Solution Manual for College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14th Edition

2 FUNCTIONS EXERCISE 2-1 2. 4. 6. 8. 10. The table specifies a function, since for each domain value there corresponds one and only one range value. 12. The table does not specify a function, since more than one range value corresponds to a given domain value. (Range values 1, 2 correspond to domain value 9.) 14. This is a function. 16. The graph specifies a function; each vertical line in the plane intersects the graph in at most one point. 18. The graph does not specify a function. There are vertical lines which intersect the graph in more than one point. For example, the y-axis intersects the graph in two points. 20. The graph does not specify a function. 22. y  4x  26. x  xy  1  0 is neither linear nor constant. 1 x is neither linear nor constant. 24. 28. 2 x  4 y  6  0 is linear. y  x 3  2x 1   1 simplifies to y  2 4 2 constant. Copyright © 2019 Pearson Education, Inc. 2-1 2-2 CHAPTER 2: FUNCTIONS 30. 32. 34. 36. 38. 3x 2 . Since the denominator is bigger than 1, we note that the values of f are between 0 and 3. x2  2 Furthermore, the function f has the property that f(–x) = f(x). So, adding points x = 3, x = 4, x = 5, we have: f(x) = x F(x) –5 –4 –3 –2 –1 0 1 2 3 4 5 2.78 2.67 2.45 2 1 0 1 2 2.45 2.67 2.78 The sketch is: 40. y = f(4) = 0 42. y = f(–2) = 3 44. f ( x )  4 at x  5. 46. f ( x )  0 at x  5, 0, 4. 48. Domain: all real numbers. 50. Domain: all real numbers except x = 2. 52. Domain: x  5 or [ 5,  ). 54. Given 6 x  7 y  21 . Solving for y we have: 7 y  21  6 x and y  6 7 x  3. This equation specifies a function. The domain is R, the set of real numbers. Copyright © 2019 Pearson Education, Inc. EXERCISE 2-1 56. Given x ( x  y )  4 . Solving for y we have: xy  x  4 and y  58. Given x  y  9. Solving for y we have: y  9  x 2 4x x This equation specifies a function. The domain is all real numbers except 0 2 2 2 2 2 . and y   9  x 2 . This equation does not define y as a function of x. For example, when x = 0, y =  3. 60. Given x  y  0. . Solving for y we have: y  3 3 1/6 x and y  x . This equation specifies a function. The domain is all nonnegative real numbers, i.e., x  0 . 62. f ( 3 x )  ( 3 x )  4  9 x  4 64. f ( x  1)  ( x  1)  4  x  2 x  1  4  x  2 x  3 66. f (x )  (x )  4  x  4 68. f ( 4 x )  x1/ 4 70. f ( 3)  f ( h )  ( 3)2  4  h 2  4  5  h 2  4  h 2  1 72. f ( 3  h )  ( 3  h ) 2  4  9  6 h  h 2  4  5  6 h  h 2 74. f ( 3  h )  f ( 3)  ( 3  h )2  4   ( 3)2  4   (9  6h  h 2  4)  (9  4)  6h  h 2     76. (A) f ( x  h)  3( x  h)  9  3 x  3h  9 (B) f ( x  h)  f ( x)   3x  3h  9   3 x  9  3h (C) f ( x  h)  f ( x) 3h   3 h h (A) f ( x  h)  3( x  h) 2  5( x  h)  8 78. 2 2 2 3 3 2 2 2 6   4  x 2 1/ 2 4 x 4  3( x 2  2 xh  h 2 )  5 x  5h  8  3 x 2  6 xh  3h 2  5 x  5h  8 (B)    f ( x  h)  f ( x)  3 x 2  6 xh  3h 2  5 x  5h  8  3 x 2  5 x  8   6 xh  3h 2  5h 80. (C) f ( x  h)  f ( x) 6 xh  3h 2  5h   6 x  3h  5 h h (A) f(x + h) = x2 + 2xh + h2 + 40x + 40h (B) f(x + h) – f(x) = 2xh + h2 + 40h (C) f ( x  h)  f ( x ) = 2x + h + 40 h Copyright © 2019 Pearson Education, Inc. 2-3 2-4 82. CHAPTER 2: FUNCTIONS Given A = l w = 81. 81 162 81  2l  . . Now P  2l  2 w  2l  2 Thus, w = l l l The domain is l > 0. 84. Given P = 2  + 2w = 160 or  + w = 80 and  = 80 – w. Now A =  w = (80 – w)w and A = 80w – w2 . The domain is 0 ≤ w ≤ 80. [Note: w ≤ 80 since w > 80 implies   0.] 86. (A) 88. (A) (B) p(11) = 1,340 dollars per computer p(18) = 920 dollars per computer R(x) = xp(x) = x(2,000 – 60x) thousands of dollars Domain: 1 ≤ x ≤ 25 (C) (B) Table 11 Revenue x(thousands) R(x)(thousands) 1 5 10 15 20 25 90. (A) $1,940 8,500 14,000 16,500 16,000 12,500 P(x) = R(x) – C(x) = x(2,000 – 60x) – (4,000 + 500x) thousand dollars = 1,500x – 60×2 – 4,000 Domain: 1 ≤ x ≤ 25 (C) (B) Table 13 Profit x (thousands) 1 5 10 15 20 25 P(x) (thousands) –$2,560 2,000 5,000 5,000 2,000 –4,000 Copyright © 2019 Pearson Education, Inc. EXERCISE 2-2 92. (A) Given 5v – 2s = 1.4. Solving for v, we have: v = 0.4s + 0.28. If s = 0.51, then v = 0.4(0.51) + 0.28 = 0.484 or 48.4%. (B) Solving the equation for s, we have: s = 2.5v – 0.7. If v = 0.51, then s = 2.5(0.51) – 0.7 = 0.575 or 57.5%. EXERCISE 2-2 Domain: [0, ) ; range: [1, ). 2. f ( x)  1  x 4. f ( x)  x 2  10 Domain: all real numbers; range: [10,  ). 6. f ( x )  5 x  3 Domain: all real numbers; range: all real numbers. 8. f ( x)  15  20 x 10. f ( x)  8  3 x Domain: all real numbers; range: all real numbers. 12. 14. 16. 18. 20. 22. Domain: all real numbers; range: ( ,15]. Copyright © 2019 Pearson Education, Inc. 2-5 2-6 CHAPTER 2: FUNCTIONS 24. 26. 28. The graph of h(x) = –|x – 5| is the graph of y = |x| reflected in the x axis and shifted 5 units to the right. 30. The graph of m(x) = (x + 3)2 + 4 is the graph of y = x2 shifted 3 units to the left and 4 units up. 32. The graph of g(x) = –6 + 3 x is the graph of y = 3 x 34. The graph of m(x) = –0.4×2 is the graph of y = x2 reflected in the x axis and vertically shifted 6 units down. contracted by a factor of 0.4. 36. The graph of the basic function y = |x| is shifted 3 units to the right and 2 units up. y = |x – 3| + 2 38. The graph of the basic function y = |x| is reflected in the x axis, shifted 2 units to the left and 3 units up. Equation: y = 3 – |x + 2| 40. The graph of the basic function 3 x is reflected in the x axis and shifted up 2 units. Equation: y = 2 – 3 x 42. The graph of the basic function y = x3 is reflected in the x axis, shifted to the right 3 units and up 1 unit. Equation: y = 1 – (x – 3)3 Copyright © 2019 Pearson Education, Inc. EXERCISE 2-2 44. g(x) = 3 x  3 + 2 46. g(x) = –|x – 1| 48. g(x) = 4 – (x + 2)2 50.  x  1 if x  1 g ( x)   2  2 x if x  1 52.  10  2 x if h( x )   40  0.5 x if 54.  4 x  20 if  h( x )   2 x  60 if  x  360 if  0  x  20 x  20 2-7 0  x  20 20  x  100 x  100 56. The graph of the basic function y = x is reflected in the x axis and vertically expanded by a factor of 2. Equation: y = –2x 58. The graph of the basic function y = |x| is vertically expanded by a factor of 4. Equation: y = 4|x| 60. The graph of the basic function y = x3 is vertically contracted by a factor of 0.25. Equation: y = 0.25×3. 62. Vertical shift, reflection in y axis. Reversing the order does not change the result. Consider a point (a, b) in the plane. A vertical shift of k units followed by a reflection in y axis moves (a, b) to (a, b + k) and then to (–a, b + k). In the reverse order, a reflection in y axis followed by a vertical shift of k units moves (a, b) to (–a, b) and then to (–a, b + k). The results are the same. Copyright © 2019 Pearson Education, Inc. 2-8 CHAPTER 2: FUNCTIONS 64. Vertical shift, vertical expansion. Reversing the order can change the result. For example, let (a, b) be a point in the plane. A vertical shift of k units followed by a vertical expansion of h (h > 1) moves (a, b) to (a, b + k) and then to (a, bh + kh). In the reverse order, a vertical expansion of h followed by a vertical shift of k units moves (a, b) to (a, bh) and then to (a, bh + k); (a, bh + kh) ≠ (a, bh + k). 66. Horizontal shift, vertical contraction. Reversing the order does not change the result. Consider a point (a, b) in the plane. A horizontal shift of k units followed by a vertical contraction of h (0 < h 700, the charge is 54 + .053(x – 700) = 16.9 + 0.053x. (B) Thus,  8.5  .065 x if 0  x  700 W ( x)   16.9  0.053 x if x  700 74. (A) Let x = taxable income. If 0 ≤ x ≤ 12,500, the tax due is $.02x. At x = 12,500, the tax due is $250. For 12,500 50,000, the tax due is 1,250 + .06(x – 50,000) = .06x – 1,250. Thus, 0.02 x if 0  x  12,500   T ( x)   0.04 x  250 if 12,500  x  50, 000 0.06 x  1, 250 if x  50, 000  Copyright © 2019 Pearson Education, Inc. EXERCISE 2-3 (B) 76. (A) The graph of the basic function y = x3 is vertically expanded by a factor of 463. 78. (C) T(32,000) = $1,030 T(64,000) = $2,590 (A) The graph of the basic function y = 3 x is reflected in the x axis and shifted up 10 units. (B) (B) EXERCISE 2-3 2. x 2  16 x (standard form) 2 x  16 x  64  64 (completing the square) 2 ( x  8)  64 4. x 2  12 x  8 (standard form) 2 ( x  12 x)  8 ( x 2  12 x  36)  8  36 (vertex form) (completing the square) ( x  6) 2  44 (vertex form) 6. 3 x 2  18 x  21 (standard form) 3( x 2  6 x )  21 3( x 2  6 x  9  9)  21 (completing the square) 3( x  3) 2  21  27 3( x  3) 2  6 (vertex form) 8. 5 x 2  15 x  11 2-9 (standard form) 5( x 2  3x )  11 5( x 2  3x  94  94 )  11 (completing the square) 5( x  23 ) 2  11  454 5( x  23 ) 2  14 (vertex form) Copyright © 2019 Pearson Education, Inc. 2-10 CHAPTER 2: FUNCTIONS 10. The graph of g(x) is the graph of y = x2 shifted right 1 unit and down 6 units; g ( x )  ( x  1) 2  6. 12. The graph of n(x) is the graph of y = x2 reflected in the x axis, then shifted right 4 units and up 7 units; n ( x )  ( x  4) 2  7. 14. (A) g (B) m (C) n (D) f 16. (A) x intercepts: –5, –1; y intercept: –5 (B) Vertex: (–3, 4) (C) Maximum: 4 (D) Range: y ≤ 4 or (–∞, 4] 18. (A) x intercepts: 1, 5; y intercept: 5 (B) Vertex: (3, –4) (C) Minimum: –4 (D) Range: y ≥ –4 or [–4, ∞) 20. g(x) = –(x + 2)2 + 3 (A) x intercepts: –(x + 2)2 + 3 = 0 (x + 2)2 = 3 x+2=± 3 x = –2 – 3 , –2 + 3 y intercept: –1 (B) 22. Vertex: (–2, 3) (C) Maximum: 3 (D) Range: y ≤ 3 or (–∞, 3] n(x) = (x – 4)2 – 3 (A) x intercepts: (x – 4)2 – 3 = 0 (x – 4)2 = 3 x–4=± 3 x=4– 3,4+ 3 y intercept: 13 (B) Vertex: (4, –3) (C) Minimum: –3 (D) Range: y ≥ –3 or [–3, ∞) 24. y = –(x – 4)2 + 2 26. y = [x – (–3)]2 + 1 or y = (x + 3)2 + 1 28. g(x) = x2 – 6x + 5 = x2 – 6x + 9 – 4 = (x – 3)2 – 4 (A) x intercepts: (x – 3)2 – 4 = 0 (x – 3)2 = 4 x – 3 = ±2 x = 1, 5 y intercept: 5 (B) Vertex: (3, –4) (C) Minimum: –4 (D) Range: y ≥ –4 or [–4, ∞) Copyright © 2019 Pearson Education, Inc. EXERCISE 2-3 30. 3 1   s(x) = –4×2 – 8x – 3 = –4  x 2  2 x   = –4  x 2  2 x  1   4 4   1   = –4  ( x  1) 2   = –4(x + 1)2 + 1 4  (A) –4(x + 1)2 + 1 = 0 4(x + 1)2 = 1 x intercepts: 1 4 1 x+1=± 2 3 1 x=– ,– 2 2 (x + 1)2 = y intercept: –3 (B) 32. Vertex: (–1, 1) (C) Maximum: 1 (D) Range: y ≤ 1 or (–∞, 1] v(x) = 0.5×2 + 4x + 10 = 0.5[x2 + 8x + 20] = 0.5[x2 + 8x + 16 + 4] = 0.5[(x + 4)2 + 4] = 0.5(x + 4)2 + 2 (A) x intercepts: none y intercept: 10 (B) Vertex: (–4, 2) (C) Minimum: 2 (D) Range: y ≥ 2 or [2, ∞) 34. g(x) = –0.6×2 + 3x + 4 (A) g(x) = –2: –0.6×2 + 3x + 4 = –2 0.6×2 – 3x – 6 = 0 (B) g(x) = 5: –0.6×2 + 3x + 4 = 5 –0.6×2 + 3x – 1 = 0 0.6×2 – 3x + 1 = 0 x = –1.53, 6.53 x = 0.36, 4.64 (C) g(x) = 8: –0.6×2 + 3x + 4 = 8 –0.6×2 + 3x – 4 = 0 0.6×2 – 3x + 4 = 0 No solution Copyright © 2019 Pearson Education, Inc. 2-11 2-12 CHAPTER 2: FUNCTIONS 36. Using a graphing utility with y = 100x – 7×2 – 10 and the calculus option with maximum command, we obtain 347.1429 as the maximum value. 38. m(x) = 0.20×2 – 1.6x – 1 = 0.20(x2 – 8x – 5) = 0.20[(x – 4)2 – 21] = 0.20(x – 4)2 – 4.2 (A) x intercepts: 0.20(x – 4)2 – 4.2 = 0 (x – 4)2 = 21 x – 4 = ± 21 x = 4 – 21 = –0.6, 4 + 21 = 8.6; y intercept: –1 (B) Vertex: (4, –4.2) 40. (C) Minimum: –4.2 (D) Range: y ≥ –4.2 or [ 4.2,  ) n(x) = –0.15×2 – 0.90x + 3.3 = –0.15(x2 + 6x – 22) = –0.15[(x + 3)2 – 31] = –0.15(x + 3)2 + 4.65 (A) x intercepts: –0.15(x + 3)2 + 4.65 = 0 (x + 3)2 = 31 x + 3 = ± 31 x = –3 – 31 = –8.6,–3 + 31 = 2.6; y intercept: 3.30 (B) Vertex: (–3, 4.65) 42. (C) Maximum: 4.65 (D) Range: x ≤ 4.65 or ( , 4.65] ( x  6)( x  3)  0 Therefore, either ( x  6)  0 and (x  3)  0 or ( x  6)  0 and ( x  3)  0. The first case is impossible. The second case implies 6  x  3. Solution set: (6,3). 44. x 2  7 x  12  ( x  3)( x  4)  0 Therefore, either ( x  3)  0 and (x  4)  0 or ( x  3)  0 and ( x  4)  0. The first case implies x  3 and the second case implies x  4. Solution set: (,  4]  [3, ). 46. 48. 50. –0.88 ≤ x ≤ 3.52 x = –1.27, 2.77 52. f is a quadratic function and max f(x) = f(–3) = –5 Axis: x = –3 Vertex: (–3, –5) Range: y ≤ –5 or (–∞, –5] x intercepts: None Copyright © 2019 Pearson Education, Inc. x 2.72 EXERCISE 2-3 54. (A) (B) f(x) = g(x): –0.7x(x – 7) = 0.5x + 3.5 –0.7×2 + 4.4x – 3.5 = 0 x= 4.4  (4.4) 2  4(0.7)(3.5) 1.4 = 0.93, 5.35 (C) f(x) > g(x) for 0.93 < x < 5.35 (D) f(x) < g(x) for 0 ≤ x < 0.93 or 5.35 g(x) for 1.08 < x < 6.35 (D) f(x) < g(x) for 0 ≤ x < 1.08 or 6.35 < x ≤ 9 58. The graph of a quadratic with no real zeros will not intersect the x–axis. 60. Such an equation will have b 2  4ac  0. 62. Such an equation will have k  0. a Copyright © 2019 Pearson Education, Inc. 2-13 2-14 64. CHAPTER 2: FUNCTIONS ax 2  bx  c  a ( x  h) 2  k  a ( x 2  2hx  h 2 )  k  ax 2  2ahx  ah 2  k Equating constant terms gives k  c  ah 2 . Since h is the vertex, we have h   b . Substituting then gives 2a 2  b2  b c 2  4a  4a  2 k  c  ah  c  a   66. 4ac  b 2 4a f(x) = –0.0117×2 + 0.32x + 17.9 (B) (A) x Mkt Share f ( x) 5 10 15 18.8 20.0 20.7 19.2 19.9 20.1 20 25 30 20.2 17.4 16.4 19.6 18.6 17 35 15.3 14.8 (C) For 2025, x = 45 and f(45) = –0.0117(45)2 + 0.32(45) + 17.9 = 8.6% For 2028, x = 48 and f(48) = –0.0117(48)2 + 0.32(48) + 17.9 = 6.3% (D) Market share rose from 18.8% in 1985 to a maximum of 20.7% in 1995 and then fell to 15.3% in 2010. 68. Verify 70. (A) (B) R(x) = 2,000x – 60×2 100   x = 60  x 2  3   100 2500 2500   = 60  x 2  x  3 9 9    50  2500  = 60  x     3  9   2 2 50  50,000  = 60  x   + 3  3  16.667 thousand computers (16,667 computers); 16,666.667 thousand dollars ($16,666,667) (C) 2000  60(50 / 3)  $1,000 Copyright © 2019 Pearson Education, Inc. EXERCISE 2-3 72. 2-15 (A)  50   50  p   = 2,000 – 60   = $1,000  3   3  (B) R(x) = C(x) x(2,000 – 60x) = 4,000 + 500x 2,000x – 60×2 = 4,000 + 500x 60×2 – 1,500x + 4,000 = 0 6×2 – 150x + 400 = 0 x = 3.035, 21.965 Break-even at 3.035 thousand (3,035) and 21.965 thousand (21,965) (C) 74. Loss: 1 ≤ x < 3.035 or 21.965 < x ≤ 25; Profit: 3.035 < x 0 for x < 5 Domain: x < 5 or (–∞, 5) (2-1) f(x) = 4×2 + 4x – 3 = 4(x2 + x)  3 1  = 4  x2  x    3  1 4  2 1  = 4  x    4 (vertex form) 2  Intercepts: y intercept: f(0) = 4(0)2 + 4(0)  3 = 3 x intercepts: f(x) = 0 2 1  4x   4 = 0 2  2 1  x 2 =1   1 = ±1 x+ 2 1 3 1 x= ±1= , 2 2 2  1  Vertex:   , 4  ; minimum: 4; range: y ≥ 4 or [4, ∞) 2   Copyright © 2019 Pearson Education, Inc. (2-3) 2-37 2-38 CHAPTER 2: FUNCTIONS 44. f(x) = ex – 1, g(x) = ln(x + 2) 2 f Points of intersection: (–1.54, –0.79), (0.69, 0.99) g -3 (2-5, 2-6) 3 -2 45. f(x) = 50 x2  1 : x 3 2 1 f ( x) 0 1 2 3 5 10 25 50 25 10 5 (2-1) 46. f(x) = 66 2  x2 x 3 f ( x ) 6 : 2 11 1 22 0 66 1 22 2 11 3 6 (2-1) For Problems 47–50, f(x) = 5x + 1. 47. f(f(0)) = f(5(0) + 1) = f(1) = 5(1) + 1 = 6 (2-1) 48. f(f(–1)) = f(5(–1) + 1) = f(–4) = 5(–4) + 1 = –19 (2-1) 49. f(2x – 1) = 5(2x – 1) + 1 = 10x – 4 (2-1) 50. f(4 – x) = 5(4 – x) + 1 = 20 – 5x + 1 = 21 – 5x (2-1) 51. f(x) = 3 – 2x (A) f(2) = 3 – 2(2) = 3 – 4 = –1 (B) f(2 + h) = 3 – 2(2 + h) = 3 – 4 – 2h = –1 – 2h (C) f(2 + h) – f(2) = –1 – 2h – (–1) = –2h f (2  h)  f (2) 2h =– = –2 (D) h h 52. f(x) = x2 – 3x + 1 (A) f(a) = a2 – 3a + 1 Copyright © 2019 Pearson Education, Inc. (2-1) CHAPTER 2 REVIEW 2-39 (B) f(a + h) = (a + h)2 – 3(a + h) + 1 = a2 + 2ah + h2 – 3a – 3h + 1 (C) f(a + h) – f(a) = a2 + 2ah + h2 – 3a – 3h + 1 – (a2 – 3a + 1) = 2ah + h2 – 3h (D) f ( a  h)  f ( a ) 2ah  h 2  3h h(2a  h  3) = = = 2a + h – 3 h h h (2-1) 53. The graph of m is the graph of y = |x| reflected in the x axis and shifted 4 units to the right. 54. The graph of g is the graph of y = x3 vertically contracted by a factor of 0.3 and shifted up 3 units. (2-2) 55. 56. (2-2) The graph of y = x2 is vertically expanded by a factor of 2, reflected in the x axis and shifted to the left 3 units. Equation: y = –2(x + 3)2 (2-2) Equation: f(x) = 2 x  3 – 1 (2-2) 57. f ( x)  n( x ) 5x  4 . Since degree n(x) = 1 1 = degree d(x), there is no horizontal asymptote. d ( x ) 100 x  1 (2-4) 60. 61. n( x ) x 2  100 x 2  100  2  . Since n( x )  x 2  100 has no real zeros and d ( x ) x  100 ( x  10)( x  10) d (10)  d ( 10)  0, x = 10 and x = 10 are the vertical asymptotes of the graph of f. f ( x)  f ( x)  (2-4) n( x ) x 2  3x x ( x  3) x  3    , x  0. x = 2 is a vertical asymptote of the graph of f. d ( x ) x 2  2 x x ( x  2) x  2 (2-4) 62. True; p(x) = p ( x) is a rational function for every polynomial p. 1 Copyright © 2019 Pearson Education, Inc. (2-4) 2-40 CHAPTER 2: FUNCTIONS 1 = x–1 is not a polynomial function. x 63. False; f(x) = 64. False; f(x) = 65. True: let f(x) = bx, (b > 0, b ≠ 1), then the positive x-axis is a horizontal asymptote if 0 < b 1. (2-5) 66. True: let f(x) = logbx (b > 0, b ≠ 1). If 0 < b 1, then the negative y-axis is a vertical asymptote. 67. True; f(x) = (2-6) x has vertical asymptote x = 1 and horizontal asymptote y = 1. x 1 68. 69. (2-2) (2-2) 70. 71. (2-4) y = –(x – 4)2 + 3 (2-2, 2-3) f(x) = –0.4×2 + 3.2x + 1.2 = –0.4(x2 – 8x + 16) + 7.6 = –0.4(x – 4)2 + 7.6 (A) y intercept: 1.2 x intercepts: –0.4(x – 4)2 + 7.6 = 0 (x – 4)2 = 19 x = 4 + 19 ≈ 8.4, 4 – 19 ≈ –0.4 (B) Vertex: (4.0, 7.6) (C) Maximum: 7.6 (D) Range: y ≤ 7.6 or (–∞, 7.6] (2-3) (A) y intercept: 1.2 x intercepts: –0.4, 8.4 72. (B) Vertex: (4.0, 7.6) (C) Maximum: 7.6 (D) Range: y ≤ 7.6 or (–∞, 7.6] 73. (2-3) log 10π = π log 10 = π 10log 2 = y is equivalent to log y = log 2 which implies y = 2 Similarly, ln e π = π ln e = π (Section 2-5, 4.b & g) and eln 2 = y implies ln y = ln 2 and y= 2 . (2-6) Copyright © 2019 Pearson Education, Inc. CHAPTER 2 REVIEW 74. 2-41 log x  log 3 = log 4  log (x + 4) 4 x log  log 3 x4 4 x  3 x4 x ( x  4)  12 x 2  4 x  12  0 ( x  6)( x  2)  0 x  6, 2 Since log(6) is not defined, 6 is not a solution. Therefore, the solution is x = 2. 75. ln(2x – 2) – ln(x – 1) = ln x  2x  2  = ln x ln   x  1  76.  2( x  1)  ln   = ln x  x 1  ln 2 = ln x x=2 77. ln(x + 3) – ln x = 2 ln 2  x 3 ln  = ln(22)  x  x3 =4 x x + 3 = 4x 3x = 3 x=1 (2-6) log 3×2 = 2 + log 9x ln y – ln c = –5t x log   = 2 3 y = e–5t c y = –5t c ln y = ce–5t (2-6) Let x be any positive real number and suppose log1x = y. Then 1y = x. But, 1y = 1, so x = 1, i.e., x = 1 for all positive real numbers x. This is clearly impossible. 80. (2-6) ln y = –5t + ln c 78. log 3×2 – log 9x = 2  3×2  log   =2  9x  x = 102 = 100 3 x = 300 (2-6) 79. (2-6) (2-6) The graph of y = 3 x is vertically expanded by a factor of 2, reflected in the x axis, shifted 1 unit to the left and 1 unit down. Equation: y = –2 3 x  1 – 1 (2-2) Copyright © 2019 Pearson Education, Inc. 2-42 81. CHAPTER 2: FUNCTIONS G(x) = 0.3×2 + 1.2x – 6.9 = 0.3(x2 + 4x + 4) – 8.1 = 0.3(x + 2)2 – 8.1 (A) y intercept: –6.9 x intercepts: 0.3(x + 2)2 – 8.1 = 0 (x + 2)2 = 27 x = –2 + (B) Vertex: (–2, –8.1) 27 ≈ 3.2, –2 – 27 ≈ –7.2 (C) Minimum: –8.1 (D) Range: y ≥ –8.1 or [–8.1, ∞) (2-3) (A) y intercept: –6.9 x intercept: –7.2, 3.2 82. (B) Vertex: (–2, –8.1) (C) Minimum: –8.1 (D) Range: y ≥ –8.1 or [–8.1, ∞) 83. (A) (2-3) S(x) = 3 if 0 ≤ x ≤ 20; S(x) = 3 + 0.057(x – 20) = 0.057x + 1.86 if 20 < x ≤ 200; S(200) = 13.26 S(x) = 13.26 + 0.0346(x – 200) = 0.0346x + 6.34 if 200 1000 if 3 0.057 x  1.86 if  Therefore, S(x) =  0.0346 x  6.34 if 0.0217 x  19.24 if 0  x  20 20  x  200 200  x  1000 x  1000 (B) (2-2) mt 84. r  A  P  1   ; P = 5,000, r = 0.0125, m = 4, t = 5.  m 4(5) 20  0.0125   0.0125  A  5000  1    5000  1    5321.95 4  4    After 5 years, the CD will be worth $5,321.95 Copyright © 2019 Pearson Education, Inc. (2-5) CHAPTER 2 REVIEW 2-43 mt 85. r  A  P  1   ; P = 5,000, r = 0.0105, m = 365, t = 5 m   365(5) 1825  0.0105   0.0105  A  5000  1   5000  1   365 365     After 5 years, the CD will be worth $5,269.51.  5269.51 (2-5) mt 86. r  A = P  1   , r = 0.0659, m = 12  m 12 t  0.0659  = 3P or (1.005492)12t = 3 Solve P  1  12   for t: 12t ln(1.005492) = ln 3 ln 3 ≈ 16.7 year. t= 12 ln(1.005492) 87. (2-5) A  Pe rt , r  0.0739 . Solve 2 P  Pe0.0739t for t. 2 P  Pe0.0739t e0.0739t  2 0.0739t  ln 2 t 88. ln 2  9.38 years. 0.0739 (2-5) p(x) = 50 – 1.25x Price-demand function C(x) = 160 + 10x Cost function R(x) = xp(x) = x(50 – 1.25x) Revenue function (A) (B) R = C x(50 – 1.25x) = 160 + 10x –1.25×2 + 50x = 160 + 10x –1.25×2 + 40x = 160 –1.25(x2 – 32x + 256) = 160 – 320 –1.25(x – 16)2 = –160 (x – 16)2 = 128 x = 16 + 128 ≈ 27.314, 16 – 128 ≈ 4.686 R = C at x = 4.686 thousand units (4,686 units) and x = 27.314 thousand units (27,314 units) R < C for 1 ≤ x < 4.686 or 27.314 C for 4.686 < x < 27.314 Copyright © 2019 Pearson Education, Inc. 2-44 CHAPTER 2: FUNCTIONS (C) Max Rev: 50x – 1.25×2 = R –1.25(x2 – 40x + 400) + 500 = R –1.25(x – 20)2 + 500 = R Vertex at (20, 500) Max. Rev. = 500 thousand ($500,000) occurs when output is 20 thousand (20,000 units) Wholesale price at this output: p(x) = 50 – 1.25x p(20) = 50 – 1.25(20) = $25 (2-3) 89. (A) P(x) = R(x) – C(x) = x(50 – 1.25x) – (160 + 10x) = –1.25×2 + 40x – 160 (B) P = 0 for x = 4.686 thousand units (4,686 units) and x = 27.314 thousand units (27,314 units) P < 0 for 1 ≤ x < 4.686 or 27.314 0 for 4.686 < x < 27.314 Maximum profit is 160 thousand dollars ($160,000), and this occurs at x = 16 thousand units(16,000 units). The wholesale price at this output is p(16) = 50 – 1.25(16) = $30, which is $5 greater than the $25 found in 88(C). (2-3) (C) 90. (A) The area enclosed by the storage areas is given by A = (2y)x Now, 3x + 4y = 840 3 so y = 210 – x 4 3   Thus A(x) = 2  210  x  x 4   3 = 420x – x2 2 (B) Clearly x and y must be nonnegative; the fact that y ≥ 0 implies 3 210 – x ≥ 0 4 3 and 210 ≥ x 4 840 ≥ 3x 280 ≥ x Thus, domain A: 0 ≤ x ≤ 280 (C) Copyright © 2019 Pearson Education, Inc. CHAPTER 2 REVIEW (D) Graph A(x) = 420x – 3 2 x and y = 25,000 together. 2 There are two values of x that will produce storage areas with a combined area of 25,000 square feet, one near x = 90 and the other near x = 190. 2-45 30,000 0 3 0 (E) (F) x = 86, x = 194 3 2 3 x = – (x2 – 280x) 2 2 Completing the square, we have 3 A(x) = – (x2 – 280x + 19,600 – 19,600) 2 3 = – [(x – 140)2 – 19,600] 2 3 = – (x – 140)2 + 29,400 2 A(x) = 420x – The dimensions that will produce the maximum combined area are: x = 140 ft, y = 105 ft. The maximum area is 29,400 sq. ft. 91. (A) Quadratic regression model, Table 1: (2-3) To estimate the demand at price level of $180, we solve the equation ax2 + bx + c = 180 for x. The result is x ≈ 2,833 sets. (B) Linear regression model, Table 2: (C) The condition is not stable; the price is likely to decrease since the supply at the price level of $180 exceeds the demand at this level. (D) Equilibrium price: $131.59 Equilibrium quantity: 3,587 cookware set. To estimate the supply at a price level of $180, we solve the equation ax + b = 180 for x. The result is x ≈ 4,836 sets. Copyright © 2019 Pearson Education, Inc. (2-3) 2-46 CHAPTER 2: FUNCTIONS (A) 92. Cubic Regression y  0.30395 x3  12.993 x 2  38.292 x  5, 604.8 y  0.30395(38)3  12.993(38) 2  38.292(38)  5,604.8  4,976 (B) The predicted crime index in 2025 is 4,976. 93. (A) N(0) = 1 1 N  = 2 2 (B) log 22t = log 109 = 9 2t log 2 = 9 9 ≈ 14.95 t= 2 log 2 N(1) = 4 = 22  3 N   = 8 = 23 2 N(2) = 16 = 24  Thus, we conclude that N(t) = 22t or N = 4t 94. We need to solve: 22t = 109 Thus, the mouse will die in 15 days. (2-6) 1 Given I = I0e–kd. When d = 73.6, I = I0. Thus, we have: 2 1 –k(73.6) I =I e 2 0 0 1 e–k(73.6) = 2 1 2 ln(0.5) ≈ 0.00942 k= 73.6 –k(73.6) = ln Thus, k ≈ 0.00942. To find the depth at which 1% of the surface light remains, we set I = 0.01I0 and solve 0.01I0 = I0e–0.00942d for d: 0.01 = e–0.00942d –0.00942d = ln 0.01 ln 0.01 ≈ 488.87 d= 0.00942 Thus, 1% of the surface light remains at approximately 489 feet. Copyright © 2019 Pearson Education, Inc. (2-6) CHAPTER 2 REVIEW 95. (A) Logarithmic regression model: Year 2023 corresponds to x = 83; y(83) ≈ 6,134,000 cows. (B) 96. ln (0) is not defined. (2-6) Using the continuous compounding model, we have: 2P0 = P0e0.03t 2 = e0.03t 0.03t = ln 2 ln 2 t= ≈ 23.1 0.03 Thus, the model predicts that the population will double in approximately 23.1 years. 97. (2-5) (A) The exponential regression model is y = 47.194(1.0768)x. To estimate for the year 2025, let x = 45  y = 47.19368975(1.076818175)45  1,319.140047. The estimated annual expenditure for Medicare by the U.S. government, rounded to the nearest billion, is approximately $1,319 billion. (This is $1.319 trillion.) (B) To find the year, solve 47.194(1.0768)x = 2,000. Note: Use 2,000 because expenditures are in billions of dollars, and 2 trillion is 2,000 billion. 47.194(1.0768)x = 2,000 2,000 1.0768x = 47.194  2,000  ln(1.0768x) = ln    47.194   2,000  xln1.0768 = ln    47.194   2,000  ln   47.194   50.6 years x=  ln1.0768 1,980 + 50.63 = 2,030.63 Annual expenditures exceed two trillion dollars in the year 2031. (2-5) Copyright © 2019 Pearson Education, Inc. 2-47