Solution Manual for Calculus and Its Applications, 11th Edition

Chapter 2 Applications of Differentiation 2. Exercise Set 2.1 1. f  x  x2  4 x  5 f ‘  x  2 x  4 f ‘  x  exists for all real numbers. Solve f  x  x  6 x  3 f ‘  x  0 2 First, find the critical points. f ‘  x  2 x  6 2x  4  0 2 x  4 x  2 The only critical value is 2 . We use 2 to divide the real number line into two intervals, A:  , 2 and B: 2,   . f ‘  x  exists for all real numbers. We solve f ‘  x  0 2x  6  0 2 x  6 x  3 The only critical value is 3 . We use 3 to divide the real number line into two intervals, A:  , 3 and B: 3,   . A: Test  3, f ‘ 3  2  3  4  2  0 We see that f  x  is decreasing on  , 2 and increasing on 2,   , there is a relative B A f ‘ 0   2 0   4  4  0 B: Test 0, minimum at x  2 . f  2   2  4  2  5  1 2 3 Thus, there is a relative minimum at  2,1 . We We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  4, f ‘  4  2  4  6  2  0 B: Test 0, sketch the graph. x f  x 5 4 3 2 1 0 1 f ‘ 0   2  0   6  6  0 We see that f  x  is decreasing on  , 3 and increasing on 3,   , and the change from decreasing to increasing indicates that a relative minimum occurs at x  3 . We substitute into the original equation to find f  3 : f  3   3  6  3  3  12 2 Thus, there is a relative minimum at  3, 12 . We use the information obtained to sketch the graph. Other function values are listed below. x f  x 6 5 4 3 2 1 0 3 8 11 12 11 8 3 3. 10 5 2 1 2 5 10 f  x   2  3x  2 x 2 First, find the critical points. f ‘  x   3  4 x f ‘  x  exists for all real numbers. We solve f ‘  x  0 3  4 x  0 3 4 The solution is continued on the next page. x Copyright © 2016 Pearson Education, Inc. 218 Chapter 2: Applications of Differentiation 3 3 . We use  to 4 4 divide the real number line into two intervals, 3   3  A:  ,   and B:   ,   .    4  4 The only critical value is  4. f  x  5  x  x2 f ‘  x   1  2 x f ‘  x  exists for all real numbers. Solve f ‘  x  0 1  2 x  0 A B 2 x  1 x 3 4 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  1, f ‘  1  3  4 1  1  0  1 1 . We use  to 2 2 divide the real number line into two intervals, 1   1  A:  ,   and B:   ,   :   2  2 The only critical value is  f ‘ 0  3  4 0  3  0 B: Test 0, A: Test  1, f ‘  1  1  2  1  1  0 3  We see that f  x  is increasing on  ,    4 1  We see that f  x  is increasing on  ,    2  1  and decreasing on   ,   , which indicates  2  there is a relative maximum at x   2 2  1 21  Thus, there is a relative maximum at   ,  .  2 4 We sketch the graph. x f  x  3 25  Thus, there is a relative maximum at   ,  .  4 8  We use the information obtained to sketch the graph. Other function values are listed below. 3 2 1 7 0 3  34 25 8 0 1 2 2 3 12 1 . 2 21  1  1  1 f    5         2  2  2 4 25  3  3  3 f    2  3   2     4  4  4 8 f  x f ‘ 0  1  2 0  1  0 B: Test 0,  3  and decreasing on   ,   , and the change  4  from increasing to decreasing indicates that a 3 relative maximum occurs at x   . We 4 substitute into the original equation to find  3 f   :  4 x 1 2 5. 3 2 1  12 1 3 5 0 1 2 5 3 1 21 4 F  x   0.5 x 2  2 x  11 First, find the critical points. F ‘  x  x  2 F ‘  x  exists for all real numbers. We solve: F ‘  x  0 x2 0 x  2 The solution is continued on the next page Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 219 We see that g ‘  x  is decreasing on  , 1 The only critical value is 2 . We use 2 to divide the real number line into two intervals, A: , 2 and B: 2,   . and increasing on  1,  , which indicates there is a relative minimum at x  1 . B A g  1  1  6  1  3  1  2 2 Thus, there is a relative minimum at  1, 2 . 2 We sketch the graph. x g  x We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  3, F ‘  3   3  2  1  0 4 3 2 1 0 1 2 F ‘ 0   0   2  2  0 B: Test 0, We see that F  x  is decreasing on  , 2 and increasing on  2,   , and the change from decreasing to increasing indicates that a relative minimum occurs at x  2 . We substitute into the original equation to find F  2 : F 2  0.5 2  2  2  11  13 2 Thus, there is a relative minimum at  2, 13 . We use the information obtained to sketch the graph. Other function values are listed below. 1 2 x  2x  5 2 First, find the critical points. g ‘  x   3x 2  x  2 g  x   x3  g ‘  x  exists for all real numbers. We solve g ‘  x  0 2 3x  x  2  0 x F  x 5 3x  2 x  1  0  172 3x  2  0 or 4 3 11  25 2 x 2 3 or 2 1 13 0 1 6. 7. 25 10 1 2 1 10 25 x  1 2 . We use them 3 to divide the real number line into three intervals, 2  2  A:  , 1 , B:  1,  , and C:  ,   .  3  3 The critical values are 1 and  25 2 11  172 g  x   1  6 x  3x 2 g ‘  x  exists for all real numbers. Solve 1 g ‘  x  0 6  6x  0 x  1 The only critical value is 1 . We use 1 to divide the real number line into two intervals, A:  , 1 and B:  1,   : C B A g ‘  x  6  6 x 2 3 We use a test value in each interval to determine the sign of the derivative in each interval. The solution is continued on the next page. A: Test  2, g ‘  2  6  6  2   6  0 B: Test 0, x 1  0 g ‘ 0   6  6 0   6  0 Copyright © 2016 Pearson Education, Inc. 220 Chapter 2: Applications of Differentiation G ‘  x  exists for all real numbers. We solve A: Test  2, g ‘  2  3  2   2  2  8  0 G ‘  x  0 2 B: Test 0, 2 3x  2 x  1  0 g ‘  0   3  0    0   2  2  0 3x  1 x  1  0 2 C: Test 1, g ‘ 1  3 1  1  2  2  0 2 We see that g  x  is increasing on  , 1 , 2  decreasing on  1,  , and increasing on  3 2  ,  . So there is a relative maximum at  3  x  1 and a relative minimum at x  We find g 1 : 2 . 3 1 2 g  1   1   1  2  1  5 2 1 13  1   2  5  2 2 2 Then we find g   : 3 3 3 2 2 2 1 2 2 g          2  5 3 3 2 3 3   13  There is a relative maximum at  1,  , and  2  2 113  there is a relative minimum at  , .  3 27  We use the information obtained to sketch the graph. Other function values are listed x g  x 8. 3 5 2 11 x 1  0 3 x  1 or x 1 1 x 3 or x 1 1 and 1 . We use them 3 to divide the real number line into three intervals, 1   1  A:  ,   , B:   ,1 , and C: 1,   .    3  3 A: Test  1, The critical values are  G ‘  1  3  1  2  1  1  4  0 2 B: Test 0, G ‘ 0  3 0  2 0  1  1  0 2 C: Test 2, G ‘  2  3  2   2  2   1  7  0 2 1  We see that G  x  is increasing on  ,   ,  3 1,  . So there is a relative maximum at x 1 and a relative minimum at x  1 . 3 3 2 59  1  1  1  1 G             2   3  3  3  3 27 G 1  1  1  1  2  1 3 2  1 59  There is a relative maximum at   ,  , and  3 27  there is a relative minimum at 1,1 . 9 2 G  x  x  x  x  2 3 or  1  decreasing on   ,1 , and increasing on  3  8 2 4 113   5  27 9 3 27 2 0 1 3x  1  0 2 G ‘  x   3x 2  2 x  1 We use the information obtained to sketch the graph. Other function values are listed below. x G  x 2 1 0 2 3 8 1 2 4 17 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 9. 221 f  x   x3  3x 2 10. f  x   x3  3 x  6 f ‘  x   3x 2  3 First, find the critical points. f ‘  x   3x 2  6 x f ‘  x  exists for all real numbers. We solve f ‘ x  0 f ‘  x  exists for all real numbers. We solve f ‘  x  0 3x 2  3  0 2 3x  6 x  0 3x  x  2  0 x0 or x2 The critical values are 0 and 2 . We use them to divide the real number line into three intervals, A: , 0 , B: 0, 2 , and C:  2,   . C B A 2 0 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  1, f ‘  1  3  1  6  1  9  0 2 B: Test 1, f ‘ 1  3 1  6 1  3  0 C: Test 3, f ‘ 3  3 3  6 3  9  0 2 2 We see that f  x  is increasing on , 0 , decreasing on 0, 2 , and increasing on 2,   . So there is a relative maximum at x  0 and a relative minimum at x  2 . We find f 0 : f 0  0  3 0  0. 3 2 Then we find f  2 : f  2   2  3  2  4. 3 2 There is a relative maximum at 0, 0 , and there is a relative minimum at 2, 4 . We use the information obtained to sketch the graph. Other function values are listed below. x f  x 2 1 1 3 4 20 4 2 0 16 3x 2  3 x2  1 x  1 The critical values are 1 and 1 . We use them to divide the real number line into three intervals, A: , 1 , B:  1,1 , and C:1,   . A: Test  3, f ‘ 3  3  3  3  24  0 2 B: Test 0, f ‘  0   3  0   3  3  0 C: Test 2, f ‘  2  3  2  3  9  0 2 2 We see that f  x  is increasing on  , 1 , decreasing on 1,1 , and increasing on 1,  . So there is a relative maximum at x  1 and a relative minimum at x  1 . f  1   1  3  1  6  1  3  6  8 3 f 1  1  3 1  6  1  3  6  4 3 There is a relative maximum at  1,8 , and there is a relative minimum at 1, 4 . We use the information obtained to sketch the graph. Other function values are listed below. x f  x 3 2 0 2 3 12 4 6 8 24 Copyright © 2016 Pearson Education, Inc. 222 11. Chapter 2: Applications of Differentiation We see that f  x  is increasing on  , 1 , f  x   x3  3 x decreasing on 1, 0 , and increasing on 0,   . First, find the critical points. f ‘  x   3x 2  3 f ‘  x  exists for all real numbers. We solve So there is a relative maximum at x  1 and a relative minimum at x  0 . 3x 2  3  0 f 0  3 0  2 0  0 f  1  3  1  2  1  1 2 f ‘  x  0 2 is a relative minimum at 0, 0 . We use the information obtained to sketch the graph. Other function values are listed below. x f  x f ‘  0  3  0   3  3  0 2 3 2 We see that f  x  is increasing on  ,  , and that there are no relative extrema. We use the information obtained to sketch the graph. Other function values are listed below. x f  x 12. 14 4 0 4 14 1 2 2 13. 27 4 1 28 F  x   1  x3 First, find the critical points. F ‘  x   3x 2 F ‘  x  exists for all real numbers. We solve F ‘  x  0 f  x   3 x 2  2 x3 3x 2  0 f ‘  x  6 x  6 x2 f ‘  x  exists for all real numbers. We solve f ‘  x  0 6 x  6 x2  0 x0 The only critical value is 0 . We use 0 to divide the real number line into two intervals, A: , 0 , and B: 0,   . 6 x 1  x   0 6x  0 3 There is a relative maximum at 1,1 , and there x 2  1 There are no real solutions to this equation. Therefore, the function does not have any critical values. We test a point 2 1 0 1 2 3 B A or x 1  0 x0 or x  1 We know the critical values are 1 and 0 . We use them to divide the real number line into three intervals, A: , 1 , B:  1, 0 , and C: 0,   . A: Test  2, f ‘  2  6  2  6  2  12  0 2 1 B: Test  , 2 2 3  1  1  1 f ‘   6    6      0  2  2  2 2 0 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  1, F ‘ 1  3  1  3  0 2 B: Test 1, F ‘ 1  3 1  3  0 2 We see that F  x  is decreasing on  , 0 and decreasing on 0,   , so the function has no relative extema. We use the information obtained to sketch the graph on the next page. C: Test 1, f ‘ 1  6 1  6 1  12  0 2 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 223 Using the information from the previous page and deteriming other function values are listed below, we sketch the graph. x F  x 2 1 0 1 2 14. 9 2 1 0 7 A C B 0 4 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  1, G ‘  1  3  1  12  1  15  0 2 g  x   2 x3  16 B: Test 1, First, find the critical points. g ‘  x  6 x2 G ‘ 1  3 1  12 1  9  0 C: Test 5, G ‘ 5  3 5  12 5  15  0 g ‘  x  exists for all real numbers. We solve g ‘  x  0 6 x2  0 x0 The only critical value is 0 . We use 0 to divide the real number line into two intervals, A: , 0 , and B: 0,   . A: Test  1, g ‘  1  6  1  6  0 2 g ‘ 1  6 1  6  0 2 B: Test 1, We see that g  x  is increasing on  , 0 and increasing on 0,   , so the function has no relative extema. We use the information obtained to sketch the graph. Other function values are listed below. x g  x 2 1 0 1 2 3 15. The critical values are 0 and 4 . We use them to divide the real number line into three intervals, A: , 0 , B: 0, 4 , and C: 4,   . 32 18 16 14 0 38 2 2 We see that G  x  is increasing on , 0 , decreasing on 0, 4 , and increasing on 4,   . So there is a relative maximum at x  0 and a relative minimum at x  4 . We find G 0 : G 0  0  6 0  10 3  10 Then we find G  4 : G  4   4  6  4  10 3  22 There is a relative maximum at 0,10 , and there is a relative minimum at  4, 22 . We use the information obtained to sketch the graph. Other function values are listed below. x G  x 2 1 1 2 3 22 3 5 6 17 First, find the critical points. G ‘  x   3x 2  12 x G ‘  x  exists for all real numbers. We solve G ‘  x  0 x  4x  0 Dividing by 3 x  x  4  0 x0 or x4 0 x0 or x4 2  64  96  10 G  x   x3  6 x 2  10 2 2 Copyright © 2016 Pearson Education, Inc. 224 16. Chapter 2: Applications of Differentiation f  x   12  9 x  3x 2  x3 17. f ‘  x  9  6 x  3x 2 f ‘  x  exists for all real numbers. Solve f ‘  x  0 First, find the critical points. g ‘  x   3 x 2  4 x3 g ‘  x  exists for all real numbers. We solve g ‘  x  0 9  6 x  3x 2  0 x2  2 x  3  0 Dividing by  3  x  3 x  1  0 x3  0 g  x   x3  x 4 x 1  0 or x  3 or x 1 The critical values are 3 and 1 . We use them to divide the real number line into three intervals, A: , 3 , B:  3,1 , and C:1,   . We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  4, f ‘  4   9  6  4  3  4  15  0 2 B: Test 0, f ‘  0   9  6  0   3 0   9  0 2 3 x 2  4 x3  0 x 2 3  4 x   0 x2  0 x0 3  4x  0  4 x  3 3 x0 or x 4 3 The critical values are 0 and . 4 We use the critical values to divide the real number line into three intervals,  3 3  A:  , 0 , B:  0,  , and C:  ,   .  4 4  or or A C: Test 2, 0 f ‘  2  9  6  2  3  2  15  0 2 We see that f  x  is decreasing on  , 3 , increasing on 3,1 , and decreasing on 1,  . So there is a relative minimum at x  3 and a relative maximum at x  1 . f  3  12  9  3  3  3   3  15 2 3 A: Test  1, g ‘  1  3  1  4  1  7  0 B: Test there is a relative maximum at 1,17  . We use the information obtained to sketch the graph. Other function values are listed below. x f  x 5 4 2 1 0 2 3 17 8 10 1 12 10 15 2 3 2 3 1 1 1 1 , g ‘   3   4   2 2 2 2 1 1 1  3   4    0 4 8 4 3 There is a relative minimum at  3, 15 , and 3 4 We use a test value in each interval to determine the sign of the derivative in each interval. f 1  12  9 1  3 1  1  17 2 C B C: Test 1, g ‘ 1  3 1  4 1  1  0 2 3 We see that g  x  is increasing on  , 0 and  3 3   0,  , and is decreasing on  ,   . So there 4 4 is no relative extrema at x  0 but there is a 3 relative maximum at x  . 4 3 We find g   : 4 3 4 27 81 27 3 3 3 g          4 4 4 64 256 256 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 225 We use the information obtained to sketch the graph. Other function values are listed below. From the previous page, we determine there is a  3 27  . We use the relative maximum at  ,  4 256  information obtained to sketch the graph. Other function values are listed below. 18. x g  x 2 1 0 24 2 0 1 2 1 16 1 2 0 8 19. f  x   x 4  2 x3 f ‘  x  4 x  6 x 3 f ‘  x  0 2 x  2 x  3  0 2 2x  3  0 3 2 3 The critical values are 0 and . We use them to 2 divide the real number line into three intervals,  3 3  A:  , 0 , B:  0,  , and C:  ,   .  2 2  x0 x or A: Test  1, f ‘  1  4  1  6  1  10  0 3 2 B: Test 1, f ‘ 1  4 1  6 1  2  0 C: Test 2, f ‘  2  4  2  6  2  8  0 3 2 3 2 Since f  x  is decreasing on both  , 0 and  3 3   0,  , and increasing on  ,   , there is no 2 2 relative extrema at x  0 but there is a relative 3 minimum at x  . 2 4 2 1 0 1 2 3 32 3 0 1 0 27 1 3 x  2 x2  4 x  1 3 First, find the critical points. f ‘  x  x2  4 x  4 f  x  f ‘  x  0 2 x  4x  4  0  x  2 2  0 4 x3  6 x 2  0 or f  x f ‘  x  exists for all real numbers. We solve 2 f ‘  x  exists for all real numbers. Solve x2  0 x 3 27 3 3 3 f       2    2 2 2 16  3 27  There is a relative minimum at  ,   .  2 16  x2 The only critical value is 2 . We divide the real number line into two intervals, A: , 2 and B:  2,   . A B 2 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test 0, f ‘ 0  0  4 0  4  4  0 2 B: Test 3, f ‘ 3  3  4 3  4  1  0 2 We see that f  x  is increasing on both  , 2 and 2,   . Therefore, there are no relative extrema. We use the information obtained to sketch the graph. Other function values are listed below. x f  x 3 2 40  59 3 1  22 3 0 1 1 2 5 3 3 2 Copyright © 2016 Pearson Education, Inc. 4 3 226 20. Chapter 2: Applications of Differentiation 1 F  x    x3  3x 2  9 x  2 3 F ‘  x   x2  6 x  9 21. F ‘  x  exists for all real numbers. Solve F ‘  x  0 f  x   3 x 4  15 x 2  12 First, find the critical points. f ‘  x   12 x3  30 x f ‘  x  exists for all real numbers. We solve f ‘  x  0 2  x  6x  9  0 12 x3  30 x  0 x2  6 x  9  0 6 x 2×2  5  0   x  32  0 x3  0 x3 The only critical value is 3 . We divide the real number line into two intervals, A: ,3 and B: 3,   . A B 3 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test 0, F ‘ 0   0  6 0  9  9  0 2 B: Test 4, F ‘  4    4  6  4  9  1  0 2 We see that F  x  is decreasing on both  ,3 and 3,   . Therefore, there are no relative extrmea. We use the information obtained to sketch the graph. Other function values are listed below. x 3 2 F  x 65 104 3  6x  0 or x0 or x0 or 2×2  5  0 5 x2  2 x 10 2 10 10 and  . We 2 2 use them to divide the real number line into four intervals,   10  10  A:  ,   , B:   2 , 0  , 2     The critical values are 0,   10  10  C:  0, , and D:  , .  2    2  C B A 0  10 2 D 10 2 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  2, f ‘  2  12  2  30  2  36  0 3 B: Test  1, 1 43 3 0 1 2  13 3 C: Test 1, 2  20 3 3 7 D: Test 2, f ‘  1  12  1  30  1  18  0 3 f ‘ 1  12 1  30 1  18  0 3 f ‘  2  12  2  30  2  36  0 3 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 227 From the previous page, we see that f  x  is  10  decreasing on  ,  , increasing on 2    10   10    2 , 0  , decreasing again on  0, 2  ,      10  ,   . Thus, there and increasing again on   2  10 , a relative 2 maximum at x  0 , and another relative is a relative minimum at x   10 . 2  10  We find f   :  2  22. x f  x 3 2 1 1 2 3 120 0 0 0 0 120 g  x   2 x 4  20 x 2  18 g ‘  x   8 x3  40 x g ‘  x  exists for all real numbers. We solve g ‘  x  0 minimum at x  3 8 x  40 x  0   8x x2  5  0 4 2  10   10   10  f   3  15       12  2   2   2  8x  0 or x2  5  0 x0 or x2  5 27 4 f Then we find 0 : x0 or x 5  The critical values are 0, 5 and  5 . We use them to divide the real number line into four intervals, f 0  3 0  15 0  12  12 4 2     C: 0, 5  , and D:  5,   . A: ,  5 , B:  5, 0 ,  10  Then we find f  :  2  4 2  10   10   10  f  3  15      12  2   2   2   0  5 27 4  10 27  ,   and There are relative minima at   4   2 C B A D 5 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  3, g ‘  3  8  3  40  3  96  0 3  10 27   2 , 4  .   B: Test  1, There is a relative maximum at  0,12 . We use the information obtained above to sketch the graph. Other function values are listed at the top of the next column. g ‘  1  8  1  40  1  32  0 3 C: Test 1, g ‘ 1  8 1  40 1  32  0 3 D: Test 3, g ‘ 3  8 3  40 3  96  0 3 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. 228 Chapter 2: Applications of Differentiation From the previoius page, we see that g  x  is  We use a test value in each interval to determine the sign of the derivative in each interval. 1 1  0 A: Test  3, G ‘  3  2 3 3 3  3  2  decreasing on ,  5 , increasing on  5, 0 , decreasing again on 0, 5  , and increasing again on  5,   . Thus, there is a B: Test  1, G ‘  1  relative minimum at x   5 , a relative maximum at x  0 , and another relative minimum at x  5 .    g  5 2  5 2 4 2 10 3 2 1 6 0,18 We use the information obtained to sketch the graph. Other function values are listed below. 23. x g  x 4 3 1 1 3 4 210 0 0 0 0 210  1 0 3 We use the information obtained to sketch the graph. Other function values are listed below. x G  x 2  5   2  5   20  5   18  32 There are relative minima at  5, 32 and  5, 32 . There is a relative maximum at g 3 , 2 and 2,  . Thus, there are no relative extrema for G  x  . g 0  2 0  20 0  18  18 4 3  1  2 2 We see that G  x  is increasing on both   20  5   18  32 4 1 24. 2 1 0 1 2 F  x   3 x  1   x  1 3 1 F ‘  x   2 1  x  1 3 1 3 1 3  x  1 3 2 F ‘  x  does not exist when 3  x  1 3  0 , which means that F ‘  x  does not 2 exist when x  1 . The equation F ‘  x   0 has no solution, therefore, the only critical value is x 1. We use 1 to divide the real number line into two intervals, A: ,1 and B: 1,   . G  x   3 x  2   x  2 3 1 First, find the critical points. 2 1 G ‘  x    x  2 3 1 3 1  2 3  x  2 3 G ‘  x  does not exist when x  2 . The equation G ‘  x   0 has no solution, therefore, the only critical value is x  2 . We use 2 to divide the real number line into two intervals, A: , 2 and B:  2,   : A B 2 We use a test value in each interval to determine the sign of the derivative in each interval. 1 1  0 A: Test 0, F ‘ 0  2 3 3 3 0  1 B: Test 2, F ‘ 2  1 3  2  1 3 2  1 0 3 We see that F  x  is increasing on both ,1 and 1,  . Thus, there are no relative extrema for F  x  . We use the information obtained to sketch the graph at the top of the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 229 Using the information from the previous page, we sketch the graph. Other function values are listed. x F  x 7 0 1 2 9 25. 2 1 0 1 2 26. f  x  1  x 3 f  x 8 1 1 8 3 0 0 3 f  x    x  3 3  5 2 f ‘  x  2 First, find the critical points. 2  13 f ‘  x  x 3 2  3 3 x f ‘  x  does not exist when  1 2  x  3 3 3 2 3  x  3 3 1 f ‘  x  does not exist when x  3 . The equation f ‘  x   0 has no solution, therefore, the only 3 3 x  0 , which means that f ‘  x  does not exist when x  0 . The equation f ‘  x   0 has no solution, therefore, the only critical value is x  0. We use 0 to divide the real number line into two intervals, A:  , 0 and B: 0,   : A x B critical value is x  3 . We use 3 to divide the real number line into two intervals, A:  , 3 and B:  3,   : A: Test  4, f ‘   4  B: Test  2, f ‘ 2  2 1  1  3   4  3 3 2 3  2  3 3 2 0 3 2 0 3 We see that f  x  is decreasing on  , 3 and increasing on  3,   . Thus, there is a relative minimum at x  3 . 0 f  3  3  3 3  5  5 : 2 We use a test value in each interval to determine the sign of the derivative in each interval. 2 2  0 A: Test  1, f ‘  1   3 3 1 3 2 2 B: Test 1, f ‘ 1   3    0 3 3 1 We see that f  x  is increasing on , 0 and decreasing on 0,   . Thus, there is a relative maximum at x  0 . We find f 0 : f 0  1  0 3  1 . 2 Therefore, there is a relative minimum at 3, 5 . We use the information obtained to sketch the graph. Other function values are listed below. x f  x 11 4 2 5 1 4 4 1 Therefore, there is a relative maximum at 0,1 . We use the information obtained to sketch the graph. Other function values are listed at the top of the next column. Copyright © 2016 Pearson Education, Inc. 230 27. Chapter 2: Applications of Differentiation 1 x x 1 First, find the critical points. G  x 3  54 2  85 1 1 2 4 4  85 3  54 8 G  x  2    8 x 2  1   2 x  G ‘  x   8  1 x 2  1 2 16 x   x  1 2 2 G ‘  x  exists for all real numbers. Setting the derivative equal to zero, we have: G ‘  x  0 16 x  x  1 2 2 28. 0 x 1    5 x2  1  A  2 x  0 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  1, G ‘  1  G ‘ 1  16  1 1  1 2 16 1 12  1 2  16  4  0 4 16  40 2 4 We see that G  x  is decreasing on  , 0 and increasing on 0,   . Thus, a relative minimum occurs at x  0 . We find G 0 : 8  0 2  1 2  x  1 2 2 F ‘  x  exists for all real numbers. We solve F ‘  x  0 10 x  x  1 2 B 1 10 x  x0 The only critical value is 0 . We use 0 to divide the real number line into two intervals, A:  , 0 and B: 0,   : G 0  2 F ‘  x   5  1 x 2  1 16 x  0 B: Test 1, 5 F  x  2 0 x0 The only critical values is 0 . We use 0 to divide the real number line into two intervals, A:  , 0 and B: 0,   : A: Test  1, F ‘  1  10  1 2  10 5  0 4 2 2  5 10  0 4 2 1  1 2 B: Test 1, F ‘ 1  10 1 1  1 2 We see that F  x  is increasing on , 0 and decreasing on 0,   . Thus, a relative maximum occurs at x  0 . We find F 0 :  8 5 Thus, there is a relative minimum at 0, 8 . F 0   We use the information obtained to sketch the graph. Other function values are listed at the top of the next column. Thus, there is a relative maximum at 0,5 .  0 2  1 5 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 231 We use the information obtained on the previous page to sketch the graph. Other function values are listed below. 29. x F  x 3 1 2 2 1 1 1 5 2 2 3 1 A: Test  2, g ‘  2  5 2 2 2 g  1   x  1 2 2  x  1 2 2 2 4  4  2 2 2 2 2  1 40  12 0 25 0 x2  1  0  2 4  2 2 information obtained to sketch the graph. Other function values are listed below. x g  x g ‘  x  0 4  4 x2  0 4  1 there is a relative maximum at 1, 2 . We use the g ‘  x  exists for all real numbers. We solve   Multiplying by x 2  1 2 Dividing by  4 x2  1 x 1 30. x  1 The critical values are 1 and 1 . We use them to divide the real number line into three intervals, A:  , 1 , B:  1,1 , and C:1,   . A 02  1 There is a relative minimum at  1, 2 , and 4  4 x2  x  1 2 1  1 Then we find g 1 : 4 1 4 g 1  2  2 1  1 2 Quotient Rule 4 x2  4  8×2 2 4  4 0  12 0 25 x  1 and a relative maximum at x  1 . We find g 1 : 2 2 2  1  1,  . So there is a relative minimum at  x  1 4  4 x 2 x g ‘  x   x  1 4  4 x2 2 2 increasing on 1,1 , and decreasing again on x2  1 First, find the critical points.  2 We see that g  x  is decreasing on  , 1 , 4x  g ‘ 2  C: Test 2, 1 2 g  x  g ‘ 0   B: Test 0, 4  4  2 B C 1 1 We use a test value in each interval to determine the sign of the derivative in each interval. 3  65 2  85 0 2 0 3 6 5 g  x  g ‘  x  g ‘  x  8 5 x2 x2  1  x  1 2 x  x 2 x  x  1 2 2 2 2 2x  x  1 2 2 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. 232 Chapter 2: Applications of Differentiation From the previous page, g ‘  x  exists for all real 31. numbers. We solve g ‘  x  0  2x 2  x 1 2 x0 The only critical values is 0 . We use 0 to divide the real number line into two intervals, A:  , 0 and B: 0,   : f ‘  x  does not exist when x  0 . The equation f ‘  x   0 has no solution, therefore, the only critical value is x  0 . We use 0 to divide the real number line into two intervals, A:  , 0 and B: 0,   : A: Test  1, 2  1 1  1 2 2  1 First, find the critical points. 1 2 f ‘  x   x 3 3 1 1   2 3 2 3 3  x 3 x 0 g ‘  1  f  x  3 x   x 3 2 1  0 4 2 A B B: Test 1, g ‘ 1   2 1  1  1 2 0 2 1   0 2 4 2 We use a test value in each interval to determine the sign of the derivative in each interval. 1 1  0 A: Test  1, f ‘  1  2 3 3 3 1 We see that g  x  is decreasing on  , 0 and   increasing on 0,   . Thus, a relative minimum B: Test 1, f ‘ 1  occurs at x  0 . We find g 0 : g 0   02 0 02  1 2 4 5 1 1 2 1 1 2 2 4 5 3 9 10 1 0 3 and 0,   . Thus, there are no relative extrema for f  x  . We use the information obtained to use the information obtained to sketch the graph. Other function values are listed below. x g  x 9 10 2   3  3 1     We see that f  x  is increasing on both  , 0 Thus, there is a relative minimum at 0, 0 . We 3 1 sketch the graph. Other function values are listed below. 32. x f  x 8 1 0 1 8 2 1 0 1 2 f  x    x  1 3 1 f ‘  x   2 1  x  1 3 1 3 1 3  x  1 3 2 f ‘  x  does not exist when x  1 . The equation f ‘  x   0 has no solution, therefore, the only critical value is x  1 . Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 233 We use 1 to divide the real number line into two intervals, A: , 1 and B:  1,   : Using the information from the previous page, we use 1 to divide the real number line into two intervals, A: , 1 and B:  1,   : 1 A: Test  2, f ‘  2   0 2 3 3 3  2  1 1 We use a test value in each interval to determine the sign of the derivative in each interval. A: Test  2, 1 1 B: Test 0, f ‘ 0   0 2 3 0  1 3 3 We see that f  x  is increasing on both B: Test 0, We use the information obtained to sketch the graph. Other function values are listed below. 33. f  x 9 2 1 0 7 2 1 0 1 2 0   1 1  0 0   2 0   5 5 We see that g  x  is decreasing on  , 1 and increasing on  1,  , and the change from g ‘ 0     1 g  1  12  2 1  5  4  2 Thus, there is a relative minimum at 1, 2 . 2 First, find the critical points.  12 1 g ‘  x  x2  2 x  5 2 x  2 2 2  x  1  1 2 2 x2  2 x  5  We use the information obtained to sketch the graph. Other function values are listed below.    x 1 2 x  2x  5 The equation x 2  2 x  5  0 has no realnumber solution, so g ‘  x  exists for all real numbers. Next we find out where the derivative is zero. We solve g ‘  x  0 x 1 2 0 2 decreasing to increasing indicates that a relative minimum occurs at x  1 . We substitute into the original equation to find g 1 : g  x  x2  2 x  5  x2  2 x  5   2   1 1  0 2   2   2  2   5 5 g ‘  2   intervals, Thus, there are no relative extrema for f  x . x B A 1 x  2x  5 x 1  0 x  1 The only critical value is 1 . 34. x g  x 4 2 0 1 3 3.61 2.24 2.24 2.83 4.47 F  x   x 1 1   x2  1 2   12   32  1 F ‘  x     x2  1 2 x   2  x  x  1 2 3 2 F ‘  x  exists for all real numbers. The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. 234 Chapter 2: Applications of Differentiation Using the derivative from the previous page, We solve F ‘  x  0 x  x  1 2 3 2 70. Answers may vary, one such graph is: 0 x0 The only critical value is 0 . We use 0 to divide the real number line into two intervals, A:  , 0 and B: 0,   : A: Test  1, F ‘  1   1 1  1 2 3 2  1 8 71. Answers may vary, one such graph is: 0 B: Test 1, F ‘ 1  1 1  1 2 3 2  1 8 0 We see that F  x  is increasing on , 0 and decreasing on 0,   . Thus, a relative maximum occurs at x  0 . 1 F 0   1  0 2  1 72. Answers may vary, one such graph is: Thus, there is a relative maximum at 0,1 . We use the information obtained to sketch the graph. Other function values are listed below. x F  x 3 2 1 1 2 3 0.32 0.45 0.71 0.71 0.45 0.32 35. – 68. 73. Answers may vary, one such graph is: Left to the student. 69. Answers may vary, one such graph is: Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 235 74. Answers may vary, one such graph is: 79. Answers may vary, one such graph is: y x -5 -4 -3 -2 -1 0 1 2 3 4 5 80. Answers may vary, one such graph is: y 75. Answers may vary, one such graph is: x -5 -4 -3 -2 -1 0 1 2 3 4 5 81. Answers may vary, one such graph is: f(x) 76. Answers may vary, one such graph is: x -5 77. Answers may vary, one such graph is: -4 -3 -2 -1 0 1 2 3 4 5 82. Answers may vary, one such graph is: g(x) x -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 83. Answers may vary, one such graph is: H(x) 78. Answers may vary, one such graph is: x -9 -8 -7 -6 -5 -4 -3 -2 -1 Copyright © 2016 Pearson Education, Inc. 0 1 2 3 4 5 6 7 8 9 236 Chapter 2: Applications of Differentiation 84. Answers may vary, one such graph is: We use them to divide the interval 0,   into K(x) three intervals: A: 0,2.108 B:  2.08,3.897  , and C:3.897  85. 0 1 2 3 4 5 6 7 8 9 The critical value of a function f is an interior value c of its domain at which the tangent to the graph is horizontal  f ‘ c   0 or the tangent is vertical  f ‘ c  does not exist  . The critical values for this graph are x1 , x3 , x4 , x5 , x6 , x7 , x8 , x10 . 86. The function is increasing on intervals a, b and c, d  . A line tangent to the curve at any point on either of these intervals has a positive slope. Thus, the function is increasing on the intervals for which the first derivative is positive. Similarly, we see that on the intervals b, c and d , e the function is decreasing. A line tangent to the curve at any point on either of these intervals has a negative slope. Thus, the function is decreasing on the intervals for which first derivative is negative. 87. Letting t be years since 2006 and E be thousand of employees, we have the function: E t   107.833t 3  971.369t 2  2657.917t  50347.83 First, we find the critical points. E ‘ t   323.499t 2  1942.738t  2657.917 E ‘ t  exists for all real numbers. Solve E ‘ t   0 323.499t 2  1942.738t  2657.917  0 Using the quadratic formula, we have: t 1942.738   1942.7382  4 323.499 2657.917  2 323.499  1942.738  334,896.970312 646.998 t  2.108 or t  3.897 There are two critical values.  C B A x -9 -8 -7 -6 -5 -4 -3 -2 -1 2.108 3.897 0 Next, we test a point in each interval to determine the sign of the derivative. A: Test 1, E ‘ 1  323.499 1  1942.738 1  2657.917 2  1038.678  0 B: Test 3, E ‘ 3  323.499 3  1942.738 3  2657.917 2  258.806  0 C: Test 4, E ‘  4  323.499  4  1942.738  4  2657.917 2  62.949  0 Since, E t  is increasing on  0, 2.108 and decreasing on 2.108,3.897  and there is a relative maximum at t  2.108 . E  2.108  107.833  2.108  971.369 2.108  3 2 2657.917  2.108  50347.83  52, 644.383 There is a relative minimum at 2.108, 52,644.383 . Since, E t  is decreasing on 2.108,3.897  and increasing 3.897,   on and there is a relative minimum at t  3.897 . E 3.897   107.833 3.897   971.369 3.897   3 2 2657.917 3.897   50347.83  52,335.73 There is a relative maxnimum at 3.897, 52,335.73 . We sketch the graph. t T t  0 1 3 5 8 50,347 52,142 52, 491 52,832 64, 654 Copyright © 2016 Pearson Education, Inc. . Exercise Set 2.1 88. 237 N  a    a 2  300a  6, 0  a  300 A: Test 20, N ‘ a  exists for all real numbers. Solve, B: Test 100, f ‘  20  0.00518 20  0.457  0.3534  0 N ‘ a   2a  300 f ‘ 100  0.00518 100  0.457  0.061  0 N ‘ a   0 We see that f t  is decreasing on 0,88.22 2a  300  0 2a  300 a  150 The only critical value is150 . We divide the interval 0,300 into two intervals, and increasing on 88.22,  , so there is a relative minimum at t  88.22 . f 88.22  0.00259 88.22  0.457 88.22  36.237 2 A: 0,150 and B: 150,300 . A: Test 100,  16.09 There is a relative minimum at about 88.22,16.08 . Thus, the latitude that is closest N ‘ 100  2 100  300  100  0 to the equator at which the full eclipse could be view is 16.08 degrees south and will occur 88.22 minutes after the start of the eclipse. We use the information obtained above to sketch the graph. Other function values are listed below. B: Test 200, N ‘  200  2  200  300  100  0 Since, N a  is increasing on 0,150 and decreasing on 150,300 , there is a relative maximum at x  150 . N 150   150  300 150  6  22,506 2 There is a relative maximum at 150, 22,506 . We sketch the graph. 89. a N a  0 100 200 300 6 20, 006 20, 006 6 90. f  x 20 30 40 60 80 28.133 24.858 22.101 18.141 16.253 T t   0.1t 2  1.2t  98.6, 0  t  12 First, we find the critical points. T ‘ t   0.2t  1.2 f t   0.00259t 2  0.457t  36.237 First, we find the critical points. f ‘ t   0.00518t  0.457 f ‘  x  exists everywhere, so we solve f ‘ t   0 0.00518t  0.457  0 t  88.22 The only critical value is about 88.22 we use it to break up the interval 0,   into two intervals A: 0,88.28 and B: 88.22,   . B A 0 x T ‘ t  exists for all real numbers. Solve T ‘ t   0 0.2t  1.2  0 t6 The only critical value is 6. We use it to divide the interval  0,12 into two intervals: A: 0,6 and B: 6,12 B A 0 6 12 The solution is continued on the next page. 88.22 Copyright © 2016 Pearson Education, Inc. 238 Chapter 2: Applications of Differentiation Next, we test a point in each interval found on the previous page to determine the sign of the derivative. A: Test 1, T ‘ 1  0.2 1  1.2  1.0  0 B: Test 7, T ‘ 7   0.2 7   1.2  0.2  0 Since, T t  is increasing on 0, 6 and decreasing on 6,12 , there is a relative 92. The derivative is positive over the interval , 2 and negative over the interval 2,   . Furthermore, it is equal to zero when x  2 . This means that the function is increasing over the interval  , 2 , decreasing over the interval 2,   and has a horizontal tangent at x  2 . A possible graph is shown below. maximum at t  6 . T 6  0.16  1.2 6  98.6  102.2 2 There is a relative maximum at 6,102.2 . We sketch the graph. t T t  0 3 5 7 8 12 98.6 101.3 102.1 102.1 101.8 98.6 93. The derivative is positive over the interval ,1 and negative over the interval 1,  . Furthermore, it is equal to zero when x  1 . This means that the function is increasing over the interval  ,1 , decreasing over the interval 91. The derivative is negative over the interval , 1 and positive over the interval 1,  . Furthermore, it is equal to zero when x  1 . This means that the function is decreasing over the interval , 1 , increasing over the 1,  and has a horizontal tangent at x  1 . A possible graph is shown below. interval 1,  and has a horizontal tangent at x  1 . A possible graph is shown below. 94. The derivative is negative over the interval , 1 and positive over the interval 1,  . Furthermore, it is equal to zero when x  1 . This means that the function is decreasing over the interval , 1 , increasing over the interval 1,  and has a horizontal tangent at x  1 . A possible graph is shown below. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 239 95. The derivative is positive over the interval 4, 2 and negative over the intervals 97. f  x    x 6  4 x5  54 x 4  160 x3  641x 2 828 x  1200 Using the calculator we enter the function into the graphing editor as follows: , 4 and 2,   . Furthermore, it is equal to zero when x  4 and x  2 . This means that the function is decreasing over the interval , 4 , then increasing over the interval 4, 2 , and then decreasing again over the interval 2,   . The function has horizontal tangents at x  4 and x  2 . A possible graph is shown below. Using the following window: The graph of the function is: 96. The derivative is negative over the interval 1,3 and intervals and positive over the intervals  , 1 and 3,   . Furthermore, it is We find the relative extrema using the minimum/maximum feature on the calculator. There are relative minima at 3.683, 2288.03 and 2.116, 1083.08 . equal to zero when x  1 and x  3 . This means that the function is increasing over the interval  , 1 , then decreasing over the There are relative maxima at 6.262,3213.8 , 0.559,1440.06 , and interval  1,3 , and then increasing again over the interval 3,   . The function has horizontal tangents at x  1 and x  3 . A possible graph is shown below. 5.054, 6674.12. 98. f  x   x 4  4 x3  36 x 2  160 x  400 Using the calculator we enter the function into the graphing editor as follows: Using the following window: The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. 240 Chapter 2: Applications of Differentiation The graph of the function is: 100. f  x   x 9  x 2 Using the calculator we enter the function into the graphing editor as follows: We find the relative extrema using the minimum/maximum feature on the calculator. There are relative minima at 5, 425 and 4, 304 . Using the following window: There is a relative maximum at 2,560 . 99. f  x  3 4  x2  1 Using the calculator we enter the function into the graphing editor as follows: The graph of the function is: Using the following window: Notice, the calculator has trouble drawing the graph. The graph should continue to the xintercepts at 3, 0 and 3,0 . Fortunately, this does not hinder our efforts to find the extrema. We find the relative extrema using the minimum/maximum feature on the calculator. There is a relative minimum at  2.12, 4.5 . The graph of the function is: There is a relative maximum at 2.12, 4.5 . 101. f  x   x  2 Using the calculator we enter the function into the graphing editor as follows: We find the relative extrema using the minimum/maximum feature on the calculator. There are relative minima at 2,1 and  2,1 . There is a relative maximum at 0, 2.587 . Using the following window: The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 241 103. f  x   x 2  1 The graph of the function is: Using the calculator we enter the function into the graphing editor as follows: We find the relative extrema using the minimum/maximum feature on the calculator. The graph is decreasing over the interval , 2 . Using the following window: The graph is increasing over the interval 2,   . There is a relative minimum at 2, 0 . The derivative does not exist at x  2 The graph of the function is: 102. f  x   2 x  5 Using the following window: We find the relative extrema using the minimum/maximum feature on the calculator. The graph is decreasing over the interval , 1 and 0,1 . The graph is increasing over the interval 1, 0 and 2,  . There are relative mimima at 1, 0 and 1,0 . The graph of the function is: There is a relative maximum at 0,1 . The derivative does not exist at x  1 and x  1. 104. f  x   x 2  3x  2 We find the relative extrema using the minimum/maximum feature on the calculator. The graph is decreasing over the interval 5   ,  . 2 The graph is increasing over the interval 5   ,   . 2 5  There is a relative minimum at  , 0  . 2  The derivative does not exist at x  Using the calculator we enter the function into the graphing editor as follows: The solution is continued on the next page. 5 2 Copyright © 2016 Pearson Education, Inc. 242 Chapter 2: Applications of Differentiation The graph of the function is: Using the following window: The graph of the function is: We find the relative extrema using the minimum/maximum feature on the calculator. The graph is decreasing over the interval , 3 and 0,3 . The graph is increasing over the interval 3, 0 and 3,  . We find the relative extrema using the minimum/maximum feature on the calculator. The graph is decreasing over the interval 3 ,1 and  , 2  . 2 The graph is increasing over the interval  3 1,  and  2,   . 2 There are relative mimima at 3, 0 and 3, 0 . There is a relative maximum at 0,9 . The derivative does not exist at x  3 and x  3. 106. f  x    x 2  4 x  4 Enter the function into the graphing editor: There are relative mimima at 1, 0 and 2, 0 . 3 1 There is a relative maximum at  ,  . 2 4 The derivative does not exist at x  1 and x  2. Using the following window: 105. f  x   9  x 2 Using the calculator we enter the function into the graphing editor as follows: The graph of the function is: Using the following window: The solution is continued on the next column. We find the relative extrema using the minimum/maximum feature on the calculator. The graph is decreasing over the interval , 2 . The graph is increasing over the interval 2,   . There is a relative minimum at 2, 0 . The derivative exists for all values of x. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.1 243 107. f  x   x3  1 The graph of the function is: Using the calculator we enter the function into the graphing editor as follows: We find the relative extrema using the minimum/maximum feature on the calculator. The graph is decreasing over the interval , 1.41 , 1, 0 and 0,1.41 . Using the following window: The graph is increasing over the interval 1.41, 0 , 0,1 and 1.41,  . There are relative mimima at 1.41, 0 , 0, 0 and 1.41, 0 . The graph of the function is: There are relative maxima at 1, 0 and 1,0 . The derivative does not exist at x  1.41 and x  1.41. 109. We find the relative extrema using the minimum/maximum feature on the calculator. The graph is decreasing over the interval , 1 . a) We enter the data into the calculator and run a cubic regression. The calculator returns The graph is increasing over the interval 1,  . There is a relative minimum at 1,0 . The derivative does not exist at x  1. 108. f  x   x 4  2 x 2 Using the calculator we enter the function into the graphing editor as follows: When we try to run a quartic regression, the calculator returns a domain error. Therefore, the cubic regression fits best. b) The domain of the function is the set of nonnegative real numbers. Realistically, there would be some upper limit upon daily caloric intake. c) The cubic regression model appears to have a relative minimum at 4316, 77.85 and it appears to have a relative maximum at 3333, 79.14 . This leads us to believe that eating too many calories might shorten life expectancy. Using the following window: The solution is continued at the top of the next column. Copyright © 2016 Pearson Education, Inc. 244 Chapter 2: Applications of Differentiation 110. a) The cubic function fits best. In fact some calculators will return an error message when an attempt is made to fit a quartic function to the data. b) The domain of the function is the set of nonnegative real numbers. Realistically, there would be some upper limit upon daily caloric intake. c) The cubic regression model does not appears to have a relative extrema. The greater the daily caloric intake, the lower the infant mortality. b) Answers will vary. When the equations are somewhat complex, the best way to determine a viewing window is to use the table screen on the calculator and observing appropriate y-values for selected x-values. You will need to set your table to accept selected x-values. Enter the table set up feature on your calculator and turn on the ask feature for your independent variable. This will allow you to enter an x-value and the calculator will return the y-value. You should make your ranges large enough so that all the data points will be easily viewed in the window. 111. a) Answers will vary. In Exercises 1-16 the function is given in equation form. The most accurate way to select an appropriate viewing window, one should first determine the domain, because that will help determine the x-range. For polynomials the domain is all real numbers, so we will typically select a x-range that is symmetric about 0. Next, you should find the critical values and make sure that your x-range contains them. Finally, you should determine the xintercepts and make sure the x-range includes them. To find the y-range, you should find the y-values of the critical points and make sure the y-range includes those values. You should also make sure that the y-range includes the y-intercept. To avoid the calculations required to find the relative extrema and the zeros as described above, we can determine a good window by using the table screen on the calculator and observing the appropriate yvalues for selected x-values. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 245 Next, find the critical points of f  x  . Since Exercise Set 2.2 1. f ‘  x  exists for all real numbers x, the only Next, find the critical points of f  x  . Since critical points occur when f ‘  x   0 . 2x 1  0 2 x  1 1 x 2 1 We find the function value at x   . 2 critical points occur when f ‘  x   0 . 1 1 5  1  1  1 f         1   1   .  2  2  2 4 2 4 2 x  0 x0 We find the function value at x  0 .  1 5 The critical point is   ,   .  2 4 Next, we apply the Second Derivative test. f ”  x   2 f  x  4  x 2 First, find f ‘  x  and f ”  x  . f ‘  x   2 x f ”  x   2 f ‘  x  exists for all real numbers x, the only 2 f ‘  x  0 f 0  4  0  4 .  1 f ”     2  0  2 Next, we apply the Second Derivative test. f ”  x   2 5  1 Therefore, f      is a relative minimum.  2 4 2 The critical point is  0, 4  . f ”  0  2  0 Therefore, f  0   4 is a relative maximum. 2. f  x  x2  x f ‘x  2x  1 f  x  5  x2 f ”  x   2 f ”  x   2 critical points occur when f ‘  x   0 . critical points occur when f ‘  x   0 . 2x 1  0 f ‘  x  exists for all real numbers. The only f ‘  x   2 x f ‘  x  exists for all real numbers. The only f ‘  x  0 f ‘  x  0 x 2 x  0 x0 We find the function value at x  0 . f 0  5  0  5 . 2 The critical point is  0,5 . 1 2 We find the function value at x  1 . 2 2 1 1 1 1 1 1 f         . 2 2 2 4 2 4 f ”  0  2  0 1 1 The critical point is  ,   . 2 4 Next, we apply the Second Derivative test. f ”  x   2 f  x  x  x  1 1 f ”    2  0 2 Next, we apply the Second Derivative test. f ”  x   2 Therefore, f  0   5 is a relative maximum. 3. 4. 2 First, find f ‘  x  and f ”  x  . f ‘  x  2x 1 1 1 Therefore, f     is a relative minimum. 2 4 f ”  x   2 Copyright © 2016 Pearson Education, Inc. 246 5. Chapter 2: Applications of Differentiation f  x   4 x 2  3x  1 We find the function value at x  First, find f ‘  x  and f ”  x  . 2 4 4 4 f    5    8    7 5 5 5 16 32 35 .    5 5 5 19  5  4 19  The critical point is  ,   . 5 5 Next, we apply the Second Derivative test. f ”  x   10 f ‘  x   8 x  3 f ”  x   8 Next, find the critical points of f  x  . Since f ‘  x  exists for all real numbers x, the only critical points occur when f ‘  x   0 . f ‘  x  0 8 x  3  0  8 x  3 3 x 8 We find the function value at x  4 f ”    10  0 5 3 . 8 19 4 Therefore, f     is a relative maximum. 5 5 2 3 3 3 f    4    3    1 8 8 8  9  9  4     1  64  8 7. 9 18 16   16 16 16 7  16 7 3 The critical point is  ,   .  8 16  Next, we apply the Second Derivative test. f ”  x   8  3 f ”    8  0 8 f  x   5 x 2  8 x  7 f ‘  x   10 x  8 f ”  x   6 x Next, find the critical points of f  x  . Since f ‘  x  exists for all real numbers x, the only critical points occur when f ‘  x   0 . f ‘  x  0 3x 2  12  0 x2  4 x 4 x  2 There are two critical values. First, we find the function value at x  2 . f ”  x   10 f  2    2   12  2   1 3 f ‘  x  exists for all real numbers. The only critical points occur when f ‘  x   0 . 10 x  8  0 4 x 5 f ‘  x   3 x 2  12 3x 2  12 7  3 Therefore, f     is a relative maximum. 8 16 f ‘  x  0 f  x   x 3  12 x  1 First, find f ‘  x  and f ”  x  . .  6. 4 . 5  8  24  1 .  15 The critical point is  2,15 . Next, we apply the Second Derivative test. f ”  x   6 x f ”  2   6  2   12  0 Therefore, f  2   15 is a relative maximum. The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 247 Next, we find the function value at x  2 . 1  The critical point is  , 1 . 2  Next, we apply the Second Derivative test. f ”  x   48 x f  2   2  12 2  1 3 .  8  24  1  17 The critical point is  2, 17  . 1 1 f ”    48    24  0 2 2 Next, we apply the Second Derivative test. f ”  x   6 x 1 Therefore, f    1 is a relative minimum. 2 f ”  2   6  2   12  0 Therefore, f  2   17 is a relative minimum. 8. f  x   x 3  27 x a) Find f ‘  x  and f ”  x  . f  x   8×3  6 x  1 f ‘  x   3x 2  27 f ‘  x   24 x 2  6 f ”  x   6 x f ”  x   48 x f ‘  x  exists for all real numbers. The only critical points occur when f ‘  x   0 . f ‘  x  0 The domain of f is . b) Find the critical points of f  x  . Since f ‘  x  exists for all real numbers x, the only critical points occur when f ‘  x   0 . 24 x 2  6  0 3 x 2  27  0 4×2  1  0 1 x2  4 x 9. x2  9 x  3 There are two critical values x  3 and x3. 1 2 1 1 There are two critical values x   and x  . 2 2 1 We find the function value at x   2 3  1  1  1 f    8    6   1  2  2  2 . 3  1  The critical point is   , 3 .  2  Next, we apply the Second Derivative test. f ”  x   48 x  1  1 f ”     48     24  0  2  2  1 Therefore, f     3 is a relative maximum.  2 1 Now, we find the function value at x  . 2 3 f  3   3  27  3  54 . 3 The critical point on the graph is  3, 54  . f  3   3  27  3  54 . 3 The critical point on the graph is  3, 54  . c) We apply the Second Derivative test to the critical points. For x  3 f ”  x   6 x f ”  3  6  3  18  0 The critical point  3, 54  is a relative maximum. For x  3 f ”  x   6 x f ” 3  6  3  18  0 The critical point  3, 54  is a relative minimum. The solution is continued on the next page. 1 1 1 f    8    6   1 . 2 2 2  1 Copyright © 2016 Pearson Education, Inc. 248 Chapter 2: Applications of Differentiation If we use the critical values x  3 and x  3 to divide the real line into three intervals, , 3 , 3, 3 , and 3,  , we know from the extrema above, that f  x  is increasing over the interval   , 3 , decreasing over the interval  3, 3 and then increasing again over the interval 3,   . d) Find the points of inflection. f ”  x  exists for all real numbers, so we solve the equation f ”  x   0 6x  0 x0 Therefore, a possible inflection point occurs at x  0 . f  0    0   27  0   0 . A possible inflection point on the graph is the point  0, 0  . e) To determine concavity, we use the possible inflection point to divide the real number line into two intervals A :  , 0  and B:  0,   . We test a point in each interval. A: Test  1: f ”  1  6  1  6  0 f ” 1  6 1  6  0 Then, f  x  is concave down on the interval  , 0 and concave up on the interval 0,   , so 0, 0  is an inflection point. f) Finally, we use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. x f x 4 1 1 4 10. 44 26 26  44 f  x   x  12 x 3 a) Find f ‘  x  and f ”  x  . f ‘  x   3 x 2  12 f ”  x   6 x The domain of f is . f ‘x  0 f ‘  x  0 3x 2  12  0 3x 2  12 x2  4 x  2 There are two critical values x  2 and x  2. We find the function value at x  2 f  2    2   12  2   16 . 3 The critical point on the graph is  2,16  . Next, we find the function value at x  2 . f  2    2   12  2   16 . 3 The critical point on the graph is  2, 16  . 3 B: Test 1: b) f ‘  x  exists for all real numbers. Solve: c) Apply the Second Derivative test to the critical points. For x  2 f ”  x   6 x f ”  2   6  2   12  0 The critical point  2,16  is a relative maximum. For x  2 f ”  x   6 x f ”  2   6  2   12  0 We determine the critical point  2, 16  is a relative minimum. If we use the critical values x  2 and x  2 to divide the real line into three intervals,  , 2  ,  2, 2  , and  2,  , we know from the extrema above, that f  x  is increasing over the interval  , 2  , decreasing over the interval  2, 2  and then increasing again over the interval  2,   . d) Find the points of inflection. f ”  x  exists for all real numbers, so we solve the equation f ”  x   0 6x  0 x0 Therefore, a possible inflection point occurs at x  0 . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 249 We evaluate the function at x  0 . f  0    0   12  0   0 . 3 Next, we find the function value at x  3 . This gives the point  0, 0  on the graph. f 3  2 3  3 3  36 3  28 e) To determine concavity, we use the possible inflection point to divide the real number line into two intervals A :  , 0  and  54  27  108  28 B:  0,   . We test a point in each interval A: Test  1: f ”  1  6  1  6  0 f ” 1  6 1  6  0 B: Test 1: Then, f  x  is concave down on the interval  , 0 and concave up on the interval 0,   , so 0, 0  is an inflection point. f) Finally, we use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. x f x 3 1 1 3 9 11  11 9 11. The critical point on the graph is  2, 72  . 3 2  53 The critical point on the graph is  3, 53 . c) Apply the Second Derivative test to the critical points. For x  2 f ”  x   12 x  6 f ”  2   12  2   6  30  0 The critical point  2, 72  is a relative maximum. For x  3 f ”  x   12 x  6 f ” 3  12 3  6  30  0 The critical point  3, 53 is a relative minimum. We use the critical values x  2 and x  3 to divide the real line into three intervals, A:  , 2  , B:  2, 3 , and C: 3,  , we know from the extrema above, that f  x  is increasing over the interval  , 2  , f  x   2 x 3  3x 2  36 x  28 a) First, find f ‘  x  and f ”  x  . decreasing over the interval 2, 3 and then increasing again over the interval 3,   . f ‘  x   6 x  6 x  36 d) Find the points of inflection. f ”  x  exists The domain of f is . for all real numbers, so we solve the equation f ”  x   0 2 f ”  x   12 x  6 b) Find the critical points of f  x  . Since f ‘  x  exists for all real numbers x, the only critical points occur when f ‘  x   0 . f ‘  x  0 6 x 2  6 x  36  0 x2  x  6  0 12 x  6  0 1 2 Therefore, a possible inflection point occurs 1 at x  . 2 x 3 2  x  3 x  2  0 1 1 1 1 f    2    3    36    28 2 2 2 2 x  3 or x  2 There are two critical values x  2 and x3. We find the function value at x  2 19 . 2 A possible inflection point on the graph is  1 19  the point  ,  . 2 2  x  3  0 or x  2  0 f  2  2  2  3  2  36  2  28 3 2   16  12  72  28  72 Copyright © 2016 Pearson Education, Inc. 250 Chapter 2: Applications of Differentiation e) To determine concavity, we use the possible inflection point to divide the real number line into two intervals 1  1  A :  ,  and B:  ,   . We test a point   2  2 in each interval A: Test 0 : f ”  0   12  0  6  6  0 B: Test 1: f ” 1  12 1  6  6  0 Then, f  x  is concave down on the interval 1   ,  and concave up on the interval 2 1   1 19   ,   , so  ,  is an inflection point. 2 2 2 f) We use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. x f x 3 1 0 1 4 55 59 28 9 36 Next, we find the function value at x  2 . f  2   3  2   36  2   3  51 . 3 The critical point on the graph is  2, 51 . c) Apply the Second Derivative test to the critical points. For x  2 f ”  x   18 x f ”  2   18  2   36  0 The critical point  2, 45 is a relative maximum. For x  2 f ”  x   18 x f ”  2   18  2   36  0 The critical point  2, 51 is a relative minimum. We use the critical values x  2 and x  2 to divide the real line into three intervals, A:  , 2  , B:  2, 2  , and C:  2,  , we know from the extrema above, that f  x  is increasing over the interval  , 2  , decreasing over the interval  2, 2  and then increasing again over the interval  2,   . d) Find the points of inflection. f ”  x  exists 12. for all real numbers, so we solve the equation f ”  x   0 f  x   3x 3  36 x  3 a) First, find f ‘  x  and f ”  x  . 18 x  0 f ‘  x   9 x 2  36 x0 Therefore, a possible inflection point occurs at x  0 . The domain of f is . f  0   3  0   36  0   3  3 . f ”  x   18 x 3 b) f ‘  x  exists for all real numbers. Solve: f ‘  x  0 This gives the point  0, 3 on the graph. e) To determine concavity, we use the possible inflection point to divide the real number line into two intervals A :  , 0  and B:  0,   . We test a point in 2 9 x  36  0 9 x 2  36 x2  4 x  2 There are two critical values x  2 and x  2. We find the function value at x  2 . f  2   3  2   36  2   3  45 . 3 The critical point on the graph is  2, 45 . each interval A: Test  1: f ”  1  18  1  18  0 B: Test 1: f ” 1  18 1  18  0 Then, f  x  is concave down on the interval  , 0 and concave up on the interval 0,   , so 0, 3 is an inflection point. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 f) 251 The critical point  0,80  is a relative We use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. x f x 3 1 1 3 24 30 36 30 maximum. We use the critical values x  6, and x  0 to divide the real line into three intervals, A : , 6 , B:  6, 0 , and C: 0,  , we know from the extrema above, that f  x  is decreasing over the interval  , 6  , increasing over the interval  6, 0  and then decreasing again over the interval  0,   . 13. d) Find the points of inflection. f ”  x  exists f  x   80  9 x 2  x 3 for all real numbers, so we solve the equation f ”  x   0 a) First, find f ‘  x  and f ”  x  . f ‘  x   18 x  3x 2 f ”  x   18  6 x 18  6 x  0 The domain of f is . x  3 Therefore, a possible inflection point occurs at x  3 . b) Next, find the critical points of f  x  . Since f ‘  x  exists for all real numbers x, the f  3  80  9  3   3  26 . 2 only critical points occur when f ‘  x   0 . A possible inflection point on the graph is the point  3, 26  . f ‘  x  0 18 x  3x 2  0 e) To determine concavity, we use the possible inflection point to divide the real number line into two intervals A :  , 3 and B:  3,   . We test a point 3 x  x  6   0 3 x  0 or x  6  0 x  0 or x  6 There are two critical values x  6 and x0. We find the function value at x  6 . in each interval A: Test  4 : f ”  4   18  6  4   6  0 B: Test 0 : f ”  0  18  6  0  18  0 Then, f  x  is concave up on the interval f  6  80  9  6   6 2 3  , 3 and concave down on the interval  3, , , so  3, 26  is an inflection point. .  80  324  216  28 The critical point on the graph is  6, 28  . f) We use the preceding information to sketch the graph of the function. Additional function values are also calculated. Next, we find the function value at x  0. f  0   80  9  0    0   80 . 2 3 The critical point on the graph is  0,80  . x c) We Apply the Second Derivative test to the critical points. For x  6 f ”  x   18  6 x f x 9 4 2 0 2 3 80 0 52 80 36 28 f ”  6  18  6  6  18  0 The critical point  6, 28  is a relative 3 minimum. For x  0 f ”  x   18  6 x f ”  0  18  6  0  18  0 Copyright © 2016 Pearson Education, Inc. 252 14. Chapter 2: Applications of Differentiation 8 3 1 x  2x  3 3 a) Find f ‘  x  and f ”  x  . 1 2 f ”  x   16 x f  x  For x  f ‘  x   8×2  2 1 1 f ”    16    8  0 2 2 The domain of f is . 1 1 The critical point  ,   is a relative  2 3 minimum. Therefore, f  x  is increasing over the f ”  x   16 x b) f ‘  x  exists for all real numbers. Solve: f ‘  x  0 8×2  2  0 1  interval  ,   , decreasing over the  2 8×2  2 x2  1 4 x  1 1 interval   ,  and then increasing again  2 2 1 2 There are two critical values x   x 1 and 2 1 . 2 We find the function value at x   1 . 2 3 1  1  The critical point on the graph is   ,1 .  2  1 . 2 3 1 8 1 1 1 f       2   2 32 2 3 1  3 d) We find the points of inflection. f ”  x  exists for all real numbers, so we solve the equation f ”  x   0 16 x  0  1 8 1  1 1 f       2     2 3 2  2 3 Next, we find the function value at x  1  over the interval  ,   . 2  . 1 1 The critical point on the graph is  ,   .  2 3 c) Apply the Second Derivative test to the critical points. 1 For x   2 f ”  x   16 x x0 Therefore, a possible inflection point occurs at x  0 . 8 3 1 1 f  0   0  2  0   . 3 3 3  1 This gives the point  0,  on the graph.  3 e) To determine concavity, we use the possible inflection point to divide the real number line into two intervals A :  , 0  and B:  0,   . We test a point in each interval A: Test  1: f ”  1  16  1  16  0 B: Test 1: f ” 1  16 1  16  0 Then, f  x  is concave down on the interval  , 0 and concave up on the interval  1 3 0,   , so  0,  is an inflection point.  1  1 f ”     16     8  0  2  2  1  The critical point   ,1 is a relative  2  maximum. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 253 We use the critical values x  1 and x  1 to divide the real line into three intervals, A :  , 1 , B:  1,1 , and C: 1,  , we f) We use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. x f x decreasing over the interval   , 1 , 17  13 2 1 1 2 15. know from the extrema above, that f  x  is increasing over the interval  1,1 and then decreasing again over the interval 1,   . 1 d) Find the points of inflection. f ”  x  exists 53 3 for all real numbers, so we solve the equation f ”  x   0 f  x    x 3  3x  2 a) First, find f ‘  x  and f ”  x  . 6 x  0 f ‘  x   3x 2  3 x0 Therefore, a possible inflection point occurs at x  0 . f ”  x   6 x The domain of f is . f  0     0   3  0   2  2 . b) Find the critical points of f  x  . Since 3 f ‘  x  exists for all real numbers x, the only A possible inflection point on the graph is the point  0, 2  . critical points occur when f ‘  x   0 . f) To determine concavity, we use the possible inflection point to divide the real number line into two intervals A :  , 0  and B:  0,   . f ‘  x  0 3 x 2  3  0   3 x 2  1  0 We test a point in each interval A: Test  1: f ”  1  6  1  6  0 2 x 1  0 B: Test 1: f ” 1  6 1  6  0 x  1 There are two critical values x  1 and x 1. We find the function value at x  1 . Then, f  x  is concave up on the interval  , 0 and concave down on the interval 0,   , so 0, 2  is an inflection point. f  1    1  3  1  2  4 . 3 e) We use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. x f x The critical point on the graph is  1, 4  . We find the function value at x  1 . f 1   1  3 1  2  0 . 3 The critical point on the graph is 1, 0  . 3 2 2 3 c) We apply the Second Derivative test to the critical points. For x  1 f ”  x   6 x f ”  1  6  1  6  0 The critical point  1, 4  is a relative minimum. For x  1 f ”  x   6 x f ” 1  6 1  6  0 16. 16 0 4 20 f  x    x 3  3x 2  4 a) First, find f ‘  x  and f ”  x  . f ‘  x   3 x 2  6 x f ”  x   6 x  6 The critical point 1, 0  is a relative The domain of f is . maximum. Copyright © 2016 Pearson Education, Inc. 254 Chapter 2: Applications of Differentiation b) f ‘  x  exists for all real numbers. Solve: line into two intervals A :   ,1 and f ‘  x  0 B: 1,   . We test a point in each interval A: Test 0 : f ” 0  6 0  6  6  0 3x 2  6 x  0 B: Test 2 : f ”  2  6  2  6  6  0 3x  x  2   0 3x  0 or Then, f  x  is concave up on the interval x20  ,1 and concave down on the interval 1,   , so 1, 2  is an inflection point. x0 or x2 There are two critical values x  0 and x  2. We find the function value at x  0 . f) We use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. x f x f  0     0   3  0   4  4 3 2 The critical point on the graph is  0, 4  . Next, we find the function value at x  2 . 2 1 3 f  2     2  3  2  4  0 . 3 2 The critical point on the graph is  2, 0  . c) Apply the Second Derivative test to the critical points. For x  0 f ”  x   6 x  6 f ”  0  6 0  6  6  0 The critical point  0, 4  is a relative minimum. For x  2 f ”  x   6 x  6 16 0 4 20 4 17. f  x   3x 4  16 x 3  18 x 2 a) First, find f ‘  x  and f ”  x  . f ‘  x   12 x 3  48 x 2  36 x f ”  x   36 x 2  96 x  36 The domain of f is . f ”  2   6  2   6  6  0 The critical point  2, 0  is a relative maximum. Therefore, f  x  is decreasing over the interval  , 0  , increasing over the interval 0, 2  and then decreasing again over the interval  2,   . d) Find the points of inflection. f ”  x  exists for all real numbers, so we solve the equation f ”  x   0 6 x  6  0 b) Next, find the critical points of f  x  . Since f ‘  x  exists for all real numbers x, the only critical points occur when f ‘  x   0 . f ‘ x  0 3 2 12 x  48 x  36 x  0   12 x x 2  4 x  3  0 12 x  x  1 x  3  0 12 x  0 or x  1  0 or x  3  0 x  0 or x  1 or x3 There are three critical values x  0 , x  1 , and x  3 . Then f  0   3  0   16  0   18  0   0 4 x 1 Therefore, a possible inflection point occurs at x  1 . f 1   1  3 1  4 3 2  1  3  4 .  2 This gives the point 1,  2  on the graph. 3 2 f 1  3 1  16 1  18 1  5 4 3 2 f  3  3  3  16  3  18  3  27 4 3 2 Thus, the critical points  0, 0  , 1, 5 , and  3, 27  are on the graph. e) To determine concavity, we use the possible inflection point to divide the real number Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 255 c) Apply the Second Derivative test to the critical points. A: Test 0 : f ”  0  36  0  96  0  36  36  0 2 f ”  0   36  0   96  0   36  36  0 2 B: Test 1: The critical point  0, 0  is a relative f ” 1  36 1  96 1  36  24  0 2 minimum. C: Test 3: f ” 1  36 1  96 1  36  24  0 2 f ”  3  36  3  96  3  36  72  0 2 The critical point 1, 5 is a relative Then, f  x  is concave up on the interval maximum.  , 0.451 and concave down on the interval  0.451, 2.215 and concave up on the interval  2.215,   , so  0.451, 2.321 and  2.215, 13.358 are inflection points. f ”  3  36  3  96  3  36  72  0 2 The critical point  3, 27  is a relative minimum. We use the critical values 0,1, and 3 to divide the real line into four intervals, A :  , 0  , B:  0,1 , C: 1, 3 and D:  3,  , we know from the extrema above, that f  x  is decreasing over the intervals f) We use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. x f x  , 0 and 1, 3 and f  x  increasing over the intervals  0,1 and  3,   . d) Find the points of inflection. f ”  x  exists 1 2 4 37 8 32 for all real numbers, so we solve the equation f ”  x   0 . f ”  x   0 2 18. 36 x  96 x  36  0   12 3x 2  8 x  3  0 f  x   3x 4  4 x 3  12 x 2  5 a) 3x 2  8 x  3  0 Using the quadratic formula, we find that 4 7 , so x  0.451 or x  2.215 are x 3 possible inflection points. f  0.451  2.321 f  2.215  13.358 f ‘  x   12 x 3  12 x 2  24 x f ”  x   36 x 2  24 x  24 The domain of f is . b) f ‘  x  exists for all real numbers. Solve f ‘  x  0 12 x 3  12 x 2  24 x  0   12 x x 2  x  2  0 So,  0.451, 2.321 and  2.215, 13.358 are two more points on the graph. e) To determine concavity, we use the possible inflection point to divide the real number line into three intervals A :  , 0.451 , B:  0.451, 2.215 , and C:  2.215,   . We test a point in each interval to determine the sign of the second derivative. 12 x  x  2  x  1  0 12 x  0 or x  2  0 x 1  0 or x  0 or x  2 or x 1 There are three critical values x  2 , x  0 , and x  1 . Then f  2   3  2   4  2   12  2   5 4 3 2  27 f  0   3  0   4  0   12  0   5  5 4 3 2 f 1  3 1  4 1  12 1  5  0 4 3 2 Thus, the critical points  2, 27  ,  0,5 , and 1, 0  are on the graph. Copyright © 2016 Pearson Education, Inc. 256 Chapter 2: Applications of Differentiation f) We use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. c) Apply the Second Derivative test to the critical points. f ”  2   36  2   24  2   24  72  0 2 The critical point  2, 27  is a relative minimum. f ”  0   36  0   24  0   24  24  0 2 The critical point  0,5 is a relative maximum. x f x 3 1 2 32 8 37 f ” 1  36 1  24 1  24  36  0 2 The critical point 1, 0  is a relative minimum. Then f  x  is decreasing over the intervals  , 2  and  0,1 and f  x  increasing over the intervals  2, 0  and 1,   . d) Find the points of inflection. f ”  x  exists for all real numbers, so we solve the equation f ”  x   0 36 x 2  24 x  24  0 Using the quadratic formula, we find that 1  7 , so x  1.215 or x  0.549 are x 3 possible inflection points. f  1.215  13.358 f  0.549   2.321 So,  1.215, 13.358  and  0.549, 2.321 are two more points on the graph. e) To determine concavity, we use the possible inflection points to divide the real number line into three intervals A :  , 1.215 , B:  1.215, 0.549  ,and C:  0.549,   . 19. f  x  x4  6×2 a) First, find f ‘  x  and f ”  x  . f ‘  x   4 x 3  12 x f ”  x   12 x 2  12 The domain of f is . b) Find the critical points of f  x  . Since f ‘  x  exists for all real numbers x, the only critical points occur when f ‘  x   0 . f ‘  x  0 3 4 x  12 x  0   4x  0 or x0 or 4 x x2  3  0 C: Test 1: f ” 1  36  0 Then, f  x  is concave up on the interval  , 1.215 and concave down on the interval  1.215, 0.549  and concave up on the interval 0.549,   , so  1.215, 13.358 and 0.549, 2.321 are inflection points. x 3 There are three critical values  3, 0, and 3. Then    f  3   3 We test a point in each interval A: Test  2 : f ”  2   72  0 B: Test 0 : f ”  0  24  0 x2  3  0   6  3  4 2  9  6  3  9 f 0   0   6 0   0 4 f 2  3   3  6 3 4 2  9  6  3  9   Thus, the critical points  3, 9 ,  0, 0  ,  3, 9 and are on the graph. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 257 A: Test  2 : c) Apply the Second Derivative test to the critical points.    f ”  3  12  3 f ”  2   12  2   12  36  0 2   12 2 B: Test 0 :  12  3  12  24  0  f ”  0  12  0  12  12  0 2  The critical point  3, 9 is a relative C: Test 2: f ”  2   12  2   12  36  0 2 minimum. f ”  0   12  0   12  12  0 2 Then, f  x  is concave up on the intervals The critical point  0, 0  is a relative  , 1 and 1,   and concave down on the interval  1,1 , so  1, 5 and 1,  5 maximum. f ”  3   12  3   12 2 are inflection points. f) We use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed.  12  3  12  24  0 The critical point  3, 9 is a relative minimum. If we use the critical values  3, 0, and 3 to divide the real line into four intervals,       A : ,  3 , B:  3, 0 , C: 0, 3 , and D:  3, Then f  x  is decreasing over the intervals ,  3  and 0, 3  , and f  x  20. increasing over the intervals  3,0 and  3,  . for all real numbers, so we solve the equation f ”  x   0 12 x 2  12  0 f x 3 2 2 3 27 8 8 27 f  x  2 x2  x4 a) d) Find the points of inflection. f ”  x  exists x f ‘  x   4 x  4 x3 f ”  x   4  12 x 2 The domain of f is . b) f ‘  x  exists for all real numbers. Solve f ‘  x  0 4 x  4 x3  0 x2  1  0   4 x 1  x2  0 x2  1 x  1 So x  1 or x  1 are possible inflection points. f  1   1  6  1  1  6  5 4 2 f 1  1  6 1  1  6  5 4 2 . So,  1, 5 and 1,  5 are two more points on the graph. e) To determine concavity, we use the possible inflection point to divide the real number line into three intervals A :  , 1 , B:  1,1 , and C: 1,   . We test a point in each interval 4 x 1  x 1  x   0 4x  0 or 1  x  0 or 1  x  0 x  0 or x  1 or x  1 There are three critical values x  1 , x  0 , and x  1 . Then f  1  2  1   1  1 2 4 f 0   2 0   0   0 2 4 f 1  2 1  1  1 2 4 Thus, the critical points  1,1 ,  0, 0  , and 1,1 are on the graph. Copyright © 2016 Pearson Education, Inc. 258 Chapter 2: Applications of Differentiation Therefore, f  x  is concave down on the c) Apply the Second Derivative test to the critical points.   1  1  ,   and intervals  ,   and   3 3  f ”  1  4  12  1  8  0 2 The critical point  1,1 is a relative f  x  is concave up on the interval maximum.  1 5  1 1   1 5 , ,  and  ,    , so    3 9  3 3 3 9 are inflection points. f) We use the preceding information to sketch the graph of the function. Additional function values can also be calculated as needed. x f x f ”  0   4  12  0   4  0 2 The critical point  0, 0  is a relative minimum. f ” 1  4  12 1  8  0 2 The critical point 1,1 is a relative maximum. Then f  x  is increasing over the intervals for all real numbers, so we solve the equation f ”  x   0 4  12 x 2  0 1 x2  3  1 5  1 5 ,  and  , are two more So,     3 9  3 9 points on the graph. e) To determine concavity, we use the possible inflection points to divide the real number  1  line into three intervals A :  ,  ,  3  1 1   1  B:   , , and C:  , .    3  3 3 We test a point in each interval A: Test  1: f ”  1  8  0 C: Test 1: f ” 1  8  0 21. f  x   x3  6 x2  9 x  1 a) First, find f ‘  x  and f ”  x  . f ‘  x   3x 2  12 x  9 1 1 x  3 3 1 1 there exists So at x   and x  3 3 possible inflection points.  1  5 f    3 9 .  1  5 f   3  9 B: Test 0 : f ”  0  4  0 8 1 0 1 8 2 1 0 1 2  , 1 and 0,1 , and f  x  is decreasing over the intervals  1, 0  and 1,   . d) Find the points of inflection. f ”  x  exists f ”  x   6 x  12 The domain of f is . b) f ‘  x  exists for all values of x, so the only critical points of f are where f ‘  x   0 . f ‘  x  0 2 3 x  12 x  9  0   3 x2  4 x  3  0 3  x  1 x  3  0 x  1  0 or x  3  0 x  1 or x3 The critical values are 1 and 3 . Then, f 1  1  6 1  9 1  1  5 3 2 f  3   3  6  3  9  3  1  1 3 2 So, 1, 5 and 3,1 are on the graph. c) Applying the Second Derivative Test, we have: f ” 1  6 1  12  6  0 f ”  3  6  3  12  6  0 Therefore, 1, 5 is a relative maximum and 3,1 is a relative minimum. The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 259 If we use the points 1 and 3 to divide the real number line into three intervals, ,1 , 1,3 , and 3,  . Using the extrema, we know f  x  is increasing on the intervals   ,1 and  3,   , and f  x  is decreasing on the interval 1, 3 . d) Find the points of inflection. f ”  x  exists for all real numbers. Solve f ”  x   0 6 x  12  0 6 x  12 f 2  2  6 2  9 2  1  3 2 Thus, the point  2, 3 on the graph is a possible inflection point. e) To determine concavity, we use 2 to divide the real number line into two intervals, A:  , 2  and B:  2,   . Then test a point in each interval. A: Test 0, f ”  0  6 0   12  12  0 B: Test 3, f ”  3  6  3  12  6  0 Thus, f  x  is concave down on the interval  , 2  and concave up on the interval  2,   and  2, 3 is an inflection point. f) We sketch the graph. Additional points may be found as necessary. 22. x f x 2 1 0 4 5 49 15 1 5 21 2  2  2  2  2 f       2    4    3  3  3  3  3 8 8 8    3 27 9 3 121  27 f  2   2  2  2  4  2  3  5 2  2 121  The critical points   , and  2, 5  3 27  are on the graph. c) Applying the Second Derivative Test, we have:  2  2 f ”     6     4  4  4  8  0  3  3  2 121    ,  is a relative maximum. 3 27  f ”  2   6  2   4  12  4  8  0  2, 5 is a relative minimum. 2  Therefore f  x  is increasing on  ,    3 and on  2,   and f  x  is decreasing on  2    , 2  . 3 inflection points occur when f ”  x   0 . 6x  4  0 4 2 x  6 3 There is a possible inflection point at x  3 2 2 2 2 2 f       2   4  3 3 3 3 3 f ”  x   6 x  4 The domain of f is . b) f ‘  x  exists for all real numbers. Solve 3x  2 x  2  0 3 2 and 2 . 3 for all values of f  x  , so the only possible f ‘  x   3x 2  4 x  4 3x 2  4 x  4  0 The critical values are  d) Find the points of inflection. f ”  x  exists f  x   x3  2 x2  4 x  3 a) or x  2  0 2 x or x  2 3 3 x2 There is a possible inflection point at x  2 . 3 3x  2  0 8 8 8   3 27 9 3 7  27  7  2 Another point on the graph is  ,   .  3 27  Copyright © 2016 Pearson Education, Inc. 2 . 3 260 Chapter 2: Applications of Differentiation 2 to divide 3 the real number line into two intervals, 2  2  A :  ,  and B:  ,   . Then test a   3  3 point in each interval. A: Test 0, f ”  0  6  0  4  4  0 e) To determine concavity we use B: Test 1, f ” 1  6 1  4  2  0 2  We see that f is concave down on  ,   3 7  2  2 and concave up on  ,   , so  ,   is 3   3 27  an inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. 23. x f x 2 1 0 1 3 4 5 4 3 2 0 19 2 The Second Derivative Test fails, we will have to use the First Derivative Test for 3  x  0 . Divide  ,  into two intervals,  2  3 A:  ,0  and B:  0,  , and test a point  2 in each interval. A: Test  1, f ‘  1  4  1  6  1  10  0 3 2 B: Test 1, f ‘ 1  4 1  6 1  2  0 Since, f is decreasing on both intervals, 0, 0  is not a relative extremum. 3 2 2  3  3  3 f ”    12    12    9  0 2 2 2  3 27  Therefore,  ,   is a relative minimum.  2 16  a) First, find f ‘  x  and f ”  x  . Thus, f  x  is increasing on the interval f ‘  x   4 x3  6 x2 3   ,   , and f  x  is decreasing on the 2 f ”  x   12 x 2  12 x The domain of f is . b) f ‘  x  exists for all values of x, so the only critical points of f are where f ‘  x   0 . f ‘  x  0 4 x3  6 x2  0 12 x  0 The critical values are 0 and 3 . 2 f  0   0  2  0  0 3 4 d) Find the points of inflection. f ”  x  exists for all real numbers. Solve f ”  x   0 12 x  x  1  0 or 2x  3  0 3 or x 2 4  3 intervals  ,0 and  0,  .  2 12 x 2  12 x  0 2 x 2  2 x  3  0 x0 f ”  0   12  0   12  0   0 We use the Second Derivative Test for 3 x . 2 f  x   x4  2 x3 2 x2  0  3 27  So,  0, 0  and  ,   are on the graph.  2 16  c) Applying the Second Derivative Test, we have: 3 or x  0 or x 1 There are a possible inflection points at x  0 and x  1 . f 0  0 found earlier f 1  1  2 1  1 4 27  3  3  3 f      2    2 2 2 16 x 1  0 3 Thus, the points  0, 0  and 1,  1 on the graph are possible inflection points. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 261 e) Use 0 and 1 to divide the real number line into two intervals, A:  , 0  , B: 0,1 , and C: 1,   . Then test a point in each interval. A: Test  1, f ”  1  12 1  12  1  24  0 2 1 , 2 B: Test 2 1 1 1 f ”    12    12    3  0 2 2 2 C: Test 2, f ” 2  12 2  12 2  24  0 2 Thus, f  x  is concave up on the intervals  , 0  and 1,   and concave down on the interval  0,1 . Also, the points  0, 0  and 1, 1 are inflection points. f) Using the preceding information, we sketch the graph. Additional points may be found as necessary. 24. x f x 2 1 2 3 32 3 0 27 f ”  1  36  1  24  1  36  24 2  12  0 So  1, 1 is a relative minimum. f ”  0   36  0   24  0   0 2 The test fails. We will use the First Derivative Test. We use 0 to divide the interval  1,   into two intervals, A:  1, 0  and B:  0,   , and test a point in each interval. 1 A: Test  , 2 3  1  1  1 f ‘     12     12     2  2  2  2 3 0 2 B: Test 2, f ‘  2   12  2   12  2   144  0 f is increasing on both intervals  1, 0 and 0,   . Therefore, 0, 0  is not a 3 2 relative extremum. Since  1, 1 is a relative minimum, we know that f is decreasing on   , 1 . d) Find the points of inflection. f ”  x  exists for all values of x, so the only possible inflection points occur when f ”  x   0 . f  x   3x 4  4 x 3 a) c) Applying the Second Derivative Test, we have: f ”  x   0 f ‘  x   12 x 3  12 x 2 36 x 2  24 x  0 12 x  3x  2   0 f ”  x   36 x 2  24 x 12 x  0 The domain of f is . b) f ‘  x  exists for all real numbers. Solve 12 x 3  12 x 2  0 12 x 2  x  1  0 12 x 2  0 or x  1  0 or 3x  2  0 2 3 There are a possible inflection points at 2 x   and x  0 . 3 x0 x or 4 3 x  0 or x  1 The critical values are 1 and 0 . 16  2  2  2 f    3   4      3  3  3 27 f  1  3  1  4  1  1 f 0   3 0   4 0  0 f 0   3 0   4 0  0  2 16  This gives one additional point   ,    3 27  on the graph. 4 4 3 3 The critical points  1, 1 and  0, 0  are on the graph. 4 Copyright © 2016 Pearson Education, Inc. 3 262 Chapter 2: Applications of Differentiation 2 and 0 to 3 divide the real number line into three 2   2  intervals, A :  ,   , B:   ,0  , and    3  3 C:  0,   . Then test a point in each interval. e) To determine concavity we use  A: Test  1, f ”  1  36  1  24  1  12  0 2 1 B: Test  , 2 2  1  1  1 f ”     36     24     2  2  2  3  0 C: Test 1, f ” 1  36 1  24 1  60  0 2 We see that f is concave up on the intervals 2   ,   and  0,   , and concave down on 3  2   2 16  the interval   , 0  , so both   ,    3   3 27  and  0, 0  are inflection points. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x 2 1 2 25. 16 7 80 3 2  125  150  675  400 f  9    9   6  9   135  9  3 2  729  486  1215  972 The critical points  5, 400  and  9, 972  are on the graph. c) Applying the Second Derivative Test, we have: f ”  5  6  5  12  30  12  42  0 The critical point  5, 400  is a relative maximum. f ”  9   6  9   12  54  12  42  0 The critical point  9, 972  is a relative minimum. If we use the points 5 and 9 to divide the real number line into three intervals  , 5 ,  5, 9  , and 9,  we see that f  x  is increasing on the intervals  , 5 and 9,   and f  x  is decreasing on the interval  5, 9  . d) Find the points of inflection. f ”  x  exists set the second derivative equal to zero and find possible inflection points. 6 x  12  0 6 x  12 x2 The only possible inflection point is 2 . f ‘  x   3x 2  12 x  135 f ”  x   6 x  12 f  2    2   6  2   135  2  3 The domain of f is . b) f ‘  x  exists for all values of x, so the only critical points of f are where f ‘  x   0 . 3x 2  12 x  135  0 2  8  24  270  286 The point  2, 286  is a possible inflection point on the graph. e) To determine concavity we use 2 to divide the real number line into two intervals, A :  , 2  and B:  2,   , Then test a point x 2  4 x  45  0  x  9 x  5  0 x9 0 f  5   5  6  5  135  5 for all values of x, so the only possible inflection points occur when f ”  x   0 . We f  x   x 3  6 x 2  135 x a) The function values are or x  5  0 x9 or x  5 The critical values are 5 and 9 . in each interval at the top of the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 263 Therefore, f  x  is increasing on the Using the information from the previous page, we test a point in each interval. A: Test 0, f ”  0   6  0  12  12  0 intervals  , 6  and 8,   and f  x  is decreasing on the interval  6,8 . B: Test 3, f ”  3  6  3  12  6  0 We see that f is concave down on the interval  , 2  and concave up on the d) Find the inflection points. f ”  x  exists for all real numbers. Solve: f ”  x   0 interval  2,   , Therefore  2, 286  is an 6x  6  0 inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x The possible inflection point 1, 286  is on the graph. e) To determine concavity, use 1 to divide the real number line into two intervals, A :  ,1 and B: 1,   and test a point in 572 250 324 0 286 700 0 400  11 10 3 0 2 5 15 16 26. x 1 f 1  286 each interval. A: Test 0, f ”  0  6  0  6  6  0 B: Test 2, f ”  2   6  2   6  6  0 Therefore, f  x  is concave down on the interval   ,1 and concave up on the f ‘  x   x 3  3x 2  144 x  140 a) interval 1,   . The point 1, 286  is an f ‘  x   3x  6 x  144 2 inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x f ”  x   6 x  6 The domain of f is . b) f ‘  x  exists for all real numbers. Solve: f ‘ x  0  12 10 3 0 4 14 15 2 3 x  6 x  144  0 x 2  2 x  48  0  x  6 x  8  0 x6 0 or x  8  0 x  6 or f  6  400 x8 Critical Values f 8  972 So the critical points  6, 400  and 8, 972  are on the graph. c) Using the Second Derivative Test, we have: f ”  6  42  0 27. 572 0 238 140 700 0 400 f  x   x 4  4 x 3  10 a) f ‘  x   4 x 3  12 x 2 f ”  x   12 x 2  24 x The domain of f is . f ” 8  42  0 So,  6, 400  is a relative maximum and 8, 972  is a relative minimum. Copyright © 2016 Pearson Education, Inc. 264 Chapter 2: Applications of Differentiation b) f ‘  x  exists for all values of x, so the only Possible points of inflection occur at x  0 and x  2. critical points of f are where f ‘  x   0 . f  0   0  4  0  10  10 4 4 x 3  12 x 2  0 f  2    2   4  2   10  6 4 4 x 2  x  3  0 4 x2  0 inflection points on the graph. e) To determine concavity we use 0 and 2 to divide the real number line into three intervals, A :  , 0  , B:  0, 2  , and C:  2,   , Then x0 or x3 The critical values are 0 and 3 . f  0   0  4  0  10  10 3 f  3   3  4  3  10  17 4 3 The points  0,10  and  2, 6  are possible or x  3  0 4 3 3 test a point in each interval. A: Test  1, The critical points  0,10  and  3, 17  are on the graph. c) Applying the Second Derivative Test, we have: f ”  1  12  1  24  1  36  0 2 B: Test 1, f ”  0   12  0   24  0   0 2 f ” 1  12 1  24 1  12  0 2 The test fails, we will use the First Derivative Test. Divide  , 3 into two intervals, C: Test 3, f ”  3  12  3  24  3  36  0 We see that f is concave up on the intervals  , 0 and  2,   and concave down on 2 A:   , 0  and B:  0, 3 , and test a point in each interval. A: Test  1, the interval  0, 2  . Therefore both  0,10  and  2, 6  are inflection points. f ‘  1  4  1  12  1  16  0 3 2 f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x B: Test 1, f ‘ 1  4 1  12 1  8  0 3 2 Since, f is decreasing on both intervals, 0,10  is not a relative extremum. We use the Second Derivative Test for x3. 2 1 1 4 5 f ”  3  12  3  24  3  36  0 2 The critical point  3, 17  is a relative minimum. When we applied the First Derivative Test, we saw that f  x  was decreasing on the intervals   , 0  and  0, 3 . Since  3, 17  is a relative minimum, we know that f  x  is increasing on  3,   . d) Find the points of inflection. f ”  x  exists for all values of x, so the only possible inflection points occur when f ”  x   0 . 12 x 2  24 x  0 12 x  x  2   0 12 x  0 or x  2  0 x  0 or x2 28. 58 15 7 10 135 4 3 x  2 x2  x 3 f ‘  x   4 x2  4 x  1 f  x  a) f ”  x   8 x  4 The domain of f is . b) f ‘  x  exists for all real numbers. Solve: f ‘  x  0 2 4x  4x 1  0  2 x  12  0 2x 1  0 1 x 2 Copyright © 2016 Pearson Education, Inc. Critical Value Exercise Set 2.2 265 Therefore, f  x  is concave down on the We evaluate the function at the critical value. 1 1 f   2 6 1  interval  ,  and concave up on the  2 1  1 1 interval  ,   . The point  ,  is an 2  2 6 inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. 1 1 So the critical point  ,  is on the graph. 2 6 c) Using the Second Derivative Test, we have: 1 f ”    0 and the test fails. We will use 2 the First Derivative Test. Divide the real number line into two 1  1  intervals, A:  ,  and B:  ,   , and  2  2 test a point in each interval. A: Test 0, f ‘ 0  4  0  4 0   1  1  0 2 B: Test 1, f ‘ 1  4 1  4 1  1  1  0 2 Since, f is increasing on both intervals, 1 1  ,  is not a relative extremum. 2 6 However, we now know that f  x  is increasing over the intervals 1  1   ,  and  ,   . 2 2 d) Find the inflection points. f ”  x  exists for all real numbers. Solve: f ”  x   0 29. x f x 6 3 1 366 57  13 3 0 1 0 3 6 21 222 1 3 f  x   x 3  6 x 2  12 x  6 a) f ‘  x   3x 2  12 x  12 f ”  x   6 x  12 The domain of f is . b) f ‘  x  exists for all values of x, so the only critical points of f are where f ‘  x   0 . 8x  4  0 3x 2  12 x  12  0 1 2 1 1 f   2 6 x2  4 x  4  0 x Dividing by 3  x  2  0 2 x20 1 1 The possible inflection point is  ,  . 2 6 1 to divide the 2 real number line into two intervals, 1  1  A:  ,  and B:  ,   and test a point  2  2 in each interval. A: Test 0, f ”  0  8  0  4  4  0 e) To determine concavity, use B: Test 1, f ” 1  8 1  4  4  0 x2 The critical value is 2 . f  2    2   6  2   12  2   6  2 3 2 The critical point  2, 2  is on the graph. c) Applying the Second Derivative Test, we have: f ”  2   6  2   12  0 The test fails, we will use the First Derivative Test. Divide the real line into two intervals, A:  , 2  and B:  2,   , and test a point in each interval on the next page. Copyright © 2016 Pearson Education, Inc. 266 Chapter 2: Applications of Differentiation b) f ‘  x  exists for all real numbers, and the Testing a point in each interval, we have: A: Test 0, equation f ‘  x   0 has no real solution. f ‘  0  3  0  12  0  12  12  0 2  3x 2  3  0 for all x  . Thus, f  x  has no   critical points. c) Since f ‘  x   0 for all real numbers, f  x  B: Test 3, f ‘  3  3  3  12  3  12  3  0 Since, f is increasing on both intervals,  2, 2  is not a relative extremum. 2 is increasing over the entire domain  ,   . d) Find the points of inflection. Since f ”  x  When we applied the First Derivative Test, we saw that f  x  was increasing on the exists for all real numbers, solve: f ”  x   0 intervals  , 2  and  2,   . 6x  0 d) Find the points of inflection. f ”  x  exists x0 f 0   1 for all values of x, so the only possible inflection points occur when f ”  x   0 . 6 x  12  0 6 x  12 x2 We have already seen that f  2   2 ,so So the point  0,1 is a possible inflection point on the graph. e) To determine concavity, we use 0 to divide the real number line into two intervals. A :  , 0  and B:  0,   , Then test a point the point  2, 2  is a possible inflection point in each interval. A: Test  1, f ”  1  6  1  6  0 on the graph. e) To determine concavity we use 2 to divide the real number line into two intervals, A :  , 2  and B:  2,   , Then test a point B: Test 1, We see that f  x  is concave down on the interval  , 0  and concave up on the in each interval. A: Test 0, f ”  0  6  0  12  12  0 interval  0,   . Therefore, the point  0,1 is an inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. B: Test 3, f ” 3  6  3  12  6  0 We see that f  x  is concave down on the interval  , 2  and concave up on the interval  2,   . Therefore, the point  2, 2  is an inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x 1 0 1 3 4 25 6 1 3 10 f ” 1  6 1  6  0 31. x f x 2 1 1 2 13 3 5 15 f  x   5 x 3  3x 5 a) f ‘  x   15 x 2  15 x 4 f ”  x   30 x  60 x 3 30. f  x   x 3  3x  1 a) The domain of f is . f ‘  x   3x 2  3 f ”  x   6 x The domain of f is . Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 267 b) f ‘  x  exists for all values of x, so the only critical points of f are where f ‘  x   0 . 15 x 2  15 x 4  0  decreasing on   , 1 . Since 1, 2  is a or 1  x 2  0 x0 or x  1 The critical values are 1, 0, and 1 . f  1  5  1  3  1  2 3 5 3 5 f 1  5 1  3 1  2 5 are on the graph. c) Applying the Second Derivative Test, we have: f ”  1  30  1  60  1  30  0 3 So, the critical point  1, 2  is a relative minimum. 3 The test fails, we will use the First Derivative Test. Divide  1,1 into two intervals, A:  1, 0  and B:  0,1 , and test a point in each interval. 1 A: Test  , 2 2  1  1  1 f ‘     15     15     2  2  2 4 45 0 16 1 , 2 2 1 1 1 f ‘    15    15   2 2 2 4 45  0 16 Since, f is increasing on both intervals, 0, 0  is not a relative extremum. We use the Second Derivative Test for x  1 . f ” 1  30 1  60 1  30  0 3 The critical point 1, 2  is a relative maximum. d) Find the points of inflection. f ”  x  exists   30 x  0 or 1  2×2  0 x0 or x2  x0 or x 30 x 1  2 x 2  0 1 2 1 1  2 2 3 f ”  0   30  0   60  0   0 B: Test decreasing on 1,   . 30 x  60 x 3  0 The critical points  1, 2  ,  0, 0  and 1, 2   relative maximum, we know that f  x  is for all values of x, so the only possible inflection points occur when f ”  x   0 . f  0  5  0  3  0  0 3 intervals  1, 0  and  0,1 . Since  1, 2  is a relative minimum, we know that f  x  is  15 x 2 1  x 2  0 15 x 2  0 When we applied the First Derivative Test, we saw that f  x  was increasing on the  1   1   1  f    5     3     2 2 2 5  1.237 f  0  5  0  3  0  0 3 5 3  1   1   1  f  5  3  2   2   2  5  1.237  1  , 1.237  ,  0, 0  and The points     2  1  ,1.237  are possible inflection points   2 on the graph. e) To determine concavity we use 1 1 to divide the real number  , 0, and 2 2  1  line into four intervals, A :  ,  ,  2  1   1  B:   ,0  , C:  0,  , and   2  2  1  D:  ,  .  2  Then test a point in each interval on the next page. Copyright © 2016 Pearson Education, Inc. 268 Chapter 2: Applications of Differentiation Testing the point in each interval, we have f ”  1  30  1  60  1 3 A: Test  1,  30  0  3 15 0 2 3 15 0 2 3  30  0 We see that f is concave up on the intervals  1   1   ,   and  0,  and concave 2 2 down on the intervals  1   1  ,0  and  ,  . Therefore, the    2  2   1  , 1.237  ,  0, 0  and points     2  1  ,1.237  are inflection points.   2 f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x 32. 2 56 f  x   20 x 3  3x 5 a) x  2 f ‘  x   60 x 2  15 x 4 f ”  x   120 x  60 x 3 The domain of f is . So the critical points  2, 64  ,  0, 0  , and  2, 64  are on the graph. c) Applying the Second Derivative Test, we have: 3 f ” 1  30 1  60 1 17 32 x  0 or f  2   64 f ”  2   120  2   60  2   240  0 D: Test 1 , 1 2 or 4  x 2  0 f  2   64 1 1 1 f ”    30    60   2 2 2 56  17 32 15 x 2  0 f 0  0 1 C: Test , 2 2  12  15 x 2 4  x 2  0  1  1  1 f ”     30     60     2  2  2  f ‘  x  0 60 x 2  15 x 4  0 1 B: Test  , 2  b) f ‘  x  exists for all real numbers. Solve: So, the critical point  2, 64  is a relative minimum. Testing the remaining critical values we have: f ”  0   120  0   60  0   0 3 The test fails, we will use the First Derivative Test. Divide  2, 2  into two intervals, A:  2, 0  and B:  0, 2  , and test a point in each interval. A: Test  1, f ‘  1  60  1  15  1  45  0 2 4 B: Test 1, f ‘ 1  60 1  15 1  45  0 Since, f is increasing on both intervals, 0, 0  is not a relative extremum. We use the Second Derivative Test for x  2. 2 4 f ”  2   120  2   60  2   240  0 3 The critical point  2, 64  is a relative maximum. c) When we applied the First Derivative Test, we saw that f  x  was increasing on the intervals  2, 0  and  0, 2  . Since  2, 64  is a relative minimum, we know that f  x  is decreasing on  , 2  . Since  2, 64  is a relative maximum, we know that f  x  is decreasing on  2,   . Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 269 d) Find the inflection points. f ”  x  exists for all real numbers. Solve: 120 x  60 x 3  0  f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x  60 x 2  x 2  0 3 1 1 3 2 60 x  0 or 2 x  0 x0 or x2  2 x0 or x 2   f  2  28 2 33. f 0  0  2   28 2 The points   2 , 28 2  ,  0, 0  and  2, 28 2  are possible inflection points on f the graph. e) To determine concavity we use  2, 0, and 2 to divide the real number line into four intervals,       2,  .  f ”  2   120  2   60  2  3  240  0 B: Test  1, f ”  1  120  1  60  1 3  60  0 C: Test 1, f ‘  x   2 x  6 x2  4 x3 f ”  x   2  12 x  12 x 2 b) f ‘  x  exists for all values of x, so the only critical points of f are where f ‘  x   0 . 2 x  6 x 2  4 x3  0 2x  0 or 1  2 x  0 x0 or x or 1  x  0 1 2 or The critical values are 0, f  0    0  1   0    0 x 1 1 , and 1 . 2 2 2 2 1  1  1   1 f      1       2  2   2 16 2 2 3 f ”  2   120  2   60  2  a) f 1  1 1  1   0 f ” 1  120 1  60 1  60  0 D: Test 2 , 2 2 x 1  x 1  2 x   0 Then test a point in each interval. A: Test  2, f  x   x 2 1  x   x 2  2 x 3  x 4 2 x 1  3 x  2 x 2   0 A : ,  2 , B:  2, 0 , C: 0, 2 , and D: 189 17 17 189 3  240  0 We see that f is concave up on the intervals ,  2  and 0, 2  and concave down on the intervals   2, 0 and  2,   . 2 1 1  The critical points  0,0  ,  ,  , and  2 16  1,0 are on the graph. c) Applying the Second Derivative Test, we have: f ”  0   2  12  0   12  0   2  0 2 So, the critical point  0,0  is a relative minimum.   Therefore, the points  2 , 28 2 ,  0, 0  and  2, 28 2  are inflection points. 2  1  1  1 f ”    2  12    12    1  0  2  2  2 1 1  So, the critical point  ,  is a relative  2 16  maximum. f ” 1  2  12 1  12 1  2  0 2 So, the critical point 1,0  is a relative minimum. The solution is continued. Copyright © 2016 Pearson Education, Inc. 270 Chapter 2: Applications of Differentiation Therefore, the points  0.211,0.028 and 1 , and 1 to divide the 2 real number line into four intervals, 1 1  ,0 ,  0,  ,  ,1 , and 1,   , we 2 2 We use the points 0,  0.789,0.028 are inflection points. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f  x know that f  x  is decreasing on the 2 1 2 3 1  intervals  ,0  and  ,1 , and f  x  is 2  increasing on the intervals  1  0, 2  and 1,   . d) Find the points of inflection. f ”  x  exists for all values of x, so the only possible inflection points occur when f ”  x   0 . 34. f  x   x 2 3  x   2 1  6 x  6 x2  0 Dividing by 2 Using the quadratic formula we have: a) 3  4  9x  6x  x f ‘  x   18 x  18 x 2  4 x 3 f ”  x   18  36 x  12 x 2 3 3 x 6 x  0.211 or x  0.789 f  0.211  0.028 The domain of f is . b) f ‘  x  exists for all values of x, so the only critical points of f are where f ‘  x   0 . f  0.789  0.028 18 x  18 x 2  4 x 3  0 The points,  0.211,0.028 and  0.789,0.028 are possible inflection points on the graph. e) To determine concavity we use 0.211 and 0.789 to divide the real number line into three intervals, A :  ,0.211 , B:  0.211, 0.789 , and C:  0.789,   Then test a point in each interval. 2x  0 or 3  2 x  0 x0 or 2 x 3  2 x 3  x   0 x or 3  x  0 3 2 or The critical values are 0, x3 3 , and 3 . 2 2 2 2 81  3   3    3  f      3       2   2    2  16 2 20 2 2 20 We see that f is concave up on the intervals  ,0.211 and  0.789,  and concave down on the interval  0.211,0.789 .  2 f ”  0   2  12  0   12  0  f ” 1  2  12 1  12 1  2 x 9  9 x  2 x2  0 f  0   0  3   0  0 1  1  1  1 B: Test , f ”    2  12    12    2  2  2 2  1  0 C: Test 1, 2  x2 9  6x  x2 2  12 x  12 x 2  0 A: Test 0, 36 4 4 36 f  3   3  3   3  0 2 2  3 81  The critical points  0, 0  ,  ,  ,and  2 16  3, 0  are on the graph. c) Applying the Second Derivative Test, we have: f ”  0   18  36  0   12  0   18  0 2 So, the critical point  0, 0  is a relative minimum. The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 271 Then test a point in each interval. A: Test 0, Applying the Second Derivative test: 2  3  3  3 f ”    18  36    12    9  0 2 2 2 f ”  0   18  36  0   12  0   3 81  So, the critical point  ,  is a relative  2 16  maximum.  18  0 B: Test 1, f ”  3  18  36  3  12  3  18  0  6  0 C: Test 3, f ” 1  18  36 1  12 1 2 So, the critical point  3, 0  is a relative 3 , and 3 to divide the 2 real number line into four intervals, 3 3  ,0 ,  0,  ,  ,3 , and 3,   . We 2 2 know that f  x  is decreasing on the down on the interval  0.634,2.366  . Therefore, the points  0.634,2.250  and  2.366,2.250 are inflection points. 3  intervals  ,0 and  ,3 , and f  x  is 2  increasing on the intervals  3  0,  and  3,   . 2 f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. d) Find the points of inflection. f ”  x  exists for all values of x, so the only possible inflection points occur when f ”  x   0 . 18  36 x  12 x 2  0 35. x f  x 2 1 1 2 4 100 16 4 4 16 f  x    x  1 3 2 a) f ‘  x  1 2 2  x  1  3  1 3 3  x  1 3 f  2.366   2.250 f ”  x     2.366,2.250 are possible inflection points The domain of f is . The points,  0.634,2.250  and on the graph. e) To determine concavity we use 0.634and 2.366 to divide the real number line into three intervals, A :  ,0.634  , B:  0.634,2.366  , and C:  2.366,   2  18  0 We see that f is concave up on the intervals  ,0.634 and  2.366,  and concave We use the points 0, 3  6 x  2×2  0 Dividing by 6 Using the quadratic formula we have: 3 3 x 2 x  0.634 or x  2.366 f  0.634   2.250 2 f ”  3  18  36  3  12  3 minimum. 2 4 2 2  x  1  3   4 9 9  x  1 3 b) f ‘  x  does not exist for x  1 . The equation f ‘  x   0 has no solution, therefore, x  1 is the only critical point. f 1  1  1 3  0 . 2 So, the critical point, 1,0  is on the graph. c) We apply the First Derivative Test. We use 1 to divide the real number line into two intervals A :  ,1 and B: 1,  and then we test a point in each interval at the top of the next page. Copyright © 2016 Pearson Education, Inc. 272 Chapter 2: Applications of Differentiation Testing a point in each interval, we have 2 2  0 A: Test 0, f ‘  0   1 3 3 3   0   1 B: Test 2, f ‘  2  2 3   2   1 1 3  2 0 3 1,0 is a relative minimum. f  x  is decreasing on the interval  ,1 and increasing on the interval 1,  . d) Find the points of inflection. f ”  x  does not exist when x  1 . The equation f ”  x   0 has no solution, so x  1 is the only possible inflection point. We know that f 1  0 . e) To determine concavity, we divide the real number line into two intervals, A :  ,1 and B: 1,  and then we test a point in each interval. A: Test 0, f ”  0    B: Test 2, f ”  2    2 9   0   1 4 3 2 9   2   1 4 3  2 0 9  2 0 9 Thus, f  x  is concave down on the intervals  ,1 and 1,  . Therefore, the point 1,0 is not an inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f  x 7 0 2 9 36. 4 1 1 4 equation f ‘  x   0 has no solution, therefore, x  1 is the only critical value. f  1   1  1 3  0 . 2 So, the critical point,  1,0  , is on the graph. c) We apply the First Derivative Test. We use 1 to divide the real number line into two intervals A :  , 1 and B:  1,   and then we test a point in each interval. A: Test  2, 2 2  0 f ‘  2   1 3 3 3   2   1 B: Test 0, f ‘  0  2 3   0   1 1  3 2 0 3 Thus,  1,0  is a relative minimum. We also know that f  x  is decreasing on the interval , 1 and increasing on the interval 1,  . d) Find the points of inflection. f ”  x  does not exist when x  1 . The equation f ”  x   0 has no solution, so x  1 is the only possible inflection point. We know that f  1  0 . e) To determine concavity, we divide the real number line into two intervals, A :  , 1 and B:  1,   and then we test a point in each interval. A: Test  2, f ”  2    2 2 2 1 f ‘  x    x  1 3  1 3 3  x  1 3 2 2 4 f ”  x     x  1 3   4 9 9  x  1 3 The domain of f is . 2 9   2   1 4 3  2 0 9 B: Test 0, f  x    x  1 3 a) b) f ‘  x  does not exist for x  1 . The f ”  0    2 9   0   1 4 3  2 0 9 Thus, f  x  is concave down on the interval  , 1 and on the interval  1,  . Therefore, the point  1,0  is not an inflection point. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 273 f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. 37. x f  x 9 2 0 7 4 1 1 4 A: Test 2, f ”  2    B: Test 4, f ”  4    3 2 9  4   3 5 3  2 0 9  2 0 9  , 3 and f  x  is concave down on the interval  3,   . Therefore, the point  3, 1 is an inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. f  x    x  3 3  1 f ‘  x  9  2   3 5 Thus, f  x  is concave up on the interval 1 a) 2 2 1 1  x  3 3  2 3 3  x  3 3 2 2 5 f ”  x     x  3 3   5 9 9  x  3 3 The domain of f is . b) f ‘  x  does not exist for x  3 . The equation f ‘  x   0 has no solution, therefore, x  3 is the only critical value. f  3   3  3 3  1  1 . 1 So, the critical point,  3, 1 is on the graph. c) We apply the First Derivative Test. We use 3 to divide the real number line into two intervals A :  , 3 and B:  3,   and then we test a point in each interval. 1 1 A: Test 2, f ‘  2    0 2 3 3 3  2   3 B: Test 4, f ‘  4   1 3  4   3 2 3  1 0 3 f  x  is increasing on both intervals  , 3 and 3,   , therefore 3, 1 is not a relative extremum. d) Find the points of inflection. f ”  x  does not exist when x  3 . The equation f ”  x   0 has no solution, so at x  3 is the only possible inflection point. We know that f  3   1 . e) To determine concavity, we divide the real number line into two intervals, A :  , 3 and B:  3,   and then we test a point in each interval. 38. x f x 5 2 4 11 3 2 0 1 f  x    x  2 3  3 1 a) f ‘  x  2 1 1  x  2  3  2 3 3  x  2 3 f ”  x    5 2 2  x  2  3   5 9 9  x  2 3 The domain of f is . b) f ‘  x  does not exist for x  2 . The equation f ‘  x   0 has no solution, therefore, x  2 is the only critical point. f  2   2  2 3  3  3 . 1 So, the critical point,  2, 3 is on the graph. c) We apply the First Derivative Test. We use 2 to divide the real number line into two intervals A :  , 2  and B:  2,   and then we test a point in each interval. 1 1 A: Test 1, f ‘ 1   0 2 3 3 3 1  2  B: Test 3, f ‘  3  1 3  3  2  2 3  1 0 3 f  x  is increasing on both intervals  , 2  and  2,   , therefore  2, 3 is not a relative extremum. Copyright © 2016 Pearson Education, Inc. 274 Chapter 2: Applications of Differentiation d) Find the points of inflection. f ”  x  does not A: Test 3, f ‘  3   exist when x  2 . The equation f ”  x   0 3  4 0 3 B: Test 5, f ‘ 5   determine concavity, we divide the real number line into two intervals, A :  , 2  Thus,  4, 5 is a relative maximum. each interval. the interval  , 4  and decreasing on the A: Test 1, f ” 1   B: Test 3, f ”  3   2 9 1  2  5 3  2 9  3  2  5 3 e) To determine concavity, we divide the real number line into two intervals, A :  , 4  and B:  4,   and then we test a point in each interval: A: Test 3, f ” 3  B: Test 5, f ” 5  1 2 4 5 4 9 3  4 3 4 4 9 5  4 4 3  4 0 9  4 0 9 Thus, f  x  is concave up on both intervals  , 4  and  4,   Therefore, the point  4, 5 is not an inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x f  x   2  x  4  3  5 2 f ”  x   4 0 3 has no solution, so x  4 is the only possible inflection point. We know that f  4   5 . f x f ‘  x    exist when x  4 . The equation f ”  x   0 2 0 9 an inflection point. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. 6 1 3 10 3 d) Find the points of inflection. f ”  x  does not  , 2  and f  x  is concave down on the interval  2,   . Therefore, the point  2, 3 is x 3 5  4  1 interval  4,   . 2 0 9  4 We also know that f  x  is increasing on Thus, f  x  is concave up on the interval a) 3  3  4  1 has no solution, so x  2 is the only possible inflection point. We know that f  2   3 . To and B:  2,   and then we test a point in 39. 4 1 4 4  x  4  3   1 3 3  x  4 3 4 3 5 12 4 4 4  x  4  3  4 9 9  x  4 3 3 3 3 3 The domain of f is . b) f ‘  x  does not exist for x  4 . The equation f ‘  x   0 has no solution, therefore, x  4 is the only critical point. f  4  2  4  4  5  5 . 2 3 So, the critical point  4, 5 , is on the graph. c) We apply the First Derivative Test. We use 4 to divide the real number line into two intervals A :  , 4  and B:  4,   and then 40. f  x   3  x  2  3  3 2 a) f ‘  x   f ”  x   1 6 2  x  2  3   1 3  x  2 3 4 6 2  x  2  3  4 9 3  x  2 3 The domain of f is . we test a point in each interval at the top of the next column: Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 275 b) f ‘  x  does not exist for x  2 . The equation f ‘  x   0 has no solution; therefore, x  2 is f  2  3  2  2  3  3 . 3 So, the critical point,  2, 3 is on the graph. c) We apply the First Derivative Test. We use 2 to divide the real number line into two intervals A :  , 2  and B:  2,   and then we test a point in each interval. 2 A: Test 1, f ‘ 1   1  2  0 1  2 3 B: Test 3, f ‘  3   2 3  2 1 3  2  0 Thus,  2, 3 is a relative maximum. We also know that f  x  is increasing on the interval  , 2  and decreasing on the interval 2,   . d) Find the points of inflection. f ”  x  does not exist when x  2 . The equation f ”  x   0 has no solution, so x  2 is the only possible inflection point. We know that f  2   3 . e) To determine concavity, we divide the real number line into two intervals, A :  , 2  and B:  2,   and then we test a point in each interval. 2 2 A: Test 1, f ” 1   0 4 3 3 3 1  2  B: Test 3, f ”  3   f  x   x 1  x2   x 1  x2 a) the only critical point. 2 41. 2 3  3  2  4 3  2 0 3 Thus, f  x  is concave up on both intervals  , 2  and  2,   Therefore, the point  2, 3 is not an inflection point. f ‘  x   x    1  x2  f x 6 1 3 10 9 0 0 9 1 2   12 1 1  x2  2 x  2   1 1 2 2 x2  1  2 x2  1 1  x  1  x  1 f ”  x    2 x  1    1  x   2 x    2 1 2 2 2 2  2 3 2 1  x  4 x  2  2 1 2 x 3  3x  1  x  2 3 2 The domain of f  x  is  1,1 . b) f ‘  x  does not exist when x  1 . However, the domain of f  x  is  1,1 . Therefore, relative extrema cannot occur at x  1 or x  1 because there is not an open interval containing 1 or 1 on which the function is defined. The other critical points occur where f ‘  x   0 . 2 x2  1 1  x2 0 2 x2  1  0 x2  1 2 x 1 2 The critical values are  1 1 . and 2 2  1   1   1     1  f        2 2 2 2  1  1  1  2  2 f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x    1  1   2  2 1 2 The solution is continued on the next page.  Copyright © 2016 Pearson Education, Inc. 276 Chapter 2: Applications of Differentiation f ”  x   0 Evaluating the second critical value, we have:  1   1   1  f   1   2   2   2  2 2 x 3  3 x   1  1    1  2  2 3 2 0 3 2 x  3 x  0   x 2 x 2  3  0  1  1   2  2   1  x2 1 2  1 1  1 1 ,  and  ,   are Therefore,     2 2 2 2 critical points on the graph. c) We use the Second Derivative Test.  1  f ”     4  0  2  1 1 ,  is a relative The critical point    2 2 maximum.  1  f ”  40  2   1 1  ,   . 2 2 d) Find the points of inflection. f ”  x  does not exist when x  1 and x  1 . However, inflection points cannot occur at those values because the domain of the function is  1,1 . The remaining possible inflection x0 or x0 or 2x 2  3  0 3 2  6 x 2 x2  6 . 2 Therefore, the only possible inflection point is x  0 . Evaluating the function we have f 0   0 1   0   0 . 2 Therefore,  0, 0  is a possible inflection point on e) the graph. To determine concavity, we use 0 to divide the interval  1,1 into two intervals, A:  1, 0  and B:  0,1 and then we test a point in each interval. 1  1  10 A: Test  , f ”     3  0  2 3 2 2 B: Test  1  10 f ”    3  0 2 3 2 1 , 2 Thus, f  x  is concave down on the interval  1   1 1   1  , ,1 ,  1,   ,    , and  2 2 2 2   we see that f  x  is increasing on the f  x  is decreasing on the interval or Note that f  x  is not defined for x    1 1 ,   is a relative The critical point   2 2 minimum. 1 1 to divide If we use the points  and 2 2 the interval  1,1 into three intervals  1   1  and  ,1 and intervals  1,     2  2 x0  1, 0 and f  x  is concave up on the interval  0,1 . Therefore, the point 0, 0  is an inflection f) point. We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x 1  12 0 1 2  43 1 0 3 4 points occur when f ”  x   0 . We set the second derivative equal to zero and solve for the possible inflection points. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 42. 277  f  x  x 4  x2  x 4  x2 a)    4  x2   1  1 2  12 1 4  x2  2 x  2 f ‘  x  x   1 4  2×2 2 Next, we simplify the derivative. 1  x2 2 2 f ‘  x  1  4 x 2 4  x2       x2  4  x2 4  x  2 1 1 2 2  2 2 1 f ”  x    0 4  2 x2  0 x 2 The critical values are  2 and 2 .       2 2  2 f  2  2 4 2  2 2  2 Therefore,   2, 2  and  2, 2 are critical   1  4  2 x2    4  x2  2 4  x  4 x  x 4  2 x  4x   4  x  4  x  4 x  2 x  4 x 4  x   4  x   2 x   3 2 2  2 1 points on the graph. c) We use the Second Derivative Test.    12  2  f ”   2   4   2      2 2 2 3 2   3 4 x  2 x  16 x  4 x 4  x  2 3 2 3 2   2 8 2 40 2 2 minimum.  2   12  2  f ”  2   4  2      3 2 2 3 2  2 x 3  12 x 4  x  2 3  2 3 4 2  12 2 3 The critical point  2, 2 is a relative 2 2 2  1 3 2  2 2 3 2 2 4  x    4  2 x  4  x  2 4  x2 f  2   2 4  2 2 4  2 x2  For this reason, we do not consider 2 or 2 in our discussion of relative extrema. The other critical points occur where f ‘  x  0 4 2  12 2 2 The critical point 2 The domain of f  x  is  2, 2  . b) First, we find the critical points. f ‘  x  does not exist when 4  x 2  0 . Solve: 4  x2  0 x2  4 x  2 Since the domain of f  x  is  2, 2  , 3 2 3  2 8 2  4  0 2 2  2, 2 is a relative maximum. If we use the points  2 and 2 to divide the interval  2, 2  into three intervals      2,  2 ,  2, 2 , and 2, 2 , we   see that f  x  is decreasing on the intervals 2,  2  and  2, 2 and f  x  is increasing on the interval   2 , 2  . relative extrema cannot occur at x  2 or x  2 because there is not an open interval containing 2 or 2 on which the unction is defined. Copyright © 2016 Pearson Education, Inc. 278 Chapter 2: Applications of Differentiation d) Find the points of inflection. f ”  x  does f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. not exist where 4  x 2  0 . We know that this occurs at x  2 and x  2 . However, just as relative extrema cannot occur at  2, 0  and  2,0  , they cannot be inflection points either. Inflection points could occur where f ”  x   0 . x f x 1  3 1 3 We set the second derivative equal to zero and solve for x. f ”  x   0 2 x 3  12 x 4  x  2 3 2 0 43. 8x x 1 f  x  2  x  1 8  2 x8x Quotient a) f ‘  x   Rule  x  1 2 2 x 3  12 x  0   2x  0 or x2  6  0 x0 or x2  6 x0 or x 6 2 2 2 x x2  6  0 8 x 2  8  16 x 2  Therefore, the only possible inflection point is x  0 . 2 Therefore,  0, 0  is a possible inflection point on the graph. e) To determine concavity, we use 0 to divide the interval  2, 2  into two intervals, f ”  x   2 2 8  8×2  Note that f  x  is not defined for x   6 . f 0  0 4  0  0 .  x  1  x  1 2 2 Next we find the second derivative 2 1 x 2  1  16 x   8  8 x 2  2 x 2  1  2 x           x2  1  2 2 A:  2, 0  and B:  0, 2  and then we test a  x  1 16 x  x  1  4 x 8  8x    x  1 point in each interval. A: Test  1,  f ”  1  2  1  12  1 3  4   12    3 2  10 3 32 0 B: Test 1, f ” 1  2 1  12 1 3 10  3 0 3 32  4  12  2   Thus, f  x  is concave up on the interval 2 2 2  2 4 16 x3  16 x  32 x  32 x3  x  1 2 3 16 x3  48 x  x  1 3 2 The domain of f is . b) f ‘  x  exists for all real numbers. Solve: f ‘  x  0 8  8×2 0  2, 0  and f  x  is concave down on the interval  0, 2  . Therefore, the point  0, 0  is  x  1 an inflection point. x2  1 x  1 The two critical values are x  1 and x  1 . The solution is continued on the next page. 2 2 8  8×2  0 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 279 We will test a point in each interval. 32 0 A: Test  2, f ”  2    125 B: Test  1, f ”  1  4  0 Evaluating the function at the critical values, we have: 8  1 8 f  1     4 2  1  1 2 f 1  8 1 1  1 2  C: Test 1, f ” 1  4  0 8 4 2 32 0 125 We see that f is concave down on the D: Test 2, f ”  2   The critical points  1, 4  and 1, 4  are on the graph. c) We use the Second Derivative Test. f ”  1  4  0 ,  3  and 0, 3  and intervals concave up on the intervals  3,0 and  3,  . Therefore the points  3, 2 3  , 0, 0 , and  3, 2 3  are So the point  1, 4  is a relative minimum. f ” 1  4  0 So the point 1, 4  is a relative maximum. inflection points. f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. f  x  is decreasing on the intervals  ,1 and 1,  , and f  x  is increasing on the interval 1,1 . d) Find the points of inflection. f ”  x  exists x f x for all real numbers, so the only possible points of inflection occur when f ”  x   0 . 3  125 2 16 x 3  48 x  16 5 2 16 5 12 5  x  1 3 2 0 3 16 x 3  48 x  0   44. 16 x x 2  3  0 or x 2  3  0 16 x  0 x0  x  1 1  x 2 x  Quotient a) f ‘  x   Rule  x  1   f  3  2 3 f 0   0  3  2 3 The points   3, 2 3  ,  0, 0  , and  3, 2 3  are three possible inflection f points on the graph. e) To determine concavity we use  3, 0, and f ”  x    1  x2  x  1 2         3,      x2  1 2 2 2 x3  6 x  x  1 2 3 The domain of f is .  intervals, A : ,  3 , B:  3, 0 , C: 0, 3 ,and D: 2 Next, we find the second derivative. 2 1 x 2  1  2 x   1  x 2  2 x 2  1  2 x    3 to divide the real number line into four  2 2 x  0 or x 3 There are three possible inflection points x   3, 0, and 3 .  x x 1 2 2 x2  3 or f  x  Copyright © 2016 Pearson Education, Inc.  280 Chapter 2: Applications of Differentiation b) Since f ‘  x  exists for all real numbers, the f ”  x   0 only critical values are where f ‘  x   0 .  1  x2  x2  1  0 2  x2  1 3 2x  6x  0 Multiplying   by x 2 1 2 x2  1 x   1  1 The two critical values are x  1 and x  1 . 1  1  1 2 1  1 2   2x x2  3  0 2x  0 or x 2  3  0 x0 or   f  3  1   1 The critical points  1,   and 1,  are    2 2 on the graph. c) We use the Second Derivative Test. f  0  1  1 2 f ”  1   2  1  6  1 3 3  4 1  0 8 2  12  1   1  So the point  1,   is a relative  2 minimum. f ” 1  2 1  6 1 3 3  1 4  0 8 2 12  1   1   So the point 1,  is a relative maximum.  2 We use 1 and 1 to divide the real number line into three intervals   , 1 ,  1,1 , and 1,   . f  x  is decreasing on the intervals  ,1 and 1,  , and f  x  is increasing on the interval 1,1 . d) Find the points of inflection. f ”  x  exists for all real numbers, so the only possible points of inflection occur when f ”  x   0 . We find the possible points of inflection at the top of the next column. x2  3 x  0 or x 3 There are three possible inflection points at x   3, 0, and 3 . 1 2 f 1  0 3 1  x2  0 f  1  2 x3  6 x f  3  3   1 2 0  0  1 2  3   3  3 1 2 3 4  0 0 1  3 4  3 The points   3,  ,  0, 0  , and 4    3  3, 4  are three possible inflection   points on the graph. e) To determine concavity we use  3, 0, and 3 to divide the real number line into four intervals,       A : ,  3 , B:  3, 0 , C: 0, 3 , and D:  3, . We test a point in each interval. 4 A: Test  2, f ”  2    0 125 1 B: Test  1, f ”  1   0 2 1 C: Test 1, f ” 1    0 2 4 0 D: Test 2, f ”  2   125 We see that f is concave down on the intervals ,  3  and 0, 3  and concave up on the intervals  3,0 and  3,  . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 281 Using the information from the previous  3 , page, we determine the points   3,  4    3 0, 0  , and  3,  are inflection points. 4   f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x 3  103 2  25 2 2 5 3 10 3 45. 2 Extended Power Rule 2  x  1 2  8  8x 2  x2  1 2 x 2   2  2  x  1  2 2 2 4 8 x 2  8  32 x 2  x  1 3 8  24 x 2  x  1 2  2 8x  0 x0 The critical value is x  0 . 4 f 0   4  0 2  1 f ”  x   0 . 8  24 x 2   x2  1 3 0 Multiplying 8  24 x 2  0 2 2  x  1  x  1 8  8x22 x   x  1   by x 2 1 d) f ”  x  exists for all real numbers. Solve 8x 2 Multiplying 0 and f  x  is increasing on the interval  0,   . Next, we find the second derivative. f ”  x   2 f  x  is decreasing on the interval  , 0  , 2   x  1 2 So the point  0, 4  is a relative minimum. 2  8x c) We use the Second Derivative Test. f ”  0   8  0   a) f ‘  x   4  1  x  1  2 x f ‘  x   8 x  x  1 x2  1 f ‘  x  0 The critical point  0, 4  is on the graph. 1 4  4 x 2  1 f  x  2 x 1  b) f ‘  x  exists for all real numbers Solve: 3 The domain of f is . x2    by x2 1 3 1 3 x 1 3 There are two possible inflection points 1 1 .  and 3 3  1  4 4 4 f      3  2 1 4  3  1   1    1 3 3 3  1   1  , 3 and  , 3 are The points      3  3 possible inflection points on the graph. 1 e) To determine concavity we use  and 3 1 to divide the real number line into three 3   1 1  1  , B:   , intervals, A :  ,   ,   3 3 3  1  ,   . The solution is continued. and C:   3  Copyright © 2016 Pearson Education, Inc. 282 Chapter 2: Applications of Differentiation Then test a point in each interval. A: Test  1, f ”  1  2  0 B: Test 0, f ”  0  8  0  1 1  , concave up on the interval   .  3 3  1  , 3 and Therefore the points     3  1  , 3 are inflection points.   3 f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. 46. 4 4  17 2  45 1 1 2 2 2  45 4 4  17 1 2 2 6 x  x  1 2 2 2 2 2 2 2 2 4 18 x 2  6  x  1 2  2 The critical point  0, 3 is on the graph. c) We use the Second Derivative Test. 18  0  6 6 f ” 0    6  0 2 2 1  0  1   So the point  0, 3 is a relative maximum. We use 0 to divide the real number line into two intervals  , 0  and  0,   . f  x  is   x2  1 3 0 2 18 x  6  0 18 x 2  6 1 x2  3 1 3 There are two possible inflection points at 1 1 . x and 3 3 x 2  x  1 6  6 x  2  x  1 2 x       x  1   x  1  x  1 6  6 x 22 x    x  1   x0 The critical value is x  0 . 3 f 0   3  0 2  1 18 x 2  6 2 2 Multiplying by x 2 1 for all real numbers, so the only possible points of inflection occur when f ”  x   0 . 2 2 6 x  0 d) Find the points of inflection. f ”  x  exists 3  3 x2  1 x 1 f ”  x  0 f  x  is decreasing on the interval 0,   . 2  2 increasing on the interval  , 0 , and   a) f ‘  x   3  1  x  1  2 x   6 x  x  1 f  x   x  1 2 We see that f is concave down on the   1  1  ,   and intervals  ,   and    3 3 f x only critical values are where f ‘  x   0 . 6 x C: Test 1, f ” 1  2  0 x b) Since f ‘  x  exists for all real numbers, the 3  1  3 f     3   1 2    1 3 3 3 9    1 4 4 1 3 3 The solution is continued on the next page. The domain of f is . Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 283 Evaluating the function at the second possible inflection point, we have:  1  3 f    3   1 2   1 3 3 3 9    1 4 4 1 3 3  1 9  1 9 ,  and  , are The points      3 4  3 4 possible inflection points on the graph. e) To determine concavity we use 1 1 to divide the real number  and 3 3 line into three intervals,   1 1  1  A :  ,  ,  , B:    , and  3 3 3  1  C:  ,   3  Then test a point in each interval. A: Test  1, f ”  1  B: Test 0, f ”  0  C: Test 1, f ” 1  18  1  6 2 1  1 2 18  0   6  3 3 0 2 f) We sketch the graph using the preceding information. Additional function values may also be calculated as necessary. x f x 3 1 1 3 3 10 3 2 3 2 3 10 47. Answers may vary, one possible graph is: 48. Answers may vary, one possible graph is: 2 02  1  6  0 3 18 1  6 2 1  1 2 3  3 0 2 We see that f is concave up on the intervals   1  1  ,   and concave  ,   and  3 3  49. Answers may vary, one possible graph is:  1 1  , down on the interval   .  3 3  1 9 ,  and Therefore the points    3 4  1 9 ,  are inflection points.  3 4 Copyright © 2016 Pearson Education, Inc. 284 Chapter 2: Applications of Differentiation 50. Answers may vary, one possible graph is: 55. Answers may vary, one possible graph is: 7y 6 5 4 3 2 1 -1 -1 0 -2 -3 -4 x 1 2 3 4 5 56. Answers may vary, one possible graph is: 3 51. Answers may vary, one possible graph is: y 2 1 -3 -2 -1 -1 x 0 1 2 3 -2 -3 57. R  x   50 x  0.5 x 2 C  x   4 x  10 P  x  R  x  C  x 52. Answers may vary, one possible graph is:    50 x  0.5 x 2   4 x  10   0.5 x 2  46 x  10 We will restrict the domains of all three functions to x  0 since a negative number of units cannot be produced and sold. First graph R  x   50 x  0.5 x 2 R ‘  x   50  x 53. Answers may vary, one possible graph is: R ”  x   1 Since R ‘  x  exists for all x  0 , the only critical points are where R ‘  x   0 . 50  x  0 50  x Critical Value Find the function value at x  50 . R 50   50 50   0.5 50  2  2500  1250 54. Answers may vary, one possible graph is:  1250 This critical point 50,1250  is on the graph. We use the Second Derivative Test: R ” 50   1  0 The point 50,1250  is a relative maximum. We use 50 to divide the interval  0,   into two intervals,  0,50  and 50,  . We know that R is increasing on  0,50  and decreasing on 50,   . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 285 Next, find the inflection points. Since R ”  x  exists for all x  0 , and R ”  x   1 , there are Sketch the graph using the preceding information. no possible inflection points. Furthermore, since R ”  x   0 for all x  0 , R is concave down over the interval  0,   . Sketch the graph using the preceding information. The x-intercepts of R are found by solving R  x   0. 50 x  0.5 x 2  0 0.5 x 100  x   0 0.5 x  0 or 100  x  0 100  x x  0 or The x-intercepts are  0, 0  and 100,0  . Next, we graph C  x   4 x  10 . This is a linear function with slope 4 and y-intercept  0,10  . C  x  is increasing over the entire domain x  0 58. R  x   50 x  0.5 x 2 C  x   10 x  3 P  x  R  x  C  x    50 x  0.5 x 2  10 x  3 2 and has no relative extrema or points of inflection. Finally, we graph P  x   0.5 x 2  46 x  10  0.5 x  40 x  3 We will restrict the domains of all three functions to x  0 . First graph R  x   50 x  0.5 x 2 as in Exercise P ”  x   1 57. Next, we graph C  x   10 x  3 . This is a linear P ‘  x    x  46 Since P ‘  x  exists for all x  0 , the only critical points occur when P ‘  x   0 .  x  46  0 46  x Critical Value Find the function value at x  46 . P  46  0.5  46  46  46  10 2  1058  2116  10  1048 The critical point  46,1048  is on the graph. We use the Second Derivative Test: P ”  46   1  0 The point  46,1048  is a relative maximum. We use 46 to divide the interval  0,   into two intervals,  0, 46  and  46,  , we know that P is increasing on 0, 46  and decreasing on  46,   . Next, find the inflection points. Since P ”  x  exists for all x  0 , and P ”  x   1 , there are no possible inflection points. Furthermore, since P ”  x   0 for all x  0 , P is concave down over the interval  0,   . function with slope 10 and y-intercept  0, 3 . C  x  is increasing over the entire domain x  0 and has no relative extrema or points of inflection. Finally, we graph P  x   0.5 x 2  40 x  3 P ‘  x    x  40 P ”  x   1 Since P ‘  x  exists for all x  0 , the only critical points occur when P ‘  x   0 .  x  40  0 40  x P  40   797 Critical Value The critical point  40, 797  is on the graph. We use the Second Derivative Test: P ”  40   1  0 The point  40, 797  is a relative maximum. We know that P is increasing on  0, 40  and decreasing on  40,   . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. 286 Chapter 2: Applications of Differentiation Next, find the inflection points. Since P ”  x  exists for all x  0 , and P ”  x   1 , there are no possible inflection points. Furthermore, since P ”  x   0 for all x  0 , P is concave down over the interval  0,   . Sketch the graph using the preceding information. The point 16.20, 7.17  is a relative minimum. If we use the point 16.20 to divide the domain into two intervals,  0,16.20  and 16.20,40 , we know that p is decreasing on 0,16.20  and increasing on 16.20, 40  . Next, we find the inflection points. p ”  x  exists for all real numbers, so the only possible inflection points are where p ”  x   0 78 x  480 0 75, 000 78 x  480  0 78 x  480 59. p  x  13x 3  240 x 2  2460 x  585,000 75,000 39 x 2  480 x  2460 p ‘  x  75,000 78 x  480 p ”  x   75,000 Since p ‘  x  exists for all real numbers, the only p  6.15 x  6.15 13  6.15  240  6.15  2460  6.15  585, 000 3  2 75, 000  7.52 The point  6.15, 7.52  is a possible inflection critical points are where p ‘  x   0 . point. To determine concavity, we use 6.15 to divide the domain into two intervals A :  0, 6.15 and B:  6.15,40 and test a point in 39 x 2  480 x  2460 0 75,000 A: Test 1, p ” 1  39 x 2  480 x  2460  0 Using the quadratic formula, we have: x  b  b 2  4ac 2a   480    4802  4 39  2460 2  39  480  614,160  78 x  3.89 or x  16.20 Critical values Since the domain of the function is 0  x  40 , we consider only x  16.20 p 16.20  13 16.20   240 16.20   2460 16.20   585, 000 3  2 75, 000  7.17 The critical point 16.20, 7.17  is on the graph. each interval. 78 1  480  0.005  0 75,000 78  7   480  0.00088  0 75, 000 Then p is concave down on  0, 6.15 and B: Test 7, p ”  7   concave up on 6.15, 40  and the point 6.15, 7.52  is a point of inflection. Sketch the graph for 0  x  40 using the preceding information. Additional function values may be calculated if necessary. x p x 0 8 12 20 24 32 40 7.8 7.42 7.25 7.25 7.57 9.15 12.46 We use the Second Derivative Test: 78 16.20   480 p ”  x    0.01  0 75,000 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 60. 287 f  x   0.025 x 2  0.71x  20.44 These are two possible points of inflection. R 3.26  18.36 f ‘  x   0.05 x  0.71 R 14.48  62.69 The two points of inflection are 3.26,18.36 and 14.48, 62.69 . However, f ”  x   0.05 f ‘  x  exists for all real numbers. Solve f ‘ x  0 t  14.48 is not in the domain of this function. The left most inflection point 3.26,18.36  0.05 x  0.71  0 0.05 x  0.71 x  14.2 implies that rate of change of the amount of rainfall is decreasing the fastest at this point. f 14.2   0.025 14.2   0.71 14.2   20.44 2  15.399 f ” 14.2   0.05 , so 14.2,15.399  is a relative   V ‘  r   k  40r  3r  62. V  r   k 20r 2  r 3 , 2 minimum. Then f  x  is decreasing on V ”  r   k  40  6r  0,14.2  and increasing on 14.2, 30 . V ‘  r  exists for all r in  0, 20  , so the only Next, find the points of inflection. Since f ”  x   0.05 exists for all real numbers and is critical points occur where V ‘  r   0 .  interval  0, 30  . 40r  3r 2  0 r  40  3r   0 r  0 or 40  3r  0 r  0 or 40  3r 40 r  0 or r 3 Using the Second Derivative Test: V ”  0  k  40  6  0  40k  0 Sketch the graph of f  x  using the information above. Additional values may be calculated as necessary. f x 0 5 10 15 20 25 30 20.44 17.52 15.84 15.42 16.24 18.32 21.64  40  Since V ”    0 , we know that there is a  3 0.615t  27.745 To find the inflection points we find the first and second derivatives. R ‘ t   0.024t 3  0.639t 2  3.404t  0.615 R ” t   0.072t 2  1.278t  3.404 Since R ” t  exists for all real values, we solve R ” t   0 By the quadratic formula. 1.278  t  3.26  k  0   40   40   V ”    k  40  6     40k  0  3  3   relative maximum at x  61. The monthly rainfall is approximated by R t   0.006t 4  0.213t 3  1.702t 2  t  k 40r  3r 2  0 always positive, f  x  is concave up on the x 0  r  20 or 40 . Thus, for an 3 40 mm or 13.33 mm , the 3 maximum velocity is needed to remove the object. object whose radius is 63. Observe that h is increasing for all values of x for which g is positive and h is decreasing for all values of x for which g is negative. Furthermore, for all values of x for which g=0, h has a horizontal tangent. Therefore, g=h’. 1.278  4  0.072  3.404  2  0.072  2 t  14.48 Copyright © 2016 Pearson Education, Inc. 288 64. 65. Chapter 2: Applications of Differentiation Observe that g is increasing for all values of x for which h is positive and g is decreasing for all values of x for which h is negative. Furthermore, for all values of x for which h=0, g has a horizontal tangent. Therefore, h  g ‘ . f  x   ax 2  bx  c, 70. False, consider f  x   3 x  3 x  2 . This function has points of inflection at x  0 , x  1 and x  2 without any critical values between the points of inflection. 71. True. a0 f ‘  x   2ax  b 72. False: The function does not switch concavity at the extreme value. Since f ‘  x  exists for all real numbers, the only 73. True. f ”  x   2a critical points occur when f ‘  x   0 . We solve: 2ax  b  0 2ax   b b x 2a b So the critical value will occur at x  . 2a Applying the second derivative test, we see that f ”  x   2a  0, for a  0 f ”  x   2a  0, for a  0 Therefore, a relative maximum occurs at b when a  0 and a relative minimum x 2a b when a  0 . occurs at x  2a 74. The rate of change is maximized at the points of inflection. Looking at the graph, we estimate the points of inflection to be 75 days after January first, and 270 days after January first. Therefore, the number of hours of daylight are increasing most rapidly approximately 75 days after January 1st or approximately March 16th and the number of hours of daylight are decreasing most rapidly approximately 270 days after January 1st or approximately September 27th. 75. f  x  4x  6x 3 Graphing the function on the calculator we have: 2 66. The point of inflection will occur when the second derivative is equal to zero or undefined. The derivatives of the function are: g  x   ax 3  bx 2  cx  d g ‘  x   3ax 2  2bx  c Using the minimum/maximum feature on the calculator, we estimate a relative maximum at 0, 0  and a relative minimum at 1, 2  . g ”  x   6ax  2b. Setting the second derivative equal to zero and solving for x, we have: 6ax  2b  0 2b b x  . 6a 3a Therefore, the point of inflection must occur at b x . 3a 67. True. 68. True. 76. f  x   3x 3  2 x 2 We estimate a relative maximum at 1,1 and a relative minimum at  0, 0  . 69. True. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.2 77. 289 f  x   x 2 1  x  3 80. f  x  x  x Graphing the function on the calculator we have: We estimate a relative minimum at  0.25, 0.25 . Using the minimum/maximum feature on the calculator, we estimate a relative maximum at 0.4, 0.035 and a relative minimum at 0, 0  . 78. f  x   x2  x  2 3 We estimate a relative maximum at  0, 0  and a 4  relative minimum at  , 1.106  . 5  79. 81. a) The cubic and quartic functions appear to be equally good fits. The cubic function will ease the computations, where as the quartic function will be a little better fit to the data. Also they quartic function declines sharply after 2011 so the cubic function is probably a better fit. b) The domain of is the set of non-negative real numbers. c) The graphs show that the cubic function does not have relative extrema on a reasonable domain. The quartic function has a relative maximum around x  10.3 . However, judging by the data, it appears that the percentage of households headed by someone with a bachelor’s degree or higher is increasing with time. f  x    x  1 3   x  1 3 2 2 Graphing the function on the calculator we have: Using the minimum/maximum feature on the calculator, we estimate a relative maximum at  1,1.587  and a relative minimum at 1, 1.587  . Copyright © 2016 Pearson Education, Inc. 290 Chapter 2: Applications of Differentiation Once the expression is in simplified form, we set the denominator equal to zero and solve. x  x  1 x  1  0 Exercise Set 2.3 1. 2. 3. x4 x2 The expression is in simplified form. We set the denominator equal to zero and solve. x2 0 x2 The vertical asymptote is the line x  2 . x  0 or x  1  0 or x  1  0 x  0 or x  1 or x 1 The vertical asymptotes are the lines x  0, x  1, and x  1 . f  x  6. 2x  3 x5 The expression is in simplified form. The vertical asymptote is the line x  5 . f  x  f  x  x  25 First, we write the function in simplified form. 5x f  x   x  5 x  5 7. 3x 2  3x  x  3 x  3 x 9 Once the expression is in simplified form, we set the denominator equal to zero and solve.  x  3 x  3  0 x  3  0 or x  3  0 x  3 or x  3 The vertical asymptotes are the lines x  3 and x  3 . 5. f  x     x3 x  x  1 x  1 x2 2 x  6x  8 First, we write the function in simplified form. x2 f  x    x  2 x  4 1 Dividing common , x  2 factors x4 Once the expression is in simplified form, we set the denominator equal to zero and solve. x4 0 x  4 The vertical asymptote is the line x  4 . 8. x6 x  7x  6 First, we write the function in simplified form. x6 1  f  x  , x  6  x  6 x  1 x  1 f  x  2 The vertical asymptote is the line x  1 . 9. x3 x3  x First, we write the function in simplified form. x3 f  x  x x2  1 f  x   x  5  0 or x  5  0 x  5 or x  5 The vertical asymptotes are the lines x  5 and x  5 . f  x   The vertical asymptotes are the lines x  0, x  2, and x  4 . Once the expression is in simplified form, we set the denominator equal to zero and solve.  x  5 x  5  0 4. 3  5x 2 x2 x  6×2  8x First, we write the function in simplified form. x2 x2  f  x  2 x  x  4 x  2 x x  6x  8 f  x  10. 7 x  49 The function is in simplified form. The equation x 2  49  0 has no real solution; therefore, the function does not have any vertical asymptotes. f  x  2 6 x  36 The function is in simplified form. The equation x 2  36  0 has no real solution; therefore, the function does not have any vertical asymptotes. f  x  2 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 11. 291 6x 8x  3 To find the horizontal asymptote, we consider lim f  x  . To find the limit, we will use some f  x  13. x  4x x  3x To find the horizontal asymptote, we consider lim f  x  . To find the limit, we will use some f  x  x  b algebra and the fact that as x  , n  0 for ax any positive integer n. 6x lim f  x   lim x  x  8 x  3 1 6x Multiplying by a  lim x form of 1 x  8 x  3 1 x 6x x  lim 3 x  8 x  x x 6  lim 3 x  8 x 6  as x  , b  0  n 80   ax algebra and the fact that as x  , 1 2 4x Multiplying by a  lim 2 x 1 form of 1 x  x  3 x x2 4x 2  lim 2 x x  x 3  x2 x2 4  lim x 3 x  1 2 x 0  as x  , b  0   n 1 0   ax 6 3  . 8 4 In a similar manner, it can be shown that 3 lim f  x   . x  4 x  3x 2 3 3 1 lim 2  lim   . 1 60 2 x  6 x  x x  6 x In a similar manner, it can be shown that 1 lim f  x   . x   2 1 The horizontal asymptote is the line y  . 2 b  0 for ax n any positive integer n. 4x lim f  x   lim 2 x  x  x  3 x  3 The horizontal asymptote is the line y  . 4 3x 2 12. f  x   2 6x  x Find lim f  x  . 2  0. In a similar manner, it can be shown that lim f  x   0. x   The horizontal asymptote is the line y  0 . 14. 2x 3x  x 2 Find lim f  x  . f  x  3 x  2 2 2x 0  0. lim 3 2  lim x  1 3 0 x  3x  x x  3 x In a similar manner, it can be shown that lim f  x   0. x   The horizontal asymptote is the line y  0 . Copyright © 2016 Pearson Education, Inc. 292 15. Chapter 2: Applications of Differentiation In a similar manner, it can be shown that lim f  x   . 2 x To find the horizontal asymptote, we consider lim f  x  . To find the limit, we will use the f  x  4  x   The function increases without bound as x   and decreases without bound as x  . Therefore, the function does not have a horizontal asymptote. x  fact that as x  , b  0 for any positive ax n integer n. 2 lim f  x   lim 4  x  x  x  40 18. as x  , b  0 . n   ax 8 x 4  5x 2 2 x3  x2 Find lim f  x  . f  x  x  5 5 lim 8 x  x  x x  . lim  lim  1 x  2 x 3  x 2 x  20 2 x In a similar manner, it can be shown that lim f  x   . 4 In a similar manner, it can be shown that lim f  x   4. 8x  8×4  5×2 x   The horizontal asymptote is the line y  4 . x   3 16. f  x   5  x Find lim f  x  . The function increases without bound as x   and decreases without bound as x  . Therefore, the function does not have a horizontal asymptote. x  3  5  0  5. x In a similar manner, it can be shown that lim f  x   5. lim 5  x  x   The horizontal asymptote is the line y  5 . 6 x3  4 x 17. f  x   3x 2  x To find the horizontal asymptote, we consider lim f  x  . To find the limit, we will use some x  algebra and the fact that as x  , b  0 for ax n any positive integer . 6 x3  4 x lim f  x   lim x  x  3 x 2  x 1 6 x3  4 x x 2  lim  1 x  3 x 2  x x2 4 6x  x  lim 1 x  3 x 4 lim 6 x  x  x   3 0 Multiplying by a form of 1 19. 4 x 3  3x  2 x3  2 x  4 To find the horizontal asymptote, we consider lim f  x  . To find the limit, we will use some f  x  x  algebra and the fact that as x  , b  0 for ax n any positive integer n. 4 x3  3 x  2 lim f  x   lim 3 x  x  x  2 x  4 1 4 x  3 x  2 x3  lim 3  1 x  x  2 x  4 x3 3 2 4 2  3 x x  lim 2 4 x  1 2  3 x x b 4   as x  , 0  n   1  ax 4 In a similar manner, it can be shown that lim f  x   4. 3 x   The horizontal asymptote is the line y  4 . Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 20. 293 6 x4  4 x2  7 2 x5  x  3 Find lim f  x  . f  x  22. x  5x 4  2 x3  x x5  x 3  8 Find lim f  x  . f  x  x  6 4 7   6 x4  4 x2  7 x x3 x5  lim lim 1 3 x  2 x5  x  3 x  2 4  5 x x 0  0 20 In a similar manner, it can be shown that lim f  x   0. 5 2 1  2 5 5×4  2 x3  x x x x  0 0. lim  lim 1 8 x  x 5  x 3  8 x  1 1 2  5 x x In a similar manner, it can be shown that lim f  x   0. x   The horizontal asymptote is the line y  0 . x   The horizontal asymptote is the line y  0 . 21. 2 x3  4 x  1 4 x3  2 x  3 To find the horizontal asymptote, we consider lim f  x  . To find the limit, we will use some f  x  x  algebra and the fact that as x  , b  0 for ax n any positive integer n. 2 x3  4 x  1 lim f  x   lim 3 x  x  4 x  2 x  3 1 2 x  4 x  1 x3  lim 3  x  4 x  2 x  3 1 x3 4 1 2 2  3 x x  lim 2 3 x  4 2  3 x x 200  as x  , b  0  n   400 ax 2 1   4 2 In a similar manner, it can be shown that 1 lim f  x   . x   2 1 The horizontal asymptote is the line y  . 2 3 23. 4  4 x 1 x a) Intercepts. Since the numerator is the constant 4, there are no x-intercepts. The number 0 is not in the domain of the function, so there are no y-intercepts. b) Asymptotes. Vertical. The denominator is 0 for x  0 , so the line x  0 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 4 f ‘  x   4 x 2   2 x 8 3 f ”  x   8 x  3 x The domain of f is  , 0    0,   as f  x  determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 0, but 0 is not in the domain of the function, so x  0 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. 4 A: Test  1, f ‘  1    4  0  12 B: Test 1, f ‘ 1   4 12  4  0 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. 294 Chapter 2: Applications of Differentiation Then f  x  is decreasing on both intervals. d) Critical Points. f ‘  x  exists for all values of Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 0, x except 0, but 0 is not in the domain of the function, so x  0 is not a critical value. The equation f ‘  x   0 has no solution, so there but because 0 is not in the domain of the function, there cannot be an inflection point at 0. The equation f ”  x   0 has no are no critical points. e) Increasing, decreasing, relative extrema. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and solution; therefore, there are no inflection points. g) Concavity. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. A: Test  1, f ”  1  B: Test 1, f ” 1  B: Test 8  13 8 1 3  8  0 80 Therefore, f  x  is concave down on  , 0 and concave up on 0,   . h) Sketch. 24. 5  5 x 1 x a) Intercepts. Since the numerator is the constant 5 , there are no x-intercepts. The number 0 is not in the domain of the function, so there are no y-intercepts. b) Asymptotes. Vertical. The denominator is 0 for x  0 , so the line x  0 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 5 f ‘  x   5 x 2  2 x 10 f ”  x   10 x 3   3 x The domain of f is  , 0    0,   as f  x   we test a point in each interval. 5 A: Test  1, f ‘  1  5 0  12 1, f ‘ 1  5 12 5 0 Then f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 0, but because 0 is not in the domain of the function, there cannot be an inflection point at 0. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. 10 A: Test  1, f ”  1    10  0  13 B: Test 1, f ” 1   10 13  10  0 Therefore, f  x  is concave up on  , 0  and concave down on  0,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. determined in step (b). Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 25. 295 g) Concavity. We use 5 to divide the real number line into two intervals A:  ,5 and B: 5,   , and we test a 2 1  2  x  5 x5 a) Intercepts. Since the numerator is the constant 2 , there are no x-intercepts. To find the y-intercepts we compute f  0  . f  x  f 0  2 0  5  point in each interval. A: Test 4, f ”  4   2 5  2 The point  0,  is the y-intercept.  5 b) Asymptotes. Vertical. The denominator is 0 for x  5 , so the line x  5 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 2 2 f ‘  x   2  x  5   x  52 f ”  x   4  x  5 3 4   x  53 The domain of f is  ,5  5,   as determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 5, but 5 is not in the domain of the function, so x  5 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 5 to divide the real number line into two intervals A:  ,5 and B: 5,   , and we test a point in each interval. 2 2 A: Test 4, f ‘  4    20 2 1  4  5 B: Test 6, f ‘ 6  2  6  5 2  2 20 1 Then f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 5, but because 5 is not in the domain of the function, there cannot be an inflection point at 5. The equation f ”  x   0 has no solution; therefore, there are no inflection points. B: Test 6, f ”  6  4  4  5 3 4 6  5 3  4 40 1  4  4  0 1 Therefore, f  x  is concave up on  ,5 and concave down on 5,   . h) Sketch. 26. 1 1   x  5 x5 a) Intercepts. Since the numerator is the constant 1, there are no x-intercepts. To find the y-intercepts we compute f  0  . f  x  f 0  1 1  0  5 5   1  The point  0,   is the y-intercept.  5 b) Asymptotes. Vertical. The denominator is 0 for x  5 , so the line x  5 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 2 f ‘  x     x  5   x  5 2 f ”  x   2  x  5 3  2  x  53 The domain of f is  ,5  5,   as determined in step (b). Copyright © 2016 Pearson Education, Inc. 296 Chapter 2: Applications of Differentiation d) Critical Points. f ‘  x  exists for all values of x except 5, but 5 is not in the domain of the function, so x  5 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 5 to divide the real number line into two intervals A:  ,5 and B: 5,   , and we test a point in each interval. 1 1 A: Test 4, f ‘  4      1  0 2 1  4  5 B: Test 6, f ‘  6  1  6  5 2 1    1  0 1 Then f  x  is decreasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 5, but because 5 is not in the domain of the function, there cannot be an inflection point at 5. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 5 to divide the real number line into two intervals A:  ,5 and B: 5,   , and we test a point in each interval. 2 2 A: Test 4, f ”  4     2  0 3 1   4  5 B: Test 6, f ”  6  2  2 20 1 27. 1 1   x  3 x3 a) Intercepts. Since the numerator is the constant 1, there are no x-intercepts. To find the y-intercepts we compute f  0  . f  x  f 0  1 0  3  1 3 1  The point  0,   is the y-intercept.  3 b) Asymptotes. Vertical. The denominator is 0 for x  3 , so the line x  3 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 2 f ‘  x     x  3   x  32 f ”  x   2  x  3 3  2  x  33 The domain of f is  , 3  3,   as determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 3, but 3 is not in the domain of the function, so x  3 is not a critical value. The equation f ‘  x   0 has no solution, so there Therefore, f  x  is concave down on are no critical points. e) Increasing, decreasing, relative extrema. We use 3 to divide the real number line into two intervals A:  , 3 and B: 3,   , and h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. we test a point in each interval. 1 A: Test 2, f ‘  2    1  0 2  2  3 6  5 3  ,5 and concave up on 5,   . B: Test 4, f ‘  4   1  4  3 2  1  0 Then f  x  is decreasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 3, but because 3 is not in the domain of the function, there cannot be an inflection point at 3. The equation f ”  x   0 has no solution; therefore, there are no inflection points. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 297 g) Concavity. We use 3 to divide the real number line into two intervals A:  , 3 and B:  3,   , and we test a point in each interval. A: Test 2, f ”  2   B: Test 4, f ”  4   2 2  3 3 2 4  3 3  2  0 20 Therefore, f  x  is concave down on  , 3 and concave up on 3,   . d) Critical Points. f ‘  x  exists for all values of x except 2 , but 2 is not in the domain of the function, so x  2 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 2 to divide the real number line into two intervals A:  , 2  and B:  2,   , and we test a point in each interval. A: Test  3, f ‘  3  h) Sketch. 1  3  2 B: Test  1, f ‘  1  2 1 1  2 2  1  0  1  0 Then f  x  is decreasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 2 , 1 1 28. f  x     x  2 x2 a) Intercepts. Since the numerator is the constant 1, there are no x-intercepts. To find the y-intercepts we compute f  0  . f 0  1 1  0  2 2  1 The point  0,  is the y-intercept.  2 b) Asymptotes. Vertical. The denominator is 0 for x  2 , so the line x  2 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 2 f ‘  x     x  2   x  2 2 f ”  x   2  x  2  3  but because 2 is not in the domain of the function, there cannot be an inflection point at 2 . The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 2 to divide the real number line into two intervals A:  , 2  and B:  2,   , and we test a point in each interval. A: Test  3, f ”  3  B: Test  1, f ”  1  2 3  2 3 2  1  2 3  2  0 20 Therefore, f  x  is concave down on  , 2  and concave up on  2,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. 2  x  2 3 The domain of f is  , 2    2,   as determined in step (b). Copyright © 2016 Pearson Education, Inc. 298 29. Chapter 2: Applications of Differentiation 2 1  2  x  5 x 5 a) Intercepts. Since the numerator is the constant 2 , there are no x-intercepts. To find the y-intercepts we compute f  0  . f  x  f 0  2 0  5  f) Inflection points. f ”  x  does not exist at 5 , but because 5 is not in the domain of the function, there cannot be an inflection point at 5 . The equation f ”  x   0 has no 2 2  5 5 solution; therefore, there are no inflection points. g) Concavity. We use 5 to divide the real number line into two intervals A:  , 5 and B:  5,   , and we test a 2  The point  0,   is the y-intercept.  5 b) Asymptotes. Vertical. The denominator is 0 for x  5 , so the line x  5 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 2 2 f ‘  x   2  x  5   x  52 f ”  x   4  x  5 3 point in each interval. A: Test  6, f ”  6  B: Test  4, f ”  4   4  6  5 3 4 4  5 3 40  4  0 Therefore, f  x  is concave up on   , 5 and concave down on  5,   . h) Sketch. 4   x  5 3 The domain of f is  , 5   5,   as determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 5 , but 5 is not in the domain of the function, so x  5 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 5 to divide the real number line into two intervals A:  , 5 and B:  5,   , and we test a point in each interval. A: Test  6, f ‘  6  2 6  5 B: Test  4, f ‘  4   20 2 2 4  5 2 20 Then f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. 30. 3 1  3  x  3 x3 a) Intercepts. Since the numerator is the constant 3 , there are no x-intercepts. To find the y-intercepts we compute f  0  . f  x  f  0  3  3 1 0  3 3 The point  0,1 is the y-intercept. b) Asymptotes. Vertical. The denominator is 0 for x  3 , so the line x  3 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 299 c) Derivatives and Domain. f ‘  x   3  x  3 2 f ”  x   6  x  3  3 h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. 3  x  32 6   x  33 The domain of f is  , 3  3,   as determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 3, but 3 is not in the domain of the function, so x  3 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 3 to divide the real number line into two intervals A:  , 3 and B:  3,   , and we test a point in each interval. 3 A: Test 2, f ‘  2    3 0 2 2  3 B: Test 4, f ‘  4   3 4  3 2  3 0 Then f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 3, but because 3 is not in the domain of the function, there cannot be an inflection point at 3. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 3 to divide the real number line into two intervals A:  , 3 and B:  3,   , and we test a point in each interval. A: Test 2, f ”  2  B: Test 4, f ”  4   6 2  3 3 6 4  3 3 60  6  0 Therefore, f  x  is concave up on  , 3 and concave down on  3,   . 31. 2x 1 x a) Intercepts. To find the x-intercepts, solve f  x  0 . f  x  2x 1 0 x 2x 1  0 1 2 This value does not make the denominator  1  0; therefore, the x-intercept is   , 0  .  2  x The number 0 is not in the domain of f  x  so there are no y-intercepts. b) Asymptotes. Vertical. The denominator is 0 for x  0 , so the line x  0 is a vertical asymptote. Horizontal. The numerator and the denominator have the same degree, so 2 y  , or y  2 is the horizontal asymptote. 1 Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 f ‘  x   2 x 2 f ”  x   2 x 3  3 x The domain of f is  , 0    0,   as determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 0, but 0 is not in the domain of the function, so x  0 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. Copyright © 2016 Pearson Education, Inc. 300 Chapter 2: Applications of Differentiation e) Increasing, decreasing, relative extrema. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. 1  1  0 A: Test  1, f ‘  1   12 B: Test 1, f ‘ 1   1 12  1  0 Then f  x  is decreasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 0, but because 0 is not in the domain of the function, there cannot be an inflection point at 0. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. A: Test  1, f ”  1  B: Test 1, f ” 1  2  1 3 2 1 3  2  0 20 Therefore, f  x  is concave down on  , 0 and concave up on 0,   . h) Sketch. 1 does not make the denominator 3 1  0, the x-intercept is  ,0  . 3  The number 0 is not in the domain of f  x  so there are no y-intercepts. b) Asymptotes. Vertical. The denominator is 0 for x  0 , so the line x  0 is a vertical asymptote. Horizontal. The numerator and the denominator have the same degree, so 3 y  , or y  3 is the horizontal asymptote. 1 Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 f ‘  x  2 x 2 f ”  x   2 x 3   3 x The domain of f is  , 0    0,   as Since x  determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 0, but 0 is not in the domain of the function, so x  0 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. 1 A: Test  1, f ‘  1  1 0  12 B: Test 1, f ‘ 1  32. 3x  1 x a) Intercepts. To find the x-intercepts, solve f  x  0 . f  x  3x  1 0 x 3x  1  0 3x  1 x 1 3 1  12 1 0 Then f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 0, but because 0 is not in the domain of the function, there cannot be an inflection point at 0. The equation f ”  x   0 has no solution; therefore, there are no inflection points. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 301 g) Concavity. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. A: Test  1, f ”  1  B: Test 1, f ” 1  2  13 2 13 20 d) Critical Points. f ‘  x  exists for all values of x except 0, but 0 is not in the domain of the function, so x  0 is not a critical value. The critical points will occur when f ‘  x   0 . 1  2  0 Therefore, f  x  is concave up on  , 0  and concave down on  0,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. 2 0 x2 2 1 2 x 2 x 2 x 2 Thus,  2 and 2 are critical values.    2   2 2 , so the critical points   2, 2 2  and  2, 2 2  f  2  2 2 and f are on the graph. e) Increasing, decreasing, relative extrema. We use  2 , 0, and 2 to divide the real number line into four intervals   B:  2,0 , C: 0, 2  , and D:  2,   . A: ,  2 33. 2 x2  2  x x a) Intercepts. The equation f  x   0 has no f  x  x  real solutions, so there are no x-intercepts. The number 0 is not in the domain of f  x  so there are no y-intercepts. b) Asymptotes. Vertical. The denominator is 0 for x  0 , so the line x  0 is a vertical asymptote. Horizontal. The degree of the numerator is greater than the degree of the denominator, so there are no horizontal asymptotes. Slant. The degree of the numerator is exactly one greater than the degree of the denominator. As x approaches  , 2 approaches x. Therefore, x y  x is the slant asymptote. c) Derivatives and Domain. 2 f ‘  x   1  2 x 2  1  2 x 4 f ”  x   4 x 3  3 x The domain of f is  , 0    0,   as f  x  x  determined in step (b). 2 A: Test  2, f ‘  2   1   2  B: Test  1, f ‘  1  1  C: Test 1, f ‘ 1  1  D: Test 2, f ‘  2   1  2 12 2  2 2 2 2  12  1 0 2  1  0  1  0  1 0 2 Then f  x  is increasing on ,  2  and  2,  and is decreasing on  2, 0 and 0, 2  . Therefore,  2, 2 2  is a relative maximum, and  2, 2 2  is a relative minimum. f) Inflection points. f ”  x  does not exist at 0, but because 0 is not in the domain of the function, there cannot be an inflection point at 0. The equation f ”  x   0 has no solution; therefore, there are no inflection points. Copyright © 2016 Pearson Education, Inc. 302 Chapter 2: Applications of Differentiation g) Concavity. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. A: Test  1, f ”  1  B: Test 1, f ” 1  4  13 4 13  4  0 40 Therefore, f  x  is concave down on  , 0 and concave up on 0,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. d) Critical Points. f ‘  x  exists for all values of x except 0, but 0 is not in the domain of the function, so x  0 is not a critical value. The solution to f ‘  x   0 is x  3 . Thus, 3and 3 are critical values. f  3  6 and f  3  6 , so the critical points  3, 6  and  3, 6  are on the graph. e) Increasing, decreasing, relative extrema. We use 3, 0, and 3 to divide the real number line into four intervals A:  , 3 B:  3,0 , C:  0, 3 , and D:  3,   . 7 0 16 B: Test  1, f ‘  1  8  0 A: Test  4, f ‘  4   C: Test 1, f ‘ 1  8  0 7 0 16 Then f  x  is increasing on D: Test 4, f ‘  4   34. 9 x a) Intercepts. The equation f  x   0 has no f  x  x  real solutions, so there are no x-intercepts. The number 0 is not in the domain of f  x  so there are no y-intercepts. b) Asymptotes. Vertical. The denominator is 0 for x  0 , so the line x  0 is a vertical asymptote. Horizontal. The degree of the numerator is greater than the degree of the denominator, so there are no horizontal asymptotes. 9 Slant. As x approaches  , f  x   x  x approaches x. Thus, y  x is the slant asymptote. c) Derivatives and Domain. 9 f ‘  x   1  9 x 2  1  2 x 18 3 f ”  x   18 x  3 x The domain of f is  , 0    0,   as , 3 and 3,   and is decreasing on 3, 0  and 0,3 . Therefore,  3, 6 is a relative maximum, and  3, 6  is a relative minimum. f) Inflection points. f ”  x  does not exist at 0, but because 0 is not in the domain of the function, there cannot be an inflection point at 0. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. A: Test  1, f ”  1  18  0 B: Test 1, f ” 1  18  0 Therefore, f  x  is concave down on  , 0 and concave up on 0,   . determined in step (b). Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 303 Then f  x  is decreasing on  , 0  and is h) Sketch. increasing on  0,   . Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 0, but because 0 is not in the domain of the function, there cannot be an inflection point at 0. The equation f ”  x   0 has no 35. 1   x 2 x2 a) Intercepts. Since the numerator is the constant 1 , there are no x-intercepts. The number 0 is not in the domain of the function, so there are no y-intercepts. b) Asymptotes. Vertical. The denominator is 0 for x  0 , so the line x  0 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 2 f ‘  x   2 x 3  3 x 6 4 f ”  x   6 x   4 x The domain of f is  , 0    0,   as determined in step (b). d) Critical Points. f ‘  x  exists for all values of solution; therefore, there are no inflection points. g) Concavity. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a f  x  x except 0, but 0 is not in the domain of the function, so x  0 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. 2 A: Test  1, f ‘  1   2  0  13 B: Test 1, f ‘ 1  2 13 20 point in each interval. A: Test  1, f ”  1   B: Test 1, f ” 1   6  14 6 14  6  0  6  0 Therefore, f  x  is concave down on both intervals. h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. 36. 2  2 x 2 x2 a) Intercepts. Since the numerator is the constant 2, there are no x-intercepts. The number 0 is not in the domain of the function, so there are no y-intercepts. b) Asymptotes. Vertical. The denominator is 0 for x  0 , so the line x  0 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. f  x  Copyright © 2016 Pearson Education, Inc. 304 Chapter 2: Applications of Differentiation c) Derivatives and Domain. 4 f ‘  x   4 x 3   3 x 12 4 f ”  x   12 x  4 x The domain of f is  , 0    0,   as determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 0, but 0 is not in the domain of the function, so x  0 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. A: Test  1, f ‘  1  4  0 B: Test 1, f ‘ 1  4  0 Then f  x  is increasing on  , 0  and is decreasing on  0,   . Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 0, but because 0 is not in the domain of the function, there cannot be an inflection point at 0. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. A: Test  1, f ”  1  B: Test 1, f ” 1  12  1 12 14 4  12  0  12  0 Therefore, f  x  is concave up on both intervals. h) Sketch. 37. x x3 a) Intercepts. To find the x-intercepts, solve f  x  0 . f  x  x 0 x3 x0 Since x  0 does not make the denominator 0, the x-intercept is  0, 0  . f  0   0 , so the y-intercept is  0, 0  also. b) Asymptotes. Vertical. The denominator is 0 for x  3 , so the line x  3 is a vertical asymptote. Horizontal. The numerator and the denominator have the same degree, so 1 y  , or y  1 is the horizontal asymptote. 1 Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 3 f ‘  x   x  32 f ”  x   6  x  3 3  6  x  33 The domain of f is  , 3  3,   as determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 3, but 3 is not in the domain of the function, so x  3 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 3 to divide the real number line into two intervals A:  , 3 and B:  3,   , and we test a point in each interval. 3 A: Test 2, f ‘  2   3  0 2 2  3 B: Test 4, f ‘ 4  3 4  3 2  3  0 Then f  x  is decreasing on both intervals. Since there are no critical points, there are no relative extrema. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 305 f) Inflection points. f ”  x  does not exist at 3, but because 3 is not in the domain of the function, there cannot be an inflection point at 3. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 3 to divide the real number line into two intervals A:  , 3 and B:  3,   , and we test a point in each interval. A: Test 2, f ” 2   B: Test 4, f ”  4  6 2  3 3 6  6  0 60 3 4  3 Therefore, f  x  is concave down on  , 3 and concave up on 3,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. c) Derivatives and Domain. 2 f ‘  x   x  2 2 f ”  x   4  x  2  3  4  x  23 The domain of f is  , 2    2,   as determined in step (b). d) Critical Points. f ‘  x  exists for all values of x except 2 , but 2 is not in the domain of the function, so x  2 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 2 to divide the real number line into two intervals A:  , 2  and B:  2,   , and we test a point in each interval. A: Test  3, f ‘  3  2  0 B: Test  1, f ‘  1  2  0 Then f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 2 , 38. x x2 a) Intercepts. The numerator is 0 for x  0 and since this value of x does not make the denominator 0, the x-intercept is  0, 0  . f  x  f  0   0 , so the y-intercept is  0, 0  also. b) Asymptotes. Vertical. The denominator is 0 for x  2 , so the line x  2 is a vertical asymptote. Horizontal. The numerator and the denominator have the same degree, so 1 y  , or y  1 is the horizontal asymptote. 1 Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. but because 2 is not in the domain of the function, there cannot be an inflection point at 2 . The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 2 to divide the real number line into two intervals A:  , 2  and B:  2,   , and we test a point in each interval. A: Test  3, f ”  3  4  0 B: Test  1, f ” 1  4  0 Therefore, f  x  is concave up on  , 2  and concave down on  2,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. Copyright © 2016 Pearson Education, Inc. 306 39. Chapter 2: Applications of Differentiation  f) Inflection points. f ”  x  exists for all real  1 1  x2  3 x 3 a) Intercepts. Since the numerator is the constant 1, there are no x-intercepts. 1 1 f 0    , so the y-intercept is 2 0   3 3 f  x  2 numbers. Solve f ”  x   0 . 6 x2  6  x  2   6 x2  6 x2  1 x  1 1 1 f  1  and f 1  4 4 1 1     So,  1,  and 1,  are possible points   4 4 of inflection. g) Concavity. We use 1 and 1 to divide the real number line into three intervals A:  , 1 B:  1,1 , and C: 1,   18 0 343 2 B: Test 0, f ”  0    0 9 18 C: Test 2, f ”  2   0 343 Therefore, f  x  is concave up on   , 1  A: Test  2, f ”  2   6 x2  6 f ”  x    x  3 2 3 The domain of f is  as determined in step (b). d) Critical Points. f ‘  x  exists for all real and 1,   , and concave down on  1,1 . numbers. Solve f ‘  x   0   2x  x2  3 2 1   1 Thus the points  1,  and 1,  are    4 4 points of inflection. h) Sketch. 0 2x  0 x0 The critical value is 0. From step (a) we  1 found  0,  is on the graph.  3 e) Increasing, decreasing, relative extrema. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. 1 A: Test  1, f ‘  1   0 8 1 B: Test 1, f ‘ 1    0 8 Then f  x  is increasing on  , 0 and is  1 decreasing on 0,   . Thus  0,  is a  3 relative maximum. 0 6 x2  6  0  1  0,  . 3 b) Asymptotes. Vertical. x 2  3  0 has no real solution, so there are no vertical asymptotes. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 2 2x f ‘  x   2 x x 2  3   2 x2  3  3 2 40.   1 1   x2  2 x 2 a) Intercepts. Since the numerator is the constant 1 , there are no x-intercepts. 1 1  f 0    , so the y-intercept is  0,   .  2 2 b) Asymptotes. Vertical. x 2  2  0 has no real solution, so there are no vertical asymptotes. f  x  2 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 307 Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 2 2x f ‘  x  2 x x2  2  2 2 x 2  f ”  x     2 2 to divide and 3 3 the real number line into three intervals   2 2 2 A:  ,  B:   ,  , 3   3 3 g) Concavity. We use   2  and C:  ,   3   A: Test  1, f ”  1   2 6 x  4  x  2 2 3 2 0 27 1 0 2 2 C: Test 1, f ” 1   0 27 Therefore, f  x  is concave down on 0, f ”  0  B: Test The domain of f is  . d) Critical Points. f ‘  x  exists for all real numbers. f ‘  x   0 for x  0 , so 0 is a critical value. From step (a) we already 1  know  0,   is on the graph.  2 e) Increasing, decreasing, relative extrema. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and   2  2  ,  3  and  3 ,   and concave up      2 2 on   ,  . Therefore the points  3 3  2 3  2 3   3 ,  8  and  3 ,  8  are points of     inflection. h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. we test a point in each interval. 2 A: Test  1, f ‘  1    0 9 2 B: Test 1, f ‘ 1   0 9 Then f  x  is decreasing on  , 0 and is 1  increasing on 0,   . Thus  0,   is a  2 relative minimum. f) Inflection points. f ”  x  exists for all real numbers. f ”  x   0 for x    2 and 3 2 , so 3 2 are possible inflection points. 3  2  2 3 3 f    and f     8 8  3  3  2 3  2 3 ,   and  ,   are points of So,    3 8  3 8 inflection. 41. x3 x3 1   , x  3 x 2  9  x  3 x  3 x  3 We write the expression in simplified form noting that the domain is restricted to all real numbers except for x  3. a) Intercepts. f  x   0 has no solution. x  3 f  x  is not in the domain of the function. Therefore, there are no x-intercepts. To find the y-intercepts we compute f  0  . f 0  1 1  0  3 3 1  The point  0,   is the y-intercept.  3 Copyright © 2016 Pearson Education, Inc. 308 Chapter 2: Applications of Differentiation f) Inflection points. f ”  x  does not exist at 3, b) Asymptotes. Vertical. In the original function, the denominator is 0 for x  3 or x  3 , however, x  3 also made the numerator equal to 0. We look at the limits to determine if there are vertical asymptotes at these points. x3 1 1 1 lim  lim   . x 3 x 2  9 x 3 x  3 3  3 6 Because the limit exists, the line x  3 is not a vertical asymptote. Instead, we have a removable discontinuity, or a “hole” at the 1  point  3,   .  6 but because 3 is not in the domain of the function, there cannot be an inflection point at 3. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 3 and 3 to divide the real number line into three intervals A:  , 3 B:  3, 3 and C:  3,   and we test a point in each interval. A: Test  4, f ”  4   B: Test 2, f ”  2   1  An open circle is drawn at  3,   to show  6 that it is not part of the graph. The denominator is 0 for x  3 and the numerator is not 0 at this value, so the line x  3 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 2 f ‘  x     x  3   x  32 f ”  x   2  x  3 3  C: Test 4, f ”  4   2  4   3 2 2  3 3 2  4  3 3 3  2 343 0  2  0 20 Therefore, f  x  is concave down on  , 3 and  3, 3 and concave up on 3,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. 2  x  33 The domain of f as determined in step (b) is , 3  3, 3  3,   . d) Critical Points. f ‘  x  exists for all values of x except 3, but 3 is not in the domain of the function, so x  3 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 3 and 3 to divide the real number line into three intervals A:  , 3 B:  3, 3 and C:  3,   . We notice that f ‘  x   0 for all real numbers, f  x  is decreasing on all three intervals  , 3 ,  3, 3 , and  3,   . Since there are no critical points, there are no relative extrema. 42. x 1 1  , x  1 2 x 1 x 1 We write the expression in simplified form noting that the domain is restricted to all real numbers except for x  1. a) Intercepts. f  x   0 has no solution, so there are no x-intercepts. To find the y-intercepts we compute f  0  f  x  f 0  1 1 0  1 The point  0,1 is the y-intercept. b) Asymptotes. Vertical. In the original function, the denominator is 0 for x  1 or x  1 , however, x  1 also made the numerator equal to 0. The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 309 g) Concavity. We use 1 and 1 to divide the real number line into three intervals A:  , 1 B:  1,1 and C: 1,   . We look at the limits to determine if there are vertical asymptotes at these points. x 1 1 1 lim 2  lim  . Because the limit x 1 x  1 x 1 x  1 2 exists, the line x  1 is not a vertical asymptote. Instead, we have a removable  1 discontinuity, or a “hole” at the point 1,  .  2 An open circle is drawn at this point to show that it is not part of the graph. The denominator is 0 for x  1 and the numerator is not 0 at this value, so the line x  1 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 2 f ‘  x     x  1   x  12 f ”  x   2  x  1 3  and we test a point in each interval. A: Test  2, f ”  2   2  0 B: Test 0, f ”  0  2  0 2 0 27 Therefore, f  x  is concave down on C: Test 2, f ”  2    , 1 and concave up on  1,1 and 1,   . h) Sketch. 43. 2  x  13 The domain of f is  , 1   1,1  1,   as determined above. d) Critical Points. f ‘  x  exists for all values of x except 1 , but 1 is not in the domain of the function, so x  1 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 1 and 1 to divide the real number line into three intervals A:  , 1 B:  1,1 and C: 1,   . We notice that f ‘  x   0 for all real numbers, f  x  is decreasing on all three intervals  , 1 ,  1,1 , and 1,   . Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 1 , but because 1 is not in the domain of the function, there cannot be an inflection point at 1 . The equation f ”  x   0 has no x 1 x2 a) Intercepts. To find the x-intercepts, solve f  x  0 . f  x  x 1 0 x2 x 1 Since x  1 does not make the denominator 0, the x-intercept is 1, 0  . 0 1 1   , so the y-intercept is 02 2 1   0,   . 2 b) Asymptotes. Vertical. The denominator is 0 for x  2 , so the line x  2 is a vertical asymptote. Horizontal. The numerator and the denominator have the same degree, so 1 y  , or y  1 is the horizontal asymptote. 1 Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. f 0  solution; therefore, there are no inflection points. Copyright © 2016 Pearson Education, Inc. 310 Chapter 2: Applications of Differentiation c) Derivatives and Domain. 3 f ‘  x   x  2 2 f ”  x    44. x2 x 1 a) Intercepts. f  x   0 for x  2 and this value f  x  does not make the denominator 0, the xintercept is  2, 0  . f  0   2 , so the y- 6  x  2 3 The domain of f is   , 2    2,   as determined in part (b). d) Critical Points. f ‘  x  exists for all values of x except 2 , but 2 is not in the domain of the function, so x  2 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 2 to divide the real number line into two intervals A:  , 2  and B:  2,   , and we test a point in each interval. A: Test  3, f ‘  3  3  0 B: Test  1, f ‘  1  3  0 Then f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 2 , but because 2 is not in the domain of the function, there cannot be an inflection point at 2 . The equation f ”  x   0 has no intercept is  0, 2  . b) Asymptotes. Vertical. The denominator is 0 for x  1 , so the line x  1 is a vertical asymptote. Horizontal. The numerator and the denominator have the same degree, so 1 y  , or y  1 is the horizontal asymptote. 1 Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 3 f ‘ x   x  12 f ”  x    6  x  13 The domain of f is  , 1   1,   as determined in part (b). d) Critical Points. f ‘  x  exists for all values of x except 1 , but 1 is not in the domain of the function, so x  1 is not a critical value. The equation f ‘  x   0 has no solution, so solution; therefore, there are no inflection points. g) Concavity. We use 2 to divide the real number line into two intervals A:  , 2  and B:  2,   , and we test a point in each interval. A: Test  3, f ”  3  6  0 there are no critical points. e) Increasing, decreasing, relative extrema. We use 1 to divide the real number line into two intervals A:  , 1 and B:  1,   . We notice that Therefore, f  x  is concave up on  , 2  there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 1 , B: Test  1, f ”  1  6  0 and concave down on  2,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. f ‘  x   0 for all real numbers; therefore, f  x  is increasing on both intervals. Since but because 1 is not in the domain of the function, there cannot be an inflection point at 1 . The equation f ”  x   0 has no solution; therefore, there are no inflection points. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 311 g) Concavity. We use 1 to divide the real number line into two intervals A:  , 1 and B:  1,   , and we test a point in each interval. A: Test  2, f ”  2   6  0 B: Test 0, f ”  0  6  0 Therefore, f  x  is concave up on   , 1 and concave down on  1,   . h) Sketch. As x approaches  , f  x  approaches x  1 , so y  x  1 is the slant asymptote. c) Derivatives and Domain. x2  2 x  9 f ‘  x   x  12 f ”  x    16  x  13 The domain of f is  , 1   1,   as determined in part (b). d) Critical Points. f ‘  x  exists for all values of x except 1 , but 1 is not in the domain of the function, so x  1 is not a critical value. f ‘  x   0 has no real solution, so there are 45. x2  9 x 1 a) Intercepts. To find the x-intercepts, solve f  x  0 . f x  x2  9 0 x 1 x2  9  0 x  3 Neither of these values make the denominator 0, so the x-intercepts are  3, 0 and 3, 0  . f  0   9 , so the y-intercept is  0, 9  . b) Asymptotes. Vertical. The denominator is 0 for x  1 , so the line x  1 is a vertical asymptote. Horizontal. The degree of the numerator is greater than the degree of the denominator, so there are no horizontal asymptotes. Slant. Divide the numerator by the denominator. 1 x 2 9 x 1 x x2  x  x9 x  1 no critical points. e) Increasing, decreasing, relative extrema. We use 1 to divide the real number line into two intervals A:  , 1 and B:  1,   We test a point in each interval. A: Test  2, f ‘  2   9  0 B: Test 0, f ‘  0  9  0 Then f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 1 , but because 1 is not in the domain of the function, there cannot be an inflection point at 1 . The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 1 to divide the real number line into two intervals A:  , 1 and B:  1,   , and we test a point in each interval. A: Test  2, f ”  2   16  0 B: Test 0, f ”  0  16  0 Therefore, f  x  is concave up on   , 1 and concave down on  1,   . h) Sketch. 8 By dividing, we get 8 f  x  x  1 x 1 Copyright © 2016 Pearson Education, Inc. 312 46. Chapter 2: Applications of Differentiation x2  4 x3 a) Intercepts. The numerator is 0 for x  2 or x  2 and neither of these values make the denominator 0, so the x-intercepts are 02  4 4   , so  2, 0  and  2, 0 . f 0  03 3 4  the y-intercept is  0,   .  3 b) Asymptotes. Vertical. The denominator is 0 for x  3 , so the line x  3 is a vertical asymptote. Horizontal. The degree of the numerator is greater than the degree of the denominator, so there are no horizontal asymptotes. Slant. By dividing the numerator by the denominator, we get 5 f  x  x  3  x3 As x approaches  , f  x  approaches f x  x  3 , so y  x  3 is the slant asymptote. c) Derivatives and Domain. x2  6x  4 f ‘  x   x  32 f ”  x   10  x  33 The domain of f is  , 3   3,   as determined in part (b). d) Critical Points. f ‘  x  exists for all values of x except 3 , but 3 is not in the domain of the function, so x  3 is not a critical value. Solve f ‘  x   0 . x2  6x  4  x  32 We test a point in each interval. 4 A: Test  6, f ‘  6   0 9 B: Test  4, f ‘  4   4  0 C: Test  2, f ‘  2   4  0 4 0 9 Then f  x  is increasing on the intervals D: Test 0, f ‘  0  , 5.236 and 0.764,   , and is decreasing on the intervals 5.226, 3 and 3, 0.764 . Therefore,  5.236, 10.472  is a relative maximum and  0.764, 1.528  is a relative minimum. f) Inflection points. f ”  x  does not exist at 3 , but because 3 is not in the domain of the function, there cannot be an inflection point at 3 . The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 3 to divide the real number line into two intervals A:  , 3 and B:  3,   , and we test a point in each interval. A: Test  4, f ”  4   10  0 B: Test  2, f ”  2   10  0 Therefore, f  x  is concave down on  , 3 and concave up on  3,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. 0 x2  6x  4  0 Using the x  3  5 Quadratic Formula x  5.236 or x  0.764 f  5.236   10.472 and f  0.764   1.528 , so  5.236, 10.472  and  0.764, 1.528  are on the graph. e) Increasing, decreasing, relative extrema. We use 5.236,  3, and  0.764 to divide the real number line into four intervals A:  , 5.236 , B:  5.236, 3 , C:  3, 0.764 , and D:  0.764,   Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 47. f  x  313 x3 2 x  2 x  15  x3 1  ,  x  3 x  5 x  5 x  3. We write the expression in simplified form noting that the domain is restricted to all real numbers except for x  5 and x  3 . a) Intercepts. f  x   0 has no solution. x  3 is not in the domain of the function. Therefore, there are no x-intercepts. To find the y-intercepts we compute f  0  : f 0  03 02  2 0  15  1 5  1 The point  0,  is the y-intercept.  5 b) Asymptotes. Vertical. In the original function, the denominator is 0 for x  5 or x  3 ; however, x  3 also made the numerator equal to 0. We look at the limits to determine if there are vertical asymptotes at these points. x3 1 1 lim 2  lim  x 3 x  2 x  15 x 3 x  5 8 Because the limit exists, the line x  3 is not a vertical asymptote. Instead, we have a removable discontinuity, or a “hole” at the  1 point  3,  . An open circle is drawn at this  8 point to show that it is not part of the graph. The denominator is 0 for x  5 and the numerator is not 0 at this value, so the line x  5 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 2 f ‘  x     x  5   x  5 2 f ”  x   2  x  5 3  d) Critical Points. f ‘  x  exists for all values of x except 5 , but 5 is not in the domain of the function, so x  5 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 5 and 3 to divide the real number line into three intervals A:   , 5 , B:  5, 3 , and C:  3,   . We notice that f ‘  x   0 for all real numbers, f  x  is decreasing on all three intervals  , 5 ,  5, 3 , and 3,   . Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 5 , but because 5 is not in the domain of the function, there cannot be an inflection point at 5 . The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 5 and 3 to divide the real number line into three intervals A:  , 5 B:  5, 3 and C:  3,   , and we test a point in each interval. A: Test  6, f ”  6   2  0 B: Test  4, f ”  4   2  0 2 0 729 Therefore, f  x  is concave down on C: Test 4, f ”  4    , 5 and concave up on  5, 3 and 3,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. 2  x  53 The domain of f as determined in part (b) is , 5  5, 3  3,   . Copyright © 2016 Pearson Education, Inc. 314 48. Chapter 2: Applications of Differentiation d) Critical Points. f ‘  x  exists for all values of x 1 x 1 1   , x  2 x  3  x  3 x  1 x  3 x  1. We write the expression in simplified form noting that the domain is restricted to all real numbers except for x  1 and x  3 a) Intercepts. f  x   0 has no solution. x  1 f  x  2 x except 3, but 3 is not in the domain of the function, so x  3 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 1 and 3 to divide the real number line into three intervals A:  , 1 B:  1, 3 and C:  3,   . is not in the domain of the function. Therefore, there are no x-intercepts. To find the y-intercepts we compute f  0  : We notice that f ‘  x   0 for all real 0 1 numbers, so f  x  is decreasing on all three 1 f 0   2 0  2  0  3 3 intervals  , 1 ,  1, 3 , and 3,   . 1  The point  0,   is the y-intercept.  3 b) Asymptotes. Vertical. In the original function, the denominator is 0 for x  1 or x  3 , however, x  1 also made the numerator equal to 0. We look at the limits to determine if there are vertical asymptotes at these points. x 1 1 1 1 lim 2  lim   x 1 x  2 x  3 x 1 x  3 1  3 4 Because the limit exists, the line x  1 is not a vertical asymptote. Instead, we have a removable discontinuity, or a “hole” at the 1  point  1,   . An open circle is drawn at  4 this point to show that it is not part of the graph. The denominator is 0 for x  3 and the numerator is not 0 at this value, so the line x  3 is a vertical asymptote. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 2 f ‘  x     x  3   x  32 f ”  x   2  x  3 3  Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  does not exist at 3, but because 3 is not in the domain of the function, there cannot be an inflection point at 3. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 1 and 3 to divide the real number line into three intervals A:  , 1 B:  1, 3 and C:  3,   , and we test a point in each interval. A: Test  2, f ”  2    2 125 0 B: Test 2, f ”  2   2  0 C: Test 4, f ”  4  2  0 Therefore, f  x  is concave down on  , 1 and  1, 3 and concave up on 3,   . h) Sketch. 2  x  33 The domain of f as determined in part (b) is , 1  1, 3  3,   . 49. 2 x2 x 2  16 a) Intercepts. The numerator is 0 for x  0 and this value does not make the denominator 0, so the x-intercept is  0, 0  . f  x  f  0   0 , so the y-intercept is  0, 0  also. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 315 f) Inflection points. f ”  x  does not exist at b) Asymptotes. Vertical. The denominator is 0 when x 2  16  0 4 and 4 , but because 4 and 4 are not in the domain of the function, there cannot be an inflection point at 4 or 4 . The equation x 2  16 f ”  x   0 has no real solution; therefore, x  4 So the lines x  4 and x  4 are vertical asymptotes. Horizontal. The numerator and the denominator have the same degree, so 2 y  , or y  2 is the horizontal asymptote. 1 Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 64 x f ‘  x   2 x 2  16  f ”  x   there are no inflection points. g) Concavity. We use 4 and 4 to divide the real number line into three intervals A:  , 4  B:  4, 4  and C:  4,   , and we test a point in each interval. 5824 A: Test  5, f ”  5  0 729 1 B: Test 0, f ”  0    0 4 5824 0 C: Test 5, f ” 5  729 Therefore, f  x  is concave up on the  192 x 2  1024 intervals  , 4  and  4,   and concave  x  16 2 3 down on the interval  4, 4  . The domain of f is h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed.  ,  4     4, 4    4,   as determined in part (b). d) Critical Points. f ‘  x  exists for all values of x except x  4 and x  4 , but 4 and 4 are not in the domain of the function, so x  4 and x  4 are not critical values. f ‘  x   0 for x  0 , so  0, 0  is the only critical point. e) Increasing, decreasing, relative extrema. We use 4, 0, and 4 to divide the real number line into four intervals A:  , 4 B:  4, 0 , C: 0, 4 , and D:  4,   . We test a point in each interval. 320 A: Test  5, f ‘ 5  0 81 64 B: Test  1, f ‘  1  0 225 64 C: Test 1, f ‘ 1   0 225 320 D: Test 5, f ‘ 5   0 81 Then f  x  is increasing on the intervals 50. x 2  x  2  x  2  x  1 x2   2  x  1 x  1 2  x  1 2×2  2 We write the expression in simplified form noting that the domain is restricted to all real numbers except for x  1. a) Intercepts. f  x   0 for x  2 ; therefore, f  x  the x-intercept is  2, 0  .  0 2   0   2  1 2 2  0  2 The point  0,1 is the y-intercept. f  0  , 4 and 4, 0 , and is decreasing on the intervals 0, 4  and  4,   . Thus, there is a relative maximum at  0, 0  . Copyright © 2016 Pearson Education, Inc. 316 Chapter 2: Applications of Differentiation f) Inflection points. f ”  x  does not exist at 1 , b) Asymptotes. Vertical. In the original function, the denominator is 0 for x  1 or x  1 , however, x  1 also made the numerator equal to 0. We look at the limits to determine if there are vertical asymptotes at these points. x2  x  2 x2 3 lim  lim  . Because x 1 2 x 2  2 x 1 2  x  1 4 the limit exists, the line x  1 is not a vertical asymptote. Instead, we have a removable  3 discontinuity, or a “hole” at the point 1,  .  4 An open circle is drawn at this point to show that it is not part of the graph. The denominator is 0 for x  1 and the numerator is not 0 at this value, so the line x  1 is a vertical asymptote. Horizontal. The numerator and the denominator have the same degree, so 1 y  is the horizontal asymptote. 2 Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 1 2 f ‘  x     x  1  2 2  x  1 f ”  x   2  x  1 3  1  x  1 The domain of f is  , 1   1,1  1,   3 as determined in part (b). d) Critical Points. f ‘  x  exists for all values of x except 1 , but 1 is not in the domain of the function, so x  1 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 1 and 1 to divide the real number line into three intervals A:  , 1 B:  1,1 and C: 1,   . We notice that f ‘  x   0 for all real numbers, f  x  is decreasing on all three intervals  , 1 ,  1,1 , and 1,   . Since there are no critical points, there are no relative extrema. but because 1 is not in the domain of the function, there cannot be an inflection point at 1 . The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 1 and 1 to divide the real number line into three intervals A:  , 1 B:  1,1 and C: 1,   . and we test a point in each interval. A: Test  2, f ”  2   1  0 B: Test 0, f ” 0  1  0 1 0 27 Therefore, f  x  is concave down on the C: Test 2, f ”  2   interval   , 1 and concave up on the intervals  1,1 and 1,   . h) Sketch. 51. 10 x 4 a) Intercepts. Since the numerator is the constant 10, there are no x-intercepts. 5  5 f 0  , so the y-intercept is  0,  .  2 2 b) Asymptotes. Vertical. x 2  4  0 has no real solution, so there are no vertical asymptotes. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 2 20 x f ‘  x   20 x x 2  4   2 2 x 4 f  x  2  f ”  x     2 20 3 x  4  x  4 2  3 The domain of f is  . Copyright © 2016 Pearson Education, Inc.   Exercise Set 2.3 317 d) Critical Points. f ‘  x  exists for all real Therefore, f  x  is concave up on numbers. f ‘  x   0 for x  0 , so 0 is a   2  2  ,   and concave  ,   and  3 3  critical value. From step (a) we already  5 know  0,  is on the graph.  2 e) Increasing, decreasing, relative extrema. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and  2 2  , down on    . Thus, the points  3 3  2 15   2 15  ,  and  , are points of    3 8  3 8 inflection. h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. we test a point in each interval. 4 A: Test  1, f ‘  1   0 5 4 B: Test 1, f ‘ 1    0 5 Then f  x  is increasing on  , 0 and is  5 decreasing on 0,   . Thus  0,  is a  2 relative maximum. f) Inflection points. f ”  x  exists for all real 2 , which 3 are possible points of inflection. We have:  2  15  2  15 f  and f   .    3  8 3 8 numbers. f ”  x   0 for x    2 15   2 15  ,  and  , are inflection So,    3 8  3 8  points. 2 2 g) Concavity. We use  to divide and 3 3 the real number line into three intervals   2 2  2  A:  ,  B:   ,   , and   3 3 3  2  C:  ,  .  3  5 0 16 5 0, f ”  0    0 4 5 0 2, f ”  2   16 A: Test  2, f ”  2   B: Test C: Test 52. 1 x2  1 a) Intercepts. Since the numerator is a constant 1, there are no x-intercepts. 1 f 0   1  0 2  1 f  x  The point  0, 1 is the y-intercept. b) Asymptotes. Vertical. The denominator x 2  1   x  1 x  1 is 0 for x  1 or x  1 , so the lines x  1 and x  1 are vertical asymptotes. Horizontal. The degree of the numerator is less than the degree of the denominator, so y  0 is the horizontal asymptote. Slant. There is no slant asymptote since the degree of the numerator is not one more than the degree of the denominator. c) Derivatives and Domain. 2 x f ‘  x  2 x2  1  f ”  x       x  1 2 2 3x  1 2 3 The domain of f is  , 1   1,1  1,   as determined in part (b). Copyright © 2016 Pearson Education, Inc. 318 Chapter 2: Applications of Differentiation d) Critical Points. f ‘  x  exists for all values of h) Sketch. x except 1 and 1 , but these values are not in the domain of the function, so x  1 and x  1 are not critical values. f ‘  x   0 for x  0 . From step (a) we know f  0   1 , so the critical point is  0, 1 . e) Increasing, decreasing, relative extrema. We use 1, 0, and 1 to divide the real number line into four intervals A:  , 1 , B:  1, 0  , C:  0,1 , and D: 1,   , and we test a point in each interval. 4 A: Test  2, f ‘  2   0 9 1  1  16 B: Test  , f ‘     0  2 9 2 1 16 1 , f ‘     0 2 2 9 4 f ‘ 2    0 D: Test 2, 9 We see that f  x  is increasing on the C: Test intervals , 1 and  1, 0  , and is decreasing on the intervals 0,1 and 1,  . Therefore,  0, 1 is a relative maximum. f) Inflection points. f ”  x  does not exist at 1 and 1 , but because these values are not in the domain of the function, there cannot be an inflection point at 1 or 1 . The equation f ”  x   0 has no real solution; therefore, there are no inflection points. g) Concavity. We use 1 and 1 to divide the real number line into three intervals A:  , 1 B:  1,1 and C: 1,   . and we test a point in each interval. 26 A: Test  2, f ”  2   0 27 B: Test 0, f ”  0  2  0 26 0 27 Therefore, f  x  is concave up on   , 1 C: Test 2, f ”  2   and 1,   , and concave down on  1,1 . 53. x2  1 x a) Intercepts. The equation f  x   0 has no f  x  real solutions, so there are no x-intercepts. The number 0 is not in the domain of f  x  so there are no y-intercepts. b) Asymptotes. Vertical. The denominator is 0 for x  0 , so the line x  0 is a vertical asymptote. Horizontal. The degree of the numerator is greater than the degree of the denominator, so there are no horizontal asymptotes. Slant. The degree of the numerator is exactly one greater than the degree of the denominator. When we divide the numerator by the denominator we have x2  1 1 f x   x  . As x approaches x x 1 , f  x   x  approaches x. Therefore, x y  x is the slant asymptote. c) Derivatives and Domain. x2  1 f ‘  x  x2 2 f ”  x   3 x The domain of f is   , 0    0,   as determined in part (b). d) Critical Points. f ‘  x  exists for all values of x except 0, but 0 is not in the domain of the function, so x  0 is not a critical value. The critical points will occur when f ‘  x   0 . x2  1 0 x2 x2  1  0 x2  1 x  1 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 319 From the previous page, we determined 1 and 1 are critical values, thus f  1  2 and f 1  2 , so the critical points  1, 2  and 1, 2  are on the graph. e) Increasing, decreasing, relative extrema. We use 1, 0, and 1 to divide the real number line into four intervals A:  , 1 B:  1, 0  , C: 0,1 , and D: 1,   . We test a point in each interval. 3 f ‘  2    0 A: Test  2, 4 1  1 B: Test  , f ‘     3  0  2 2 C: Test 1 , 2 1 f ‘    3  0 2 3 0 4 Then f  x  is increasing on D: Test 2, f ‘  2  , 1 and 1,   and is decreasing on 1, 0 and 0,1 . Therefore,  1, 2  is a relative maximum, and 1, 2  is a relative minimum. f) Inflection points. f ”  x  does not exist at 0, but because 0 is not in the domain of the function, there cannot be an inflection point at 0. The equation f ”  x   0 has no solution; therefore, there are no inflection points. g) Concavity. We use 0 to divide the real number line into two intervals A:  , 0  and B:  0,   , and we test a point in each interval. A: Test  1, f ”  1  2  0 B: Test 1, f ” 1  2  0 Therefore, f  x  is concave down on  , 0 and concave up on 0,   . h) Sketch. Use the preceding information to sketch the graph. Compute additional function values as needed. 54. x3 x 1 a) Intercepts. The numerator is 0 for x  0 and this value does not make the denominator 0, the x-intercept is  0, 0  . f  x  2 f  0   0 , so the y-intercept is  0, 0  also. b) Vertical. The denominator x 2  1   x  1 x  1 is 0 for x  1 or x  1 , so the lines x  1 and x  1 are vertical asymptotes. Horizontal. The degree of the numerator is greater than the degree of the denominator, so there are no horizontal asymptotes. Slant. The degree of the numerator is exactly one greater than the degree of the denominator. When we divide the numerator by the denominator we have x3 x f  x  2  x 2 . As x x 1 x 1 x approaches  , f  x   x  2 approaches x 1 x. Therefore, y  x is the slant asymptote. c) Derivatives and Domain. x 4  3x 2 f ‘  x  2 x2  1  f ”  x    3 2x  6x  x  1 2 3 The domain of f as determined in part (b) is  , 1   1,1  1,   . d) Critical Points. f ‘  x  exists for all values of x except 1 and 1 , but these values are not in the domain of the function, so x  1 and x  1 are not a critical values. f ‘  x   0 for x   3 , x  0, and x  3 . Therefore,   3 3 , f  0   0 , and 2 3 3 . The critical points f 3  2   3 3 3 3   3,  2  ,  0, 0  and  3, 2  are     on the graph. f  3    Copyright © 2016 Pearson Education, Inc. 320 Chapter 2: Applications of Differentiation e) Increasing, decreasing, relative extrema. We use  3,  1, 0, 1, and 3 to divide the real number line into six intervals 1 208 1 , f ”     0 2 2 27 28 D: Test 2, f ”  2  0 27 Therefore, f  x  is concave down on C: Test   B:  3, 1 , C: 1,0 , D:  0,1 , E: 1, 3  , and F:  3,   . We A: ,  3  , 0  and 0,1 and concave up on  1, 0  and 1,   . Thus, the point 0, 0  is a test a point in each interval. 4 A: Test  2, f ‘  2   0 9 3 27  3 B: Test  , f ‘      0  2 2 25 1 C: Test  , 2 11  1 f ‘      0  2 9 D: Test 1 , 2 11 1 f ‘     0 2 9 E: Test 3 , 2 27 3 0 f ‘    2 25 4 0 9 Then f  x  is increasing on the intervals F: Test 2, f ‘  2  ,  3  and  3,  and is decreasing on the intervals   3, 1 ,  1, 0 , 0,1 , and 1, 3  . Therefore, 0, 0  is not a  3 3  is a relative extremum,   3, 2    3 3 relative maximum, and  3, is a 2   relative minimum. f) Inflection points. f ”  x  does not exist at 1 and 1 , but because these values are not in the domain of the function, there cannot be an inflection point at 1 or 1 . f ”  x   0 when x  0 . We know that f  0   0 , so there is a possible inflection point at  0, 0  . g) Concavity. We use  1, 0, and 1 to divide the real number line into four intervals A:  , 1 B:  1, 0  , C: 0,1 , and D: 1,  , and we test a point in each interval. 28 A: Test  2, f ”  2   0 27 1  1  208 B: Test  , f ”     0  2  27 2 point of inflection. h) Sketch. 55. x 2  16  x  4  x  4    x  4, x  4 x4 x4 Notice that f  x   x  4 for all values of x f  x  except x  4 , where it is undefined. The graph of f  x  will be the graph of y  x  4 except at the point x  4 . a) Intercepts. f  x   0 when x  4 , so the xintercept is  4, 0  . f 0    4. The point  0,  4  is the y-intercept. b) Asymptotes. In simplified form f  x   x  4 , a linear function everywhere except x   4 . So there are no asymptotes of any kind. In the original function, the denominator is 0 for x  4 ; however, x  4 also made the numerator equal to 0. We look at the limits to determine if there are vertical asymptotes at these points. x 2  16 lim  lim  x  4   8 . Because x 4 x  4 x 4 the limit exists, the line x   4 is not a vertical asymptote. Instead, we have a removable discontinuity, or a “hole” at the point  4, 8 . An open circle is drawn at  4, 8 to show that it is not part of the graph. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 321 c) Derivatives and Domain. f ‘  x   1, x  4 f 0  f ”  x   0, x  4 d) Critical Points. There are no critical points. e) Increasing, decreasing, relative extrema. We use 4 to divide the real number line into two intervals A:  , 4  and B:  4,   . We notice that f ‘  x   0 for all real numbers in the domain, f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  is constant; therefore, there are no points of inflection. g) Concavity. f ”  x  is 0; therefore, there is no concavity. h) Sketch. Use the preceding information to sketch the graph. 0 2  9 9  3  0   3 3 The point  0, 3 is the y-intercept. b) Asymptotes. In simplified form f  x   x  3 , a linear function everywhere except x  3 . So there are no asymptotes of any kind. In the original function, the denominator is 0 for x  3 ; however, x  3 also made the numerator equal to 0. We look at the limits to determine if there is a vertical asymptote at this point. x2  9 lim  lim  x  3  6 . x3 x  3 x3 The limit exists; therefore, the line x  3 is not a vertical asymptote. Instead, we have a removable discontinuity, or a “hole” at the point  3, 6  . An open circle is drawn at this point to show that it is not part of the graph. c) Derivatives and Domain. f ‘  x   1, x  3 f ”  x   0, x  3 The domain of f is  , 3  3,   as Note: In the preceding problem, we could have noticed that the graph of x 2  16 is the graph of f  x   x  4 f  x  x4 with the exception of the point 4, 8 which is a removable discontinuity. We simply need to graph f  x   x  4 with a hole at the point 4, 8 and determine all other aspects of the graph of f  x  from the linear graph. 56. x2  9 x3 We write the expression in simplified form:  x  3 x  3  x  3, x  3 f  x  x3 Note that the domain is restricted to all real numbers except for x  3. a) Intercepts. The numerator is 0 when x  3 or x  3 however, x  3 is not in the domain of f  x  , so the x-intercept is f x  determined in part (b). d) Critical Points. f ‘  x  exists for all values of x except 3, but 3 is not in the domain of the function, so x  3 is not a critical value. The equation f ‘  x   0 has no solution, so there are no critical points. e) Increasing, decreasing, relative extrema. We use 3 to divide the real number line into two intervals A:  , 3 and B: 3,   . We notice that f ‘  x   0 for all real numbers in the domain. f  x  is increasing on both intervals. Since there are no critical points, there are no relative extrema. f) Inflection points. f ”  x  is constant; therefore, there are no points of inflection. g) Concavity. f ”  x  is 0; therefore, there is no concavity. h) Sketch.  3, 0 . Copyright © 2016 Pearson Education, Inc. 322 57. Chapter 2: Applications of Differentiation C  x   3x 2  80 C  x  3 x 2  80 80   3x  x x x b) Using the techniques of this section we find the following information. We will only consider the values of x is  0,   . a) A  x   Intercepts. None. Asymptotes. x  0 is the vertical asymptote. There is no horizontal asymptote. As x 80 approaches x 3x. Therefore, y  3 x is the slant asymptote. Increasing, decreasing, relative extrema. 80 A ‘  x   3  2 . A ‘  x  is not defined for x x  0 , however that value is outside the domain of the function. A ‘  x   0 when approaches  , A  x   3x  x  80  80 , and A    2 240 . 3  3  Using x  80 to divide the interval  0,   3   80  80  and  ,  , into two intervals,  0,  3    3  and testing a point in each interval, we find  80  that A  x  is decreasing on  0, and 3    80  ,   . Therefore, the increasing on   3   80  , 2 240  is a relative minimum. point   3  Inflection points, concavity. 160 A ”  x   3 exists for all values of t in x 0,   . The equation A ”  x   0 has no real solution, so there are no possible points of inflection. Furthermore, A ”  x   0 for all x in the domain, so A  x  is concave up on 0,   . We use this information to sketch the graph. Additional values may be computed as necessary. c) The degree of the numerator is exactly one greater than the degree of the denominator. When we divide the numerator by the denominator we have 3 x 2  80 80 . As x A x   3x  x x 80 approaches  , A  x   3x  approaches x 3x. Therefore, y  3 x is the slant asymptote. This means that when a large number of pairs of rocket skates are produced, the average cost can be estimated by multiplying the number of pairs produced by 3 thousand dollars. 58. V  t   50  25t 2  t  2 2 a) V  0   50  25  0  2  0  2  2  50  0  50 The inventory’s value after 0 months is $50 hundreds or $5000. V 5  50  25 5 2   5  2  2  50  625  37.24 49 The inventory’s value after 5 months is $37.24 hundreds or $3724. V 10  50  25 10  2 10  2 2  50  2500  32.64 144 The inventory’s value after 10 months is $32.64 hundreds or $3264. V  70  50  25  70  2 70  2  2  50  122, 500 5184  26.37 The inventory’s value after 70 months is $26.37 hundreds or $2637. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 323 b) Find V ‘  t  and V ”  t  . V ‘ t    Inflection points, concavity. V ”  t  exist for all values of t in  0,   . V ” t   0 , when t  22 50t   25t 2 2 t  21 t  1 . We use this to split the domain into two intervals. A:  0,1 and B: 1,   . t  2  2 2 Testing points in each interval, we see V  t  t  2 t  250t   25t 2 2  t  24  V ” t    is concave down on  0,1 and concave up on 1,   . 100t We use this information to sketch the graph. Additional values may be computed as necessary. t  23 t  23 100  100t 3 t  22 1 t  2  3 2 t  22 100 t  2  100t 3  t  26   200t  200 t  2 this from the horizontal asymptote on the graph. 200t  200 t  24 59. Solve V ‘  t   0 . 100t t  23 Yes; The value below which V will never fall is lim V  t   25 . We observed t  4 V ‘  t  exists for all values of t in  0,   .  d) 0 100t  0 t0 Since t  0 is an endpoint of the domain, there cannot be a relative extrema at t  0 . We notice that V ‘  t   0 for all x in the domain, therefore, V  t  is decreasing over the interval 0,   . Since V  t  is decreasing, the absolute maximum value of the inventory will be $5000 when t  0 . c) Using the techniques of this section, we find the following additional information: Intercepts. There are no t-intercepts in 0,   . The V-intercept is the point 0,50 Asymptotes. There are no vertical asymptotes in  0,   . The line V  25 is a horizontal asymptote. There are no slant asymptotes. Increasing, decreasing, relative extrema. We have already seen that V  t  is C  x   5000  600 x 1 2 x  1000 x 2 a) Profit is revenue minus cost, therefore, the total profit function is: P  x  R  x  C  x R  x   1 2 x  1000 x  5000  600 x  2 1   x 2  400 x  5000 2 P  x  b) A  x   x 1 2 x  400 x  5000  2 x 1 5000   x  400  2 x c) As x approaches  , A  x  approaches  1 1 x  400 . Therefore, y   x  400 is 2 2 the slant asymptote. This represents the average profit for x items, when x is a large number of items.  decreasing over the interval 0,   . There are no relative extrema. Copyright © 2016 Pearson Education, Inc. 324 Chapter 2: Applications of Differentiation d) Using the techniques of this section we find the following additional information. Intercepts. The x-intercepts are 12.70, 0  c) Using the techniques of this section we find the following additional information. Intercepts. No p-intercepts. The point  0, 480  is the C-intercept. and  787.30, 0  . There is no P-intercept. Vertical: p  100 Horizontal : C  0 Slant: None Increasing, decreasing, relative extrema. C  p  is increasing over the interval 0,100  . There are no relative extrema. Inflection points, concavity. C  p  is Asymptotes. x0 Asymptotes. Vertical: Horizonta: None 1 x  400 2 Increasing, decreasing, relative extrema. A  x  is increasing over the interval  0,100 Slant: y and decreasing over the interval 100,   . concave up on the interval  0,100  . there The point 100, 300  is a relative maximum. are no inflection points. We use this information and compute other function values as necessary to draw the graph. Inflection points, concavity. A  x  is concave down on the interval  0,   . There are no inflection points. We use this information and compute other function values as necessary to sketch the graph. d) 60. C  p  48, 000 100  p 48,000 48, 000   480 100  0 100 The cost of removing 0% of the pollutants from a chemical spill is $480. 48,000 48,000 C  20    600 100   20 80 The cost of removing 20% of the pollutants from a chemical spill is $600. 48,000 48, 000 C 80    2400 100  80 20 The cost of removing 80% of the pollutants from a chemical spill is $2400. 48, 000 48,000 C  90    4800 100   90 10 The cost of removing 90% of the pollutants from a chemical spill is $4800. b) The domain of C is 0  p  100 since it is not possible to remove less than 0% or more than 100% of the pollutants, and C  p  is not a) C  0  61. From the result in part (b) we see that there is a vertical asymptote at p  100 . This means that the cost of cleaning up the spill increases without bound as the amount of pollutants removed approaches 100%. The company will not be able to afford to clean up 100% of the pollutants. P  x  a) 1 1  0.0362 x P 5  1  0.84674 1  0.0362 5 In 1995, the purchasing power of a dollar was $0.85. 1 P 10   0.73421439 1  0.0362 10 In 2000, the purchasing power of a dollar was $0.73. 1 P  25   0.5249343 1  0.0362  25 In 2015, the purchasing power of a dollar was $0.52. defined for p  100 . Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 325 b) Solve P  x   0.50 1  0.50 1  0.0362 x 1  0.50 1  0.0362 x  1  0.50  0.0181x 0.50  0.0181x 27.624  x 27.6 years after 1990, or in 2017, the purchasing power of a dollar will be $0.50. c) Find lim P  x  . x  1 lim P  x   lim x  x  1  0.0362 x 1 1  lim x x  1  0.0362 x 1 x 1 x  lim x  1  0.0362 x 0  0 0  0.0362 lim P  x   0 . x  62. A0 A t   2 t 1 If we assume 100 cc is the initial amount 100 . injected, then A  t   2 t 1 100 a) A  0    100  0 2  1 After 0 hours (at the time of injection) there are 100 cc’s of medication in the bloodstream. 100 A 1  2  50 1  1 After 1 hour, there are 50 cc’s of the medication in the bloodstream. 100 A 2   20 2  2  1 After 2 hours, there are 20 cc’s of the medication in the bloodstream. 100 A 7  2  7 2  1 After 7 hours, there are 2 cc’s of the medication in the bloodstream. 100 A 10  102  1  0.99009901 After 10 hours, there is approximately 0.9901 cc’s of the medication in the bloodstream. 200t . Notice that A ‘  t   0 for b) A ‘  t   2 t2 1   all t in the interval  0,   , so there are no relative extrema, and A  t  is decreasing on the interval 0,   . Therefore, the maximum value of A  t  is the initial value of 100 cc at the time of injection. c) Using the techniques of this section we find the following additional information. Intercepts. There are no t-intercepts. The A-intercept is  0,100  . Asymptotes. Vertical: None Horizontal: y  0 Slant: None Increasing, decreasing, relative extrema. A  t  is decreasing over the interval 0,   . There are no relative extrema. Inflection points, concavity. A ” t     exists for all t in the 200 3t 2  1 t  1 2 3 interval  0,   . A ” t   0 when 3t 2  1  0 . The only solution to the equation on  0,   is t 1 . We divide the interval  0,   into 3  1   1  and  ,   and test two intervals  0,    3  3 a point in each interval. A  t  is concave  1  down on the interval  0,  and is concave  3  1  ,  . The point up on the interval   3   1  ,75 is an inflection point.   3 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. 326 Chapter 2: Applications of Differentiation We use the information on the previous page and compute other function values as necessary to draw the graph. d) 4 36  lim  n 0 n 0 n n 0 n If the pitcher gives up one or more runs but gets no one out, the pitcher would be credited with zero innings pitched. b) lim E  n   lim 9  64. Vertical asymptotes occur at values of the variable for which the function is undefined. Thus, they cannot be part of the graph. 65. Asymptotes can be thought of as “limiting lines” for the graph of a function. The graphs and limits in Examples 1, 3 and 6 in section 2.3 of the text book illustrate vertical, horizontal and slant asymptotes. No, the medication does not completely leave the bloodstream. We notice that lim A  t   0 . This means that as t t  approaches  , A approaches 0, but does not actually reach the value 0. We also notice that the equation A  t   0 has no solution, so according to this model, the medication will never completely leave the bloodstream. 63. 4 n a) Calculate each value for the given n. 4 E  9   9   4.00 9 4 E  6  9   6.00 6 4 E  3  9   12.00 3 4 E 1  9   36.00 1 4 2  3 E    9   9  4    54.00 2  3  2 3 4 1  3 E    9   9  4    108.00 1  3  1 3 We complete the table. Earned-Run Innings average  E  Pitched  n  66. E n   9  9 6 3 1 2 3 1 3 4.00 6.00 12.00 36.00 54.00 108.00 67. 5 3x  3x 2  5 x   lim lim x  2  x x  2 1 x lim  3 x   0 x    lim  3 x    x  1 lim x x 0 x   x, for x  0 , we have Using x    x , for x  0 x x lim  1 and lim  1 . x 0 x x 0 x x Therefore, lim does not exist. x 0 x 68.   x  2 x  2 x  4 x3  8  lim 2 x 2 x  4 x 2  x  2  x  2  lim 2 x2  2 x  4 x 2 x2  lim  22  2 2  4 . 2  2  3 69. 7 6 x  6 x 3  7 x x  lim lim 3 10 x  2 x 2  3 x  10 x  2  2 x x lim  6 x   0  x  200  Copyright © 2016 Pearson Education, Inc.  Exercise Set 2.3 70. 327    x  1 x  x  1 x3  1  lim 2 x 1 x  1 x 1  x  1 x  1 lim 2 75. f  x  x 3  2 x 2  3x x 2  25 76. f  x  1 2 x 77. f x  x2  3 2x  4 x2  x  1 x 1 x 1  lim 12  1  1  1  1  71. f  x  x2  72. f  x  3 2 1 x2 x x2  1 a) Solve f  x   0. 73. f  x  x3  4 x2  x  6 x2  x  2 x2  3 0 2x  4 x2  3  0 x 1 2 x2  3 x   3  1.732 The x-intercepts are:  1.732, 0 and 1.732, 0 . 74. x 3  2 x 2  15 x f  x  2 x  5 x  14 b) Evaluate at x  0. 02  3 3  2 0   4 4 The y-intercept is  0, 0.75 . f 0   Copyright © 2016 Pearson Education, Inc. 328 Chapter 2: Applications of Differentiation c) Vertical. x2 Horizontal. None Slant. Dividing the numerator by the denominator we have x2  3 f x  2x  4 1 1  x 1 2 2x  4 As x approaches  , We verify this algebraically. lim  x 2 lim x 1 79. x 2  3x  2 x3 15 y  6 x3  0 x2  x 5 -5 x 2 4 6 -15 Therefore, the nonlinear asymptote is y  x2  6 . Using a calculator, we graph the function, f, and the asymptote, y, below. a) lim f  x   1; lim f  x   1 x   As x increases without bound, f  x  +36 x 37x  9 Therefore, x5  x  9 37 x  9 f  x  3  x2  6  3 . x  6x x  6x As x gets large, f  x  approaches x 2  6 x . 8 -10 x  x5  x  9 x3  6 x Using long division we have: x2 6 3 5 x  6x x  x9 f  x  6 x 3 0 12  3 1  2 1  3 x5  6 x 3 10 -8 -6 -4 -2 x 2  3x  2  x3 0 x  1  0.5 x  1. 2 The slant asymptote is y  0.5 x  1 f  x   2  2  3  2   2  2   3 0 f  x  approaches 78. x 2  3x  2  x3 x 2  3x  2 approaches x3 x x2   1. x x As x decreases without bound, f  x  x 2  3x  2 approaches x3 x x2   1. x x b) The domain of the function appears to be  , 2    1, 3  3,   . We must throw out values that make the radicand negative, or the denominator 0. c) From the graph it appears that lim  f  x   0 and lim  f  x   0 . x 2 x 1 80. a) Visually inspecting the graph, there appears  1 to be relative maxiumum at  0,  .  6 However, noticing the graph dips below the horizontal asymptote around x  5 , we would think there is a relative minimum for some value x  5. This graph does not give us enough detail to visually determine that point. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 329 b) Calculating the first derivative, we have: x 2  10 x  1 f ‘  x  . 2 x2  x  6   f ‘  x  does not exist when x  3 or x  2 ; however, those values are not in the domain of f  x  . So we set f ‘  x   0 and solve for x. x 2  10 x  1  x  x  6 2 2 0 x 2  10 x  1  0 By the quadratic formula, we have: x   10  102  4 11 2 1 81. One possible rational function would be 2 x f  x  . x2 Using the techniques in this section, we sketch the graph. Intercepts. f  x   0 when x  0 . It turns out the x-intercept and the y-intercept is  0, 0  . Asymptotes. x  2 is the vertical asymptote. The degree of the numerator equals that of the denominator, the line y  2 is the horizontal asymptote. Increasing, decreasing, relative extrema. 4 . f ‘  x  is not defined for f ‘ x   x  2 2 x  2 , however that value is outside the domain of the function. f ‘  x   0 so f  x  is increasing 10  100  4 2 10  96  2 Using the calculator, f ‘  x   0 when on the intervals  , 2  and  2,   . There are x  0.10 and x  9.90 . Substititing these values back into the function, we find the critical values occur approximately at  0.101, 0.168  and The equation f ”  x   0 has no real solution, so  9.899, 0.952  . c) Entering the graph in to the calculator and using the maximum feature we see: no relative extrema. Inflection points, concavity. 8 does not exist when x  2 . f ”  x    x  23 there are no possible points of inflection. Furthermore, f ”  x   0 for all x in  , 2  , so f  x  is concave up on  , 2  , and , f ”  x   0 for all x in  2,   , so f  x  is concave down on  2,   . We use this information to sketch the graph. Additional values may be computed as necessary. 8 6 4 2 The result is confirmed. d) Entering the graph in to the calculator and using the minimum feature we see: e) The result is confirmed. Our estimates from part (a) are not very close. Even the “obvious” maximum was not correct. It is doubtful that we would have identified the relative minimum without calculus. -5 -4 -3 -2 -1 -2 -4 -6 -8 f(x) x 0 1 2 3 4 5 82. One possible rational function would be 3x  1 f  x  . x Using the techniques in this section, we sketch the graph. 1 Intercepts. f  x   0 when x  . The x3 1  intercept is  ,0  . When x  0 , f  x  is 3  undefined, so there is no y-intercept. Copyright © 2016 Pearson Education, Inc. 330 Chapter 2: Applications of Differentiation Asymptotes. x  0 is the vertical asymptote. The degree of the numerator equals that of the denominator, the line y  3 is the horizontal asymptote. Increasing, decreasing, relative extrema. 1 f ‘  x   2 . f ‘  x  is not defined for x  0 , x however that value is outside the domain of the function. f ‘  x   0 so f  x  is increasing on the interval  , 0  and  0,   . There are no relative extrema. Inflection points, concavity. 2 f ”  x   3 does not exist when x  0 . The x equation f ”  x   0 has no real solution, so there are no possible points of inflection. Furthermore, f ”  x   0 for all x in  , 0  , so f  x  is concave up on  , 0  , and , f ”  x   0 for all x in  0,   , so f  x  is -4 -3 -2 -1 -2 -4 -6 -8 x 1 2 3 4 5   so there is a critical value at  0, 2  . We notice that g ‘  x   0 when x  1 so g  x  is decreasing on the interval   , 1 . g ‘  x   0 when 1  x so g  x  is increasing on the interval 1,   . g ‘  x   0 when 1  x  0 so g  x  is decreasing on the interval  1, 0  . g ‘  x   0 when 0  x  1 so g  x  is increasing on the interval  0,1 . There are is a relative minimum at  0, 2  . Inflection points, concavity. 2 3x 2  1 . g ”  x  does not exist g ”  x   3 x2  1    g  x  is concave up on  1,1 , and , g ”  x   0 for all x in   , 1 and 1,   , so g  x  is concave down on   , 1 and 1,   . 83. One possible rational function would be x2  2 g  x  2 . x 1 Using the techniques in this section, we sketch the graph. Intercepts. g  x   0 when x   2 . The x- intercepts are  2,0 and x  1 , however that value is outside the domain of the function. g ‘  x   0 , when x  0 , real solution, so there are no possible points of inflection. Furthermore, g ”  x   0 for all x in  1,1 , so f(x) 0  when x  1 . The equation g ”  x   0 has no We use this information to sketch the graph. Additional values may be computed as necessary. -5   concave down on  0,   . 8 6 4 2 Increasing, decreasing, relative extrema. 2x g ‘  x  . g ‘  x  is not defined for 2 x2  1 We use this information to sketch the graph. Additional values may be computed as necessary. 8 6 4 2 -5 -4 -3 -2  2,0 When x  0 , g  x   2 , so the y-intercept is  0, 2  . Asymptotes. x  1 and x  1 are the vertical asymptote. The degree of the numerator equals that of the denominator, the line y  1 is the horizontal asymptote. -1 -2 -4 -6 -8 g(x) x 0 1 2 3 4 5 84. One possible rational function would be 3x 2  15 g  x  2 . x  2x Using the techniques in this section, we sketch the graph. Intercepts. g  x   0 when x   5 . The x-    5, 0. When x  0, intercepts are  5, 0 and g  x  is undefined, so there is no y-intercept. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 331 Asymptotes. x  2 and x  0 are the vertical asymptote. The degree of the numerator equals that of the denominator, the line y  3 is the horizontal asymptote. Increasing, decreasing, relative extrema. g ‘  x   6 x 2  5 x  5 x  x  2 2 2  . g ‘  x  is not defined for x  2 or x  0 , however that value is outside the domain of the function. g ‘  x   0 , 5  5 5  5 and x  , so there 2 2   5  5 3 5  5  are critical values at  , 2 2   when x    5  5 3 and  , 2    5  5  g ”  x    6 2 x 3  15 x 2  30 x  20 3 3 We find that g ”  x   0 when x  4.816 and g ”  x   0 when 4.816  x  2 , so a point of inflection exists at  4.816, 4.025 Furthermore, g ”  x   0 for all x in  0,   , so g  x  is concave up on 0,   , and , g ”  x   0 for all x in  2, 0  , so g  x  is concave down on  2, 0  . We use this information to sketch the graph. Additional values may be computed as necessary. g(x) 5 x 5  5  x  2 so g  x  is 2  5  5  , 2  . increasing on the interval   2  g ‘  x   0 when 5  5 so g  x  is 2  5  5  increasing on the interval  2, . 2   There is a relative minimum at   5  5 3 5  5  . , 2 2   g ‘  x   0 when 2  x   5  5  x0 2 so g  x  is increasing on the interval We find that g ‘  x   0 when -18-16-14-12-10 -8 -6 -4 -2 0 2 4 6 8 10 12 14 16 18 -5 -10 -15 85. One possible rational function would be 8 h  x  2 . x  x6 Using the techniques in this section, we sketch the graph. Intercepts. h  x   0 has no real solution. The are no x-intercepts. When x  0 , h  x   4 , so 3  4 the y-intercept is  0,  .  3 Asymptotes. x  3 and x  2 are the vertical asymptotes. The degree of the numerator is less than that of the denominator, the line y  0 is the horizontal asymptote.  5  5   2 ,0  . There is a relative maximum at     . g ”  x  does x  x  2 not exist when x  2 and x  0 . The equation g ”  x   0 has a solution when x  4.816 .  2 5  5 so g  x  is g ‘  x   0 when x  2  5  5  decreasing on the interval  , . 2    Inflection points, concavity.    5  5 3 5  5  . , 2 2   Copyright © 2016 Pearson Education, Inc. 332 Chapter 2: Applications of Differentiation Increasing, decreasing, relative extrema. 8  2 x  1 h ‘  x  . h ‘  x  is not defined  x  2 2  x  32 for x  3 or x  2 , however those values are outside the domain of the function. h ‘  x   0 , 1 so there is a critical value at 2 when x  are no x-intercepts. When x  0 , h  x   3 , so the y-intercept is  0, 3 .  1 32   ,  2 25 Asymptotes. x   1 so h  x  is 2 1   decreasing on the interval  3,  .  2  h ‘  x   0 when 3  x  1  x  2 so h  x  is 2  1  increasing on the interval  , 2  . There is a  2  We find h ‘  x   0 when  1 32  relative minimum at  ,  . Furthermore,  2 25  h ‘  x   0 when x  3 so h  x  is decreasing on the interval   , 3 . h ‘  x   0 when 2  x so h  x  is increasing on the interval  2,   . Inflection points, concavity. h ”  x   86. One possible rational function would be 3 h  x  2 . 4x 1 Using the techniques in this section, we sketch the graph. Intercepts. h  x   0 has no real solution. The  16 3x 2  3x  7  x  2  x  3 3 3  . h ”  x  does not exist when x  2 and x  0 . The equation h ”  x   0 has no real solution. Therefore h  x  has no points of inflection. h ”  x   0 when asymptotes. The degree of the numerator is less than that of the denominator, the line y  0 is the horizontal asymptote. Increasing, decreasing, relative extrema. 24 x h ‘  x  . h ‘  x  is not defined for 2 4 x2  1  intervals   , 3 and  2,   . h ”  x   0 when 3  x  2 , so h  x  is concave up on  3, 2  . We use this information to sketch the graph. Additional values may be computed as necessary. h(x) 5 1 1 x   or x  , however those values are 2 2 outside the domain of the function. h ‘  x   0 , when x  0 so there is a critical value at  0, 3 1  x  0 so h  x  is 2  1  increasing on the interval  ,0  . h ‘  x   0  2  h ‘  x   0 when  1 so h  x  is decreasing on the 2  1 interval  0,  . There is a relative maximum at  2 when 0  x  1 so h  x  2 1  is increasing on the interval  ,   and  2 1 h ‘  x   0 when  x so h  x  is decreasing on 2 1   the interval  ,   . 2  Furthermore, h ‘  x   0 when x   x -6 -4 -2 0 2 4 6  0, 3 . x  3 and x  2 . Therefore, h  x  is concave down on the -8 1 1 and x  are the vertical 2 2 8 -5 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.3 333 Inflection points, concavity. h ”  x     . h ”  x  does not exist 24 12 x 2  1 4 x  1 2 3 1 1 and x  . The equation 2 4 h ”  x   0 has no real solution. when x   Therefore h  x  has no points of inflection. 1 1  x  and x  2 . 2 2 Therefore, h  x  is concave down on the interval h ”  x   0 when  1  1 1  , . h ”  x   0 when x   and when  2 2  2 1  x , so h  x  is concave up on the intervals 2 1  1  ,   and  ,   .  2  2 We use this information to sketch the graph. Additional values may be computed as necessary. h(x) 5 x -4 -2 0 2 4 -5 Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 334 c) The critical value and the endpoints are: 1 0, , and 2 . 2 d) Evaluate f  x  for each value in step (c). Exercise Set 2.4 1. 2. a) The absolute maximum gasoline mileage is obtained at a speed of 55 mph. b) The absolute minimum gasoline mileage is obtained at a speed of 5 mph. c) At 70 mph, the fuel economy is 25 mpg. f 0  4   0   0  4 2 2 17 1 1 1 f    4     4.25 2 2 2 4 f 2  4   2  2  2 2 Over the interval  30, 70 , the vehicle’s On the interval  0, 2 , the absolute absolute maximum fuel economy is 30 mpg at 55 mph. The vehicle’s absolute minimum fuel economy is 25 mpg at 70 mph. 3. f  x  5  x  x2 ; a) Find f ‘  x  0, 2 f ‘x  1 2x 5. b) Find the Critical values. The derivative exists for all real numbers. Thus, we solve f ‘  x  0 1 2x  0 1  2x 1 x 2 c) List the critical values and endpoints. These 1 values are 0, , and 2 . 2 d) Evaluate f  x  at each value in step (c). f  0  5   0   0  5 2 2 21 1 1 1 f    5        5.25 2 2 2 4 f 2  5  2  2  3 2 21 , is the 4 1 absolute maximum, it occurs at x  . The 2 smallest of these values, 3, is the absolute minimum, it occurs at x  2 . The largest of these values, 4. f  x  4  x  x2 ; a) f ‘x  1 2x 17 1 , which occurs at x  . 4 2 The absolute minimum is 2, which occurs at x  2. maximum is 0, 2  2,1 1 2 x  2 x  5; 2 a) Find f ‘  x  f  x   x3  f ‘  x   3x 2  x  2 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘  x  0 3x 2  x  2  0 3x  2 x  1  0 3x  2  0 or x  1  0 2 x   or x  1 3 c) List the critical values and endpoints. These 2 values are are: 2,  , and 1 . 3 d) Evaluate f  x  for each value in step (c). f  2    2   3 3 1 22  2 2  5  9 2 2 85  2  2 1  2  2 f          2    5   3.148  3  3 2  3  3 27 f 1  1  3 1 2 15 1  2 1  5   7.5 2 2 On the interval  2,1 , the absolute b) f ‘  x  exists for all real numbers. Solve: 1 2x  0 1 x 2 15 , which occurs at x  1 . The 2 absolute minimum is 9 , which occurs at x  2. maximum is Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 6. 335 f  x   x 3  x 2  x  2; a) f ‘  x   3x 2  2 x  1  1, 2 f  0   0   3 maximum is 5.5 , which occurs at x  1 . The absolute minimum is 2, which occurs at x  2. f ‘  x  0 3x 2  2 x  1  0 8. 1 x or x 1 3 c) The critical value and the endpoints are: 1 1,  , 1, and 2 . 3 d) Evaluate f  x  for each value in step (c). f ‘  x  0 2 3x  2 x  1  0 3x  1 x  1  0 2 3 1 or x 1 3 c) List the critical values and endpoints. The critical value x  1 is not in the interval, so we exclude it. We will test the values 1 1,  , and 0 . 3 d) Evaluate f  x  for each value in step (c). x 2 f 1  1  1  1  2  1 3 2 f 2   2  2   2  2  4 3 2 On the interval  1, 2  , the absolute maximum is 4 , which occurs at x  2 . The absolute minimum is 1, which occurs at x  1 and x  1. 1 2 x  2 x  4; 2 a) Find f ‘  x  f  x   x3  f  1   1   1   1  3  2 3 3  2, 0 f  0   0   0   0  3  3 3 3x  2 x  1  0 3 x  2  0 or x  1  0 2 x x  1 or 3 2 c) The critical value x  is not in the 3 interval, so we exclude it. We will test the values: 2,  1, and 0 . d) Evaluate f  x  for each value in step (c). f  1   1  1 22  2 2  4  2 2 1 2 2 On the interval  1, 0  , the absolute 86 1 , which occurs at x   . 27 3 The absolute minimum is 2, which occurs at x  1. maximum is 3x 2  x  2  0 3 2 86  1   1   1   1     3  3.2  3   3   3   3  27 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f  2   2  2 f  f ‘  x   3x 2  x  2 3 f ‘  x   3x 2  2 x  1  1, 0 b) f ‘  x  exists for all real numbers. Solve: 59  1    1    1    1  f         2  2.2  3  3  3  3 27 7. f  x   x 3  x 2  x  3; a) f  1   1   1   1  2  1 3 2 02  2 0  4  4 On the interval  2, 0 , the absolute b) f ‘  x  exists for all real numbers. Solve: 3x  1 x  1  0 1  12  2  1  4  11 2 9. f  x   2 x  4; a) Find f ‘  x   1,1 f ‘ x  2 b) and c) The derivative exists and f ‘  x   2 for all real numbers. Note that the derivative is never 0. Thus, there are no critical values for f  x  , and the absolute maximum and absolute minimum will occur at the endpoints of the interval.  5.5 Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 336 d) Evaluate f  x  at the endpoints. 12. f  1  2  1  4  2 a) f 1  2 1  4  6 maximum is 6, which occurs at x  1 . The absolute minimum is 2, which occurs at x  1. f  x   5 x  7; a) f ‘x  5 f  10  2  3  10  28  2, 3 f 10  2  3 10  32 On the interval  10,10 , the absolute b) and c) The derivative exists and is 5 for all real numbers. Note that the derivative is never 0. Thus, there are no critical values for f  x  , and the absolute maximum and absolute minimum will occur at the endpoints of the interval. d) Evaluate f  x  at the endpoints. maximum is 28, which occurs at x  10 . The absolute minimum is 32 , which occurs at x  10. 13. absolute maximum is 5 for 1  x  1 and the absolute minimum is 5 for 1  x  1 . f  3  5  3  7  8 On the interval  2, 3 , the absolute 14. maximum is 8, which occurs at x  3 . The absolute minimum is 17 , which occurs at x  2. f  x   7  4 x; a) Find f ‘  x  f ‘ x  2x  6  1,5 b) f ‘  x  exists for all real numbers. Solve: 2x  6  0 2x  6 x3 c) The critical value and the endpoints are 1, 3, and 5 . d) Evaluate f  x  for each value in step (c). f ‘  x   4 b) and c) The derivative exists and is 4 for all real numbers. Note that the derivative is never 0. Thus, there are no critical values for f  x  , f  1   1  6  1  3  4 2 f  3   3  6  3  3  12 2 and the absolute maximum and absolute minimum will occur at the endpoints of the interval. d) Evaluate f  x  at the endpoints. f 5  5  6 5  3  8 2 On the interval  1, 5 , the absolute maximum is 4, which occurs at x  1 . The absolute minimum is  12 , which occurs at x  3. f  2   7  4  2   15 f 5  7  4 5  13 maximum is 15, which occurs at x  2 . The absolute minimum is 13 , which occurs at x  5. f  x   x 2  6 x  3; a)  2, 5 On the interval  2,5 , the absolute  1,1 f  x   5; Note for all values of x, f  x   5 . Thus, the f  2   5  2   7  17 11. f ‘  x   3 b) and c) f ‘  x   3 for all real numbers; therefore, there are no critical points. The absolute maximum and minimum values occur at the endpoints. d) Evaluate f  x  at the endpoints. On the interval  1,1 , the absolute 10.  10,10 f  x   2  3 x ; 15. g  x   24;  4,13 Note for all values of x, g  x   24 . Thus, the absolute maximum is 24 for 4  x  13 and the absolute minimum is 24 for 4  x  13 . Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 16. 337 f  x   3  2 x  5x 2 ; a) f ‘  x   2  10 x b) f ‘  x  exists for all real numbers. Solve:  3,3 3x 2  6 x  0 3x  x  2   0 3x  0 or x20 x  0 or x2 c) The critical value and the endpoints are 0, 2, and 5 . Note, since 0 is an endpoint of the interval, x  0 is included in this list as an endpoint, not as a critical value. d) Evaluate f  x  for each value in step (c). b) f ‘  x  exists for all real numbers. Solve: 2  10 x  0 1 x 5 c) The critical value and the endpoints are 1 3,  , and 3 . 5 d) Evaluate f  x  for each value in step (c). f 0   0  3  0  0 3 f  3  3  2  3  5  3  36 2 f  2    2   3  2   4 3 2 16  1  1  1 f    3  2    5     3.2  5  5  5 5 3 2 is 50, which occurs at x  5 . The absolute minimum is 4 , which occurs at x  2. maximum is f  x   x 2  4 x  5; a) f ‘ x  2x  4 19. f  0   1  6  0  3  0   1 2 f 1  1  6 1  3 1  4 2 f  1   1  4  1  5  10 2 f  4   1  6  4   3  4   23 2 f  2   2  4  2  5  1 2 On the interval  0, 4 , the absolute maximum f  3   3  4  3  5  2 2 is 4, which occurs at x  1 . The absolute minimum is 23 , which occurs at x  4. On the interval  1, 3 , the absolute f  x   x 3  3x 2 ; a) f ‘  x   3x 2  6 x 0,5 f ‘ x  6  6x b) f ‘  x  exists for all real numbers. Solve: 6  6x  0 x 1 c) The critical value and the endpoints are 0, 1, and 4 . Note, since 0 is an endpoint of the interval, x  0 is included in this list as an endpoint, not as a critical value. d) Evaluate f  x  for each value in step (c). b) f ‘  x  exists for all real numbers. Solve: 2x  4  0 2x  4 x2 c) The critical value and the endpoints are 1, 2, and 3 . d) Evaluate f  x  for each value in step (c). 18. 0, 4 f  x   1  6 x  3x 2 ; a)  1,3 maximum is 10, which occurs at x  1 . The absolute minimum is 1, which occurs at x  2. 2 On the interval  0,5 , the absolute maximum On the interval  3, 3 , the absolute 17. 2 f 5  5  3 5  50 f  3  3  2  3  5  3  48 16 1 , which occurs at x   . 5 5 The absolute minimum is 48 , which occurs at x  3. 2 20. f  x   x 3  3x  6; a) f ‘  x   3x  3  1,3 2 b) f ‘  x  exists for all real numbers. Solve: 3x 2  3  0 x2  1  0 x  1 Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 338 d) Evaluate f  x  for each value in step (c). c) The critical value and the endpoints are 1, 1, and 3 . Note, since 1 is an endpoint of the interval, x  1 is included in this list as an endpoint, not a critical value. d) Evaluate f  x  for each value in step (c). f  5  3  5  2  5  325 2 f  0   3  0  2  0   0 2 2 f  1   1  3  1  6  8 3 maximum is 325, which occurs at x  5. The absolute minimum is 0, which occurs at x  0. f  3   3  3  3  6  24 3 On the interval  1, 3 , the absolute f  x   x 3  3 x; a) f ‘  x   3x 2  3 23. f ‘  x   3x 2 b) f ‘  x  exists for all real numbers. Solve:  5,1 3 x 2  0 x0 c) The critical value and the endpoints are 8, 0, and 8 . d) Evaluate f  x  for each value in step (c). 3x 2  3  0 x2  1  0 f  8  1   8  513 3 x  1 c) The critical value and the endpoints are 5,  1, and 1 . Note, since 1 is an endpoint of the interval, x  1 is included in this list as an endpoint, not a critical value. d) Evaluate f  x  for each value in step (c). f  0  1   0  1 3 f 8  1  8  511 3 On the interval  8,8 , the absolute maximum is 513, which occurs at x  8. The absolute minimum is 511 , which occurs at x  8. f  5   5  3  5  110 3 f  1   1  3  1  2 3 24. f 1  1  3 1  2 3 On the interval  5,1 , the absolute maximum is 2, which occurs at x  1. The absolute minimum is 110 , which occurs at x  5. f  x   3x 2  2 x 3 ; a) 8,8 f  x   1  x3; a) b) f ‘  x  exists for all real numbers. Solve: 22. 3 On the interval  5,1 , the absolute f 1  1  3 1  6  4 21. 3 f 1  3 1  2 1  1 3 maximum is 24, which occurs at x  3. The absolute minimum is 4, which occurs at x  1. 3 f ‘  x  6x  6×2  5,1 b) f ‘  x  exists for all real numbers. Solve: 6x  6×2  0  10,10 f  x   2 x3; a) f ‘  x  6x 2 b) f ‘  x  exists for all real numbers. Solve: 6×2  0 x0 c) The critical value and the endpoints are 10, 0, and 10 . d) Evaluate f  x  for each value in step (c). f  10  2  10  2000 3 f  0  2  0   0 3 6 x 1  x   0 x  0 or x  1 c) The critical value and the endpoints are 5, 0, and 1 . Note, since 1 is an endpoint of the interval, x  1 is included in this list as an endpoint, not a critical value. f 10  2 10  2000 3 On the interval  10,10 , the absolute maximum is 2000, which occurs at x  10. The absolute minimum is 2000 , which occurs at x  10. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 25. 339 0, 4 f  x   x 3  6 x 2  10; a) f ‘  x   3 x  12 x 27. 2 a) b) f ‘  x  exists for all real numbers. Solve: x 2 3  4 x   0 x2  0 3  4x  0 3 x  0 or x 4 c) The critical values and the endpoints are 3 1, 0, , and 1 . 4 d) Evaluate f  x  for each value in step (c). 3 2 3 2 On the interval  0, 4 , the absolute maximum 3 4 f 1  1  1  0 3 4 e) On the interval  1,1 , the absolute  3,1 f ‘  x   9  6 x  3x 2 4 27  3  3  3 f         0.105 4 4 4 256 is 10, which occurs at x  0. The absolute minimum is 22 , which occurs at x  4. a) 4 f  0   0   0  0 f  4    4   6  4   10  22 f  x   12  9 x  3 x 2  x 3 ; or f  1   1   1  2 f  0   0  6  0  10  10 26. 3 3x 2  4 x 3  0 3x  x  4  0 3x  0 or x  4  0 x  0 or x4 c) The critical values and the endpoints are 0 and 4 . Note, since the possible critical values are the endpoints of the interval, they are included in this list as endpoints, not as critical values. d) Evaluate f  x  for each value in step (c). 3 f ‘  x   3x  4 x 2 b) f ‘  x  exists for all real numbers. Solve: 3x 2  12 x  0 3 1,1 f  x   x3  x4 ; 27 3 , which occurs at x  . 256 4 The absolute minimum is 2 , which occurs at x  1 . maximum is b) f ‘  x  exists for all real numbers. Solve: 9  6 x  3×2  0 x2  2 x  3  0 28.  x  3 x  1  0 x  3 or x 1 c) The critical values and the endpoints are 3 and 1 . Note, since the possible critical values are the endpoints of the interval, they are included in this list as endpoints, not as critical values. d) Evaluate f  x  for each value in step (c). f  3  12  9  3  3  3   3  15 2 3 f 1  12  9 1  3 1  1  17 2 3 On the interval  3,1 , the absolute maximum is 17, which occurs at x  1. The absolute minimum is 15 , which occurs at x  3.  2, 2 f  x   x4  2 x3; a) f ‘  x  4x  6x 3 2 b) f ‘  x  exists for all real numbers. Solve: 4 x3  6 x2  0 2 x 2  2 x  3  0 3 2 c) The critical values and the endpoints are 3 2, 0, , and 2 . 2 d) Evaluate f  x  for each value in step (c). x0 x or f  2    2   2  2   32 4 3 f  0   0  2  0  0 4 3 4 3 27  3  3 3 f      2     1.6875 2 2 2 16 f 2   2  2  2  0 The solution is continued on the next page. 4 Copyright © 2016 Pearson Education, Inc. 3 Chapter 2: Applications of Differentiation 340 From the previous page, we determine on the interval  2, 2  , the absolute maximum is f  2    2   8  2   3  13 4 f  3   3  8  3  3  12 4 32, which occurs at x  2. The absolute 27 3 minimum is  , which occurs at x  . 16 2 29. a) f ‘  x   4 x3  4 x b) f ‘  x  exists for all real numbers. Solve: maximum is 12, which occurs at x  3 and x  3. The absolute minimum is 13 , which occurs at x  2 and x  2. 31. 8,8 f ‘ x   2  13 2 x  1 3 3x 3 a) 4 x x2  1  0 b) f ‘  x  does not exist for x  0 . The   4x  0 or equation f ‘  x   0 has no solution, so x  0 x2  1  0 is the only critical value. c) The critical values and the endpoints are 8, 0, and 8 . d) Evaluate f  x  for each value in step (c). f  8  1   8 3  3 2 f  2    2   2  2   5  13 4 2 f  1   1  2  1  5  4 f 0  1  0 3  1 f  0   0  2  0  5  5 f 8  1  8 3  3 4 2 2 4 2 2 On the interval  8,8 , the absolute f 1  1  2 1  5  4 4 2 maximum is 1, which occurs at x  0. The absolute minimum is 3 , which occurs at x  8 and x  8. f  2    2   2  2   5  13 4 2 On the interval  2, 2  , the absolute maximum is 13, which occurs at x  2 and x  2. The absolute minimum is 4 , which occurs at x  1 and x  1.  3, 3 f  x   x 4  8 x 2  3; a) 2 f  x  1 x 3 ; 4 x3  4 x  0 x  0 or x  1 c) The critical values and the endpoints are 2,  1, 0, 1, and 2 . d) Evaluate f  x  for each value in step (c). 30. 2 On the interval  3, 3 , the absolute 2, 2 f  x   x 4  2 x 2  5; 2 f ‘  x   4 x  16 x 3 b) f ‘  x  exists for all real numbers. Solve: 4 x 3  16 x  0   4x  0 or 4 x x2  4  0 x2  4  0 f  3   3  8  3  3  12 2 f  2    2  8  2  3  13 4 2 f  0   0  8  0  3  3 4 2  4,5 f  x    x  3 3  5; 2 a) f ‘  x  1 2 2  x  3 3  1 3 3  x  3 3 b) f ‘  x  does not exist for x  3 . The equation f ‘  x   0 has no solution, so x  3 is the only critical value. c) The critical values and the endpoints are 4,  3, and 5 . d) Evaluate f  x  for each value in step (c). f  4    4  3 3  5  4 2 x  0 or x  2 c) The critical values and the endpoints are 3,  2, 0, 2, and 3 . d) Evaluate f  x  for each value in step (c). 4 32. f  3   3  3 3  5  5 2 f 5  5  3 3  5  1 2 On the interval  4,5 , the absolute maximum is 1 , which occurs at x  5. The absolute minimum is 5 , which occurs at x  3. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 33. f  x  x  341 4 ; x d) Evaluate f  x  for each value in step (c).  8, 1 1 f 1  1   2 1 1 401 f  20  20    20.05 20 20 On the interval 1, 20 , the absolute 4 a) f ‘  x   1  4 x  1  2 x b) f ‘  x  does not exist for x  0 . However, 2 x  0 is not in the interval. Solve f ‘  x   0 . 1 4 0 x2 4 1 2 x 2 x 4 35. x  2 The only critical value in the interval is at x  2 . c) The critical values and the endpoints are 8,  2 and  1 . d) Evaluate f  x  for each value in step (c). f  8    8   f  2    2   f  1   1  4  2   4 1 ; x 2 2  x  1 2 x   x 2 x  Quotient Rule a) f ‘  x    x  1 2 2 2 x3  2 x  2 x3   x  1 2 2 2x   x  1 2 2x 1,20 f ‘  x   1  x 2  1  x  0 is not in the interval. Solve f ‘  x   0 . 1  2, 2 2 f ‘  x  0 4  5  1 1 x2 b) f ‘  x  does not exist for x  0 . However, a) x2 ; x 1 2 b) f ‘  x  exists for all real numbers. Solve: maximum is 4 , which occurs at x  2. 17 The absolute minimum is  , which 2 occurs at x  8. f  x  x  f  x  4 17   8.5  8 2   On the interval  8, 1 , the absolute 34. maximum is 20.05, which occurs at x  20. The absolute minimum is 2, which occurs at x  1. 1 0 x2 1 1 2 x x2  1  x  1 2 2 0 2x  0 x0 c) The critical values and the endpoints are 2, 0, and 2 . d) Evaluate f  x  for each value in step (c).  2  2  4  2  2  1 5  0 2  0 f  0   0 2  1  2 2  4 f  2   2 2  1 5 f  2   On the interval  2, 2  , the absolute 4 , which occurs at x  2 and 5 x  2. The absolute minimum is 0, which occurs at x  0. maximum is x  1 The critical value x  1 is not in the interval, and the other critical value is an endpoint. c) The endpoints are 1 and 20 . Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 342 36. f  x  a) 4x ; x 1 d) Evaluate f  x  for each value in step (c). 3,3 2 f  2   2  1 3  1 1  x  1 4  4 x 2 x   x  1 2 f ‘  x  f  1   1  1 3  0 1 2 2 f  26   26  1 3  3 1 4  4×2   x  1 2 On the interval  2, 26  , the absolute 2 minimum is 1 , which occurs at x  2. The absolute maximum is 3, which occurs at x  26. b) f ‘  x  exists for all real numbers. Solve: 4  4 x2  x  1 2 0 2 38. 2 4  4x  0 x  1 c) The critical values and the endpoints are 3,  1, 1, and 3 . d) Evaluate f  x  for each value in step (c). 4  3  1 On the interval 8, 64  , the absolute maximum is 4, which occurs at x  64. The absolute minimum is 2, which occurs at x  8. 39. – 48. 49. maximum is 2, which occurs at x  1. The absolute minimum is 2 , which occurs at x  1. 1 a) f ‘  x  1 f 64   64  3  4 On the interval  3, 3 , the absolute f  x    x  1 3 ; f ‘  x  f  8   8  3  2 6 5  3  1 4  1  2 f  1   12  1 4 1 2 f 1  2 1  1 4  3 6  f  3  2 3  1 5 2 8,64 1 23 1 x  2 3 3x 3 b) There are no critical values in the interval. c) The endpoints are 8 and 64 . d) Evaluate f  x  for each value in step (c). x 1  0 37. 1 a) 2 f  3  f  x  3 x  x 3 ;  2, 26 2 1 1  x  1 3  2 3 3  x  1 3 b) f ‘  x  does not exist for x  1 . The equation f ‘  x   0 has no solution, so x  1 is the only critical value. c) The critical values and the endpoints are 2,  1, and 26 . Left to the student. f  x   30 x  x 2 When no interval is specified , we use the real line  ,   . a) Find f ‘  x  f ‘  x   30  2 x b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . f ‘  x  0 30  2 x  0 2 x  30 x  15 The only critical value is x  15. c) Since there is only one critical value, we can apply Max-Min Principle 2. First we find f ”  x  . f ”  x   2 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 343 Since the second derivative is positive at 10, we have a minimum at x  10. Next, we find the function value at x  10. On the previous page, we determined the second derivative is constant, so f ” 15  2 . Since the second derivative is f 10   2 10   40 10   270  70 2 negative at 15, we have a maximum at x  15. Next, we find the function value at x  15. Therefore, the absolute minimum is 70, which occurs at x  10. The function has no maximum value. f 15  30 15  15  225 2 Therefore, the absolute maximum is 225, which occurs at x  15. There is no minimum value. 50. f  x   12 x  x 2 ; a) 52. a) f ‘  x  0 4 x  20  0 b) f ‘  x  exists for all real numbers. Solve f ‘  x  0 c) 12  2 x  0 Thus, we have a minimum at x  5. f 5  2 5  20 5  340  290 2 Therefore, the absolute minimum is 290, which occurs at x  5. There is no maximum value. f ” 6   2  0 . Thus, we have a maximum at x  6. f  6   12  6    6   36 2 51. x5 f ”  x   4 f ” 5  4  0 6 x The only critical value is x  6 . c) f ”  x   2 . Therefore, the absolute maximum is 36, which occurs at x  6. The function has no minimum value. f ‘  x   4 x  20  ,   b) f ‘  x  exists for all real numbers. Solve ,   f ‘  x   12  2 x f  x   2 x 2  20 x  340; 53. f  x   16 x  4 3 x ; 3 0,   a) Find f ‘  x  f ‘  x   16  4 x 2 f  x   2 x 2  40 x  270 When no interval is specified , we use the real line  ,   . a) Find f ‘  x  b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . 16  4 x 2  0 f ‘  x   4 x  40 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . 4 x  40  0 4 x  40 x  10 The only critical value is x  10. . c) Since there is only one critical value, we can apply Max-Min Principle 2. First we find f ”  x  . f ”  x   4 . The second derivative is constant, so f ” 10   4 . 4 x 2  16 x2  4 x  2 There are two critical values; however, x  2 is the only critical value on the interval  0,   . c) Since there is only one critical value in the interval, we can apply Max-Min Principle 2. First we find f ”  x  . f ”  x   8 x . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 344 Thus, we solve f ‘  x   0 . Next evaluate the second derivative at x  2. f ”  2   16  0 Since the second derivative is negative at 2 , we have a maximum at x  2. 4 3 64 f  2   16  2    2   3 3 64 Therefore, the absolute maximum is , 3 which occurs at x  2. There is no minimum value. 4 54. f  x   x  x 3 ; 3 a) f ‘  x   1  4 x 2 f ‘  x  0 60  2 x  0 60  2 x 30  x The only critical value is x  30 . c) Since there is only one critical value, we can apply Max-Min Principle 2. First we find f ”  x  . f ”  x   2 . The second derivative is constant, so f ”  30   2 . Since the second derivative is negative at 30, we have a maximum at x  30. Next, we find the function value at x  30. f  30   30  60  30   900 Therefore, the absolute maximum is 900, which occurs at x  30. The function has no minimum value. 0,  b) f ‘  x  exists for all real numbers. Solve 1  4 x2  0 4 x 2  1 x2  1 4 1 x 2 There are two critical values; however, the 1 only critical value in  0,   is x  . 2 c) f ”  x   8 x . 56. a) f ‘ x  0 25  2 x  0 c) 1 . 2  25  f ”    2  0  2 3 1 1 1 4 1 f        2 2 3 2 3 Therefore, the absolute maximum is Thus, we have a maximum at x  1 , 3 1 . The function has no 2 Therefore, the absolute maximum is which occurs at x  f  x   x  60  x   60 x  x 2 25 . There is no 2 minimum value. When no interval is specified , we use the real line  ,   . 57. f ‘  x   60  2 x . b) Find the critical values. The derivative exists for all real numbers. 25 . 2 25  625  25   25   f       25     156.25  2   2  2 4 minimum value. a) Find f ‘  x  . 25 2 f ”  x   2 x Thus, we have a maximum at x  55. f ‘  x   25  2 x  ,   b) f ‘  x  exists for all real numbers. Solve 1 f ”     4  0 . 2 which occurs at x  f  x   x  25  x   25 x  x 2 ; 1 3 x  5 x; 3 a) Find f ‘  x  . f  x  f ‘  x  x2  5 . Copyright © 2016 Pearson Education, Inc. 3, 3 625 , 4 Exercise Set 2.4 345 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . 1 10 23  3 2   3.33 3 3 3 1 f  3   3  3  3  2 3  3.464 3 3 1 f 3  3  3 3  2 3  3.464 3 1 3 10 f 2  2  3 2    3.33 3 3 Thus, the absolute maximum over the interval  2, 2  , is 2 3 , which occurs at f 2   x2  5  0 x2  5 x   5  2.236 Both critical values are in the interval  3, 3 . c) The interval is closed and there is more than one critical value, so we use Max-Min Principle 1. The critical points and the endpoints are 3,  5, 5, and 3. Next, we find the function values at these points. 1 3 f  3   3  5  3  6 3 3 1 10 5 f  5   5 5  5   7.454 3 3 3 1 10 5 f 5  5 5 5    7.454 3 3 1 3 f  3   3  5  3  6 3 Thus, the absolute maximum over the 10 5 interval  3, 3 , is , which occurs at 3 x   5 , and the absolute minimum over          3, 3 is  58. a) x   3 , and the absolute minimum over  2, 2 is 2 3 , which occurs at x  3 . 59. f  x   0.001x 2  4.8 x  60 When no interval is specified , we use the real line  ,   .  a) Find f ‘  x  f ‘  x   0.002 x  4.8   b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . 0.002 x  4.8  0 0.002 x  4.8 x  2400 The only critical value is x  2400. c) Since there is only one critical value, we can apply Max-Min Principle 2. First we find f ”  x  . 10 5 , which occurs at x  5 . 3 1 3 x  3 x; 3 f ‘  x  x2  3 f  x   f ”  x   0.002 . The second derivative is constant, so f ”  2400   0.002 . Since the second derivative is negative at 2400, we have a maximum at x  2400. Next, we find the function value at x  2400.  2, 2 b) f ‘  x  exists for all real numbers. Solve f ‘  x  0 x2  3  0 f  2400   0.001  2400   4.8  2400   60 2 x   3  1.732. Both critical values are in the interval  2, 2 . c) The critical points and the endpoints are 2,  3, 3, and 2. Next, we find the function values at these points.             5700 Therefore, the absolute maximum is 5700, which occurs at x  2400. The function has no minimum value. 60. f  x   0.01x 2  1.4 x  30 a) f ‘  x   0.02 x  1.4 Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 346 b) f ‘  x  exists for all real numbers. Solve f ‘  x  0 62. 0.02 x  1.4  0 c) 1 f  x    x 3  6 x 2  11x  50; 3 a) f ‘  x    x 2  12 x  11 b) f ‘  x  exists for all real numbers. Solve x  70 f ”  x   0.02 f ‘  x  0 f ”  70   0.02  0 2  x  12 x  11  0 Thus, we have a maximum at x  70. x 2  12 x  11  0 f  70   0.01  70   1.4  70   30  19 Therefore, the absolute maximum is 19, which occurs at x  70. There is no minimum value. 2 61. f  x    x 3  x 2  5 x  1;  x  11 x  1  0 x  11  0 c) b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . 3 x 2  2 x  5  0 63. 5 x or x  1 3 5 x  is the only critical value on the 3 interval  0,   . c) The interval 0,   is not closed. The only 5 critical value in the interval is x  . 3 Therefore, we can apply Max-Min Principle 2. First, we find the second derivative. f ”  x   6 x  2 5 5 f ”    6    2  8  0  3  3 Since the second derivative is negative when 5 5 x  , there is a maximum at x  . 3 3 Next, find the function value: 2 148 5 5 5 5 f           5  1   3  3  3  3 27 Thus, the absolute maximum over the 148 , which occurs at interval  0,   is 27 5 x  . The function has no minimum value. 3 x 1  0 x  11 or f ”  x   2 x  12 x 1 Therfore, there is a minimum at x  1. 1 3 166 2 f 1   1  6 1  111  50   3 3 Thus, the absolute minimum over the 166 , which occurs at interval  0, 3 is  3 x  1. f ‘  x   3x 2  2 x  5 .  3 x  5 x  1  0 3 x  5  0 or x  1  0 or f ” 1  2 1  12  10  0 . 0,   a) Find f ‘  x  . 3 0,3 1 3 x ; 2 a) Find f ‘  x  . f  x   15 x 2  0, 30 3 2 x . 2 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . f ‘  x   30 x  3 2 x 0 2 60 x  3x 2  0 30 x  3x  20  x   0 3x  0 or 20  x  0 x  0 or x  20 Both critical values are in the interval 0, 30 . c) Since the interval is closed and there is more than one critical value, we apply the MaxMin Principle 1. The critical values and the endpoints are 0, 20, and 30 . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 347 Next, we find the function values. 1 3 2 f  0   15  0    0  0 2 1 2 3 f  20   15  20    20   2000 2 1 2 3 f  30  15  30    30  0 2 The largest of these values, 2000, is the maximum. It occurs at x  20. The smallest of these values, 0, is the minimum. It occurs at x  0 and x  30. Thus, the absolute maximum over the interval  0, 30  , is 2000, which occurs at 65. 72 ; x f  x   2 x  72 x 1 f  x  2x  a) Find f ‘  x  . 72 . x2 b) Find the critical values. f ‘  x  does not exist for x  0 ; however, 0 is not in the interval 0,   . Therefore, we solve f ‘  x   2  72 x 2  2  f ‘  x  0 72 0 x2 72 2 2 x 2 x 2  72 2 x  20 , and the absolute minimum over 0, 30 is 0, which occurs at x  0 and x  30. 64. 1 3 x ; 0,8 2 3 a) f ‘  x   8 x  x 2 2 b) f ‘  x  exists for all real numbers. Solve: f  x  4 x2  x  6 c) The interval  0,   is not closed. The only critical value in the interval is x  6. Therefore, we can apply Max-Min Principle 2. First, we find the second derivative. 144 f ”  x   144 x 3  3 x Evaluating the second derivative at x  6 , we have: 144 2 f ”  6    0. 63 3 3 2 x 0 2 16 x  3 x 2  0 8x  x 16  3 x   0 x0 or 16 x  3  0 3 x0 or x 16 c) The critical values and the endpoints are 3 0, , and 8. 16 d) Find the function values. 1 3 2 f 0  4  0   0  0 2 2 2 Multiplying by x , since x  0. x 2  36 f ‘  x  0 Since the second derivative is positive when x  6 , there is a minimum at x  6. Next, find the function value at x  6. 72 f  6  2  6   24 6 Thus, the absolute minimum over the interval  0,   is 24, which occurs at x  6. 3 1  16  1024  16   16  f    4       37.93  3  3 2 3  27 1 3 2 f 8   4 8   8   0 2 Therefore, the absolute maximum over the 1024 25 , which or 37 interval  0,8 is 27 27 16 . The absolute minimum is occurs at x  3 0, which occurs at x  0 and x  8. 0,   The function has no maximum value over the interval  0,   . 66. 3600 ; x 3600 f ‘  x  1 2 x f  x  x  a) Copyright © 2016 Pearson Education, Inc. 0,   Chapter 2: Applications of Differentiation 348 b) Find the critical values. f ‘  x  does not First, we find the second derivative. 864 f ”  x   2  864 x 3  2  3 x Evaluating the second derivative at x  6 , we have: 864 f ”  6   2   6 0. 63 Since the second derivative is positive when x  6 , there is a minimum at x  6. Next, find the function value at x  6. 432 2 f 6  6   108 6 Thus, the absolute minimum over the interval  0,   is 108, which occurs at x  6. exists for x  0 ; however, 0 is not in the interval  0,   . Therefore, we solve f ‘  x  0 1 3600 0 x2 x 2  3600 x   60 c) The interval  0,   is not closed. The only critical value in the interval is x  60. 7200 f ”  x   3 x 7200 1 f ”  60    0. 3 30 60 Since the second derivative is positive when x  60 , there is a minimum at x  60. 3600 f  60   60   120 60 Thus, the absolute minimum over the interval  0,   is 120, which occurs at x  60. The function has no maximum value over the interval  0,   . 67. 432 ; x f  x   x 2  432 x 1 f  x  x2  250 ; 0,   x 250 a) f ‘  x   2 x  2 x b) Find the critical values. f ‘  x  does not exist f  x  x2  for x  0 ; however, 0 is not in the interval 0,   . Therefore, we solve 2x  250 x2 0 2x  432 f ‘  x   2 x  432 x  2 x  2 . x b) Find the critical values. f ‘  x  does not exist 2 for x  0 ; however, 0 is not in the interval 0,   . Therefore, we solve f ‘  x  0 432 0 x2 432 2x  2 x 3 2 x  432 68. f ‘  x  0 0,   a) Find f ‘  x  . 2x  The function has no maximum value over the interval  0,   . 2 Multiplying by x , since x  0. x 3  216 x6 c) The interval  0,   is not closed. The only critical value in the interval is x  6. Therefore, we can apply Max-Min Principle 2. 250 x2 x3  125 x5 c) The interval  0,   is not closed. The only critical value in the interval is x  5. 500 f ”  x   2  3 x 500 f ” 5  2  3  6  0 .  5 There is a minimum at x  5. 250 2 f 5  5   75 5 Thus, the absolute minimum over the interval  0,   is 75, which occurs at x  5. The function has no maximum value over the interval  0,   . Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 69. 349 1,1 f  x   2 x 4  x; a) Find f ‘  x  . b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . f ‘  x   8×3  1 . 8×3  1  0 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . 8×3  1 1 x3  8 1 x 2 8×3  1  0 8 x 3  1 1 x3   8 1 x 2 The only critical value x  interval  1,1 . c) The interval is closed, and we are looking for both the absolute maximum and absolute minimum values, so we use Max-Min Principle 1. The critical values and the endpoints are 1 1, , and 1. 2 Next, we find the function values at these points. 1 The only critical value x   is in the 2 interval  1,1 . c) The interval is closed, and we are looking for both the absolute maximum and absolute minimum values, so we use Max-Min Principle 1. The critical points and the endpoints are 1 1,  , and 1. 2 Next, we find the function values at these points. f  1  2  1   1  3 4 4 3 1 1 1 f    2       2 2 2 8 f 1  2 1  1  1 4 f  1  2  1   1  1 4 The largest of these values, 3, is the maximum. It occurs at x  1. 3 The smallest of these values,  , is the 8 1 minimum. It occurs at x  . 2 Thus, the absolute maximum over the interval  1,1 , is 3, which occurs at x  1 , 4 3  1  1  1 f    2         2  2  2 8 f 1  2 1  1  3 4 The largest of these values, 3, is the maximum. It occurs at x  1. The smallest of 3 these values,  , is the minimum. It occurs 8 1 at x   . 2 Thus, the absolute maximum over the interval  1,1 , is 3, which occurs at x  1 , and the absolute minimum over  1,1 is 3 1  , which occurs at x   . 8 2 70. f  x   2 x  x; 4 a) f ‘  x   8×3  1 1,1 1 is in the 2 and the absolute minimum over  1,1 is 3 1  , which occurs at x  . 8 2 71. f  x  3 x  x 3 ; 1 a) Find f ‘  x  f ‘  x  0,8 1 2 3 1 x  3 2 3 3 x b) Find the critical values. f ‘  x  does not exist for x  0 . The equation f ‘  x   0 has no solution, so the only critical value is 0, which is also an endpoint. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 350 c) The interval is closed, and we are looking for both the absolute maximum and absolute minimum values, so we use Max-Min Principle 1. The only critical value is an endpoint. The endpoints are 0 and 8 . Next, we find the function values at these points. f  0  3 0  0 73. a) Find f ‘  x  . f ‘  x   3  x  1 . 2 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . 3  x  1  0 2 x 1  0 x 1 The only critical value is x  1. c) Since there is only one critical value, we can apply Max-Min Principle 2. First we find f ”  x  . and the absolute minimum over  0,8 is 0, which occurs at x  0. f  x  x  x 2 ; 1 a) f ‘ x  f ”  x   6  x  1 . Now, f ” 1  6 1  1  0 , so the Max-Min 0,4 1  12 1 x  2 2 x Principle 2 fails. We cannot use Max-Min Principle 1, because there are no endpoints. b) Find the critical values. f ‘  x  does not exist We note that f ‘  x   3  x  1 is never 2 for x  0 . The equation f ‘  x   0 has no negative. Thus, f  x  is increasing solution, so the only critical value is 0, which is also an endpoint. c) The interval is closed, and we are looking for both the absolute maximum and absolute minimum values, so we use Max-Min Principle 1. The only critical value is an endpoint. The endpoints are 0 and 4 . Next, we find the function values at these points. f  0  0  0 f  4  4  2 The largest of these values, 2, is the maximum. It occurs at x  4. The smallest of these values, 0, is the minimum. It occurs at x  0. Thus, the absolute maximum over the interval  0, 4 is 2, which occurs at x  4 , and the absolute minimum over  0, 4 is 0, which occurs at x  0. 3 When no interval is specified , we use the real line  ,   . f 8  3 8  2 The largest of these values, 2, is the maximum. It occurs at x  8. The smallest of these values, 0, is the minimum. It occurs at x  0. Thus, the absolute maximum over the interval  0,8 , is 2, which occurs at x  8 , 72. f  x    x  1 everywhere except at x  1 . Therefore, the function has no maximum or minimum over the interval  ,   . Notice f ”  0  6  0 f ”  2   6  0 and f 1  0 Therefore, there is a point of inflection at 1, 0 . 74. f  x    x  1 3 When no interval is specified , we use the real line  ,   . a) f ‘  x   3  x  1 2 b) f ‘  x  exists for all real numbers. Solve: f ‘ x  0 3  x  1  0 2 x  1 The only critical value is x  1. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 c) 351 f ”  x   6  x  1 Therefore, the absolute maximum over the interval  10,10 is 59, which occurs at f ”  1  6  1  1  0 . x  10 , and the absolute minimum over the interval  10,10 is  41 , which occurs The Max-Min Principle 2 fails. We cannot use Max-Min Principle 1, because there are no endpoints. We note that f ‘  x   3  x  1 is always positive, except at x  10. 2 at x  1 . Thus, f  x  is increasing 77. a) Find f ‘  x  everywhere except at x  1 . Therefore, the function has no maximum or minimum over the interval  ,   . f ‘ x  2 b) and c) The derivative exists and is 2 for all real numbers. Therefore, f ‘  x  is never 0. Thus, Notice f ”  2   6  0 there are no critical values. We apply the Max-Min Principle 1. There is only one endpoint, x  1 . We find the function value at the endpoint. f  1  2  1  3  5 f ”  0  6  0 and f  1  0 Therefore, there is a point of inflection at  1, 0 . 75. f  x   2 x  3; a) Find f ‘  x  . We know f ‘  x   0 over the interval, so the function is increasing over the interval 1, 5 . Therefore, the minimum value will  1,1 be the left hand endpoint. The absolute minimum over the interval  1, 5 is 5 , f ‘ x  2 . . which occurs at x  1. Since the right endpoint is not included in the interval, the function has no maximum value over the interval  1, 5 . b) and c) The derivative exists and is 2 for all real numbers. Therefore, f ‘  x  is never 0. Thus, there are no critical values. We apply the Max-Min Principle 1. The endpoints are 1 and 1. We find the function values at the endpoints. f  1  2  1  3  5 f 1  2 1  3  1 Therefore, the absolute maximum over the interval  1,1 is 1 , which occurs at x  1 , and the absolute minimum over the interval  1,1 is 5 , which occurs at x  1. 76. f  x   9  5 x; a) f ‘  x   5  1,5 f  x   2 x  3;  10,10 b) and c) The derivative exists for all real numbers and is never 0. There are no critical values, so the maximum and minimum occur at the endpoints, 10 and 10. We find the function values at the endpoints. f  10  9  5  10  59 f 10  9  5 10  41 78.  1,1 f  x  x 3 ; 2 a) f ‘ x  2  13 2 x  3 3 3 x b) Find the critical values. f ‘  x  does not exist for x  0 . The equation f ‘  x   0 has no solution, so the only critical value is 0. c) The interval is closed, and we are looking for both the absolute maximum and absolute minimum values, so we use Max-Min Principle 1. The critical value and the endpoints are 1, 0, and 1. Next, we find the function values at these points. f  1   1 3  1 2 f  0   0 3  0 2 f 1  1 3  1 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. 2 Chapter 2: Applications of Differentiation 352 From the previous page, the largest of these values, 1, is the maximum. It occurs at x  1 and x  1. The smallest of these values, 0, is the minimum. It occurs at x  0. Thus, the absolute maximum over the interval  1,1 , is 1, which occurs at x  1 and x  1 , and the absolute minimum over  1,1 is 0, which occurs at x  0. 79. f  x   9  5 x; a) Find f ‘  x  . We determine that the function has no absolute extrema over the interval  ,   .  2, 3 f ‘  x   5 . 81. b) and c) The derivative exists for all real numbers and is never 0. There are no critical values. The only endpoint is the left endpoint 2 . f ‘  x   0 over the interval, so the function g  x  x 3 When no interval is specified , we use the real line  ,   . 2 a) Find g ‘  x  . g ‘  x  is decreasing and a maximum occurs at x  2 . We find the function value at x  2 . f  2   9  5  2   19 . b) Find the critical values. g ‘  x  does not exist for x  0 . The equation g ‘  x   0 has no solution, so the only critical value is 0. c) We apply the Max-Min Principle 2. 2 g ”  x    4 9x 3 g ”  0  does not exist. The absolute maximum over the interval  2, 3 is 19, which occurs at x  2 . The function has no minimum value over the interval  2, 3 . Note that g ‘  x   0 for x  0 and g ‘  x   0 1 2 80. f  x   x 3  x  3 3 When no interval is specified , we use the real line  ,   . a) for x  0 , so g  x  is decreasing on  , 0  and increasing on  0,   . Therefore, the absolute minimum over the interval  ,   is 0, which occurs at x  0. The function has no maximum value. f ‘  x  x 1 2 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . 2 x 1  0 x  1 There are two critical values 1 and 1. c) The interval  ,   is not closed, so the Max-Min Principle 1 does not apply. Since there is more than one critical value, the Max-Min Principle 2 does not apply. A quick sketch of the graph at the top of the next column will help us determine whether absolute or relative extrema occur at the critical values. 2  13 2 x  3 . 3 3 x 82. 1 3 x  2 x 2  x; 3 f ‘  x  x2  4 x  1 f  x  a) 0, 4 b) f ‘  x  exists for all real numbers. Solve: f ‘  x  0 x2  4x  1  0 We use the quadratic formula to solve the equation. x   4  42  4 11 2 1 4  12  2 3 2 The solution is continued on the next page.  Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 353 c) The interval  ,   is not closed, so the Both critical values x  2  3  0.27 and Max-Min Principle 1 does not apply. Since there is more than one critical value, the Max-Min Principle 2 does not apply. A quick sketch of the graph will help us determine absolute or relative extrema occur at the critical values. x  2  3  3.73 are in the closed interval 0, 4 . c) The interval is closed, and there is more than one critical value in the interval, so we use Max-Min Principle 1. The critical values and the endpoints are 0, 2  3, 2  3, and 4. Next, we find the function values at these points. 1 3 2 f 0  0  2 0   0  0 3   13 2  3   2 2  3   2  3  3 f 2 3    2 10  2 3  0.131 3  3 10 3 2  2 3  6.797 1 3  4   2  4 2   4  3 20   6.666 3 Thus, the absolute maximum over the 10 interval  0, 4 , is   2 3 , which occurs 3 at x  2  3 , and the absolute minimum 10 over  0, 4 is   2 3 , which occurs at 3 x  2 3 . f 4  83. We determine that the function has no absolute extrema over the interval  ,   .  3 2  3   2 2  3   2  3  f 2 3  1 1 3 1 2 x  x  2x 1 3 2 When no interval is specified , we use the real line  ,   . f  x  a) Find f ‘  x  f ‘  x  x2  x  2 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘ x  0 . x2  x  2  0  x  2 x  1  0 x  2 or x  1 There are two critical values 1 and 2. 84. 1 3 x  2 x 2  x; 3 a) g ‘  x   x 2  4 x  1 g  x  4, 0 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve g ‘  x  0 . x2  4 x  1  0 We use the quadratic formula to solve the equation. The solution to the quadratic equation is x  2  3 . Both critical values x  2  3  3.73 and x  2  3  0.27 are in the closed interval  4, 0 . c) The interval is closed, and there is more than one critical value in the interval, so we use Max-Min Principle 1. The critical values and the endpoints are 4,  2  3,  2  3, and 0. Next, we find the function values at the critical values and the endpoints. 20 g  4    6.666 3 10 g 2  3   2 3  6.797 3 10 g 2  3   2 3  0.131 3 g  0  0 The solution is continued on the next page.   Copyright © 2016 Pearson Education, Inc.   Chapter 2: Applications of Differentiation 354 b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ‘x  0 . From the previous page, we determine the absolute maximum over the interval  4, 0 , 10  2 3 , which occurs at x  2  3 , 3 and the absolute minimum over  4, 0 is is 8×3  8x  0 10  2 3 , which occurs at x  2  3 . 3 85. t  x  x  2x 4   x0 or 8x x2  1  0 x2  1  0 x  0 or x  1 There are three critical values 1, 0, and 1. c) The interval  ,   is not closed, so the 2 When no interval is specified , we use the real line  ,   . Max-Min Principle 1 does not apply. Since there is more than one critical value, the Max-Min Principle 2 does not apply. A quick sketch of the graph will help us determine absolute or relative extrema occur at the critical values. a) Find t ‘  x  t ‘  x   4 x3  4 x b) Find the critical values. The derivative exists for all real numbers. Thus, we solve t ‘  x  0 . 4 x3  4 x  0   x0 or 4 x x2  1  0 x2  1  0 x  0 or x  1 There are three critical values 1, 0, and 1. c) The interval  ,   is not closed, so the Max-Min Principle 1 does not apply. Since there is more than one critical value, the Max-Min Principle 2 does not apply. A quick sketch of the graph will help us determine whether absolute or relative extrema occur at the critical values. We determine that the function has no absolute maximum over the interval  ,   . The function’s absolute minimum is 0, which occurs at x  1 and x  1. 87. – 96. 97. Left to the student. M  t   2t 2  100t  180, 0  t  40 a) M ‘  t   4t  100 b) M ‘  t  exists for all real numbers. We solve M ‘  t   0. 4t  100  0 4t  100 We determine that the function has no absolute maximum over the interval  ,   . The function’s absolute minimum is 1 , which occurs at x  1 and x  1. 86. f  x  2 x4  4 x2  2 When no interval is specified , we use the real line  ,   . a) Find f ‘  x  f ‘  x   8×3  8x t  25 c) Since there is only one critical value, we apply the Max-Min Principle 2. First, we find the second derivative. M ” t   4 . The second derivative is negative for all values of t in the interval, therefore, a maximum occurs at t  25. M  25  2  25  100  25  180  1430 2 The maximum productivity for 0  t  40 is 1430 units per month, which occurs after t  25 years of employment. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 98. 355 N  a    a 2  300a  6, 0  a  300 a) N ‘  a   2 a  300 b) N ‘  a  exists for all real numbers. Solve: N ‘ a   0 2a  300  0 a  150 c) Since there is only one critical value, we apply the Max-Min Principle 2. First, we find the second derivative. N ”  a   2 . The second derivative is negative for all values of t in the interval, therefore, a maximum occurs at a  150. N 150    150   300 150   6  22, 506 2 The maximum number of units that can be sold is 22,506. In order to achieve this maximum, $150,000 must be spent on advertising. 99. p  x   0.039 x3  0.594 x 2  1.967 x  7.555 The absolute maximum occurs when x  8.07. According to this model, the maximum percentage of unemployed workers in service occupations occurred in 2011. 100. f  x   0.0135 x 2  0.265 x  74.6 We restrict our attention to the years 1992 to 2012. That is, we will look at the x-values 0  x  20 . a) f ‘  x   0.027 x  0.265 b) f ‘  x  exists for all real numbers. Solve: f ‘  x  0 0.027 x  0.265  0 x  9.81 The critical value 9.81 is in the interval. c) The critical value and the endpoints are 0, 9.81, and 20. d) Find the function values. f 0  0.0135 0  0.265 0  74.6 2 We restrict our attention to the years 2003 to 2013. That is, we will look at the x-values 0  x  10 . a) Find p ‘  x  . f 9.81  0.0135 9.81  0.265 9.81  74.6 p ‘  x   0.117 x  1.188 x  1.967 . f  20  0.0135  20  0.265  20  74.6 2 b) Find the critical values. p ‘  x  exists for all real numbers. Therefore, we solve p ‘ x  0 . 0.117 x 2  1.188 x  1.967  0 Using the quadratic formula, we have: x  1.188  1.1882  4 0.1171.967 2  0.117  1.188  0.490788 0.234 x  2.08 or x  8.07 The critical values are x  2.08 and x  8.07 is in the interval  0,10  .  c) The critical values and the endpoints are 0, 2.08, 8.07, and 10. d) Using a calculator, we find the function values. p 0  7.555 p 2.08  5.683 p 8.07   9.869 p 10  8.285  74.6 2  75.9 2  74.5 The maximum occurs when x  9.81. According to this model for the period 1992 to 2012, the percentage of women aged 2154 in the U.S. Civilian labor forced was at maximum in 2001. 101. We use the model P t   2.69t 4  63.941t 3  459.895t 2  688.692t  24,150.217 We consider the interval  0,   , where t  0 corresponds to the year 2000. a) Find P ‘  t  . P ‘ t   10.76t 3  191.823t 2  919.79t  688.692 b) P ‘  t  exists for all real numbers. Solve: P ‘ t   0 10.76t 3  191.823t 2  919.79t  688.692  0 Using a graphing calculator, we approximate the zeros of P ‘  t  . We find the solutions: t  0.914 t  7.232 t  9.681 Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 356 c) The critical values and the endpoints are: 0, 0.914, 7.232, and 9.681. d) Find the function values. P 0  24,150 P 0.914  23,858 P 9.681  39,533 The absolute minimum production of world wide oil was 23,858,000 barrels. The world achieved this production 0.914 years after 2000, or approximately the year 2001. 1500 x  6 x  10 We will restrict our analysis to the nonnegative real numbers  0,   , since you cannot produce and sell a negative number of amplifiers. 3000  x  3 a) P ‘  x    2 x 2  6 x  10 2   b) P ‘  x  exists for all real numbers. Solve P ‘  x  0  3000  x  3  x  6 x  10 2 2 0 x3 c) Since the interval is open, we apply the Max-Min Principle 2. P ”  x    3000 3x 2  18 x  26  x  6 x  10 2 P ‘  x  exists for all real numbers. Solve: P ‘  x  0  x  400  0 P 7.232  26,396 102. P  x   b) First, we find the critical values. P ‘  x    x  400 3  x  400 The critical value is 400 and the endpoints are 0 and 600. Using the Max-Min Principle 1. We evaluate the function at the endpoints and critical values: 1 2 P  0    0  400  0  5000 2  5000 1 2 P  400     400   400  400  5000 2  75, 000 1 2 P  600     600   400  600   5000 2  55, 000 The total profit is maximized when 400 items are produced. 104. From Exercise 103, we know that 1 P  x    x 2  400 x  5000 2 P  x 1 5000   x  400  , a) A  x   x x 2 0  x  600 . 1 5000 b) A ‘  x     2 2 x A ‘  x  exists everywhere on 0  x  600 . P ”  3  3000 Solve A ‘  x   0 . Therefore, a maximum occurs at x  3 . We find the function value at x  3 . 1500 P  3   1500 2 3  6 3  10 Producing and selling 3 amplifiers will result in a maximum weekly profit of $1500. 1 5000   2 0 2 x 103. C  x   5000  600 x 1 2 x  1000 x , 0  x  600 2 a) P  x   R  x   C  x  R  x   1 2 x  1000 x  5000  600 x  2 1   x 2  400 x  5000 2  x 2  10,000 x  100 x  100 is the only critical value in the interval 0  x  600 . The critical value and the endpoint are 100 and 600. A 100  300 A  600  91.67 The average profit is maximized when 100 items are produced. 105. B  x   305 x 2  1830 x 3 , a) B ‘  x   610 x  5490 x Copyright © 2016 Pearson Education, Inc. 0  x  0.16 2 Exercise Set 2.4 357 b) B ‘  x  exists for all real numbers. Solve: We sketch a graph of the function. B ‘  x  0 610 x  5490 x 2  0 610 x 1  9 x   0 x  0 or 1  9 x  0 1  0.11 9 c) The critical points and the endpoints are 1 0, , and 0.16. 9 d) We find the function values. x  0 or x B  0  305  0  1830  0  0 2 3 2 1 1 1 B    305    1830   9 9 9  g  2    2   4 2 B  0.16  305 0.16   1830  0.16 3  0.312 The maximum blood pressure is approximately 1.255, which occurs at a dose 1 of x  cc , or about 0.11 cc of the drug. 9 106. x  0 , which is also the endpoint of this part of the domain. For 0  x  2 , g ‘  x   5 . Therefore, there are no critical values of g  x  . The absolute extrema will occur at one of the endpoints. The function values are: 3 305  1.255 243 2 108. We look at the derivative on each piece of the function to determine any critical values. For 2  x  0 , g ‘  x   2 x . g ‘  x   0 when g 0  0  0 2 g  2   5  2   10 On the interval  2, 2  , the absolute minimum is 0 , which occurs at x  0. The absolute maximum is 10, which occurs at x  2 A sketch of the graph is shown below. In finding the absolute extrema of a function, the second derivative is used when there is exactly one critical value interior to an interval. If there is exactly one critical value, the second derivative will determine the concavity of the function on the interval, and thus determine if there is an absolute maximum or absolute minimum on the interval. 107. We look at the derivative on each piece of the function to determine any critical values. For 3  x  1 , f ‘  x   2 so there are no critical values for this part of the function. For 1  x  2 , f ‘  x   2 x . f ‘  x   0 when x  0 , which is outside the domain of this piece of the function. Therefore, there are no critical values of f  x  . The absolute extrema will occur at one of the endpoints. The function values are: f  3  2  3  1  5 f 1  2 1  1  3 f 2  4   2  0 2 109. We look at the derivative on each piece of the function to determine any critical values. For 4  x  0 , h ‘  x   2 x . h ‘  x   0 when x  0 , which is also the endpoint of this part of the domain. For 0  x  1 , h ‘  x   1 . Therefore, there are no critical values of h  x  on this part of the domain. For 1  x  2 , h ‘  x   1 . Therefore, there are no critical values of h  x  on this part of the domain. The absolute extrema will occur at one of the endpoints. The function values are determined on the next page. On the interval  3, 2  , the absolute minimum is 5 , which occurs at x  3. The absolute maximum is 3, which occurs at x  1. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 358 We determine the function values from the previous page: 111. a) The sketch of the graph is shown below: h  4   1   4   15 2 h  0  1   0  1 h 1  1  1  0 h  2   2  1  1 On the interval  4, 2  , the absolute minimum is 15 , which occurs at x  4. The absolute maximum is 1, which occurs at x  0 and x  2. A sketch of the graph is shown below. 110. We look at the derivative on each piece of the function to determine any critical values. For 2  x  0 , F ‘  x   2 x . F ‘  x   0 when x  0 , which is also the endpoint of this part of the domain. For 0  x  3 , F ‘  x   1 . Therefore, there are no critical values of F  x  on this part of the domain. For 3  x  67 , 1 . Therefore, there are no F ‘  x  2 x2 critical values of F  x  on this part of the domain. The absolute extrema will occur at one of the endpoints. The function values are: F  2    2   4  8 2 b) From the graph, the absolute maximum is 6 and occurs at x  2 . c) The absolute minimum value for this function is 2. This value occurs over the range 0  x  4 . 112. a) According to the graph, the maximum elevation is 2290 feet, this elevation occurs at 6.5 miles from the western edge of Rogers Dry Lake. b) According to the graph, the minimium elevation occurs at 2269 feet. This elevation occurs over the range of 0.5 miles to 5.5 miles from the western edge of Rogers Dry Lake.  3, 3 113. g  x   x x  3; a) Find g ‘  x  . 1 1  1 g ‘  x   x   x  3 2 1  1 x  3 2 2  1 x 2  1   x  3 2 2  x  3  x  3  2  x  3   1 2  x  3 2  x  3 1 x 1 F  0  4   0  4 2 2 1 1 2 2 Multiplying by a form of 1 F  3  3  2  1 F  67   67  2  65  8.062 On the interval  2, 67  , the absolute minimum is 1 , which occurs at x  3. The absolute maximum is approximately 8.062, which occurs at x  67. A sketch of the graph is shown below.   x 2  x  3 2 1 3x  6 2  x  3 2 Copyright © 2016 Pearson Education, Inc. 1  2  x  3 2  x  3 2 , or 1 3x  6 2 x3 Exercise Set 2.4 359 b) Find the critical values. g ‘  x  exists for all d) Find the function values. values in  3, 3 except 3 . This is a h  0   0  1   0   0 critical value as well as an endpoint. To find the other critical values, we solve g ‘  x  0 2 2 2 2 2 1 h      1      3  3  3 3 3 3 3 3x  6 0 2 x3 3x  6  0 3 x  6 x  2 The second critical value on the interval is 2 . c) On a closed interval, the Max-Min Principle 1 can always be used. The critical values and the endpoints are 3,  2, and 3 . d) Find the function value at each value in step (c).. g  3   3  3  3  0 g   2    2   2   3  2 g  3   3  3  3  3 6 Thus, the absolute maximum over the interval  3, 3 is 3 6 , which occurs at x  3 , and the absolute minimum is 2 , which occurs at x  2. 114. h  x   x 1  x ; a) Find h ‘  x  . h 1  1 1  1  0 Thus, the absolute maximum over the 2 , which occurs at interval  0,1 is 3 3 2 x  , and the absolute minimum over the 3 interval is 0, which occurs at x  0 and x  1.  2  115. C  x    2 x  4    ,  x  6   2 x  4  2  x  6 1 1 1 h ‘  x   x   1  x  2  1  11  x  2 2 x   1 x 2 1 x 2 1  x  x   2 1 x 2 1 x 3 x  2  2 1 x b) h ‘  x  does not exist for x  1. Solve: h ‘ x  0 3x  2 0 2 1 x 3 x  2  0 2 x 3 c) The critical values and the endpoints are 2 0, , and 1. 3 1 a) Find C ‘  x  . C ‘  x   2  2  x  6 2 1  2  2  x  6 2 b) Find the critical values. C ‘  x  does not exist for x  6 ; however, this value is not in the domain interval, so it is not a critical value. Solve C ‘  x   0 . 2 0,1 x6 2  x  6 2 0 2 2  x  6 2 2  x  6  2 2 Multiplying by  x  6 Since x  6. 2  x  6 2  1 x  6  1 Taking the square root of both sides. x  6 1 x  5 or x  7 The only critical value in  6,   is 7. c) Since there is only one critical value, we apply the Max-Min Principle 2. 4 3 C ”  x   4  x  6    x  6 3 C ”  7   4 7  63 40 Therefore, since C ”  7   0 , there is a minimum at x  7 . The Katie’s Clocks should use 7 “quality units” to minimize its total cost of service. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 360 116. y   x  a    x  b  2 118. 2 dy  2  x  a   2  x  b   4 x  2 a  2b dx b) The derivative exists for all real numbers. Solve: dy 0 dx 4 x  2 a  2b  0 4 x  2 a  2b ab x 2 c) There is only one critical value, we apply the Max-Min Principle 2. d2y  4  0 for all values of x. dx 2 ab Thus, y is a minimum for x  . 2 117. From exercise 101, we know that the first derivative is R t   P ‘ t   10.76t 3  191.823t 2  a) 919.79t  688.692 a) We find the maximum rate of change, by finding the critical values of the derivative.  0,8 . Taking the derivative we have: P ” t   32.28t 2  383.646t  919.79 b) P ” t  exists for all real numbers. Solve: P ” t   0 2 32.28t  383.646t  919.79  0 Using the quadratic formula, we find the zeros of P ” t  . t   383.646    383.646   4  32.28  919.79  2  32.28  2 The two solutions are t  3.33 and t  8.554. Only one of the critical values, t  3.33 , is in the interval. c) Since there is only one critical value, we apply the Max-Min Principle 2. P ”’  t   64.56t  383.646 P ”’  3.33  64.56  3.33  383.646  168.6  0 Therefore, the absolute maximum over the interval  0,8  occurs at t  3.33 . P ‘  3.33  644.427  640. In the year 2003 – 2004, worldwide oil production was increasing most rapidly. It was increasing at a rate of approximately 640,000 barrels per year. The first derivative is used to find the critical values of the function on an interval. For closed intervals, we know that absolute extrema occur at a critical value in the interval, or at an endpoint of the interval. For open intervals, absolute extrema, if they exists, will occur at a critical value in the interval. 119. P  t   0.0000000219t 4  0.0000167t 3  0.00155t 2  0.002t  0.22, a) Find P ‘  t  . 0  t  110 P ‘  t   0.0000000876t 3  0.0000501t 2  0.0031t  0.002 P ‘  t  exists for all real numbers. Solve P ‘  t   0 . We use a calculator to find the zeros of P ‘  t  . We estimate the solutions to be: x  0.639 x  71.333 x  501.223 Only one of the critical values, x  71.333 , is in the interval  0,110 . We apply the Max-Min Principle 1, to find the absolute maximum. The critical values and the endpoints are 0, 71.333, and 110. The function values at these points are P  0  0.22 P  71.333  2.755 P 110  0.174 Thus, the absolute maximum oil production for the U.S. after 1910 was 2.755 billion barrels per year. This production level occurred 71.333 year after 1910, or in 1981. b) In 2010, t  2010  1910  100 .We plug this value into the first derivative to obtain: P ‘ 100   0.0000000876 100   0.0000501 100   3 2 0.0031 100   0.002  0.1014. The rate of oil was declining at approximately 0.1014 billion of barrels per year. The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.4 361 In 2015, t  2015  1910  105 .We plug this value into the first derivative to obtain: Then, using the window: P ‘ 105  0.0000000876 105   0.0000501 105  3 2 0.0031 105   0.002  0.1234. The rate of oil was declining at approximately 0.1234 billion of barrels per year. 120. f  x   x 3  x  5 ; 2 We get the graph: 1, 4 Using a calculator, we enter the equation into the graphing editor: Using the table feature, we locate the extrema. We estimate the absolute minimum to be 0, which occurs at x  1. There is no absolute maximum. 4 x  122. f  x   x   5 ;  2  Using a calculator, we enter the equation into the graphing editor: Then, using the window: We get the graph: Then, using the window: Using the table feature, we locate the extrema. We estimate the absolute maximum to be 2.520 , which occurs at x  4 , and the absolute minimum to be 4.762 , which occurs at x  2.   2 3 2 3  12 ,  x 1 ; 4 Using a calculator, we enter the equation into the graphing editor: 121. f  x   We get the graph:  Looking at the graph, it is clear to see that there are no absolute extrema over the real numbers. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 362 123. a) Using a graphing calculator, we fit the linear equation y  x  8.857 . This corresponds to the model P  t   t  8.857 . Where P is the pressure of the contractions and t is the time in minutes. We substitute 7 for t to find the pressure at 7 minutes. P  7   7  8.857  15.857 . The pressure at 7 minutes is 15.857 mm of Hg. b) Rounding the coefficients to 3 decimal places, we find the quartic regression y  0.117 x 4  1.520 x 3  6.193 x 2  7.018 x  10.009 Changing the variables we get the model P  t   0.117t 4  1.520t 3  6.193t 2  7.018t  10.009 Using the table feature, when x  7 , y  24.857 . So the pressure at 7 minutes is 24.86 mm of mercury. (If we use the rounded coefficients above, we get 23.897 mm of Hg.) Using the trace feature, we estimate the smallest contraction on the interval  0,10 was about 7.62 mm of Hg. This occurred when x  0.765 min. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 363 3. Exercise Set 2.5 1. Express Q  xy as a function of one variable. First, we solve x  y  70 for y. x  y  70 y  70  x Next, we substitute 70  x for y in Q  xy . Q  xy Q  x 70  x  Substituting 2  70 x  x Now that Q is a function of one variable we can find the maximum. First, we find the critical values. Q ‘  x   70  2 x . Since Q ‘  x  exists for all real numbers, the only critical value will occur when Q ‘  x   0 . We solve: 70  2 x  0 70  2 x 35  x There is only one critical value. We use the second derivative to determine if the critical value is a maximum. Note that: Q ”  x   2  0 . The second derivative is Q  x   x  x  6  x 2  6 x . Finding the deriviative, we have: Q ‘ x  2x  6 The derivative exists for all values of x; thus, the only critical values are where Q ‘  x   0 . 2x  6  0 2x  6 x3 There is only one critical value. We can use the second derivative to determine whether we have a minimum. Q ”  x   2  0 for all values of x. Therefore, a negative for all values of x. Therefore, a maximum occurs at x  35. minimum occurs at x  3. Q  3   3  6  3   9 2 Now, Q  35  70  35   35  1225 2 Therefore, the maximum product is 1225, which occurs when x  35. If x  35 , then y  70  35  35 . The two numbers are 35 and 35. 2. x  y  50, so y  50  x. Q  x   xy  x 50  x   50 x  x 2 Q ‘  x   50  2 x Q ‘  x  exists for all real numbers. Solve: Q ‘  x  0 Let x be one number and y be the other number. Since the difference of the two numbers must be 6, we have x  y  6. The product, Q, of the two numbers is given by Q  xy , so our task is to minimize Q  xy , where x  y  6. First, we express Q  xy as a function of one variable. Solving x  y  6 for y, we have: x y  6 y  6 x y  x6 Next, we substitute x  6 for y in Q  xy. Thus, the minimum product is 9 when x  3 , and y  3  6  3 . 4. Let x be one number and y be the other number. The product, Q, of the two numbers is given by Q  xy , so our task is to minimize Q  xy , where x  y  4. First, we express Q  xy as a function of one variable. Solving x  y  4 for y, we have: y  x  4 Next, we substitute x  4 for y in Q  xy. Q  x  x  4  x 2  4 x 50  2 x  0 x  25 Q ”  x   2  0 for all values of x, so a maximum occurs at x  25 . Q  25  50  25   25  625 2 Thus, the maximum product is 625 when x  25 and y  50  25  25. Q ‘ x  2x  4 The derivative exists for all values of x; thus, the only critical values are where Q ‘  x   0 . 2x  4  0 2x  4 x2 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 364 There is only one critical value. We can use the second derivative to determine whether we have a minimum. Q ”  x   2  0 for all values of x. Therefore, a 2  y2  4 y2  4  2 y2  2 minimum occurs at x  2. y 2 Q  2    2   4  2   4 2 Thus, the minimum product is 4 when x  2. Substitute 2 for x in y  x  4 to find y. y  2  4  2 . The two numbers which have the minimum product are 2 and  2 . 5. 2 Maximize Q  xy , where x and y are positive 2 numbers such that x  y  4 . Express Q  xy 2 as a function of one variable 2 2 First, we solve x  y  4 for y . 2 x y 4 y2  4  x Next, we substitute 4  x for y 2 in Q  xy 2 . Q  xy 2 Q  x 4  x   4x  x2 Now that Q is a function of one variable we can find the maximum. First, we find the critical values. Q ‘ x  4  2x . Since Q ‘  x  exists for all real numbers, the only critical value will occur when Q ‘  x   0 . We set Q ‘  x   0 and solve for x: 4  2x  0 2 x  4 x  2. There is only one critical value. We use the second derivative to determine if the critical value is a maximum. Note that: Q ”  x   2  0 . The second derivative is negative for all values of x. Therefore, a maximum occurs at x  2 . Now, Q  2    4  2    2   4 2 Substitute 2 in for x in x  y 2  4 and solve for y. y 2 x and y are positive Then Q is a maximum when x  2 and y  2 . 6. Maximize Q  xy 2 , where x and y are positive numbers such that x  y 2  1 . Express Q  xy 2 as a function of one variable. x  y2  1 y2  1 x Q  xy 2 Q  x 1  x   x  x 2 Substituting Now that Q is a function of one variable we can find the maximum. First, we find the critical values. Q ‘  x   1  2 x . Since Q ‘  x  exists for all real numbers, the only critical value will occur when Q ‘  x   0 . We solve: 1  2x  0 1  x. 2 There is only one critical value. We use the second derivative to determine if the critical value is a maximum. Note that: Q ”  x   2  0 . The second derivative is negative for all values of x. Therefore, a 1 maximum occurs at x  . 2 2 1 1 1 1 Now, Q          2 2 2 4 1 When x  , we have: 2 1 y2  1  2 1 1  y 2 2 1 y x and y must be positive , 2 1 1 The maximum value of Q is when x  and 4 2 1 . y 2 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 7. 365 There is one critical value, we use the second derivative to determine if it is a minimum. Q ”  x   10  0 for all values of x, therefore, Minimize Q  x 2  2 y 2 , where x  y  3 . Express Q as a function of one variable. First, solve x  y  3 for y. x y 3 y  3 x 2 Q ” 3  0 and a minimum occurs when x  3 . When x  3 , Q 3  2 3  3 5  3  30. 2 2 Then substitute 3  x for y in Q  x  2 y . Q  x  2 3  x    x2  2 9  6 x  x2   3x 2  12 x  18 Find Q ‘  x  , where Q  x   3x 2  12 x  18 . Q ‘  x   6 x  12 This derivative exists for all values of x; thus the only critical values are where Q ‘ x  0 6 x  12  0 6 x  12 x2 Since there is only one critical value, we can use the second derivative to determine whether we have a minimum. Note that: Q ”  x   6 , which is positive for all real numbers. Thus Q ” 2   0 , so a minimum occurs when x  2. The value of Q is Q  2   2 2  2 3  2  2  42  6. Substitute 2 for x in y  3  x to find y. y  3 x y  3 2 y 1 Thus, the minimum value of Q is 6 when x  2 and y  1. 8. When x  3 , y  5  3  2 . Therefore, the minimum value of Q is 30 when x  3 and y  2. 2 2 9. Maximize Q  xy , where x and y are positive 4 2 y  1. 3 Express Q as a function of one variable. First, 4 solve x  y 2  1 for x. 3 4 2 x y 1 3 4 x  1  y2 3 4 Then substitute 1  y 2 for x in Q  xy . 3  4 2 Q  xy  1  y  y  3  4  y  y3 3 4 Find Q ‘  y  , where Q  y   y  y 3 . 3 Q ‘  y  1 4 y2 numbers such that x  This derivative exists for all values of y; thus the only critical values are where Q ‘ y  0 1 4 y2  0 4 y 2  1 y2  x  y  5 , so y  5  x. 1 4 Q  2 x 2  3 y 2  2 x 2  3 5  x  y  5 x 2  30 x  75 Q ‘  x   10 x  30 y 2 Q ‘  x  exists for all real numbers. Solve: Q ‘  x  0 10 x  30  0 x3 2 1 4 1 2 1 y must be positive 2 Since there is only one critical value, we can use the second derivative to determine whether we have a maximum. y The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 366 Only 2 is in the domain of Q. Q ”  x   8 x Note that: Q ”  y   8 y and 1 1 Q ”    8    4  0 . 2 2 When x  2 , Q ”  2  8  2  16 , so a 1 Since Q ”   is negative, a maximum occurs at 2 y 1 . 2 Evaluating the function at y  1 1 4 1 Q      2 2 3 2 1 we have: 2 x  2 and y  1 4 1   2 3 8 1 1   2 6 1  . 3 1 4 for y in x  1  y 2 to find x. Substitute 2 3 4 2 x  1 y 3 2 x  1 Thus, the maximum value of Q is x 10. 1 when 3 2 1 and y  . 3 2 4 2 4 x  y  16 , so y  16  x 2 . 3 3 4 2 4  Q  xy  x 16  x   16 x  x3   3 3 Q ‘  x   16  4 x . 2 Q ‘  x  exists for all real numbers. Solve: Q ‘  x  0 16  4 x 2  0 x  2 x2 Therefore, the maximum value of Q is 3  4 1   3 2 1 x  1 3 2 x 3 maximum occurs when x  2. When x  2 , 4 2 16 32 y  16   2   16   . 3 3 3 and 32 64 Q 2  2   . 3 3 x must be positive 64 when 3 32 . 3 11. Let x represent the width and y represent the length of the area. It is helpful to draw a picture. y x x y x y River Since the rancher has 240 yards of fence, the perimeter is 3 x  y  240 . Solving this equation for y, we have y  240  3 x . Since x and y must be positive, we are restricted to the interval 0  x  80 . The objective is to maximize area, which is given by A lw Substituting y  240  3 x for the width and x for the length, we have: A  x   x  240  3x   240 x  3x 2 . We will maximize the area over the restricted interval by first finding the derivative of the area function. A  x   240 x  3 x 2 A ‘  x   240  6 x The derivative exists for all values of x in the interval 0,80  . The only critical values are where: A ‘  x  0 240  6 x  0 6 x  240 x  40 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 367 Since there is only one critical value in the interval, we use the second derivative to determine whether we have a maximum. Note, A ”  x   6  0 for all values of x, so there is a maximum at x  40. Next we find the dimensions and the area. When x  40 , y  240  3 x  240  3  40   120 and A  40   240  40   3  40   4800. 2 2 Therefore, the maximum area is 4800 yd when the overall dimensions are 40 yd by 120 yd. 12. Let x represent the length and y represent the width of the swimming area. Since the life guard has 180 yd of rope and floats, the perimeter of the swimming area is x  2 y  180 , or x  180  2 y The objective is to maximize area, which is given by A lw Substituting x  180  2 y for the length and y for the width, we have: A  180  2 y  y  180 y  2 y 2 . We will maximize the area over the interval 0  y  90 , because y is the length of one side, and cannot be negative. We find the derivative: A ‘  y   180  4 y . This derivative exists for all values of y in 0, 90 . Thus the only critical values are where A ‘ y  0 180  4 y  0 y  45 There is one critical value on the interval. A ”  y   4  0 for all values of y, so a maximum occurs at y  45 . Next, find the dimensions and the area. When y  45 , we have x  180  2  45  90 and A  45  180  45  2  45  4050. 2 Therefore, the maximum area is 4050 yd 2 when the overall dimensions are 45 yd by 90 yd. 13. Let x represent the width and y represent the length of the area. It is helpful to draw a picture. y x x x x y Since 1200 yards of fence is available, the perimeter is 4 x  2 y  1200 . Solving this equation for y, we have y  600  2 x . Since x and y must be positive, we are restricted to the interval 0  x  300 . The objective is to maximize area, which is given by A lw Substituting y  600  2 x for the width and x for the length, we have: A  x   xy  x 600  2 x   600 x  2 x 2 . We will maximize the area over the restricted interval by first finding the derivative of the area function. A  x   600 x  2 x 2 A ‘  x   600  4 x The derivative exists for all values of x in the interval 0,300  . The only critical values are where: A ‘  x  0 600  4 x  0 4 x  600 x  150 Since there is only one critical value in the interval, we use the second derivative to determine whether we have a maximum. Note, A ”  x   4  0 for all values of x, so there is a maximum at x  150. Next we find the dimensions and the area. When x  150 , y  600  2 x  600  2 150   300 and A 150   600 150   2 150  2  45, 000. Therefore, the maximum area is 45, 000 yd 2 when the overall dimensions are 150 yd by 300 yd. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 368 14. a) Let x represent the width and y represent the length of the area. If the three areas are parallel to the stone wall, only one of the areas will utilize the stone wall as a side. With 600 ft of fencing, the perimeter of the total area will be 3 3 x  2 y  600 , or y  300  x . Since x 2 and y must be positive, we are restricted to the interval 0  x  200 . Using this information, we find the area as a function of x. 3  3  A  x   xy  x  300  x   200 x  x 2 .  2  2 Therefore, A ‘  x   300  3 x This derivative exists for all values of x. Solve: A ‘  x  0 300  3 x  0 x  100 There is one critical value on the interval. A ”  x   3  0 for all values of x. So a maximum occurs at x  100. When x  100 , 3 y  300  100   150 2 and 3 2 A 100   300 100   100   15, 000. 2 Therefore, the maximum area is 15,000 ft 2 when the overall dimensions are 100 ft by 150 ft. b) Let x represent the width and y represent the length of the area. If the three areas are perpendicular to the stone wall, all three areas will utilize the stone wall as a side. With 600 ft of fencing, the perimeter of the total area will be 4 x  y  600 , or y  600  4 x . Since x and y must be positive, we are restricted to the interval 0  x  150 . Using this information, we find the area as a function of x. A  x   xy  x 600  4 x   600 x  4 x 2 . Therefore, A ‘  x   600  8 x This derivative exists for all values of x. Solve: A ‘  x  0 600  8 x  0 x  75 There is one critical value on the interval. A ”  x   8  0 for all values of x. So a maximum occurs at x  75. When x  100 , y  600  4 75  300 and A 75  600 75  4 75  22, 500. 2 Therefore, the maximum area is 22,500 ft 2 when the overall dimensions are 75 ft by 300 ft. 15. Let x represent the length and y represent the width. It is helpful to draw a picture. x y y x The perimeter is found by adding up the length of the sides. Since it is fixed at 42 feet, the equation of the perimeter is 2 x  2 y  42 . The area is given by A  xy . First, we solve the perimeter equation for y. 2 x  2 y  42 2 y  42  2 x y  21  x Then we substitute for y into the area formula. A  xy  x  21  x   21x  x 2 We want to maximize the area on the interval 0, 21 . We consider this interval because x is the length of the rectangle and cannot be negative. Since the perimeter cannot exceed 42 feet, x cannot be greater than 21, also if x is 21 feet, the width of the rectangle would be 0 feet. We begin by finding A ‘  x  . A ‘  x   21  2 x . This derivative exists for all values of x in 0, 21 . Thus, the only critical values occur where A ‘  x   0 . We solve the equation on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 369 Solving A ‘  x   0 , we have: 21  2 x  0 2 x  21 21 x  10.5 2 Since there is only one critical value in the interval, we can use the second derivative to determine whether we have a maximum. Note that A ”  x   2  0 for all values of x. Thus, 17. When squares of length h on a side are cut out of the corners, we are left with a square base of length x. A picture will help. x x x A 10.5  2110.5  10.5 2  110.25 The maximum area is 110.25 ft2 . Note: when x  10.5 , y  21  10.5  10.5 , so the overall dimensions that will achieve the maximum area are 10.5 ft by 10.5 ft. 16. The perimeter is 2l  2 w  54 , so l  27  w Since l and w must be positive, we are restricted to the interval 0  w  27. A  w  l  w   27  w w  27 w  w2 A ‘  w   27  2 w This derivative exists for all values of w. Solve: A ‘ x  0 27  2 w  0 27  13.5 2 There is one critical value on the interval. We find the second derivative. A ”  w   2  0 for all values of w. So a w maximum occurs at w  13.5 . When w  13.5 , l  27  13.5  13.5 and A 13.5  13.5 27  13.5  182.25. Therefore, the maximum area is 182.25 ft 2 when the overall dimensions are 13.5 ft by 13.5 ft. h h A ” 10.5  0 , so a maximum occurs at x  10.5. Now, A  x   21x  x 2 h x x x The resulting volume of the box is V  lwh  x  x  h  x 2 h . We want to express V in terms of one variable. Note that the overall length of a side of the cardboard is 20 in. We see from the drawing, that h  x  h  20, or x  2 h  20. Solving for h we get: 2 h  20  x 1 1 h  20  x   10  x. 2 2 Substituting h into the volume equation, we have: 1  1  V  x 2 10  x   10 x 2  x 3 . The objective   2 2 is to maximize V  x  on the interval  0, 20  . First, we find the derivative. 3 V ‘  x   20 x  x 2 2 This derivative exists for all x in the interval 0, 20 . We set the derivative equal to zero and solve for the critical values V ‘  x  0 3 2 x 0 2 3   x  20  x   0  2  20 x  3 x0 2 3  x  20 x  0 or 2 40  13 13 x  0 or x 3 The solution is continued on the next page. x  0 or 20  Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 370 From the previous page, the only critical value 40 or about 13.33. Therefore, we in  0, 20  is 3 can use the second derivative V ”  x   20  3 x to determine if we have a maximum. We have  40   40  V ”    20  3    20  0 .  3  3 40 Therefore, there is a maximum at . 3 2 1  40   40   40  V    10       3  3 2 3  3 100 , or 3 about 33.33. Therefore, we can use the second derivative V ”  x   50  3 x to determine if we The only critical value in  0,50  is have a maximum. We have  100   100  V ”   50  3   50  0 .  3   3  100 Therefore, there is a maximum at . 3 2 1  100   100   100  V  25      3   3  2 3  3 16, 000  592 16 27 27 Now, we find the height of the box. 1  40  10 h  10      3 13 . 2 3  3 Therefore, a box with dimensions 13 13 in. by 13 13 in. by 3 13 in. will yield a 250, 000 7  9259 27 27 Now, we find the height of the box. 1  100  25 h  25    8 13 .  2 3  3 Therefore, a box with dimensions 33 13 cm by 33 13 cm by 8 13 cm will yield a in 3. maximum volume of 592 16 27 7 cm3 . maximum volume of 9259 27  18. Using the picture drawn in Exercise 17, the resulting volume of the box is V  lwh  x  x  h  x 2 h . We want to express V in terms of one variable. Note that the overall length of a side of the aluminum is 50 cm. We see from the drawing, that h  x  h  50, or x  2 h  50. Solving for h we get: 1 1 h  50  x   25  x. 2 2 Substituting h into the volume equation, we have: 1  1  V  x 2  25  x   25 x 2  x 3 . The objective  2  2 is to maximize V  x  on the interval  0,50  . First, we find the derivative. 3 V ‘  x   50 x  x 2 2 This derivative exists for all x in the interval 0,50 , so the critical values will occur when V ‘  x   0 . Solving this equation, we have: 3 2 x 0 2 3   x  50  x   0  2  50 x  x  0 or x  100  33 13 . 3  19. First, we make a drawing. y x x The surface area of the open-top, square-based, rectangular tank is found by adding the area of the base and the four sides. x 2 is the area of the base, xy is the area of one of the sides and there are four sides, therefore the surface area is given by S  x 2  4 xy . The volume must by 32 cubic feet, and is given by V  l  w  h  x 2 y  32 . To express S in terms of one variable, we solve x 2 y  32 for y: y 32 . x2 Substituting, we have:  32  S  x  x2  4x  2  x  128  x 2  128 x 1 x Now S is defined only for positive numbers, so we minimize S on the interval  0,   .  x2  The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 371 First, we find S ‘  x  . We restrict the analysis to the interval  0,   . S ‘  x   2 x  128 x 2 S ‘  x   2 x  250 x 2  2 x   2x  128 x2 Since S ‘  x  exists for all x in  0,   , the only critical values are where S ‘  x   0 . We solve the following equation: S ‘  x  0 128 0 x2 x 3  64 x4 Since there is only one critical value, we use the second derivative to determine whether we have a minimum. Note that 256 S ”  x   2  256 x 3  2  3 . x 256 S ”  4  2  3  6  0 . Since the second 4 derivative is positive, we have a minimum at x  4 . We find y when x  4 . 32 y 2 x 32  2 4 2 The surface area is minimized when x  4 ft and y  2 ft. We find the minimum surface area by substituting these values into the surface area equation. S  x 2  4 xy 2x   4  4  42 2  16  32  48 S  4   42  4  4  2  48. Therefore, when the dimensions are 4 ft by 4 ft by 2 ft, the minimum surface area will be 48 ft2. 20. Using the drawing in Exercise 19, we see that S  x 2  4 xy and V  l  w  h  x 2 y  62.5 .  62.5  , and S  x   x 2  4 x  2  , or  x  x2 250 S  x  x2   x 2  250 x 1 . x Then y  62.5 250 x2 S ‘  x  exists for all values of x in  0,   . Solve: S ‘  x  0 2x  250 x2 0 x3  125 x5 There is one critical value in the interval. We use the second derivative to determine if it is a minimum. 500 S ”  x   2  500 x 3  2  3 x 500 S ” 5  2  3  6  0 , so a minimum occurs 5 when x  5 . 62.5 When x  5 , y  2  2.5 . 5 Therefore, S 5  52  4  5  2.5  75. Therefore, when the dimensions are 5 in by 5 in by 2.5 in, the minimum surface area will be 75 in2. 21. First, we make a drawing. y x 2x The surface area of the open-top, rectangular dumpster is found by adding the area of the base and the four sides. 2x 2 is the area of the base, xy is the area of two of the sides, while 2 xy is the area of the other two sides. Therefore the surface area is given by S  2 x 2  2 xy  2  2 xy   2 x 2  6 xy . The volume must by 12 cubic yards, and is given by V  l  w  h  2 x  x  y  2 x 2 y  12 . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 372 To express S in terms of one variable, we solve 2 x 2 y  12 for y: 6 x2 Then y  6  S  x  2×2  6x  2  x  22. Let x be the width of the container, y be the length of the container and 2x be the height of the container. Since we are including the top and the bottom of the container, the surface area is given by: S  4 x 2  6 xy and the volume is given by V  y  x  2 x  2 x 2 y  18 36  2 x 2  36 x 1 x Now S is defined only for positive numbers, so we minimize S on the interval  0,   . Then, y  First, we find S ‘  x  . 9 , and x2  9  S  4 x 2  6 x  2   4 x 2  54 x 1 . x  We are restricted to the interval  0,   . S ‘  x   4 x  36 x 2 S ‘  x   8 x  54 x 2  8 x  36 x2 Since S ‘  x  exists for all x in  0,   , the only S ‘  x  0  2×2  54 x2 S ‘  x  exists for all x in the interval. Solve:  4x  54 0 x2 8 x 3  54 critical values are where S ‘  x   0 . We solve 8x  the following equation: 36 4x  2  0 x 36 4x  2 x 3 x 9 27 4 x  1.89 Since there is only one critical value, we use the second derivative to determine if it is a minimum. 108 S ”  x   8  3 x S ” 1.89   24  0 x3  x  3 9  2.08 Since there is only one critical value, we use the second derivative to determine whether we have a minimum. 72 Note that S ”  x   4  72 x 3  4  3 . x 72 S ” 3 9  4   12  0 . Since the second 3 3 9   Therefore, there is a minimum at x  1.89 . The height is twice that of the width, therefore, the height is 2 1.89   3.78 . We solve for the   length y using: 9 y  2.52 1.892 Therefore, the dimensions of the compost container with minimal surface area are 1.89 ft. by 2.52 ft. by 3.78 ft. derivative is positive, we have a minimum at x  3 9  2.08 . The width is 2.08 yd.; therefore, the length is 2  2.08  4.16 . We find the height y 6 y 2 x 6   2.082 23.  1.387 The overall dimensions of the dumpster that will minimize surface area are 2.08 yd by 4.16 yd by 1.387 yd. R  x   50 x  0.5 x 2 ; C  x   4 x  10 Profit is equal to revenue minus cost. P  x  R  x  C  x  50 x  0.5 x 2   4 x  10  0.5 x 2  46 x  10 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 373 Because x is the number of units produced and sold, we are only concerned with the nonnegative values of x. Therefore, we will find the maximum of P  x  on the interval  0,   . 25. Profit is equal to revenue minus cost. P  x  R  x  C  x   2 x  0.01x 2  0.6 x  30 First, we find P ‘  x  . P ‘  x    x  46  2  0.01x  1.4 x  30 Because x is the number of units produced and sold, we are only concerned with the nonnegative values of x. Therefore, we will find the maximum of P  x  on the interval  0,   . The derivative exists for all values of x in  0,   . Thus, we solve P ‘  x   0.  x  46  0  x  46 x  46 There is only one critical value. We can use the second derivative to determine whether we have a maximum. P ”  x   1  0 The second derivative is less than zero for all values of x. Thus, a maximum occurs at x  46. First, we find P ‘  x  . P ‘  x   0.02 x  1.4 The derivative exists for all values of x in  0,   . Thus, we solve P ‘  x   0. 0.02 x  1.4  0 0.02 x  1.4 x  70 There is only one critical value. We can use the second derivative to determine whether we have a maximum. P ”  x   0.02  0 The second derivative is less than zero for all values of x. Thus, a maximum occurs at x  70. P  46  0.5  46  46  46  10 2  1058  2116  10  1048 The maximum profit is $1048 when 46 units are produced and sold. 24. R  x   2 x; C  x   0.01x 2  0.6 x  30 R  x   50 x  0.5 x 2 ; C  x   10 x  3 P  70   0.01  70  1.4  70  30 2 P  x  R  x  C  x  49  98  30  50 x  0.5 x  10 x  3  19 The maximum profit is $19 when 70 units are produced and sold. 2  0.5 x 2  40 x  3, P ‘  x    x  40 0 x The derivative exists for all values of x in  0,   . Thus, we solve P ‘  x   0.  x  40  0 x  40 There is only one critical value. P ”  x   1  0 The second derivative is less than zero for all values of x. Thus, a maximum occurs at x  40. P  40   0.5  40   40  40   3  797 2 The maximum profit is $797 when 40 units are produced and sold. 26. R  x   5 x; C  x   0.001x 2  1.2 x  60 P  x  R  x  C  x   5 x  0.001x 2  1.2 x  60  0.001x 2  3.8 x  60, P ‘  x   0.002 x  3.8  0 x The derivative exists for all values of x in  0,   . Thus, we solve P ‘  x   0. 0.002 x  3.8  0 x  1900 There is only one critical value. P ”  x   0.002  0 The second derivative is less than zero for all values of x. The maximum occurs at x  1900. P 1900  0.001 1900   3.8 1900  60  3550 2 The maximum profit is $3550 when 1900 units are produced and sold. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 374 27. R  x  9 x  2 x2 5 thousand is approximately 3 148 1.667 thousand or 1667 units, and that 27 thousand is approximately 5.481 thousand or 5481. Thus, the maximum profit is approximately $5481 when approximately 1667 units are produced and sold. Note that x  C  x   x 3  3x 2  4 x  1 R  x  and C  x  are in thousands of dollars and x is in thousands of units. Profit is equal to revenue minus cost. P  x  R  x  C  x    9 x  2 x 2  x 3  3x 2  4 x  1   x3  x 2  5x  1 Because x is the number of units produced and sold, we are only concerned with the nonnegative values of x. Therefore, we will find the maximum of P  x  on the interval  0,   . First, we find P ‘  x  . P ‘  x   3 x 2  2 x  5 The derivative exists for all values of x in  0,   . Thus, we solve P ‘  x   0. 3x 2  2 x  5  0 3x  5 x  1  0 or or is in thousands of units. P  x  R  x  C  x 1   100 x  x 2   x3  6 x 2  89 x  100  3  0 x Thus, we solve P ‘  x   0. 5 or x  1 3 There is only one critical value in the interval 0,   . We can use the second derivative to determine whether we have a maximum. P ”  x   6 x  2 Therefore, 5 5 P ”    6    2  10  2  8  0  3  3 The second derivative is less than zero for 5 5 x  . Thus, a maximum occurs at x  . 3 3 2 5 5 5 5 P           5  1  3  3  3  3 125 25 25   1 27 9 3 125 75 225 27     27 27 27 27 148  27  1 3 x  6 x 2  89 x  100 3 R  x  and C  x  are in thousands of dollars and x C  x  The derivative exists for all values of x in  0,   . x 1  0 x  1 x 3 R  x   100 x  x 2 1   x3  5 x 2  11x  100, 3 P ‘  x    x 2  10 x  11 3 x 2  2 x  5  0 3x  5  0 3x  5 28.  x 2  10 x  11  0 x 2  10 x  11  0  x  11 x  1  0 x  11  0 or x  1  0 x  11 or x  1 There is only one critical value in the interval 0,   . We can use the second derivative to determine whether we have a maximum. P ”  x   2 x  10 Therefore, P ” 11  2 11  10  12  0 The second derivative is less than zero for x  11. Thus, a maximum occurs at x  11. 1 3 2 P 11   11  5 11  11 11  100 3  182.333 Note that x  11 thousand is 11,000 units, and that 182.333 thousand is 182,333. Thus, the maximum profit is approximately $182,333 when 11,000 units are produced and sold. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 29. 375 p  280  0.4 x Price per unit. 30. C  x   5000  0.6 x 2 Cost per unit. a) Revenue is price times quantity. Therefore, revenue can be found by multiplying the number of unit sold, x, by the price of the unit, p . Substituting 280  0.4x for p, we have: R  x  x  p  x  280  0.4 x  p  150  0.5 x , C  x   4000  0.25 x 2 a) R  x   x  p  x 150  0.5 x   150 x  0.5 x 2 b) P  x   R  x   C  x    150 x  0.5 x 2  4000  0.25 x 2 P  x   0.75 x 2  150 x  4000,  0 x c) P ‘  x   1.5 x  150 P ‘  x  exists for all real numbers in the interval  0,   . Solve: R  x   280 x  0.4 x 2 b) Profit is revenue minus cost. Therefore, P  x  R  x  C  x   280 x  0.4 x 2  5000  0.6 x 2  0x   x 2  280 x  5000, Since x is the number of units produced and sold, we will restrict the domain to the interval 0  x  . c) To determine the number of suits required to maximize profit, we first find P ‘  x  . P ‘  x   2 x  280 . The derivative exists for all real numbers in the interval  0,   . Thus, we solve P ‘  x  0 2 x  280  0 2 x  280 x  140 Since there is only one critical value, we can use the second derivative to determine whether we have a maximum. P ”  x   2  0 The second derivative is negative for all values of x; therefore, a maximum occurs at x  140. Riverside Appliances must sell 140 refrigerators to maximize profit. d) The maximum profit is found by substituting 140 for x in the profit function. P 140   140  280 140  5000 2  19, 600  39, 200  5000  14, 600 The maximum profit is $14,600. e) The price per refrigerator is given by: p  280  0.4 x Substituting 140 for x, we have: p  280  0.4 140   224 1.5 x  150  0 x  100 Since there is only one critical value, we can use the second derivative to determine whether we have a maximum. P ”  x   1.5  0 The second derivative is negative for all values of x; therefore, a maximum occurs at x  100. Raggs, Ltd. must sell 100 suits to maximize profit. d) Substitute 100 for x in the profit function. P 100  0.75 100   150 100  4000 2  3500 The maximum profit is $3500. e) p  150  0.5 100   150  50  100 The price per suit will be $100. 31. Let x be the amount by which the price of $80 should be increased. First, we express total revenue R as a function of x. There are two sources of revenue, revenue from tickets and revenue from concessions.  Number of   Price of  R  x     Rooms   Rooms  Note, when the price increases x dollars, the number of rooms occupied falls by x rooms. Thus, the number of rooms is 300  x when price increases x dollars. Therefore, the total revenue function is R  x   300  x 80  x    x 2  220 x  24, 000 The solution is continued on the next page. The price per refrigerator will be $224. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 376 The cost of maintining each occupied room is $22. Therefore the total cost function is:  Number of  C  x    22  Rooms  = 300  x  22  6600  22 x. Now we can find the profit function for the hotel. P  x  R  x  C  x   x 2  220 x  24, 000  6600  22 x    x 2  242 x  17, 400. To find x such that P  x  is a maximum, we first find P ‘  x  : P ‘  x   2 x  242 This derivative exists for all real numbers x. thus, the only critical values are where P ‘  x   0 ; so we solve that equation: P ‘  x  0 10, 000   R  x    40, 000  x  18  x     3 10, 000   x   4.50  40, 000   3 10, 000 2 x  35, 000 x  900, 000 3 Therefore, the total revenue function is 10,000 2 R  x   x  35,000 x  900,000 3 20,000 R ‘  x   x  35,000 3 This derivative exists for all real numbers x. thus, the only critical values are where R ‘  x   0 ; so we solve that equation:   20,000 x  35,000  0 3 20,000  x  35,000 3 x  5.25 20, 000 0 3 is negative, for all values of x, therefore a maximum occurs at x  5.25 . In order to maximize revenue, the university should charge $18  $5.25 , or $12.75. We can find the attendance using 10, 000 40,000  5.25  57,500 3 The average attendance when ticket price is $12.75 is 57,500 people. The second derivative R ”  x    2 x  242  0 2 x  242 x  121 The second derivative P ”  x   2  0 is negative for all values of x, therefore a maximum occurs at x  121 . The charge per unit should be $80 + $121, or $201. 32. Let x be the amount by which the price of $18 should be decreased (if x is negative, the price would be increased to maximize revenue). First, we express total revenue R as a function of x. There are two sources of revenue, revenue from tickets and revenue from concessions.  Revenue from   Revenue from  R  x      concessions  tickets   Number of   Ticket   Number of     4.50   People   Price   People  33. Let x be the amount the number of new officers Oak Glen should place on patrol. The total number of parking tickets written per day is  Number of   Avg. Tickets  p  x     officers   per day   8  x  24  4 x   4 x 2  8 x  192. To find x such that p  x  is a maximum, we first find p ‘  x  : Note, the increase in ticket sales is 10,000 x, when price drops 3x dollars. Therefore, the 10, 000 x when price increase in ticket sales is 3 drops x dollars. p ‘  x   8 x  8. This derivative exists for all real numbers x. thus, the only critical values are where P ‘x  0 . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 377 We solve the equation: P ‘  x  0 b) First, we express total revenue R as a function of x. Revenue is price times quantity demanded, therefore, R x  x  q x 8 x  8  0 8 x  8 x  1 The second derivative P ”  x   8  0 is negative for all values of x, therefore a maximum occurs at x  1 . This means that Oak Glean should place one fewer officer on patrol in order to maximize the number of parking tickets written in a day. 34. Let x equal the number of additional trees per acre which should be planted. The total yield per acre is equal to the yield per tree times the number of trees so, we have: Y  x    30  x  20  x   600  10 x  x 2 Find Y ‘  x  : Y ‘  x   10  2 x. This derivative exists for all real numbers x. Thus, the only critical values are where Y ‘  x   0 ; so we solve that equation: 10  2 x  0 x5 This corresponds to planting 5 trees. Since this is the only critical value, we can use the second derivative, R ”  x   2  0, which is negative for all values of x. A maximum occurs at x  5. Therefore, in order to maximize yield, the apple farm should plant 20 + 5, or 25 trees per acre. 35. a) First find the slope of the line. 1.12  1 12 m  0.59  1 41 12 y  1    x  1 41 12 53 y   x 41 41 So the demand function is: 12 53 q  x   x  41 41 53   12 R  x  x  x    41 41  12 2 53 x  x 41 41 To find x such that R  x  is a maximum, we  first find R ‘  x  : 24 53 x 41 41 R ‘  x  exists for all real numbers. Solve: R ‘  x   R ‘  x  0  24 53 x 0 41 41 53 x 24 x  2.21 24  0, 41 is negative, for all values of x, therefore a maximum occurs at x  2.21 . To maximize revenue, the price of nitrogen should increase 221% from the January 2001 price. The second derivative R ”  x    36. a) When x  25 , q  2.13 . When x  25  1 , or 26, then q  2.13  0.04  2.09. We use the points  25, 2.13 and  26,2.09  to find the linear demand function q  x  . First, we find the slope: 2.13  2.09 0.04 m   0.04 . 25  26 1 Next, we use the point-slope equation: q  2.13  0.04  x  25 q  2.13  0.04 x  1 q  0.04 x  3.13 Therefore, the linear demand function is: q  x   0.04 x  3.13. b) Revenue is price times quantity; therefore, the revenue function is: R  x  x  q  x  x  0.04 x  3.13  0.04 x 2  3.13x The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 378 Find R ‘  x  : 38. The volume of the box is given by V  x  x  y  x 2 y  320. R ‘  x   0.08 x  3.13 . This derivative exists for all values of x. So the only critical values occur when R ‘  x   0 ; so we solve that equation: 0.08 x  3.13  0 0.08 x  3.13 x  39.125 Since this is the only critical value, we can use the second derivative, R ”  x   0.08  0, to determine whether we have a maximum. Since R ”  39.125 is negative, R  39.125 is The area of the base is x 2 . The cost of the base is 15x 2 cents. The area of the top is x 2 . The cost of the top is 10x 2 cents. The area of each side is xy . The cost of the four a maximum. In order to maximize revenue, the State of Maryland should charge $39.125 or rounding up to $39.13 per license plate. y 37. Let x be the number of $0.10 increase that should be made. Then, R  x    Attendance    Admission Price   180  x 5  0.1x   0.1x 2  13x  900 To find x such that R  x  is a maximum, we first find R ‘  x  : R ‘  x   0.2 x  13 R ‘  x  exists for all real numbers. Solve: R ‘  x  0 0.2 x  13  0 x  65 Since there is only one critical value, we can use the second derivative, R ”  x   0.2  0, to determine whether we have a maximum. Since R ”  65 is negative, a maximum occurs at x  65 . When x  65 the admission price that will maximize revenue is given by: $5  0.165  11.50 Therefore, the theater owner should charge $11.50 per ticket. sides is 2.5  4xy  cents. The total costs in cents is given by C  15 x 2  10 x 2  2.5  4 xy   25 x 2  10 xy. To express C in terms of one variable, we solve x 2 y  320 for y: 320 . x2 Then, 3200  320  C  x   25 x 2  10 x  2   25 x 2   x  x The function is defined only for positive numbers, so we are minimizing C on the interval  0,   . First, we find C ‘  x  . 3200 x2 Since C ‘  x  exists for all x in  0,   , the only C ‘  x   50 x  3200 x 2  50 x  critical values are where C ‘  x   0 . Thus, we solve the following equation: 3200 50 x  2  0 x 3200 50 x  2 x x3  64 x4 This is the only critical value, so we can use the second derivative to determine whether we have a minimum. 6400 C ”  x   50  6400 x 3  50  3 x Note that the second derivative is positive for all positive values of x, therefore we have a minimum at x  4. We find y when x  4. 320 320 320 y 2    20 x  4 2 16 The cost is minimized when the dimensions are 4 ft by 4 ft by 20 ft. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 379 We find y when x  79.93. 5000 5000 y   62.55. x 79.93 39. The area of the parking area is given by A  x  y  5000. Since three sides are chain link fencing, the total cost of the chain link fencing is given by 4.50  2x  y  dollars. The cost is minimized when the dimensions are 62.55 ft by 79.92 ft. (Note, the wooden fence side should be 62.55 feet.) Substituting these dimensions into the cost function we have: C  9 x  11.5 y One side is wooden fencing, the cost of the wooden fence is 7 y dollars. The total costs in dollars is given by C  4.50  2 x  y   7 y  9 79.93  11.5 62.55  9 x  11.5 y. To express C in terms of one variable, we solve xy  5000 for y: 5000 . x Then, y 40. 57,500 x The function is defined only for positive numbers, so we are minimizing C on the interval  0,   .  9x  First, we find C ‘  x  . 57,500 x2 Since C ‘  x  exists for all x in  0,   , the only critical values are where C ‘  x   0 . Thus, we solve the following equation: 57,500 0 9 x2 57,500 9 x2 9 x 2  57,500 Cost of Wood  28  2 y  Therefore the total cost function is: C  70 x  56 y. To express C in terms of one variable, we solve xy  1200 for y: 1200 . x Then, y 67, 200  1200   70 x  C  x   70 x  56   x  x The function is defined only for positive numbers, so we are minimizing C on the interval  0,   . First, we find C ‘  x  . C ‘  x   70  67, 200 x2 Since C ‘  x  exists for all x in  0,   , the only critical values are where C ‘  x   0 . Thus, we 57,500 x  9 2 57,500 9 x  79.93 This is the only critical value, so we can use the second derivative to determine whether we have a minimum. 172,500 C ”  x   172,500 x 3  x3 Note that the second derivative is positive for all positive values of x, therefore we have a minimum at x  79.93. x A  x  y  1200 Cost of Stone  35  2 x   5000  C  x   9 x  11.5   x  C ‘  x   9  57,500 x 2  9   1438.695  1439. The total cost of fencing the parking area is approximately $1439. solve the following equation: 67, 200 0 70  x2 70 x 2  67, 200 x  30.98 This is the only critical value, so we can use the second derivative to determine whether we have a minimum. 134, 400 C ”  x   x3 Note that the second derivative is positive for all positive values of x, therefore we have a minimum at x  30.98. The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 380 We find y when x  30.98. 1200 y  38.73. 30.98 Therefore, the cost will be minimized when the stone wall is 30.98 yards long and the wooden fencing is 38.73 yards long. The minimum cost is given by: C  70 30.98  56 38.73  4337.48. The minimum cost will be approximately $4337. 41. Let x and y represent the outside length and width, respectively. 0.75 x  1.5 x 0.5 0.5  109.6875   0 , so A  109.6875  is a maximum. When x  109.6875  10.47, 73.125  6.98. 109.6875 Therefore, the outside dimensions should be approximately 10.47 in. by 6.98 in. to maximize the print area. y C x  y 1 y 73.125 . x We want to maximize the print area: A   x  1.5 y  1 We know that xy  73.125, so y   xy  x  1.5 y  1.5. Substituting for y, we get:  73.125   73.125  A x  x   x  1.5   1.5  x   x  109.6875  1.5 x 109.6875  74.625  x  x 109.6875 A ‘  x   1  109.6875 x 2  1  x2 A ‘  x  exists for all values in the domain of A.  73.125  x  1  A ” 42. Let x equal the lot size. Now the inventory costs are given by: 0.75 Solve: Therefore, we can use the second derivative to determine if it is a maximum. 219.375 A ”  x   219.375 x 3   . x3 Yearly carrying costs: x Cc  x   20  2  10 x Yearly reorder costs:  100  Cr  x    40  16 x    x  4000  1600 x Hence, the total inventory cost is: C  x   Cc  x   Cr  x   4000  1600. x We want to find the minimum value of C on the interval 1,100  . First, we find C ‘  x  :  10 x  4000 . x2 The derivative exists for all x in 1,100  , so the C ‘  x   10  4000 x 2  10  only critical values are where C ‘  x   0 . C ‘  x  0 10  A’  x  0 109.6875 0 x2 x 2  109.6875 x   109.6875 The only critical value in the domain of A is x  109.6875 ; Yearly Carrying Yearly reorder  Cost Cost 4000 0 x2 4000 10  2 x 2 10 x  4000 x 2  400 x  20 The only critical value in the interval is x  20 . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 381 We can use the second derivative to determine whether we have a minimum. 8000 C ”  x   8000 x 3  3 x Notice that C ”  x  is positive for all values of x in 1,100  , we have a minimum at x  20. Thus, to minimize inventory costs, the store should 100  5 times per year. The order pool tables 20 lot size will be 20 tables. 43. Let x be the lot size. Yearly Carrying Yearly reorder C x   Cost Cost We consider each cost separately. Yearly carrying costs, C c  x  : Can be found by multiplying the cost to store the items by the number of items in storage. The average x amount held in stock is , and it cost $4 per 2 bowling ball for storage. Thus: x Cc  x   4  2  2x Yearly reorder costs, C r  x  : Can be found by multiplying the cost of each order by the number of reorders. The cost of each order is 1  0.5x , and the number of orders per year is 200 . Therefore, x  200  Cr  x   1  0.5 x    x  200  100 x The total inventory cost is given by: C  x   Cc  x   Cr  x   200  2x   100, 1  x  200 x We want to find the minimum value of C on the interval 1, 200  . First, we find C ‘  x  : 200 x2 C ‘  x  exists for all x in 1, 200 . 1,100  , so the C ‘  x   2  200 x 2  2  only critical values are where C ‘  x   0 . Solving the equation, we have: C ‘  x  0 200 0 x2 x  10 The only critical value in the domain is x  10 . Therefore, we use the second derivative, 400 C ”  x   400 x 3  3 x to determine whether we have a minimum. C ” 10   0.4  0 , so C 10  is a minimum. 2 In order to minimize inventory costs. The store 200 should order  20 times per year. The lot 10 size will be 10 bowling balls. 44. Let x equal the lot size. C x  Yearly Carrying Yearly reorder  Cost Cost Yearly carrying costs, C c  x  : x x 2 Yearly reorder costs, C r  x  : Cc  x   2   720  3600   1800 Cr  x   5  2.50 x    x  x Hence, the total inventory cost is: C  x   Cc  x   Cr  x   x 3600  1800. x Then, 3600 . x2 The derivative exists for all x in 1, 720 . Solve: C ‘  x   1  3600 x 2  1  C ‘  x  0 1 3600 x2 0 x 2  3600 x  60 The only critical value in the interval is x  60 , so we can use the second derivative to determine whether we have a minimum. 7200 C ”  x   7200 x 3  3 . x The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 382 Notice that C ”  x  is positive for all values of x in 1, 720 , we have a minimum at x  60. Thus, to minimize inventory costs, the store should 720  12 times per year. The order calculators 60 lot size will be 60 calculators. 45. Let x equal the lot size. Now the inventory costs are given by: C x  Yearly Carrying Yearly reorder  Cost Cost We consider each cost separately. Yearly carrying costs, C c  x  : Can be found by multiplying the cost to store the items by the number of items in storage. The average x amount held in stock is , and it cost $2 per 2 calculator for storage. Thus: x Cc  x   8  2  4x Yearly reorder costs, C r  x  : Can be found by multiplying the cost of each order by the number of reorders. The cost of each order is 10  5x , and the number of orders per year is 360 . Therefore, x  360  Cr  x   10  5 x    x  3600  1800 x Hence, the total inventory cost is: C  x   Cc  x   Cr  x   3600  4x   1800, 1  x  360 x We want to find the minimum value of C on the interval 1, 360 . First, we find C ‘  x  : 3600 C ‘  x   4  3600 x  4  2 x The derivative exists for all x in 1, 360 , so the 2 only critical values are where C ‘  x   0 . We solve the equation at the top of the next column. C ‘  x  0 4 3600 x2 0 4 3600 x2 4 x 2  3600 x 2  900 x  30 The only critical value in the domain is x  30 . Therefore, we use the second derivative, 7200 C ”  x   7200 x 3  3 x to determine whether we have a minimum. C ”  30   0.27  0 , so C  30  is a minimum. In order to minimize inventory costs. The store 360  12 times per year. The lot should order 30 size will be 30 surf boards. 46. Let x equal the lot size. Yearly carrying costs: x Cc  x   2  2 x Yearly reorder costs:  256  Cr  x    4  2.50 x    x  1024   640 x Hence, the total inventory cost is: C  x   Cc  x   Cr  x  1024  640. x We want to find the minimum value of C on the interval 1, 256 . First, we find C ‘  x  :  x 1024 . x2 The derivative exists for all x in 1, 256 , so the C ‘  x   1  1024 x 2  1  only critical values are where C ‘  x   0 . We solve that equation: 1024 1 2  0 x 1024 1 2 x 2 x  1024 x  32 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 383 The only critical value in the interval is x  32 , so we can use the second derivative to determine whether we have a minimum. 2048 C ”  x   2048 x 3  3 x Notice that C ”  x  is positive for all values of x in 1, 256 , so we have a minimum at x  32. Thus, to minimize inventory costs, the store 256 should order calculators  8 times per year. 32 The lot size will be 32 calculators. 47. Let x be the lot size. x  4x 2  360  Cr  x   10  6 x    x  Yearly carrying cost: Cc  x   8  Yearly reorder cost:  3600  2160 x Therefore, C  x   Cc  x   Cr  x  3600  2160, 1  x  360 x 3600 C ‘  x   4  3600 x 2  4  2 x C ‘  x  exists for all x in 1, 360 . Solve:  4x  C ‘  x  0 3600 0 x2 x  30 The only critical value in the domain is x  30 . Therefore, we use the second derivative, 7200 C ”  x   7200 x 3  3 x to determine whether we have a minimum. C ”  30   0.27  0 , so C  30  is a minimum. 4 In order to minimize inventory costs. The store 360  12 times per year. The lot should order 30 size will be 30 surf boards. 48. The volume of the container must be 250 in3. Therefore, we use formula for the volume cylinder to obtain 250   r 2 h . Solving for h we have: 250 h 2 . r The container consists of a circular top and a circular bottom. The area for each of the top and bottom of the container is given by A   r 2 . The side material that when laid out is a rectangle of height h and whose length is the same as the circumference of the circular ends, 2 r . Therfore, the surface area of the side material is A  2 rh . Therefore, the total surface area is the sum of the area of the top and the bottom plus the the side material: A  2 r 2  2 rh . Substituting for h, we have area as a function of the radius, r.  250  A  r   2 r 2  2 r  2  r  500 A  r   2 r 2  r The nature of this problem requires r  0 . We differentiate the area function with respect to r: 500 A ‘  r   4 r  2 . r We find the critical values by setting the derivative equal to zero and solving for r. Remember, r  0 . A ‘ r  0 500 0 r2 500 4 r  2 r 3 4 r  500 4 r  r3  r 500 4 125 3  r  3.414 This is the only critical value in the interval r  0 . We will calculate the second derivative to determine the concavity of the function. 1000 A ”  r   4  3 . r Evaluating the second derivative at the critical value, we have: 1000 A ” 3.414  4  3.4143  37.697  0. Since the second derivative is positive at the critical value, the critical value represents a relative minimum. The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 384 We determine the height of the container by substituting back into 250 h 2 . r 250  6.828 h 2   3.414  Therefore, the dimensions of the container that will minimize the surface area are a height of 6.828 inches and a radius of 3.414 inches. 49. The volume of the container must be 400 cm3. Therefore, we use formula for the volume cylinder to obtain 400   r 2 h . Solving for h we have: 400 h 2 . r Therefore, the total surface area is the sum of the area the bottom plus the the side material: A   r 2  2 rh . Substituting for h in area formula we have area as a function of the radius, r.  400  A  r    r 2  2 r  2  r  800 r The nature of this problem requires r  0 . We differentiate the area function with respect to r: 800 A ‘  r   2 r  2 . r We find the critical values by setting the derivative equal to zero and solving for r. Remember, r  0 . A ‘ r  0 Ar   r2  2 r  800 0 r2 800 2 r  2 r 800 r3  2 400 r3 A ” 5.03  2  1600 5.033  18.855  0. Since the second derivative is positive at the critical value, the critical value represents a relative minimum. We determine the height of the container by substituting back into 400 h 2 . r 400  5.03 h 2  5.03 Therefore, the dimensions of the container that will minimize the surface area are a height of 5.03 cm and a radius of 5.03 cm. 50. The cost function for the container would be is given by:  area of the  C  0.005   top and bottom   area of the   0.003   side material  Using the information from problem 48, we know the total area for the top and bottom of the can was given by 2 r 2 and the area of the side 500 material was given by r . Therefore, the cost r function is:  500  C  r   0.005 2 r 2  0.003   r    1.5 . r The nature of the problem still requires r  0. Calculating the derivative of the cost function we have:  0.01 r 2     C ‘  r   0.01 2 r 2 1  1.5 1r 11  0.02 r   1.5 r2 Find the critical values by setting the derivative equal to zero and solving for r. C ‘ r   0  r  5.03 This is the only critical value in the interval r  0 . We will calculate the second derivative to determine the concavity of the function. 1600 A ”  r   2  3 . r We evaluate the second derivative at the critical value at the top of the nexg column. 0.02 r  1.5 r2 0 r3  75  r3 75  r  2.879 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 385 From the previous page, we determine r  2.879 is the only critical value in the interval r  0. Next, we find the second derivative: 3 C ” r   0.02  3 . r Substituting the critical value into the second derivative, we have: 3 C ”  2.879  0.022  2.8793  0.19  0. Since the second derivative is positive at the critical value, the critical value represents a relative minimum. We substitute the critical 250 value back into h  2 to determine the r value for h. Notice, we substitute before the rounding of r. 250 h 2   2.879411  9.598 The dimensions of the container that will minimize the cost of the container are a radius of 2.879 in and a height of 9.589 in. 51. The cost function for the container would be is given by:  area of  C  0.0015   the base   area of   0.0008   the side  Using the information from problem 49, the cost function is:  800  C  r   0.0015  r 2  0.0008   r  0.64  0.0015 r 2  . r The nature of the problem still requires r  0. Calculating the derivative of the cost function we have: d  0.64  C ‘ r  0.0015 r 2   dr  r  0.64  0.003 r  2 r   Find the critical values by setting the derivative equal to zero and solving for r. C ‘ r  0 0.64 0 r2 0.64 0.003 r  2 r 0.64 r3  0.003 0.64 r3 0.003 r  4.08 This is the only critical value in the interval r  0. Next, we find the second derivative: d  0.64  C ”  r    0.003 r  2  dr  r 1.28  0.003  3 . r Substituting the critical value into the second derivative, we have: 1.28 C ”  4.08  0.003   4.083 0.003 r   0.028  0. Since the second derivative is positive at the critical value, the critical value represents a relative minimum. We substitute the critical 400 value back into h  2 to determine the r value for h. Notice, we substitute before the rounding of r. 400 h 2   4.08  7.65 The dimensions of the container that will minimize the cost of the container are a radius of 4.08 cm and a height of 7.65 cm. 52. Case I. If y is the length, the girth is x  x  x  x, or 4 x. Case II. If x is the length, the girth is x  y  x  y , or 2 x  2 y. Case I. The combined length and girth is y  4 x  84. The volume is V  x  x  y  x 2 y . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 386 We want express V in terms of one variable. We solve y  4 x  84 , for y. y  4 x  84 y  84  4 x Thus, V  x 2 84  4 x   84 x 2  4 x 3 . To maximize V  x  we first find V ‘  x  . V ‘  x   168 x  12 x 2 This derivative exists for all x, so the critical values will occur when V ‘  x   0 ; therefore, 9 2 x 0 2 3   3x  28  x   0  2  84 x  3x  0 or x0 or x0 or 3 x0 2 3  x  28 2 56 x 3 28  56 3 we solve that equation. Since x  0 , the only critical value is x  168 x  12 x 2  0 We can use the second derivative, V ”  x   84  9 x , to determine whether we have a maximum.  56   56  V ”    84  9    84  0  3  3 56 Therefore, we have a maximum at x  . 3 56 3  56  If x   18.67 , then y  42     14. 3 2 3  Therefore, the dimensions that will maximize the volume of the package are 18.67 in. by 18.67 in. by 14 in. The volume is 56 56   14  4878.2 in 3 . 3 3 Comparing Case I and Case II, we see that the maximum volume is 5488 in3 when the dimensions are 14 in. by 14 in. by 28 in. 12 x 14  x   0 12 x  0 or 14  x  0 x  0 or x  14 Since x  0 , the only critical value is x  14. We can use the second derivative, V ”  x   168  24 x , to determine whether we have a maximum. V ” 14   168  24 14   168  0 Therefore, we have a maximum at x  14 . If x  14 , then y  84  4 14   28. Therefore, the dimensions that will maximize the volume of the package are 14 in. by 14 in. by 28 in. The volume is 14  14  28  5488 in 3 . Case II. The combine length and girth is x  2 x  2 y  3 x  2 y  84. The volume is V  x  x  y  x 2 y . We want express V in terms of one variable. We solve 3 x  2 y  84 , for y. 3 x  2 y  84 2 y  84  3 x 3 y  42  x 2 Thus, 3  3  V  x 2  42  x   42 x 2  x 3 .  2  2 To maximize V  x  we first find V ‘  x  . 9 2 x 2 This derivative exists for all x, so the critical values will occur when V ‘  x   0 . We solve the V ‘  x   84 x  equation at the top of the next column. 53. Let y represent the dimension on the lot line and let x represent the other dimension. Then the length of fencing that the person must pay for is 1 3 y  x  y  x  2x  y . 2 2 48 We know xy  48 ; therefore, y  . x The length of fencing as a function of x is: 3  48  72 F  x  2x     2x  , 0  x   . 2 x  x 72 x2 F ‘  x  exists for all x in  0,   . Solve: F ‘  x   2  72 x 2  2  F ‘  x  0 72 0 x2 x  6 The solution is continued on the next page. 2 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 387 Only the critical value x  6 is in  0,   . Since Then, there is only one critical value, we use the second derivative to determine whether we have a minimum. 144 F ”  x   144 x 3  3 x 2 F ”  6   0 , so F  6  is a minimum. 3 48  8. The dimensions that When x  6, y  6 minimize the cost are 6 yd by 8 yd. Where the longer side of the lot is adjacent to the neighbor’s yard. A x  54. Use the figure in the text book. Since the radius of the window is x, the diameter of the window is 2x , which is also the length of the base of the window. The circumference of a circle whose radius is x is given by: C  2 x. C  2 r  Therefore, the perimeter of the semicircle is 1 2 x   x. C 2 2 The perimeter of the three sides of the rectangle which form the remaining part of the total perimeter of the window is given by: 2 x  y  y  2 x  2 y. The total perimeter of the window is:  x  2 x  2 y  24 . Maximizing the amount of light is the same as maximizing the area of the window. The area of the circle with radius x is: A   x2 , A  r . 1 2     x  2 x 12  x  x   2 2  1   x 2  24 x  2 x 2   x 2 2  1       2  x 2  24 x  2  We maximize A on the interval  0, 24  . We first find A ‘  x  . A ‘  x      4  x  24 . Since A ‘  x  exists for all x in  0, 24  , the only critical points are where A ‘  x   0 . Thus, we solve the following equation: A’  x  0    4  x  24  0    4 x  24 x 24    4  x 24    4  24  3.36  4 This is the only critical value, so we can use the second derivative to determine whether we have a maximum. A ”  x     4  0 x Since A ”  x  is negative for all values of x, we have a maximum at x  2 Therefore, the area of the semicircle is: 1 1 A   x2. 2 2 The area of the rectangle is 2 x  y . The total area of the Norman window is 1 A   x 2  2 xy. 2 To express A in terms of one variable, we solve  x  2 x  2 y  24 for y:  x  2 x  2 y  24 2 y  24  2 x   x  y  12  x  x 2 x 24  4  y  12  24  4 . We find y when : 2 x x   24  24   2  4  4 24    4  12  12       4    4   4 12  48  12  24   4  12  24  3.36  4 The solution is continued on the next page.  Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 388 To maximize the amount of light through the 24 ft window, the dimensions must be x   4 24 ft, or approximately and y   4  x  3.36 ft and y  3.36 ft. 55. Since the stained glass transmits only half as much light as the semicircle in Exercise 54, we express the function A as: 1 1 A    x 2  2 xy 2 2 1   x 2  2 xy 4 The perimeter is still the same, so we can substitute 12   2 x  x for y to get: 1 2     x  2 x 12  x  x    4 2 1   x 2  24 x  2 x 2   x 2 4  3       2  x 2  24 x, 0  x  24.  4  A  x  y  12   12   2 xx 48   48    2  3  8  3  8 12  48  4.92 3  8 To maximize the amount of light through the 48 window, the dimensions must be x  ft 3  8 12  48 ft or approximately and y  3  8 x  2.75 ft and y  4.92 ft.  56-110. Left to the Student. 1 is x the reciprocal of the number, and x 2 is the square of the number. The sum, S, of the reciprocal and five times the square is given by: 1 S  x    5x 2 . x We want to minimize S  x  on the interval 111. Let x represent a positive number. Then, Find A ‘  x  . 0,   . First, we find S ‘  x   3  A ‘  x       4  x  24  2  1  10 x x2 Since S ‘  x  exists for all values of x in  0,   , S ‘  x    x 2  10 x   Since A ‘  x  exists for all x in  0, 24  , the only critical points are where A ‘  x   0 . Thus, we solve the following equation:  3      4  x  24  0 2 3  8 x  48  0 48  2.75 3  8 This is the only critical value, so we can use the second derivative to determine whether we have a maximum. 3 A ”  x      4  0 2 Since A ”  x  is negative for all values of x, we x have a maximum at x  when x  48 . We find y 3  8 48 at the top of the next column. 3  8 the only critical values occur when S ‘  x   0 . We solve the following equation: 1  2  10 x  0 x 1 10 x  2 x 10 x 3  1 1 x3  10 1 1  x3 10 3 10 Since there is only one critical value, we use the second derivative, 2 S ”  x   2 x 3  10  3  10 , x to determine whether it is a minimum. The second derivative is positive for all x in  0,   ; therefore, the sum is a minimum when x  3 Copyright © 2016 Pearson Education, Inc. 1 . 10 Exercise Set 2.5 389 112. Let x represent a positive number. The sum, S, of the reciprocal and four times the square is given by: 1 S  x   4 x2. x We want to minimize S  x  on the interval 0,   . First, we find S ‘  x  1  8x x2 S ‘  x  exists for all values of x in  0,   . Solve: S ‘  x    x 2  8 x   S ‘  x  0  We solve the following equation: 0.18k  2ki  0 2ki  0.18k 0.18k i 2 k i  0.09 Since there is only one critical point, we can use the second derivative to determine whether we have a minimum. Notice that P ” i   2 k , which is a negative constant  k  0  . Thus, P ” 0.09  is negative, so P  0.09  is a maximum. To maximize profit, the bank should pay 9% on its savings accounts. 1  8x  0 x2 1 1  8 2 Since there is only one critical value, we use the second derivative, 2 S ”  x   2 x 3  8  3  8 , x to determine whether it is a minimum. The second derivative is positive for all x in  0,   ; x3 therefore, the sum is a minimum when x  1 . 2 113. Let A represent the amount deposited in savings account and i represent the interest rate paid on the money deposited. If A is directly proportional to i , then there is some positive constant k such that A  ki . The interest earned by the bank is represented by 18% A, or 0.18 A . The interest paid by the bank is represented by iA . Thus the profit received by the bank is given by P  0.18 A  iA. We express P as a function of the interest the bank pays on the money deposited, i , by substituting ki for A. P  0.18  ki   i  ki   0.18ki  ki 2 We maximize P on the interval  0,   . First, we find P ‘ i  . P ‘ i   0.18k  2 ki Since P ‘ i  exists for all i in  0,   , the only critical values are where P ‘ i   0 . 114. The circumference of the circle is x  2 r . x . Solving the equation for r we get, r  2 The area of the circle is A   r 2 , thus, 2 x2  x  . substituting for r we have A      2  4 The perimeter of the square is 24  x . 24  x The length of a side of the square is . 4 Therefore, the area of the square is: 2 2 x  48 x  576  24  x  .  As    4  16 The total area is: A  Ac  As x 2 x 2  48 x  576  4 16 x2 1 2   x  3x  36 4 16 1  1    x 2  3x  36  4 16    4  2  x  3x  36  16  We minimize the area on the interval  0, 24  . First, we find A ‘  x  .  4  A’  x  2  x3  16   4   x3  8  A ‘  x  exists for all x in  0, 24  ; therefore, the only critical value occurs when A ‘  x   0. The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 390 Setting the derivative equal to zero, we solve for the critical value.  4    x3 0 8   8  24   10.56 x  3  4    4   Since there is only one critical value, we use the second derivative, 4 A ”  x   0 8 to determine whether we have a minimum. The second derivative is positive for all values of x in the interval; therefore a minimum occurs at 24  10.56 . x 4 The wire should be cut to x  10.56 in. in order to form the circle and 24  10.56  13.44 in. in order to form the square. There is no maximum if the string is to be cut. One would interpret the maximum to be at the endpoint, with the string uncut and used to form a circle. 115. Using the drawing in the text, we write a function that gives the cost of the power line. The length of the power line on the land is given by 4  x , so the cost of laying the power line underground is given by: C L  x   3000  4  x   12, 000  3000 x. The length of the power line that will be under water is 1  x 2 , so the cost of laying the power line underwater is given by: CW  x   5000 1  x 2 Therefore, the total cost of laying the power line is: C  x   C L  x   CW  x   12, 000  3000 x  5000 1  x 2 . We want to minimize C  x  over the interval 0  x  4. First, we find the derivative.  12 1 C ‘  x   3000  5000   1  x 2 2 x 2    3000  5000 x 1  x 2  3000  5000 x 1  x2    12 Since the derivative exists for all x, we find the critical values by solving the equation: C ‘ x  0 5000 x 3000  0 1  x2 3000 1  x 2  5000 x  0 5000 x  3000 1  x 2 5 x  1  x2 3 2 5   x   3  1 x  2 2 25 2 x  1  x2 9 16 2 x 1 9 9 x2  16 9 x 16 3 x 4 The only critical value in the interval  0, 4 is 3 , so we can use the second derivative to 4 determine if we have a minimum. x 1  x  5000  5000 x  12 1  x   2 x  2 C ”  x   1 2 2  5000 1  x 2      1  x 2 12    2 2 5000 x 2 1  x  1  x2 2 5000  5000 x 2 1  x  5000 1  x   5000 x  1  x  1  x2 2 2 2  1 3 3 2 2 2 5000 1  x  2 3 2 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 391 C ”  x  is positive for all x in  0, 4 ; therefore, a minimum occurs at x  3 3 . When x  , 4 4 3 13   3.25 . 4 4 Therefore, S should be 3.25 miles down shore from the power station. 4 Note: since we are minimizing cost over a closed interval, we could have used Max-Min Principle 1 to determine the minimum, and avoided finding the second derivative. The 3 critical value and the endpoints are 0, , and 4. 4 The function values at these three points are: 2 C  0  12, 000  3000  0  5000  1   0     17, 000 2   3  3  3 C    12, 000  3000    5000  1     4 4 4     16, 000 2 C  4   12, 000  3000  4   5000  1   4      20,615.53 Therefore, the minimum occurs when x  3 , or 4 when S is 3.25 miles down shore from the power station. 116. The distance from point C to point S is given by 9  x 2 , and the distance from point S to point A is 8  x . Therefore, the total energy expended by the pigeon is: E  x   1.28r 9  x 2  r 8  x  , where r is a positive constant measuring the rate of energy the pigeon uses. We want to minimize E  x  on the interval 0,8 . E ‘  x   1.28rx 9  x2  1 2 r E ‘  x  exists for all x in  0,8 . Solve: E ‘  x  0 1.28rx 9  x  2 1 2 r0 x  3.755 The only critical value in  0,8 is x  3.755 , so we can use the second derivative to determine whether we have a maximum. 11.52r and that Note that E ”  x   3 2 9  x2   E ”  x   0 for all x in the interval  0,8 . Therefore, a minimum occurs when x  3.755 . The pigeon should reach land about 8  3.755 or 4.245 miles down shore from A. 117. The objective is to maximize the parallel areas subject to the constraint of having k units of fencing. Using the figure in the book, we can let x equal the length of the two parallel sides and y equal the length of the three parallel sides. Since the two areas are the identical, we can either maximize the area of one of the smaller areas or maximize the area of the entire field. That is, we can either maximize A1  x  y or A2  2 x  y. First we will maximize A1. The perimeter of the fields is given by P  4 x  3 y . Since we have k units of fencing, we know that 4 x  3 y  k. Solve for y and substitute into the area function to get the area as a function of one variable. 4x  3 y  k 3y  k  4x k  4x 3 1 4 y k x 3 3 Substituting into the area function we have: 4  1 4 1 A1  x   k  x   kx  x 2 . 3 3  3 3 We find the first derivative with respect to x. dA1 1 8  k  x. dx 3 3 The deriviative exists for all values of x. We find the critical values by solving: dA1 0 dx 1 8 k x0 3 3 8 1  x k 3 3 1 x  k. 8 The solution is continued on the next page. y Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 392 There is only one critical value, so we can use the second derivative to determine if it is a maximum. Notice that the second derivative, d 2 A1 8   , is negative for all values of x. 2 3 dx Therefore the maximum area will occur when 1 the width of the fields are k units long. We 8 substitute in to find the length of the fields y. 1 4 y k x 3 3 1 4 1   k   k 3 3 8  1 1  k k 3 6 1 k  k . 6 6 Therefore the length of the rectangular areas that will maximize the area given k units of k fencing is y  units. 6 Notice, if we maximized A2  2 x  y . The constraint is the same: 1 4 4 x  3 y  k  y  k  x. 3 3 4  2 8 1 A2  2 x   k  x   kx  x 2 . 3 3  3 3 dA2 2 16  k  x. 3 3 dx dA2 0 dx 2 16 k x0 3 3 1 x  k. 8 There is only one critical value, so we can use the second derivative to determine if it is a maximum. Notice that the second derivative, d 2 A1 16   , is negative for all values of x. 2 3 dx Therefore the maximum area will occur when 1 the width of the fields are k units long. We 8 substitute in to find the length of the fields y. 1 4 1  k y  k   k  . 3 3 8  6 Therefore the length of the rectangular areas that will maximize the area given k units of k fencing is y  units as determined 6 previously. 118. Using the drawing in the text, we write a function which gives the total distance between the cities. The distance from C1 to the bridge can be given a 2   p  x  . The distance over the bridge 2 by is r. The distance from the bridge to C2 can be given by b2  x 2 . Therefore, the total distance between the two cities is given by: D  x   a 2   p  x   r  x 2  b2 2 To minimize the distance, we find the derivative of the function first.  12 1 2 D ‘  x    a 2   p  x    2  p  x  1   2 1  1 2 2 b  x 2   2 x  2 x p x   2 2 b  x2 a2   p  x The derivative exists for all values of x in the interval  0, p  . Therefore, the only critical values occur when D ‘  x   0. We solve this equation. x p a   p  x 2 2  x 2 b  x2 0. The solution to this equation is bp bp x or x  . ba ba bp is in  0, p  . Only x  ba Since there is only one critical value, we can use the second derivative to determine if there is a minimum. The second derivative is given by: a2 b2 D ”  x    . 3 3  a 2   p  x 2  2  x 2  b 2  2     D ”  x   0 for all values of x; therefore, a bp . The bridge ba should be located such that the distance x is bp units. ba minimum occurs at x  Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 393 x3 100 a) To determine the average cost, we divide the total cost function by the number of units produced: C  x A x  . x x3 8 x  20  100 A x  x 119. C  x   8 x  20  20 x 2  x 100 b) Taking the derivative of the total cost function and the average cost function we have: d  x3  C ‘  x  8 x  20  dx  100   8 3x 2 . 100 d  20 x 2  A ‘  x  8   dx  x 100   8 Find the function value when x  10 : 20 102   11. 10 100 The minimum average cost is $11 when 10 units are produced. 3 C ‘ 10  8  102  11 . 100 The marginal cost is $11 when 10 units are produced. d) The two values are equal: A 10   C ‘ 10   11. A 10   8    C  x x a) Taking the derivative of A  x  we have: 120. A  x   A’  x    d  1 2 1 x   8  20 x  dx 100  1  20 x 2  2 x 100 20 x  2  . 50 x c) The derivative exists for all x in  0,   ;  therefore, the critical values occur when A ‘  x   0 . Solve: 20 x  2 0 50 x x 20  50 x 2 d  C  x    dx  x  x  C ‘  x   C  x  1 x2 x  C ‘  x  C  x Quotient Rule x2 b) The derivative exists for all x in  0,   , therefore, the critical values will occur when A ‘  x0   0 . We solve the equation: x0  C ‘  x0   C  x0  x02 0 x0  C ‘  x0   C  x0   0 Multiplying by x02  0. x0  C ‘  x0   C  x0  C ‘  x0   C  x0  x0  A  x0  121. Express Q as a function of one variable. First, solve x  y  1 for y. We have: y  1 x . Substituting we have: x  x 2  20  50 x3  1000 x  10 There is only one critical value, so we use the second derivative to determine whether we have a minimum. 40 1 A ”  x   3  x 50 3 A ” 10   0. Thus A 10  is a minimum. 50 Q  x 3  2 1  x  3   x 3  2 1  3x  3x 2  x 3    x3  6 x 2  6 x  2 Next, we find Q ‘  x  . Q ‘  x   3x 2  12 x  6 The derivative exists for all values of x in the interval  0,   . The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 394 Note the constraint y  1  x actually limits us to look at the interval 0,1 . The only critical values are where Q ‘  x   0 . The derivative exists for all values of x in the interval   2, 2  ; thus, the only critical values are where Q ‘  x   0 . We solve: Q ‘  x  0 We solve the equation: 3 x 2  12 x  6  0 2 x  4x  2  0 Using the quadratic formula, we have: x  2 2 . 3  3x 2  x 2  0 3  3 x 2  x 2 1   x 2  x2   When x  2  2 , y  1   2  2   1  2. When x  2  2 , y  1  2  2  1  2.   x4  2 x2  1  0   x  1  0 Q ” 2  2  6 2  2  12  8.48  0 , so we have a minimum at x  2  2 . The minimum value of Q is found by substituting. Q  x3  2 y3   2 2   2  1  2  3  2 1  2 x2  x4 Note, Q ”  x   6 x  12 and   1  x2 2  x2 Since x and y must be positive, we only consider x  2  2 and y  1  2.   12   x 2  x 2 3 2 2 x2  1  0 x2  1 x  1 We notice that x  1 is an extraneous solution which does not work. 3  3 1 2  1  3  3  6  0 .  64 2 2 122. Express Q as a function of one variable. First, solve x 2  y 2  2 for y. We have: y 2  2  x2 y   2  x2 y is a real number if x is in the interval   2, 2  .   Therefore, the only critical value is x  1 . The critical value and the endpoints are  2,  1, and 2.       3 2 2 Q  2  3  2   2   2   3 2    Q  1  3  1  2   1 2   4 3 2 3  2   3 2    2   2    3 2 2 2 If y   2  x 2 , we substitute for y to get: Q Q  3x  y 3 The minimum value of Q is 3 2 and occurs  Q  3x   2  x   3x  2  x 2  3 2  3  when x   2 and y   2   2 2 Next, we find Q ‘  x  .   1 2  3 Q ‘  x   3    2  x 2  2 x  2   3  3x 2  x 2   3  3x 2  x 2 1 2  0 2 Next, we repeat the process for y  2  x 2 . We notice that: Q  3x  y 3 Q  3x   2  x2   3x  2  x 2  3  3 2 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 395 Next, we find Q ‘  x  . 123. Let x be the lot size.     3  3x 2  x 2  1 2  3  3x 2  x 2 The derivative exists for all values of x in the interval   2, 2  ; thus, the only critical values are where Q ‘  x   0 . We solve the equation: 3  3x 2  x 2  0 3  3x 2  x 2 1  x 2  x2  12  x 2  x 2 2  1 x 2 x 2  2  C ‘  x  0 a bQ  0 2 x2 a bQ  2 x2 x4  2 x2  1  0  x  1  0 2 x2  1  0 x2  1 x  1 We notice that x  1 is an extraneous solution which does not work. 3  3  1 2   1  3  3  6  0 . 2 Therefore, the only critical value is x  1 . The critical value and the endpoints are  2, 1, and 2. We evaluate the function:       3 2 2 Q  2  3  2   2   2   3 2    Q 1  3 1  2  1 Q  2   3  2  4 2   2   3 2 3  2    3 2 2 2 The minimum value of Q is 3 2 occurs when  ax bQ   cQ, 1  x  Q 2 x To find the minimum, we take the first deriviative: a a bQ C ‘  x    bQx 2   2 2 2 x C ‘  x  exists for all x in 1,Q  . So the only  critical values occur when C ‘  x   0 . Solve: 1  2 x2  x4 2 x ax  2 2 Q  Cr  x    b  cx    x Yearly reorder cost: bQ   cQ x Then, C  x   Cc  x   Cr  x  Yearly carrying cost: Cc  x   a  1  3 2 Q ‘  x   3    2  x 2  2 x  2 x   2 and y   2   2  0. 2 Regardless of what value of y we chose, we see that the minimum of Q, is 3 2 , when x   2 and y  0. ax 2  2bQ 2bQ x2  a 2bQ x a The only critical value in the domain is 2bQ . a Since there is only one critical value in the domain, we use the second derivative, 2bQ C ”  x   2bQx 3  3 x to determine whether we have a minimum. C ”  x   0 for all x in 1,Q  , so a minimum x 2bQ . a In order to minimize inventory costs. The store Q aQ times per year. should order  2b 2bQ a occurs at x  The lot size will be Copyright © 2016 Pearson Education, Inc. 2bQ units. a Chapter 2: Applications of Differentiation 396 124. From Exercise 123, we know that the store 2bQ aQ should order a lot size of units, a 2b times per year. When Q  2500, a  10, b  20, c  9 , the store should order: 126. Using a spreadsheet we numerically estimate the maximum: Starting Value: Step Size: 10  2500 aQ   25 times per year. 2b 2  20 x 3.22 3.23 3.24 3.25 3.26 3.27 3.28 3.29 3.3 The lot size of each order should be: 2  20 2500  100 units. 10 125-128. The starting value and step size were chosen to limit the amount of space used . Approaches can vary. 125. Using a spreadsheet we numerically estimate the maximum: Starting Value: Step Size: x 24 24.25 24.5 24.75 25 25.25 25.5 25.75 26 3.22 0.01 y  16  x 2 Q  x2  y 2.373099239 2.35947028 2.345719506 2.331844763 2.317843826 2.303714392 2.289454083 2.275060439 2.260530911 24.60524215 24.61611748 24.62442508 24.63011031 24.63311704 24.63338762 24.63086281 24.6254817 24.61718162 We determine the maximum of Q  24.63 to occur when x  3.27 and y  2.30. 24 0.25 127. Using a spreadsheet we numerically estimate the maximum: 100  2 x y 3 q  x y 17.33333333 17.16666667 17 16.83333333 16.66666667 16.5 16.33333333 16.16666667 16 416 416.2917 416.5 416.625 416.6667 416.625 416.5 416.2917 416 We determine the maximum of Q  416.67 to occur when x  25 and y  16.67. Starting Value: Step Size: 2.2 0.01 y  10  0.5 x 2 x 2.2 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.75317998 2.749172603 2.745141162 2.741085551 2.737005663 2.732901389 2.728772618 2.724619239 2.720441141 Q  x  y3 45.91202934 45.91962105 45.92477319 45.92748186 45.92774328 45.9255538 45.92090987 45.91380804 45.90424501 We determine the maximum of Q  45.93 to occur when x  2.24 and y  2.74. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.5 397 128. Using a spreadsheet we numerically estimate the maximum: Starting Value: Step Size: 4.6 0.1 x 4.6 4.7 4.8 4.9 5 5.1 5.2 5.3 5.4 y  50  x 2 Q  x2  y2 5.370288633 5.282991577 5.192301994 5.098038839 5 4.897958759 4.79165942 4.680811895 4.565084884 610.2544 616.5319 621.1584 624.0199 625 623.9799 620.8384 615.4519 607.6944 We determine the maximum of Q  625 to occur when x  5 and y  5. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 398 2. Exercise Set 2.6 1. a) Total profit is revenue minus cost. P  x  R  x  C  x  R  x   50 x  0.5 x 2 ; C  x   4 x  10  5 x  0.001x 2  1.2 x  60 a) Total profit is revenue minus cost. P  x  R  x  C  x  5 x  0.001x  1.2 x  60  0.001x 2  3.8 x  60 b) R 100  5 100   500  50 x  0.5 x 2  4 x  10 The total revenue from the sale of the first 100 units is $500.  0.5 x  46 x  10 b) Substituting 20 for x into the three functions, we have: 2 C 100   0.001 100   1.2 100   60  190 2 The total cost of producing the first 100 units is $190. R  20   50  20   0.5  20   800 2 P 100   0.001 100   3.8 100   60  310 The total revenue from the sale of the first 20 units is $800. C  20   4  20   10  90 2 The total profit is $310 when the first 100 units are produced and sold. c) R ‘  x   5 The total cost of producing the first 20 units is $90. P  20   R  20   C  20  C ‘  x   0.002 x  1.2 P ‘  x   0.002 x  3.8  800  90  710 The total profit is $710 when the first 20 units are produced and sold. Note, we could have also used the profit function, P  x  , from part (a) to find the d) R ‘ 100   5 Once 100 units have been sold, the approximate revenue for the 101st unit is $5. C ‘ 100  1.4 Once 100 units have been produced, the approximate cost for the 101st unit is $1.40. P ‘ 100   0.002 100   3.8  3.6 profit. P  20   0.5  20   46  20   10  710 2 Once 100 units have been produced and sold, the approximate profit from the sale of the 101st unit is $3.60. e) In part (b), we are observing the total revenue, cost and profit from the production and sale of the first 100 items. In part (d), we are observing the approximate revenue, cost and profit from the production and sale of the 101st unit only. These quantities are also known as the marginal revenue, marginal cost and marginal profit. c) Finding the derivative for each of the functions, we have: R ‘  x   50  x C ‘  x  4 P ‘  x    x  46 d) Substituting 20 for x in each of the three marginal functions, we have: R ‘  20   50  20  30 Once 20 units have been produced, the approximate cost for the 21st unit is $4. P ‘  20   20  46  26 Once 20 units have been produced and sold, the approximate profit from the sale of the 21st unit is $26.  2 P  x   50 x  0.5 x 2   4 x  10  Once 20 units have been sold, the approximate revenue for the 21st unit is $30. C ‘  20   4 R  x   5 x; C  x   0.001x 2  1.2 x  60 3. C  x   0.002 x 3  0.1x 2  42 x  300 a) Substituting 40 for x into the cost function, we have: C  40   0.002  40   0.1 40   42  40   300 3 2  2268 (hundreds of dollars) The current daily cost of producing 40 security systems is $226,800. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.6 399 c) C ‘  x   0.003 x 2  0.14 x  19. b) In order to find the additional cost of producing 41 chairs monthly, we first find the total cost of producing 41 security systems in a day. C ‘  25  0.003  25  0.14  25  19 2  24.375 The marginal cost when 25 daypacks have been produced is $24.38. d) Using the marginal cost from part (c), the additional cost required to produce 2 additional daypacks monthly is: 2  24.375  48.75. C  41  0.002  41  0.1 41  42  41  300 3 2  2327.94 (hundreds of dollars) Next, we subtract the cost of producing 40 security systsems daily found in part (a) from the cost of producing 41 security systems daily. C  41  C  40   2327.94  2268  59.94 Therefore, the difference in cost between producing 25 and 27 daypacks per month is approximately $48.75. e) In part (a) we found that it cost $1234.38 to produce 25 daypacks per month. In part (d) we found that the difference in cost between 25 daypacks and 27 daypacks per month was $48.75. Therefore, the approximate total cost of producing 27 daypacks per month is C  27  1234.38  48.75  1283.13. The additional daily cost of increasing production to 41 security systems daily is $5994. c) First, we find the marginal cost function, C ‘  x   0.006 x 2  0.2 x  42 Next, substituting 40 for x, we have: C ‘  40   59.6 The marginal cost when 40 security systems are produced daily is $5960. d) In part (a) we found that it cost $2268 hundred to produce 40 security systems per day. The additional cost of producing 2 additional security systems is 2($59.60)  $119.20 hundreds. Therefore, the estimated daily cost of producing 42 security systems per day is C  42   $2268  $119.20  $2387.20 hundreds or $238,720 per day. 4. C  x   0.001x 3  0.07 x 2  19 x  700 a) Substituting 25 for x into the cost function, we have: C  25  0.001  25  0.07  25  19  25  700 3 2  1234.375 The current monthly cost of producing 25 daypacks is $1234.38. b) We first find the total cost of producing 26 daypacks in a month. C  26   0.001  26   0.07  26   19  26   700 3 2  1258.896 Therefore, C  26   C  25  1258.896  1234.375  24.521 The additional cost of increasing production to 26 daypacks monthly is $24.52. We predict the cost of producing 27 daypacks monthly will be $1283.13. 5. R  x   0.005 x 3  0.01x 2  0.5 x a) Substituting 70 for x, we have: R  70   0.005  70   0.01 70   0.5  70  3 2  1715  49  35  1799 The currently daily revenue from selling 70 lawn chairs per day is $1799. b) Substituting 73 for x, we have: R  73  0.005  73  0.01 73  0.5  73 3 2  2034.875  2034.88 Therefore, the increase in revenue from increasing sales to 73 chairs per day is: R  73  R  70   2034.88  1799  235.88 Revenue will increase $235.88 per day if the number of chairs sold increases to 73 per day. c) First we find the marginal revenue function by finding the derivative of the revenue function. R ‘  x   0.015 x 2  0.02 x  0.5 Substituting 70 for x, we have: R ‘  70   0.015  70   0.02  70   0.5 2  75.40 The marginal revenue when 70 lawn chairs are sold daily is $75.40. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 400 b) First we find the total montly revenue for selling 28 suitcases 3 2 R  28   0.007  28   0.5  28   150  28  d) In part (a) we found that selling 70 lawn chairs per day resulted in a revenue of $1799. In part (c) we found that the marginal revenue when 70 chairs were sold is $75.40. Using these two numbers, we estimate the daily revenue generated by selling 71 chairs is R  71  R  70   R ‘  70   3961.66 The difference in monthly revenue from selling 26 suitcases and 28 suitcases a month is: R  28  R  26   $3961.66  $3685.03  $1799  $75.40  $1874.40. Similarly, the daily revenue generated by selling 72 chairs, or 2 additional chairs, daily is approximately R  72   R  70   2  R ‘  70   $276.63. If sales increased from 26 to 28 suitcases, monthly revenue would increase $276.63 c) First we find marginal revenue by taking the derivative of the revenue function. R ‘  x   0.021x 2  x  150  $1799  2  $75.40   $1949.80. The daily revenue generated by selling 73 chairs, or 3 additional chairs, daily is approximately R  73  R  70   3  R ‘  70  Next we substitute 26 in for x. R ‘  26   0.021  26    26   150  138.196 2 Marginal revenue is 138.20 when 26 suitcases are sold. d) From part (a), we know that when 26 suitcases are sold, total monthly revenue is $3685.03. From part (c), we know that when 26 suitcases are sold, marginal revenue is $138.20. Therefore, we estimate: R  27  R  26  R ‘  26  $1799  3  $75.40   $2025.20. 6. P  x   0.006 x 3  0.2 x 2  900 x  1200 a) P  60   $50,784 b) P  60   P  59   50, 784  49, 971.53  812.47 The dealership would lose $812.47 per week if it were only able to sell 59 cars weekly. c) P ‘  x   0.018 x 2  0.4 x  900 P ‘  60   0.018  60   0.4  60   900 R  27  $3685.03  $138.20  $3823.23 We estimate the revenue from selling 27 suitcases per month to be $3823.23. 2  811.20 The marginal profit is $811.20 when 60 cars are sold each week. d) P  61  P  60   P ‘  60   $51,595.20 The estimated profit of selling 61 cars per week is $51,595.20. 7. R  x   0.007 x 3  0.5 x 2  150 x a) Substituting 26 for x, we have: R  26   0.007  26   0.5  26   150  26  3 2  123.032  338  3900  3685.03 The current monthly revenue is $3685.03. 8. P  x   0.004 x 3  0.3 x 2  600 x  800 a) Substituting 9 for x, we have: P  9   0.004  9   0.3  9   600  9   800 3 2  4572.784 The currently weekly profit is $4572.78. b) The difference in weekly profit from selling 8 laptops and 9 laptops per week is P  9  P  8  4572.78  3978.75  594.03 Therefore, Crawford Computing would lose $594.03 each week if 8 laptops were sold each week instead of 9. c) P ‘  x   0.012 x 2  0.6 x  600 P ‘  9  0.012  9  0.6  9  600 2  593.628 The marginal profit is $593.63 when 9 laptops are sold weekly. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.6 401 d) We estimate: P 10   P  9  P ‘  9 12. C  C  x  x   C  x   4572.78  593.63 C  C  70  1  C  70   5166.41 The total weekly profit is approximately $5166.41 when 10 laptops are built and sold weekly. 9.  C  71  C  70   0.01  71  0.6  71  30  2  0.01  70  2  0.6  70   30     2.01 The additional cost of producing the 71st unit is $2.01. C ‘  x   0.02 x  0.6 N 1000   500,000 means that 500,000 computers will be sold annually when the price of the computer is $1000. N ‘ 1000  100 means that when the price is increased $1 to $1001, sales will decrease by 100 computers per year. 10. C ‘  70   0.02  70   0.6  2.00 The marginal cost when 70 units are produced is $2.00. N 1025  N 1000   25  N ‘ 1000 N 1025  500,000  25  100  497,500 13. R  x  2 x R  R  x  x   R  x  We estimate that 497,500 computers will be sold annually if the price is increased to $1025. 11. C  x   0.01x 2  0.6 x  30 Substituting x  70, and x  1 we have R  R  70  1  R  70  C  x   0.01x  1.6 x  100 2  R  71  R  70  C  C  x  x   C  x   2  71   2  70   Substituting x  80, and x  1 we have C  C  80  1  C  80  2 The additional revenue from selling the 71st unit is $2.00. Finding the derivative of R  x  we have:  0.01  81  1.6  81  100  2  0.01  80  2  1.6  80   100     3.21 The additional cost of producing the 81st unit is $3.21. Finding the derivative of C  x  we have: R ‘  x  2 . The derivative is constant; therefore, R ‘  70   2 The marginal revenue when 70 units are produced is $2.00. C ‘  x   0.02 x  1.6 Substituting 80 for x, we have: C ‘  80   0.02  80   1.6  3.20 The marginal cost when 80 units are produced is $3.20. 14. R  x   3x R  R  x  x   R  x  R  R  80  1  R  80   3  81   3  80   3 The additional revenue from selling the 81st unit is $3.00. R ‘  x  3 . R ‘  80  3 The marginal revenue when 80 units are produced is $3.00. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 402 15. C  x   0.01x 2  1.6 x  100; R  x   3 x The additional profit of producing and selling the 71st unit is $0.01. P ‘  x   0.02 x  1.4 a) Finding the profit function we have: P  x  R  x  C  x   3 x  0.01x 2  1.6 x  100 P ‘  70  0.02  70  1.4  0.00  The marginal profit when 70 units are produced and sold is $0.00.  0.01x 2  1.4 x  100 b) P  P  x  x   P  x  Substituting x  80, and x  1 we have P  P  80  1  P  80  17.  0.01  81  1.4  81  100  2  0.01 80  2  1.4  80   100     0.21 The additional profit of producing and selling the 81st unit is $0.21. Finding the derivative of P  x  we have: S  0.007  25   0.5  25   150  25  3 Substituting 80 for x, we have: P ‘  80  0.02  80  1.4  0.20 The marginal profit when 80 units are produced and sold is $0.20. Note: We notice that P  R  C and P ‘  x   R ‘  x   C ‘  x  . We could have used This result implies when the price is $25, a $1 increase in price will lead to an increase in supply of approximately 138 pens. d) We would expect the rate of change of quantity with respect to price to be positive. All things being equal, it is reasonable to assume as the price of a good or service increases, the supply for that good or service will increase. this knowledge and our work from Exercises 11 and 14 to simplify our work. We have: P  R  C  3  3.21  0.21 P ‘  80   R ‘  80   C ‘  80   3  3.20  0.20 C  x   0.01x 2  0.6 x  30; R  x   2 x a) Finding the profit function we have: P  x  R  x  C  x   2 x  0.01x 2  0.6 x  30   0.01x 2  1.4 x  30 b) P  P  x  x   P  x  P  P  70  1  P  70  18. 13 x  100 x x 13  13 x  100 1 A ‘ x  x2 100  2 x Therefore, A  A ‘  x  x A  x   A ‘ 100  x  P  71  P  70   0.01  71  1.4  71  30  2  0.01  70  2  1.4  70   30     0.01 2  109.375  312.50  3750  3546.875 Producers will want to supply 3547 units when price is $25 per unit. c) Substituting 25 for p into the answer from part (a) we have: dS 2  0.021 25   25  150  138.125 dp p  25 P ‘  x   0.02 x  1.4 16. S  0.007 p 3  0.5 p 2  150 p a) We take the derivative of the supply function with respect to price. dS  0.021 p 2  p  150 dp b) Substituting 25 for p in the supply function we have:  100 100  2 1  x  100   x  1  0.01 The average cost changes by about $0.01 . (We see an approximate decrease in average cost of one cent.) Copyright © 2016 Pearson Education, Inc. Exercise Set 2.6 19. 403 M  t   2t 2  100t  180 21. a) Substituting for t, we have:  567  36 x  104 x x is the number of years since 1960; therefore, the year 2014 corresponds to x  2014  1960  54 , and the year 2015 corresponds to x  2015  1960  55 . To estimate the increase in gross domestic product from 2014 to 2015, we establish that x  54 and x  1 . Next, we find the derivative of P  x  : 2 M 10   2 10   100 10   180  980 2 M  25  2  25  100  25  180  1430 2 M  45  2  45  100  45  180  630 b) To find marginal productivity, we take the derivative of the productivity function: M ‘  t    4t  100 . 2 P ‘  x   36 1.6 x 0.6  104  57.6 x 0.6  104 . Substituting for t, we have: M ‘  5  4  5  100  80 Therefore, P  P ‘  x  x M ‘ 10   4 10   100  60  P ‘  54  x M ‘  25  4  25  100  0 M ‘  45  4  45  100  80 We see that the additional monthly output per years of service decreases each year the employee is with the company. d) The employees marginal productivity is at its highest point when the employee is new to the company. The employee is still learning how to do the job and will make the greatest gains. As the employee gains experience, the marginal productivity begins to decrease. The employee is still being more productive each month, but just doesn’t increase total output as much as the previous month’s increase. Eventually, age catches up to the employee and they cannot produce the output they once did. Marginal productivity becomes negative as total output starts to fall. 20. S  p   0.08 p3  2 p2  10 p  11 p  18.00, p  18.20  18.00  0.20 S ‘  p   0.24 p2  4 p  10 S  S ‘  p  p  S ‘ 18.00  p    0.24 18  4 18  10  0.2  2  31.952 The supplier will supply approximately 32 more units.  1.6 M  5  2  5  100  5  180  630 c)  P  x   567  x 36 x 0.6  104   57.6  54  0.6   x  54  104 x   526.7521497 1  x  1  526.75 The gross domestic product should increase about $526.75 billion between 2014 and 2015. 22. N  x    x 2  300 x  6 x is in thousands of dollars so, x  100, x  1 . N ‘  x   2 x  300 N  N ‘  x  x  N ‘ 100  x   2 100   300  1  100 Norris will sell approximately 100 more units by increasing its advertising expenditure from $100,000 to $101,000. 23. Yes, the taxation in 2014 was progressive. The 25,001st dollar is taxed at a rate of 15%, the 80,001st dollar is taxed at a rate of 25%, and the 140,001st dollar is taxed at a rate of 33%. 24. Marcy’s marginal tax rate is 28%, while Tyrone marginal tax rate is also 28%. However, an increase in $5000 will push Tyrone into the next tax bracket putting his marginal tax at 33%.Therefore, since Marcy is in a lower marginal tax bracket, she will keep more of the $5000 after taxes. 25. Alan’s marginal tax rate is currently 25%. If he earns another $10,000, dollars, this will push him into the 28% tax bracket and he will pay about $0.28 per dollar earned in taxes. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 404 26. The marginal tax rate at $50,000 is 25%. By earning an extra $3000 it will not push her out of the 25% tax bracket. Therefore, her tax liability will not grow if she takes the extra work. 27. 30. y  f  x  x   f  x   f  3  0.02   f  3 2 2   3.02    3.02      3   3       0.1004 f ‘  x  x  1  2 x   x y  f  x   x 3 , x  2, and x  0.01 y  f  x  x   f  x  f ‘  3 x  1  2  3   0.02  2 for x and  f  2  0.01  f  2  Substituting 0.01 for x  f  2.01  f  2    2.01   2  3  0.10 3  0.1206 f ‘  x  x  3x 2  x  f  x   x3 ; f ‘  x   3x 2  f ‘  2  x  3  2   0.01 2  12  0.01 31.  f 1.2   f 1  1  1      1.2  1   0.1667 y  f  x   x 2 , x  2, and x  0.01 y  f  x  x   f  x  f ‘  x  x    x 2   x  f  2  0.01  f  2    2.01   2  2  2 f ‘ 1 x   1  0.0401 f ‘  x  x  2 x  x 32.  0.04 y  f  x   x  x 2 , x  3, and x  0.04 y  f  x  x   f  x   f  3.04   f  3 Substituting 3 for x and 0.04 for x 2 2   3.04    3.04     3   3       12.2816  12  0.2816 f ‘  x  x  1  2 x   x   0.2  f  x   x 1 ;    2  f ‘  x    x  Substituting 1 for x and 0.2 for x  0.2  4  0.01  f  3  0.04   f  3 2 Substituting 1 for x and 0.2 for x  1 0.2  f ‘  2  x  2  2    0.01 29. 1  x 1 , x  1, and x  0.2 x y  f  x  x   f  x  y  f  x   f 1  0.2   f 1 Substituting 2 for x and 0.01 for x  0.12 28. y  f  x   x  x 2 , x  3, and x  0.02  f  x   x  x 2 ;    f ‘  x   1  2 x  f ‘  3 x  1  2  3    0.04  1  x 2 , x  1, and x  0.5 x2 y  f  x  x   f  x  y  f  x   f 1  0.5   f 1  1   1    2  2  1.5    1   0.5556 f ‘  x  x  2 x 3  x f ‘ 1 x   2 1    0.5      2  0.5  3  1 Substituting 3 for x and 0.04 for x   7   0.04   0.28 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.6 33. 405 y  f  x   3 x  1, x  4, and x  2 y  f  x  x   f  x   f  4  2  f  4  f  6  f  4 6 f ‘  x  x   3  x  f  x   3 x  1; f ‘  x   3 Substituting 4 for x and 2 for x 6 y  f  x  x   f  x   f  8  0.5  f  8   2  8.5  3   2  8  3 1 f ‘  x  x   2   x 1 2 x 1  x  1 2 9 1   0.167 6 We can now approximate  8;  3  y  3  0.167  2.833 To five decimal places 8  2.82843. Thus, our approximation is reasonably accurate. 1 35. We first think of the number closest to 26 that is a perfect square. This is 25. What we will do is approximate how y  x , changes when x changes from 25 to 26. Let y  f  x  x  x 2 1 1 1  12 1  x 2 2 x y  f ‘  x  x  x 37. We first think of the number closest to 102 that is a perfect square. This is 100. What we will do is approximate how y  x , changes when x changes from 100 to 102. Let y  f  x  x  x 2 Using, y  f ‘  x  x , we have 1 y  f ‘  x   x 8  9  y f ‘  8 x   2    0.5  36. Let y  f  x   x , x  8, x  1  y  f  x   2 x  3, x  8, and x  0.5 Then f ‘  x    5  0.100  5.100 To five decimal places 26  5.09902. Thus, our approximation is reasonably accurate.  17   11 34. 26 ;  5  y Substituting 4 for x and 2 for x  3  6   1  3  4   1 f ‘  4  x   3   2  We can now approximate 26  25  y Then f ‘  x   1  12 1  x 2 2 x Using, y  f ‘  x  x , we have y  f ‘  x  x 2 x We are interested in y as x changes from 25 to 26, so 1 y   x 2 x 1  1 Replacing x with 25 and x with 1 2 25 1  25 1   0.100 10  1  x 2 x We are interested in y as x changes from 100 to 102, so 1 y   x 2 x 1   2 Replacing x with 100 and x with 2 2 100 1  2 2 10 1   0.100 10 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 406 We can now approximate 102  100  y We are interested in y as x changes from 1000 to 1005, so 1 y   x 3 2 3 x 102 ;  10  y  10  0.100 Replacing x with 1000 and x with 5, we have  10.100 To five decimal places 102  10.09950. Thus, our approximation is reasonably accurate.   1  x   3 3 2 100 3   0.150 20 We can now approximate  10.017 To five decimal places 3 1005  10.01664 Thus, our approximation is reasonably accurate. 103 ; 40. Let y  f  x   3 x , x  27, x  1  10.150 y  f ‘  x  x To five decimal places 103  10.14889. Thus, our approximation is reasonably accurate.  39. We first think of the number closest to 1005 that is a perfect cube. This is 1000. What we will do is approximate how y  3 x , changes when x changes from 1000 to 1005. Let  1 23 1 x  3 2 3 3 x Using, y  f ‘  x  x , we have y  f ‘  x  x 3 x2 3  x2 1  x 3  3  27  2  1 1  0.037 27 We can now approximate 3 28 ; 3 28  3 27  y  3  0.037  3.037 To five decimal places 3 28  3.03659 Thus, our approximation is reasonably accurate. 1 1 1 3  y  f  x  3 x  x 3 3 1005  3 1000  y  10  0.017  10  0.15  5  10  y 103  100  y Then f ‘  x   2  y  f ‘  x  x 2 x 1 3  3 1000  1 5 3 100 1   0.017 60 We can now approximate 3 1005 ; 38. Let y  f  x   x , x  100, x  3  1  x 41. y  3 x  2   3 x  2 2 1 First, we find dy : dx dy 1 3 1   3 x  2 2  3  . dx 2 2 3x  2 Then 3 dy  dx . 2 3x  2 Note that the expression for dy contains two variables x and dx. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.6 42. 407 y  x  1   x  1 2 1 First, we find 45. dy : dx   dy  dx dy  1 dy  9 x 2 2 x 3  1 dx . Note that the expression for dy contains two variables x and dx. First, we find 46. 2 2 2 . 2 x4  8×2  4 x  3  x  3 2 2 dx . y  5 x  27   x  27 5 1 dy : dx dy 1 1 4   x  27 5 1  . 4 dx 5 5  5  x  27 2 Then dy : dx  4 x 3  2 x  5  3 x 2  2 x  5 dy  2     4 x  6 x  15 x   2 x  5  10 x  15 x   2 x  5  4 x  3 x  2 x  5  2 x  5 2 3 3  x  3 First, we find dy 2  x 3  2  2 x  5  2    3 x 2  2 x  5 dx 3 4 Note that the expression for dy contains two variables x and dx.  9 x 2 2 x 3  1. Then y  x  2 x  5 2 Then  9 x2 2 x3  1 2 3  2 x4  8×2  4 x  3     2 3   2    By the extended power rule 3  4 y  2 x3  1 2  dy . By the quotient rule we have: dx x2  3 3×2  1  x3  x  2  2 x  x  3  3x  10 x  3   2 x  2 x  4 x  .   x  3 dy : First, we find dx 1 dy 3  2 x3  1 2 6 x2 dx 2 44. x3  x  2 x2  3 First, we find dy 1 1 1   x  1 2 1  . dx 2 2 x 1 Then 1 dy  dx . 2 x 1 43. y 47. . 1 5  5  x  27 4 dx . y  x 4  2 x3  5×2  3x  4 First, we find 2 dy : dx dy  4 x 3  6 x 2  10 x  3 . dx Then dy  4 x 3  6 x 2  10 x  3 dx . 2  5 x 2  2 x  3  2 x  5 .  Then dy  5 x 2  2 x  3  2 x  5 dx .  Note that the expression for dy contains two variables x and dx. 48. y   7  x 8 First, we find dy : dx dy 7 7  8  7  x   1  8  7  x  . dx Then dy  8  7  x  dx . 7 Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 408 53. Let y  f  x   x 4  x 2  8 49. From Exercise 48, we know: dy  8  7  x  dx , 7 when x  1 and dx  0.01 , we have: dy  8  7  1  0.01 7 7   22,394.88.  To approximate f  5.1 , we will use  When x  2 and dx  0.1 we have:  dy  4  2   6  2   10  2   3  0.1 2   31 0.1  3.1 51. x  5 and dx  0.1 to determine the differential dy. Substituting 5 for x and 0.1 for dx we have: dy  f ‘  5 dx  y   3 x  10    490   0.1 dy : dx  15  3 x  10  Then  49 Next, we find f  5   5   5  8 4 4 When x  4 and dx  0.03 we have: Now, f  5.1  f  5  f ‘  5 dx  608  49 dy  15  3  4   10   0.03 4  657  15  2   0.03 4 54. Let y  f  x   x 3  5 x  9  7.2 First we find f ‘  x  : y  x5  2 x3  7 x f ‘  x   3x 2  5 . dy : First, we find dx dy  5×4  6 x2  7 . dx Then dy  5 x 4  6 x 2  7 dx .  Then dy  f ‘  x  dx  To approximate f  3.2 , we will use When x  3 and dx  0.02 we have:  dy  5  3  6  3  7  0.02  2   3 x 2  5 dx  4 2  625  25  8  608 4 dy  15  3 x  10  dx .  6.88. 3   500  10   0.1 5 dy 4  5  3 x  10   3 By the extended power rule dx    4  5  2  5  0.1   4 125  10   0.1 First, we find 52.   4 x 3  2 x dx 50. From Exercise 47, we know: dy  4 x 3  6 x 2  10 x  3 dx . 3 f ‘  x   4 x3  2 x . Then dy  f ‘  x  dx  8  6   0.01  First we find f ‘  x  : x  3 and dx  0.2 to determine the differential dy. dy  f ‘  3 dx    3  3  5  0.2  2   27  5  0.2    22   0.2   4.4 The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.6 409 Next, we find 3 f  3   3   5  3  9 The differential is: dN  N ‘  t  dt  27  15  9   21 Now, f  3.2   f  3  f ‘  3 dx  25.4 S  0.02235h0.42246 w0.51456 We begin by noticing that we are wanting to estimate the change in surface area due to a change in weight w; therefore, we will first find dS . Since h  160 , we have: dw S  0.02235 160  0.42246   0.02235  8.53399783 w 0.51456   Now that we have the differential, we can use her weight of 60 kg to approximate how much her surface area changes when her weight drops 1 kg. We substitute 60 for w and 1 for dw to get:  dS  0.09814452  60  0.48544   1  0.01345 The patient’s surface area will change by 0.01345 m 2 . 56. 0.8t  1000 5t  4 First we find N ‘  t  by the quotient rule. N t  N ‘t    5t  4  0.8   0.8t  1000  5  5t  4 2 4t  3.2  4t  5000   5t  4 2 4996.8  5t  4 2  dt 4996.8 9 2  0.1  6.16889 Next, we approximate the change in bodily concentration from 2.8 hr to 2.9 hr by using 2.8 for t and 0.1 for dt. 4996.8  0.1 dN   2   5  2.8  4  w 0.51456  0.19073485w 0.51456 Now we can take the derivative of S with respect to w. dS  0.19073485  0.51456  w 0.48544 dw  0.09814452w 0.48544 Therefore, dS  0.09814452 w 0.48544 dw  5t  4  2 We approximate the change in bodily concentration from 1.0 hr to 1.1 hr by using 1.0 for t and 0.1 for dt. 4996.8  0.1 dN   2   5 1.0   4   21  4.4 55. 4996.8  4996.8 18 2  0.1  1.54222 The concentration changes more from 1.0 hr to 1.1 hr. 57. p  x   0.06 x3  0.5 x 2  1.64 x  24.76 First we find p ‘  x  . p ‘  x   0.06  3x 2   0.5  2 x   1.64  0.18 x 2  x  1.64. The differential is: dp  p ‘  x  dx   0.18 x 2  x  1.64  dx Since x is the number of years since 2008, we have 2010 implies x  2 and 2012 implies x  4 . To estimate the change in ticket prices from 2010 and 2012, we substitute 2 for x and 2 for dx.   dp  0.18  2    2   1.64  2  2   0.36  2   0.72 To estimate the change in ticket prices from 2014 and 2016, we substitute 6 for x and 2 for dx.   dp  0.18  6    6   1.64  2  2   2.12  2   4.24 Ticket prices will increase more between 2014 and 2016. Copyright © 2016 Pearson Education, Inc. 410 Chapter 2: Applications of Differentiation 58. The circumference of the earth, which is the original length of the rope, is given by C  r   2 r , where r is the radius of the earth. b) The possible extra area is 628 square feet. Since each additional can will cover 300 square feet, they will need to bring 3 extra cans to account for any extra area. c) 3 additional cans will cost the painters $90. We need to find the change in the length of the radius, r , when the length of the rope is increased 10 feet. 61. The volume of the spherical cavern is given by: V r   4 3 r 3 First we find the derivatve of the volume function. r 4 V ‘  r    3r 2   4 r 2 . 3 r Therefore, the differential is: dV  4 r 2 dr Substituting the given information, we have: dV  4 3.14400 2  4, 019, 200. 2 Using differentials, C  C ‘  r  r represents the change in the length of the rope. Therefore, C  10 , and we have: 10  C ‘  r  r Noticing that C ‘  r   2 , we have: The enlarged cavern will contain an additional 4,019,200 cubic feet. 62. R  x  p  x 10  2  r 10  r 2 Therefore, the rope is raised approximately 5 r   1.59 feet above the earth.  59. A  x   3  100 x  x 2 . 3 1 3 x . R ‘  x   100  x 2  100  2 2 x To find Marginal Average Cost, we take the derivative of the average cost function. By the quotient rule we have: x  C ‘  x   C  x  1 A ‘  x  x2 x  C ‘  x  C  x  . x2 60. a) The surface area of the water tank is given by A  2 r . Calculating the differential we have: 2 dA  A ‘ r  dr  4 rdr. The tolerance is in feet is 0.5 . dA  4 3.14100 0.5  628   100  x x 63. C  x p  100  x p  400  x Since revenue is price times quantity, the revenue function is given by: R  x  p  x   400  x  x  400 x  x 2 To find the marginal revenue, we take the derivative of the revenue function. Thus: R ‘  x   400  2 x 64. p  500  x R  x  p  x   500  x  x  500 x  x 2 R ‘  x   500  2 x . The approximate difference in surface area when the tolerance is taken into consideration is 628 square feet. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.6 65. 411 4000 3 x Since revenue is price times quantity, the revenue function is given by: R  x  p  x p  4000    3 x  x   4000  3 x To find the marginal revenue, we take the derivative of the revenue function. Thus: R ‘  x  3 66. 3000 5 x R  x  p  x p  3000    5 x  x   3000  5 x . R ‘  x   5. 67. Answers will vary. Calculus in its present form was essentially developed independently in the 17th century by Isaac Newton and Gottfried Wilhelm von Leibniz. During the 18th century calculus was challenged by some philosophers and religious leaders who argued that the infinitely small quantities represented by differentials were meaningless. These critics were silenced when the concept of “limit” was introduced. In it, a differential was not thought of as an infinitely small quantity; rather, a derivative was considered to be the limit approached by two differentials as each becomes infinitely small. 68. For a function y  f  x  , the differential, dy, can be used to approximate the true change in the value of f  x  when a small change is made in the value of x. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 412 c) Solve E  x   1 Exercise Set 2.7 1. x 1 500  x x  500  x 2 x  500 x  250 A price of $250 will maximize total revenue. a) The demand function is q  D  x   400  x . The definition of the elasticity of demand is x  D ‘  x . In order to given by: E  x    D  x find the elasticity of demand, we need to find the derivative of the demand function first. dq d  D ‘  x   400  x   1 . dx dx Next, we substitute 1 for D ‘  x  , and 3. a) D ‘  x    4 E  x   x 50  x b) Substituting x  46 into the expression found in part (a) we have: 46 46 23 E  46     11.5 50  46 4 2 elasticity. x   1 x  400  x 400  x b) Substituting x  125 into the expression found in part (a) we have: 125  125  5 E 125  400  125 275 11 E  x   Since E  46   1 , demand is elastic. c) We solve E  x   1 x 1 50  x x  50  x 2 x  50 x  25 A price of $25 will maximize total revenue. 5 Since E 125  is less than one, the 11 demand is inelastic. c) The values of x for which E  x   1 will maximize total revenue. We solve: E  x  1 2. q  D  x   500  x; x  38 a) D ‘  x   1 E  x   x  D ‘  x x   1 x   D  x 500  x 500  x 38 19  b) E  38  500  38 231 E  38   1 , so demand is inelastic. x  D ‘ x x   4  4x   D  x 200  4 x 200  4 x  400  x for D  x  into the expression for x 1 400  x x  400  x 2 x  400 x  200 A price of $200 will maximize total revenue. q  D  x   200  4 x; x  46 4. q  D  x   500  2 x; x  57 a) D ‘  x   2 E  x   x  D ‘  x D  x  x   2  500  2 x  2x 500  2 x x 250  x 57 57  b) E 57   250  57 193 E 57   1 , so demand is inelastic.  c) Solve E  x   1 x 1 250  x x  250  x 2 x  250 x  125 A price of $125 will maximize total revenue. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.7 5. 413 Next, we take the derivative of the demand function, using the Chain Rule: 1 1 d D ‘  x    600  x  2   600  x  dx 2 1  2 600  x Making the appropriate substitutions into the elasticity function, we have   1 x x  D ‘ x  2 600  x  E  x     D  x 600  x 400 ; x  50 x a) First, we rewrite the demand function. 400 D  x   400 x 1 . x Next, we take the derivative of the demand function, using the Power Rule. D ‘  x   400  1 x 2  400 x 2 q  D  x  Making the appropriate substitutions into the elasticity function, we have x  400 x 2 x  D ‘ x   E  x   D  x 400 x 1   x x  2 600  x  2  600  x  600  x 400 x 1 1 400 x 1 Therefore, E  x   1 for all values of x .  x 1200  2 x b) Substituting x  100 into the expression found in part (a) we have: 100 100 1 E 100     1200  2 100 1000 10  b) E 50   1 , so demand is unit elastic. c) E  x   1 for all values of x . Therefore, total revenue is maximized for all values of x . In other words, total revenue is the same regardless of the price. 6. c) Solve E  x   1 3000 ; x  60 x 3000 a) D  x    3000 x 1 x D ‘  x   3000 x 2 q  D  x  x 1 1200  2 x x  1200  2 x 3x  1200 Making the appropriate substitutions into the elasticity function, we have x  3000 x 2 x  D ‘  x   E  x   D  x 3000 x 1   3000 x 1  1 3000 x 1 Therefore, E  x   1 for all values of x . b) E  60   1 , so demand is unit elastic. c) E  x   1 for all values of x . Therefore, total revenue is maximized for all values of x . In other words, total revenue is the same regardless of the price. 7. Since E 100   1 , demand is inelastic. q  D  x   600  x ; x  100 a) First rewrite the demand function: D  x    600  x  2 . 1 x  400 A price of $400 will maximize total revenue. 8. q  D  x   300  x ; x  250 a) D  x    300  x  2 . 1 1 1 d 300  x  2  300  x   2 dx 1 D ‘  x   2 300  x E  x   x  D ‘ x  D  x   1 x  2 300  x  300  x x x x  2 300  2  300  x  300  x  x 600  2 x Copyright © 2016 Pearson Education, Inc.  Chapter 2: Applications of Differentiation 414 b) E  250   250 5  600  2  250  2 10. Since E  250   1 , demand is elastic. c) Solve E  x   1 x 1 600  2 x x  600  2 x  x  200 A price of $200 will maximize total revenue. q  D  x  100  x  32 ; x8  2 x  12 2 2 a) D  x   500  2 x  12  . 3 D ‘  x   500  2  2 x  12   2  3  2000  2 x  12  3x  600 9. 500 q  D  x  2000  2 x  123 Making the appropriate substitutions into the elasticity function, we have  2000  x   3 x  D ‘  x   2 x  12   E  x     500 D  x ; x 1 a) First, we rewrite the demand function:  2 x  12 2 D  x   100  x  3 . 2  2000   2 x  12 2 2x  x  3 500 x6   2 x  12   Next, we take the derivative of the demand function, using the Chain Rule:  3  d D ‘  x   100  2  x  3   x  3   dx   200  x  3  b) E 8  3 2  8 16 8   8  6 14 7 Since E 8   1 , demand is elastic. 200 c) Solve E  x   1  x  33 Making the appropriate substitutions into the elasticity function, we have:  200  x   3 x  D ‘  x   x  3  E  x    100 D  x  x  32  200   x  32  x 3   x  3  100 2x x3 b) Substituting x  1 into the expression found in part (a) we have: 2  1 2 1 E 1    1 3 4 2 Since E 1  1 , demand is inelastic.  c) Solve E  x   1 2x 1 x6 2x  x  6 x6 A price of $6 will maximize total revenue. 11. q  D  x   50, 000  300 x  3 x 2 a) D ‘  x   300  6 x E  x      x  D ‘  x D  x x  300  6 x  50, 000  300 x  3 x 2 300 x  6 x 2 50, 000  300 x  3 x 2 6 x 2  300 x 50, 000  300 x  3 x 2 2x 1 x3 2x  x  3 x3 A price of $3 will maximize total revenue. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.7 415 f) Substituting the answer found in part (e) into the demand function we have: b) Substituting in to the elasticity of demand we have: E 75  6 75  300 75 D 114.98  50, 000  300 114.98  3 114.98 2 50, 000  300 75  3 75 2  44,832.80. The demand for oil is about 44,832 million barrels per day or 44.8 billion barrels per day at a price of $114.98 a barrel. At the time this solution was created, the price of oil was $100.83 per barrel. Thus according to this model, by increasing the price, oil producers could increase revenue. g) The demand for oil is inelastic at $110 a barrel. Therefore an increase in price will result in an increase in total revenue. 11, 250 55, 625  0.20. Since E 75   1 , the demand for oil is  inelastic at $75 a barrel. c) Substituting in to the elasticity of demand we have: E 100  6 100  300 100 2 50, 000  300 100  3 100 2 30, 000 50, 000  0.6. Since E 100   1 , the demand for oil is  inelastic at $100 a barrel. d) Substituting in to the elasticity of demand we have: E 125  6 125  300 125 2 50, 000  300 125  3 125 2 56, 250 40, 625  1.38. Since E 125  1 , the demand for oil is  elastic at $125 a barrel. e) Revenue will be maximized when E  x   1 . Therefore, we solve: 6 x 2  300 x 1 50, 000  300 x  3 x 2 6 x 2  300 x  50, 000  300 x  3 x 2 9 x 2  600 x  50, 000  0 Using the quadratic formula, we find that the solutions to the equation. x   600  6002  4 950000 2 9  600  2,160, 000 18 x  48.316 or x  114.983. The only solution that is feasible is x  114.98 . Thus, oil revenues will be maximized when price is $114.98 a barrel. x 2 12. q  D  x   967  25 x a) D ‘  x   25 E  x   x  D ‘ x x   25  D  x 967  25 x 25 x 967  25 x b) We set E  x   1 and solve for x.  25 x 1 967  25 x 25 x  967  25 x 50 x  967 x  19.34 Demand is unitary elastic when price is 19.34 cents. c) Demand is elastic when E  x   1 . Testing a value on each side of 19.34 cents, we have: 25  19 E 19    0.97  1 967  25  19 25  20 E  20   1.07  1 967  25  20 Therefore, the demand for cookies is elastic for prices greater than 19.34 cents. d) Demand is inelastic when E  x   1 . Using the calculations from part (c), we see that the demand for cookies is inelastic for prices less than 19.34 cents. e) Total revenue is maximized when E  x   1 . In part (b) we showed that E  x   1 when price was 19.34 cents. Therefore, revenue will be maximized when price is 19.34 cents. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 416 f) We have shown that the demand for cookies is elastic when the price of cookies is 20 cents. Therefore a small increase in price will cause total revenue to decrease. 13. 14. q  D  x   200  x 3 2 x  300 10 x  11 a) Using the Quotient Rule, we have: 10 x  11  2   2 x  300  10 D ‘  x  10 x  112 q  D  x  a) First, we rewrite the demand function to make it easier to find the derivative.  D  x   200  x 3   . 1     b) E  3      200  x  3  3x 3 2 200  x 3 x  D ‘  x  D  x x 2978 10 x  112 2 x  300 10 x  11 2978 x 10 x  11   10 x  112 2 x  300  b) E  3  2 1489 x 10 x  11 x  150 1489  3 10 3  11 3  150 1489 2091  0.7121 Since E  3  1 , the demand for tomato   3x 400  2 x 3 plants is inelastic when the price is $3 per plant. c) We determined in part (b) that the demand for tomato plants was inelastic at $3. Therefore, an increase in the price of tomato plants will lead to an increase in revenue. 3 400  2  3 10 x  112 E  x   3 3  3 2978 Therefore, 3x 3 2 10 x  112 2 Next, using the Chain Rule, we have: 1 1 2 D ‘  x   200  x 3 3 x 2 2 3 x 2  2 200  x 3 Now, substituting in the elasticity function we get:  3 x 2  x  3 x  D ‘  x  2 200  x  E  x    D  x 200  x 3  20 x  22   20 x  3000 3 81 346  0.2341 Since E  3  1 , the demand for computer  games is inelastic when price is $3. c) From part (b) we know that the demand for computer games is inelastic at a price of $3. Therefore an increase in the price of computer games will lead to an increase in the total revenue. 15. q  D  x   180  10 x a) First we find the derivative of the demand function. D ‘  x   10 Therefore, the elasticity of demand is: x  D ‘  x E  x   D  x  x   10 180  10 x 10 x  180  10 x 10 x  180  10 x Copyright © 2016 Pearson Education, Inc. Exercise Set 2.7 417 b) Substituting in to the elasticity of demand we have: 10 8 E 8  180  10 8 Substituting into the elasticity function we have:  80 100  0.8. c) Revenue will be maximized when E  x   1 .  Therefore, we solve: 10 x 1 180  10 x 10 x  180  10 x 20 x  180 x  9. When the price of sunglass cases is $9 a case, revenue will be maximized. d) A 20% increase in the price would increase the price from $8 a case to $9.60 a case. At $8 a case, demand is: q  D 8  180  10 8  100 cases. Therefore revenue is price times quantity or R  p  q  $8 100  $800. At $9.60 a case, demand is: q  D 9.60  180  10 9.60  84 cases. Therefore revenue is price times quantity or R  p  q  $9.60  84  $806.40. Therefore, revenue will increase if Tipton Industries raises the price 20%. We know increasing from $8 to $9 would increase revenue, and increasing from $9 to $9.60 would decrease revenue, we just didn’t know how much the increase and decrease would be. Therefore, the answer to part (d) does not contradict the answers to parts (b) and (c). In fact it confirms them. 16.  n 1 x   nkx x  D ‘  x E  x    D  x kx  n    nk x  x  n 1 kx n     nk x  n kx  n n b) No, the elasticity of demand is constant for all prices. E  x   n . c) Total revenue is maximized when E  x   1 . Since E  x   n , total revenue will be maximized when n  1 . 17. Answers will vary. The elasticity of demand is a measure of the responsiveness of quantity demanded to changes in price. This measure allows economists to determine how sensitive quantity demanded is to price changes and help predict the effect of price changes on total revenue. 18. Answers will vary. In general, the greater the availability of substitutes and the better the substitutes are will cause goods to have a higher elasticity of demand. For example, the demand for tea is relatively elastic because of the wide range of substitutes available, such as coffee, soda, or water. However, a diabetic’s demand for insulin is very inelastic because there are no close substitutes for insulin. k xn a) First, we rewrite the demand function to make it easier to find the derivative. D  x   k  xn Using the Power Rule, we have: D ‘  x   k    n  x  n 1   nkx  n 1 q  D  x  Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 418 3. Exercise Set 2.8 1. Differentiate implicitly to find  dy . dx We have 3 x 3  y 2  8. Differentiating both sides with respect to x yields: d d 3×3  y 2  8 dx dx d 3 d 2 d 3x  8 y  dx dx dx dy dy Next, we isolate 9 x2  2 y  . 0 dx dx dy 2 y  9 x 2 dx dy 9 x 2  dx 2 y Find the slope of the tangent line to the curve at  2, 4  .   2 dy 3x 2  dx 4y Find the slope of the tangent line to the curve at  2, 1 . Replacing x with 2 and y with 1 , we have: dy 3 x 2 3  2 12     3. dx 4y 4  1 4 2 The slope of the tangent line to the curve at  2, 1 is 3. 4. The slope of the tangent line to the curve at 9  2, 4  is . 2 2. x3  2 y3  6 Differentiating both sides with respect to x yields: d 3 d x  2 y3  6  dx dx dy dy 0 Next, we isolate 3×2  6 y 2  dx dx dy  3x 2 6 y2  dx   dy  x 2  dx 2 y 2 Find the slope of the tangent line to the curve at  2, 1 .  dy 6 x 2  dx 8y Replacing x with 2 and y with 4 , we have: dy 9 x 2 9 2 36 9     . dx 2 y 2  4 8 2 2 x 3  4 y 2  12 Differentiating both sides with respect to x yields: d d 2 x3  4 y 2  12 dx dx dy dy 0 Next, we isolate 6 x2  8 y  . dx dx dy  6 x 2 8y  dx 2 x2  3 y3  5 Differentiating both sides with respect to x yields: d d 2 x2  3 y3  5  dx dx dy dy 0 Next, we isolate 4x  9 y2  . dx dx dy 9 y 2   4 x dx dy 4x  dx 9 y 2   Find the slope of the tangent line to the curve at  2,1 . dy 4  2  8 8    . 2 dx 9 1 9 9 The slope of the tangent line to the curve at 8  2,1 is  . 9 4 dy   2     2 . dx 2  12 2 2 The slope of the tangent line to the curve at  2, 1 is 2. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.8 5. 419 x2  y2  1 Differentiating both sides with respect to x yields: d 2 d x  y2  1 dx dx dy dy 2x  2 y  . Next, we isolate 0 dx dx dy 2y   2 x dx dy 2 x  2y dx  7.   4 x3 y  dy 3y  dx 2x Find the slope of the tangent line to the curve at  1, 3 . 1 3 and y with , we have: 2 2 x2  y2  1 Differentiating both sides with respect to x yields: d 2 d x  y2  1 dx dx dy dy 0 2x  2 y  . Next, we isolate dx dx dy 2 y   2 x dx dy x  dx y Find the slope of the tangent line to the curve at    3, 2  . 3 3 dy   . 2 dx 2 The slope of the tangent line to the curve at 3 3, 2 is . 2   dy dy  6 x 2 y 2  0 Next, we isolate . dx dx dy 3 2 2  6 x y 4x y  dx dy 6 x 2 y 2  dx 4 x3 y 1 dy x 1   2  . dx y 3 3 2 The slope of the tangent line to the curve at 1 3 1 .  2 , 2  is  3   6.      dy x  dx y Find the slope of the tangent line to the curve at 1 3 2, 2 .   Replacing x with 2 x 3 y 2  18 Differentiating both sides with respect to x yields: d d 2 x3 y 2  18 dx dx d 2 d y  y2 Product Rule 2 x3 2 x3  0 dx dx dy   2 x3  2 y    y 2 6 x 2  0  dx  Replacing x with 1 and y with 3, we have: 3 3 9 dy 3y    dx 2x 2  1 2 The slope of the tangent line to the curve at 9  1, 3 is . 2 3 x 2 y 4  12 Differentiating both sides with respect to x yields: d d 3x 2 y 4  12 dx dx dy   Product Rule 3 x 2  4 y 3    y 4  3  2 x   0  dx  8.   dy dy  6 xy 4  0 Next, we isolate . dx dx dy 12 x 2 y 3   6 xy 4 dx dy y  2x dx Find the slope of the tangent line to the curve at  2, 1 . 12 x 2 y 3   1  1 dy  . dx 2  2 4 The slope of the tangent line to the curve at 1  2, 1 is . 4 Copyright © 2016 Pearson Education, Inc. 420 Chapter 2: Applications of Differentiation x 4  x 2 y 3  12 Differentiating both sides with respect to x yields: d 4 d x  x2 y3  12 dx dx d 4 d 2 3 d x  x y  12 dx dx dx    dy  4 x3   x 2  3 y 2    y 3  2 x   0 Product Rule   dx   Find the slope of the tangent line to the curve at 3, 2  . 9.     4 x3  3 x 2 y 2  dy 3  3  2  2  98 1    . dx 12 2  3 2  12 The slope of the tangent line to the curve at 1 3, 2  is  . 12 2   11. dy  2 xy 3  0 dx Next, we isolate 3 x 2 y 2  dy dx    . dy  2 xy 3  4 x3 dx dy 2 xy 3  4 x3  dx 3 x 2 y 2 x Replacing x with 2 and y with 1, we have: dy 4 x 2  2 y 3 4 2  2 1 7    2 dx 3 3 xy 2 3  21 3 The slope of the tangent line to the curve at 7  2,1 is  . 3 10. x 3  x 2 y 2  9 Differentiating both sides with respect to x yields: d 3 d x  x2 y2   9  dx dx    dy  3 x 2   x 2  2 y    y 2  2 x   0 dx      Replacing x with 1 and y with 2 , we have: 2  2 dy 2 y 4 4     dx x  2 y 1  2  2 3 3 The slope of the tangent line to the curve at 4 1, 2  is  . 3  dy  2 xy 2  0 dx dy 2 x 2 y   2 xy 2  3 x 2 dx 3x 2  2 x 2 y  dy 2 xy 2  3 x 2  dx 2 x 2 y dy 3 x  2 y  2 xy dx 2  dy dy  y  2y   2  0 dx dx dy dy x   2y   2 y dx dx dy x  2 y   2  y dx 2 y dy  dx x  2 y Find the slope of the tangent line to the curve at 1, 2  . dy 4 x 2  2 y 3  dx 3 xy 2 Find the slope of the tangent line to the curve at  2,1 . 2 xy  y 2  2 x  0 Differentiating both sides with respect to x yields: d d xy  y 2  2 x  0  dx dx d d d d  xy   y 2  2 x   0 dx dx dx dx   dy   dy  x    y 1  2 y   2 1  0 dx  dx  12. xy  x  2 y  3 Differentiating both sides with respect to x yields: d d  xy  x  2 y   3 dx dx   dy   dy 0  x    y 1  1  2  dx dx   dy dy  y 1 2  0 dx dx dy at the top of the next page. We isolate dx x Copyright © 2016 Pearson Education, Inc. Exercise Set 2.8 421 Continued from the previous page. dy  x  2   1  y dx dy 1  y  dx x  2 Find the slope of the tangent line to the curve at 2   5,  . 3 x2 y  2 x3  y 3  1  0 Differentiating both sides with respect to x yields: d 2 d x y  2 x3  y 3  1  0  dx dx  2  dy   dy 2 2 0  x    y  2 x   2 3 x  3 y  dx dx   14.      x2  2 1 1    3 1 dy   3  . dx  5  2 3 9 The slope of the tangent line to the curve at 2 1   5,  is  . 3 9 4 x3  y 4  3 y  5x  1  0 Differentiating both sides with respect to x yields: d d 4 x3  y 4  3 y  5 x  1  0 dx dx d d 4 d d d y  3 y  5x  1  0 4 x3  dx dx dx dx dx dy dy 12 x 2  4 y 3   3   5  0  0 dx dx dy dy 4 y 3   3   12 x 2  5 dx dx dy 4 y 3  3   12 x 2  5 dx  dy 12 x 2  5  dx 4 y 3  3 dy 12 x 2  5  dx 4 y 3  3 Find the slope of the tangent line to the curve at 1, 2  . Replacing x with 1 and y with 2 , we have: 17 dy 12 x 2  5 12 1  5    29 dx 4 y 3  3 4 23  3 2 The slope of the tangent line to the curve at 17 1, 2  is  . 29  Find the slope of the tangent line to the curve at  2, 3 . dy 6  2   2  2  3 24  12 36 36     2 2 dx 4  27  23 23  2  3  3 2   dy dy  2 xy  6 x 2  3 y 2  0 dx dx dy  6 x 2  2 xy x2  3 y 2  dx dy 6 x 2  2 xy  2 dx x  3y2  13.   The slope of the tangent line to the curve at 36  2, 3 is  . 23 15. x 2  2 xy  3 y 2 Differentiating both sides with respect to x yields: d 2 d 3y2 x  2 xy  dx dx d 2 d d 3y2 x  2 xy  dx dx dx   dy   dy   2 x  2  x    y 1  3  2 y      dx dx        dy dy  2y  6y  dx dx dy dy 2x   6 y   2 x  2 y dx dx dy 2 x  6 y    2  x  y  dx dy 2  x  y   dx 2x  6 y 2x  2x  2  x  y  dy  dx 2   x  3 y  dy x y  dx 3 y  x Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 422 16. 2 xy  3  0 Differentiating both sides with respect to x yields: d d 2 xy  3  0 dx dx   dy   2   x    y 1  0  0   dx   2x  17. dy  2y  0 dx dy 2x   2 y dx dy y  dx x 18. x 2  y 2  25 Differentiating both sides with respect to x yields: d 2 d x  y2   25 dx dx dy 0 2x  2 y  dx dy  2 x 2y dx dy 2 x  dx 2 y   dy x  dx y   dy 5 x 4  dx 3 y 2 y5  x3 Differentiating both sides with respect to x yields: d 5 d 3 y  x dx dx dy  3x 2 5 y4  dx    dy x  dx y y 3  x5 Differentiating both sides with respect to x yields: d 3 d 5 y  x dx dx dy  5x 4 3y2  dx   20. x 2  y 2  16 Differentiating both sides with respect to x yields: d 2 d x  y2  16 dx dx d 2 d 2 d x  y  16 dx dx dx dy 0 2x  2 y  dx dy 2 y   2 x dx dy 2 x  dx 2 y  19.   dy 3x 2  dx 5 y 4 21. x 2 y 3  x 3 y 4  11 Differentiating both sides with respect to x yields: d 2 3 d x y  x3 y 4  11 dx dx d 2 3 d 3 4 x y  x y 0 dx dx Notice: d 2 3 dy   x y  x2  3 y 2    y 3 2 x   dx dx           3x 2 y 2  dy  2 xy 3 dx and d 3 4 dy   x y  x 3  4 y 3    y 4 3x 2  dx dx  dy  4 x 3 y 3   3x 2 y 4 dx Therefore, d 2 3 d 3 4 x y  x y 0 dx dx dy dy 3 x 2 y 2   2 xy 3  4 x 3 y 3   3 x 2 y 4  0 dx dx dy , on the next page. We isolate dx   Copyright © 2016 Pearson Education, Inc.       Exercise Set 2.8 423 Continued from the previous page. dy 4 x 3 y 3  3x 2 y 2   3 x 2 y 4  2 xy 3 dx dy 3x 2 y 4  2 xy 3  dx 4 x 3 y 3  3x 2 y 2  3 3 p  1  dp dx 2    dp 3  dx 3 p 2  1   2 2 dy xy 3 xy  2 y  dx xy 2 4 x 2 y  3x 25.   2 y  3xy 2 dy  2 dx 4 x y  3x 22. x 3 y 2  x 5 y 3  19 Differentiating both sides with respect to x yields: d 3 2 d x y  x5 y 3   19 dx dx  2 x3 y  dy dx  3x 2 y 2  3x 5 y 2  5 2 3xp 2   dy dx  5x 4 y 3  0  5 x y  3x y 2 x y  3x y  dy dx 3 xp 3  24 Differentiating both sides with respect to x yields: d d xp 3   24 dx dx dp   Product Rule x  3 p 2    p 3 1  0  dx  4 3 2 2 4 3 2 2 dy 5 x y  3x y  5 2 dx 3x y  2 x 3 y 26. x 3 p 2  108 Differentiating both sides with respect to x yields: dy 5 x 2 y 2  3 y  dx 3x 3 y  2 x 23. p 2  p  2 x  40 Differentiating both sides with respect to x yields: d d p2  p  2 x   40 dx dx dp dp 2p   2 1  0 dx dx dp  2 p  1   2 dx 2 dp  dx 2 p  1  24.  p 3  p  3 x  50 Differentiating both sides with respect to x yields: d 3 d p  p  3x  50 dx dx dp dp 3 p2    3 1  0 dx dx dp We isolate at the top of the next column. dx   d 3 2 d x p  108 dx dx   dp   x 3  2 p    p 2 3x 2  0  dx  2 x3 p    dp   p3 dx dp  p 3  dx 3 xp 2 dp p  3x dx 27. dp  3x 2 p 2 dx dp 3x 2 p 2  dx 2 x3 p dp 3p  dx 2x x 2 p  xp  1 1 2x  p Multiply both sides by 2 x  p to clear the fraction.  x 2 p  xp  1  2 x  p     1 2 x  p   2x  p  x 2 p  xp  1  2 x  p The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 424 Differentiating both sides of the equation on the previous page with respect to x yields: d 2 d x p  xp  1  2 x  p  dx dx d 2 d d d d x p xp  1  p 2x  dx dx dx dx dx dp dp dp  p  2x  x   p 1  0  2  x2  dx dx dx dp dp 2 dp  x  1  2  2 xp  p x  dx dx dx dp x2  x  1  2  2 xp  p dx dp 2  2 xp  p  2 dx x  x 1 xp 28. 2 x p Multiply both sides by x  p to clear the fraction.  xp    2  x  p   x  p   x  p      xp  2 x  2 p Differentiate both sides with respect to x. d d  xp    2 x  2 p  dx dx dp dp  p 1  2 1  2  x dx dx dp dp  2  2 p x dx dx dp  x  2   2  p dx dp 2  p  dx x  2 29.  p  4  x  3  48 Expanding the left hand side of the equation we have: px  3 p  4 x  12  48 px  3 p  4 x  36 Differentiating both sides with respect to x yields: d d  px  3 p  4 x   36 dx dx dp dp p 1  x   3  4 1  0 dx dx dp  x  3    p  4 dx dp  p  4  dx x3 30. 1000  300 p  25 p 2  x Differentiating both sides with respect to x yields: d d 1000  300 p  25 p 2   x dx dx dp dp 300   25  2 p  1 dx dx dp 50 p  300   1 dx 1 dp  dx 50 p  300  31.  G 2  H 2  25 We differentiate both sides with respect to t. d d G 2  H 2   25 dt dt dG dH 2G   2H 0 dt dt dH dG 2H   2G  dt dt dH G dG   dt H dt We find H when G  0 :   02  H 2  25 H 2  25 H  5, H is nonnegative Next, we substitute 5 in for H , 0 in for G , and dG dH to determine 3 in for . dt dt dH G dG   dt H dt 0    3  0 5 We find H when G  1 : 12  H 2  25 H 2  24 H  24  2 6, H is nonnegative Next, we substitute 2 6 in for H , 1 in for G , dG dH and 3 in for to determine . dt dt dH G dG   dt H dt 1 3   3   2 6 2 6   The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.8 425 We find H when G  3 : Next, we substitute 10 for x and 5 for dx/dt. dR  50  10  5   40  5  200 dt The rate of change of total revenue with respect to time is $200 per day. C  x   10 x  3 Differentiating with respect to time we have: d d C  x   10 x  3 dt dt dC dx  10  dt dt Next, we substitute 10 for x and 5 for dx/dt. dC  10  5  50 dt The rate of change of total cost with respect to time is $50 per day. Profit is revenue minus cost. Therefore; P  x  R  x  C  x 3  H  25 2 2 H 2  16 H  4, H is nonnegative Next, we substitute 4 in for H , 3 in for G , and dG dH to determine 3 in for dt dt. dH G dG   dt H dt 3 9   3   4  4 32. A3  B 3  9 We differentiate both sides with respect to t. d 3 d A  B 3  9 dt dt dA dB  3B 2  0 3 A2  dt dt dA dB  3 B 2  3 A2  dt dt    50 x  0.5 x 2  10 x  3  0.5 x 2  40 x  3 Differentiating with respect to time we have: d d P  x  0.5 x 2  40 x  3 dt dt dP dx dx   x   40 dt dt dt dP dx   40  x   dt dt Next, we substitute 10 for x and 5 for dx/dt. dP   40  10  5   305  150 dt The rate of change of total profit with respect to time is $150 per day.  dA 3B 2 dB   dt 3 A2 dt dA  B 2 dB  2  dt dt A We find B when A  2 : A3  B 3  9  2 3  B 3  9 8  B3  9 B3  1 B 1 Next, we substitute 2 for A, 1 for B, and 3 for dB dA into the formula for : dt dt dA  1 3   3   2 4 dt 2 2 33. R  x   50 x  0.5 x 2 Differentiating with respect to time we have: d d R  x  50 x  0.5 x 2 dt dt dR dx dx  50   x  dt dt dt dR dx  50  x   dt dt   34.  R  x   50 x  0.5 x 2 Differentiating with respect to time we have: d d R  x  50 x  0.5 x 2 dt dt dR dx dx  50   x  dt dt dt dR dx  50  x   dt dt Next, we substitute 30 for x and 20 for dx/dt. dR  50  30  20   20  20  400 dt The rate of change of total revenue with respect to time is $400 per day.   The solutioin is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 426 Next, we substitute 200 for x and 300 for dx/dt. dC  1.2  200  300  72, 000 dt The rate of change of total cost with respect to time is $72,000 per day. Profit is revenue minus cost. Therefore; P  x  R  x  C  x Looking at the cost function C  x   4 x  10 . Differentiating with respect to time we have: d d C  x    4 x  10  dt dt dC dx  4 dt dt Next, we substitute 30 for x and 20 for dx/dt. dC  4   20  80 dt The rate of change of total cost with respect to time is $80 per day. Profit is revenue minus cost. Therefore; P  x  R  x  C  x   280 x  0.4 x 2  5000  0.6 x 2   x  280 x  5000 Differentiating with respect to time we have: d d P  x   x 2  280 x  5000 dt dt dP dx dx  2 x   280 dt dt dt dP dx   280  2 x   dt dt Next, we substitute 200 for x and 300 for dx/dt. dP   280  2  200   300  36,000 dt The rate of change of total profit with respect to time is $36, 000 per day.   50 x  0.5 x 2   4 x  10  0.5 x 2  46 x  10 Differentiating with respect to time we have: d d P  x  0.5 x 2  46 x  10 dt dt dP dx dx   x   46 dt dt dt dP dx   46  x   dt dt Next, we substitute 30 for x and 20 for dx/dt. dP   46   30   20  16 20  320 dt The rate of change of total profit with respect to time is $320 per day.  35.  R  x   280 x  0.4 x 2 Differentiating with respect to time we have: d d R  x  280 x  0.4 x 2 dt dt dR dx dx  280   0.8 x  dt dt dt dR dx   280  0.8 x   dt dt Next, we substitute 200 for x and 300 for dx/dt. dR   280  0.8  200  300  36,000 dt The rate of change of total revenue with respect to time is $36,000 per day. C  x   5000  0.6 x 2   Differentiating with respect to time we have: d d C  x  5000  0.6 x 2 dt dt dC dx  1.2 x  dt dt    2 36.  R x  2x Differentiating with respect to time we have: d d R  x   2 x  dt dt dR dx  2 dt dt Next, we substitute 20 for x and 8 for dx/dt. dR  2  8  16 dt The rate of change of total revenue with respect to time is $16 per day. C  x   0.01x 2  0.6 x  30 Differentiating with respect to time we have: d d C  x  0.01x 2  0.6 x  30 dt dt dC dx dx  0.02 x   0.6  dt dt dt dC dx   0.02 x  0.6  dt dt Next, we substitute 20 for x and 8 for dx/dt. dC   0.02  20  0.6  8  8 dt The rate of change of total cost with respect to time is $8 per day.   The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.8 427 Profit is revenue minus cost. Therefore; P  x  R  x  C  x  2  2 x  0.01x  0.6 x  30 38. From Exercise 37, we know that   0.01x 2  1.4 x  30 Differentiating with respect to time we have: d d P  x  0.01x 2  1.4 x  30 dt dt dP dx dx  0.02 x   1.4 dt dt dt dP dx  1.4  0.02 x   dt dt Next, we substitute 20 for x and 8 for dx/dt. dP  1.4  0.02  20  8  8 dt The rate of change of total profit with respect to time is $8 per day.  37. Total revenue is changing at a rate of $1.39 per day. 39. 5 p  4 x  2 px  60 First, we take the derivative of both sides of the equation with respect to t. d d 5 p  4 x  2 px    60  dt dt   dp dx dx dp   5  4  2 p  x 0 dt dt dt dt       Product Rule  dx  5  2  3   1.5 dt  4  2  5   11  1.5 14 16.5  14  1.18 Sales are changing at a rate of 1.18 sales per day.  Constant Multiple Rule Chain Rule dr  4.7 dt miles per year. Substituting these values into the derivative, we have: dA  2 792 4.7  dt  23,388.52899  23,389 Therefore, in 2013 the Arctic ice cap was changing at a rate of 23, 389 mi 2 per year. Another way of stating this is to say that the Arctic ice cap was shrinking at a rate of 23,389 mi2/yr. In 2013, the r was 792 miles, and dp dx dx dp  4  2 p   2x  0 dt dt dt dt dx Next, we solve for . dt dx dx dp dp  5  2 x  4  2p dt dt dt dt dx dp  4  2 p    5  2 x   dt dt dx  5  2 x  dp   dt  4  2 p  dt have: A   r2 To find the rate of change of the area of the Arctic ice cap with respect to time, we take the derivative of both sides of the equation with respect to t. d d A   r 2  dt dt dA d    r 2  dt dt dA dr      2r   dt dt   dA dr  2 r  dt dt 5 dp , we dt dx 16.5  dt 14 dR dp dx  x  p dt dt dt Substituting the appropriate values, we have: dR  16.5    31.5  5   1.39  14  dt  Substituting 3 for x¸ 5 for p, and 1.5 for R  xp 40. A   r2 d d A   r 2  dt dt dA dr  2 r  dt dt Substituting, we have: dA  2  25 1  50  157.0796 dt The area of the wound is decreasing at a rate of 157.08 mm2/day. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 428 41. b) Using the derivative found in part (a), and substituting the values for R and dR/dt, we have: dV 500  0.075    0.0002  0.48 dt  0.0156 The speed of the blood is changing at a rate of 0.0156 mm / sec 2 . hw 60 First, we substitute 165 for h, and then we take the derivative of both sides with respect to t. 165w 165 1 2  w S 60 60 d d  165 1   S   dt  60  w 2  dt   S dS 165 1  1 2 dw   w  dt 60 2 dt 165 1 dw  1  120 w 2 dt  p  dR   0 2R  4 Lv  dt  pR dR   2 Lv dt Substituting 70 for L, 400 for p and 0.003 for v , we have: 400 R dV dR   dt 2  70 0.003 dt 400 R dR  0.42 dt dR  952.38 R  dt b) Using the derivative in part (a), we substitute 0.00015 for dR/dt and 0.1 for R to get: dV  952.38  0.1   0.00015 dt  0.0143 The speed of the person’s blood will be increasing at a rate of 0.0143 mm/sec2.    p  dR  2R  4 Lv  dt  pR dR   2 Lv dt Substituting, we have: 500 R dV dR   dt 2 80 0.003 dt   500 R dR  0.48 dt   p R2  r2 4 Lv We assume that r, p, L and v are constants. a) Taking the derivative of both sides with respect to t, we have: dV d  p   R2  r2  dt dt  4 Lv  p  d 2 d 2  R  r  4 Lv  dt dt     165 dw  120 w dt Now, we will substitute 70 for w and dw 2 for . dt dS 165    2 dt 120 70  0.0256 Therefore, Kim’s surface area is changing at a rate of 0.0256 m2 /month. We could also say that Kim’s surface area is decreasing by 0.0256 m2/month.  42. V   p R2  r2 4 Lv We assume that r, p, L and v are constants. a) Taking the derivative of both sides with respect to t, we have: dV d  p   R2  r2  dt dt  4 Lv  p  d 2 d 2  R  r  4 Lv  dt dt  43. V  44. D2  x2  y2 After 1 hour, D 2  252  602 D 2  4225 D  65 dx dy  25 and  60 dt dt The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Exercise Set 2.8 429 Differentiating both sides of the distance equation on the previous page with respect to t, we have: d 2 d  2 D   x  y 2  dt dt dD dx dy 2D   2x   2 y  dt dt dt dx dy x  y dD dt dt  dt D Substituting the appropriate information, we have: dD  25   25   60   60   65 dt 65 One hour after the cars leave, the distance between the two cars is increasing at a rate of 65 mph. 45. Since the ladder forms a right triangle with the wall and the ground, we know that: x 2  y 2  262 x 2  y 2  676 dy . dt Differentiating both sides of the equation with respect to t, we have: d  2 d x  y 2    676  dt dt dx dy 2x  2 y 0 dt dt dy dx 2y Subtracting  2 x dt dt dy 2 x dx Dividing by 2 y   2 y dt dt dy  x dx   dt y dt The lower end of the wall is being pulled away from the wall at a rate of 5 feet per second; dx  5. therefore, dt When the lower end is 10 feet away from the wall, x  10 , we substitute and solve for y at the top of the next column. We are looking for 102  y 2  676 100  y 2  676 y 2  676  100 y 2  576 y   576 y  24 y  24 y must be positive We substitute 10 for x, 24 for y and 5 for dx dt into the derivative to get: dy  x dx   dt y dt  10  5   24 25 12  2 121  When the lower end of the ladder is 10 feet from the wall, the top of the ladder is moving down the wall at a rate of 2 121 feet per second. 46. First we draw a picture. D x 2x First, from the picture we know that: x2  2 x   D2 2 5x 2  D2 When x is 440 m wide, we have D 2  5  440 2 D 2  968,000 D  440 5 Differentiating both sides of 5x 2  D 2 with respect to t, we have: dx dD 10 x  2D dt dt dx D dD  dt 5 x dt The solution is continued on the next page. Copyright © 2016 Pearson Education, Inc. Chapter 2: Applications of Differentiation 430 dD  90 , x  440 , and dt D  440 5 we have: Now, when  48.   5 90 5  18 5 We are looking to find how fast the area is changing. A  2 x2 dA dx  4x dt dt Substituting, we have: dA  4  440 18 5 dt  dy  x  2  1 dx y 2 1 1 dy  y 2  y  1  dx x x2   31,680 5  70,838.63 The area is changing at a rate of 70,838.63 m2/yr. 49. 1 1  2 5 2 x y x 2  y 2  5 Differentiating both sides with respect to x, we have: d  2  d  2  d x  y   5 dx   dx  dx dy 2 x 3  2 y 3 0 dx dy 2 y 3  2 x 3 dx 4 3 r 3 Differentiating both sides with respect to t, we have: dV d  4 3  r   dt dt  3  4 d     r 3  3 dt 47. V  4  2 dr   3r 3  dt  dr  4 r 2  dt Next, substituting 0.7 for dr/dt and 7.5 for r, we have: dV 2  4  7.5  0.7  dt  4 56.25 0.7  dy x 3   3 dx y dy y3  3 dx x   157.5  494.8 The cantaloupe’s volume is changing approximately at the rate of 494.8 cm3/week.  1 d 12 d x  y 2  1 dx dx 1  1 2 1  1 2 dy x  y  0 dx 2 2 1  1 2 dy 1 1 y    x 2 dx 2 2 1  12 dy  2 x  1  12 dx y 2 440 5 dx  90 dt 5  440  x  y 1 50. y3  x 1 x 1 d  3  d  x  1 y  dx   dx  x  1  3y2 dy  x  11   x  11  dx  x  12 3y2 2 dy  dx  x  12 2 dy  dx 3 y 2  x  12 Copyright © 2016 Pearson Education, Inc. Exercise Set 2.8 51. 431 dy  dy  2 2 3  x  y  1    3  x  y  1     dx   dx  x2  1 x2  1 Differentiating both sides with respect to x, we have: d  2  d  x2  1 y    dx   dx  x 2  1  y2     5×4  5 y 4 3 x  y  3 x  y 2  x2  1 2 x   x2  1 2 x  dy 2y  2 dx x2  1 2y   2  54.  2 dy . dx d d  xy  x  2 y   dx  4 dx dy dy x  y 1  1  2 0 dx dx dy  x  2   1  y dx dy 1  y 1  y   2 x dx x2 d2y dy implicitly to find . Differentiate dx dx 2 d  dy  d  1  y   dx  dx  dx  2  x  2 d  32 d x  y 3  1  dx   dx  3 1 2 2 13 dy x  y 0 dx 2 3 2 13 dy 3 1 y  x 2 dx 3 2 3 12  x dy  2 2 1 3 dx y 3 1 dy 9 x 2 9 1 2 1 3   1  x y 4 y 3 4 dx 53. 2 xy  x  2 y  4 Differentiate implicitly to find x 2  y 3 1 d2y  x  y 3   x  y 3  x 5  y 5 dx 2 Differentiating both sides with respect to x, we have: d  d  x  y 3   x  y 3    x5  y 5   dx dx d d d d 3 3  x  y    x  y   x5  y5 dx dx dx dx 2 d 2 d 3 x  y   x  y  3 x  y x  y  dx dx dy 5×4  5 y 4 dx Continued at the top of the next column. dy dx 2 Simplification will yield: dy 5 x 4  6 x 2  6 y 2  dx 12 xy  5 y 4 2x dy  dx y x 2  1 2 3 dx  5×4  5 y 4 4 dy 5 x  3  x  y   3  x  y   dx 3  x  y 2  3  x  y 2  5 y 4 4x dy  dx 2 y x 2  1 2 52. 2 dy 2   2 5×4  3  x  y   3  x  y     3x  y  3  x  y   3  x  y   5 y 4  dy    dx 2 4x dy 2y  2 2 dx x 1  dx 3 x  y dy 2 x 3  2 x  2 x 3  2 x  2 dx x2  1  2 dy dy dx     2  x   dy    1  y  1 dx  2  x 2 2  x   1 y    1  y  2 x  2  x 2 1 y 1 y  2  x 2 2  2y  2  x 2 Copyright © 2016 Pearson Education, Inc. Substituting dy 1 y for dx 2 x 2 Chapter 2: Applications of Differentiation 432 55. y 2  xy  x 2  5 56. dy . dx Differentiate implicitly to find Differentiate implicitly to find Differentiate d2y dy implicitly to find . dx dx 2 d y  dx 2  2 y  x    dy   dy   2    y  2 x   2  1   dx  dx  2 y  x 2  d y  dx 2 57. dy dx 2 y  x  3 y  3 x      y  2x dy for 2y  x dx y  2x 2y  x  2 y  x 2 3 y 2y  x y  2x  3x  2y  x 2y  x  2 y  x 2 2 6 y  3xy  3xy  6 x  2 y  x 3 6 y 2  6 xy  6 x 2 2 y  x  2 6 y 2  xy  x 2 2 y  x  3 dy dx y2  x2 y3 x3  y3  8 dy . dx d  3 d x  y 3   8 dx  dx dy 3x 2  3 y 2 0 dx dy 3 y 2  3x 2 dx dy 3x 2  dx 3 y 2 dy x 2  dx y 2 The solution is continued on the next page. 3  Substituting for Differentiate implicitly to find 2 Substituting d2y  dx 2 y2 dy dx y 2 x2  y y  2 y dy  dy  2  y  4 x  2 x    2 y  x   2 y  dx dx  3 y  3 x y 1  x  x y x   y  2 y Simplifying the numerator we have: d 2 y   dy dy   2 y  4 y  x  2x  2  dx dx   dx 2 d2y dy implicitly to find dx dx 2 d2y d  x     dx 2 dx  y  d  dy  d  y  2 x   dx  dx  dx  2 y  x  2 dy . dx d  2 d x  y 2   5 dx  dx dy 0 2x  2 y dx dy x  dx y d  2 d y  xy  x 2   5 dx  dx dy  dy   x  y  1  2 x  0 2y dx  dx  dy 2 y  x   y  2 x  0 dx dy 2 y  x   y  2 x dx dy y  2 x  dx 2 y  x Differentiate x2  y2  5  Copyright © 2016 Pearson Education, Inc. Exercise Set 2.8 433 60. Using the calculator, we have: d2y dy implicitly to find . dx dx 2 d 2 y d  x2     dx 2 dx  y 2  dy y 2  2 x   x2  2 y dx  Differentiate y  2 2  x2  2 xy 2  2 x 2 y  2  y   4 y 2 xy 2   61. Using the calculator, we have: Substituting for dy dx 2 x4 y y4 Simplifying the derivative, we have: 2 xy 2 y 2 x 4   2 d y 1 y y  2 4 dx y 62. Using the calculator, we have: 2 xy 3  2 x 4 y  y4   58. 59. 2 xy 3  2 x 4 y5  3 2x y  x y 5 3 63. Using the calculator, we have:  Given an equation in x and y where y is a function of x but where it is difficult or impossible to express y in terms of x, implicit differentiation allows us to find the derivative of y with respect to x. 64. Using the calculator, we have: One dictionary defines “implicit” as “capable of being understood from something else though unexpressed” or “involved in the nature or essence of something though not revealed, expressed, or developed.” When a function y of x is defined implicitly it is written as an equation in x and y where y is not expressed in terms of x but where it is understood or implied that y is indeed a function of x. Copyright © 2016 Pearson Education, Inc.