Solution Manual for Precalculus, 11th Edition

Chapter 2 Functions and Their Graphs 16. explicitly Section 2.1 1. 17. a.  1,3 2. 3  2   5  2   2 1 1  3  4   5  2   2  2   12  10   43 or 21 12 or 21.5 2 4. 3  2 x  5 2 x  2 x  1 Solution set:  x | x  1 or  , 1 5.  c. {(0, 1.411), (22, 1.305), (40, 1.229), (70, 1.121), (100, 1.031)} 18. a. Domain: {1.80, 1.78, 1.77} Range: {87.1, 86.9, 92.0, 84.1, 86.4} b. c. {(1.80, 87.1), (1.78, 86.9), (1.77, 83.0), (1.77, 84.1), (1.80, 86.4)} 52 19. Domain: {Elvis, Colleen, Kaleigh, Marissa} Range: {Jan. 8, Mar. 15, Sept. 17} Function 6. radicals 7. independent; dependent 20. Domain: {Bob, John, Chuck} Range: {Beth, Diane, Linda, Marcia} Not a function 8. a 9. c 10. False; g  0 11. False; every function is a relation, but not every relation is a function. For example, the relation x 2  y 2  1 is not a function. 12. verbally, numerically, graphically, algebraically 13. False; if the domain is not specified, we assume it is the largest set of real numbers for which the value of f is a real number. 14. False; if x is in the domain of a function f, we say that f is defined at x, or f(x) exists. 15. difference quotient b. 1 2 3. We must not allow the denominator to be 0. x  4  0  x  4 ; Domain:  x x  4 .  Domain: {0,22,40,70,100} Range: {1.031, 1.121, 1.229, 1.305, 1.411} 21. Domain: {20, 30, 40} Range: {200, 300, 350, 425} Not a function 22. Domain: {Less than 9th grade, 9th-12th grade, High School Graduate, Some College, College Graduate} Range: {$18,120, $23,251, $36,055, $45,810, $67,165} Function 23. Domain: {-3, 2, 4} Range: {6, 9, 10} Not a function 24. Domain: {–2, –1, 3, 4} Range: {3, 5, 7, 12} Function 65 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs 25. Domain: {1, 2, 3, 4} Range: {3} Function 34. Graph y  x . The graph passes the vertical line test. Thus, the equation represents a function. 26. Domain: {0, 1, 2, 3} Range: {–2, 3, 7} Function 27. Domain: {-4, 0, 3} Range: {1, 3, 5, 6} Not a function 35. x 2  8  y 2 Solve for y : y   8  x 2   28. Domain: {-4, -3, -2, -1} Range: {0, 1, 2, 3, 4} Not a function For x  0, y  2 2 . Thus, 0, 2 2 and 29. Domain: {–1, 0, 2, 4} Range: {-1, 3, 8} Function since a distinct x-value corresponds to two different y-values. 30. Domain: {–2, –1, 0, 1} Range: {3, 4, 16} Function 36. y   1  2 x For x  0, y  1 . Thus, (0, 1) and (0, –1) are on the graph. This is not a function, since a distinct xvalue corresponds to two different y-values.  0, 2 2  are on the graph. This is not a function, 31. Graph y  2 x 2  3x  4 . The graph passes the vertical line test. Thus, the equation represents a function. 37. x  y 2 Solve for y : y   x For x  1, y  1 . Thus, (1, 1) and (1, –1) are on the graph. This is not a function, since a distinct x-value corresponds to two different y-values. 38. x  y 2  1 3 32. Graph y  x . The graph passes the vertical line test. Thus, the equation represents a function. Solve for y : y   1  x For x  0, y  1 . Thus, (0, 1) and (0, –1) are on the graph. This is not a function, since a distinct xvalue corresponds to two different y-values. 39. Graph y  3 x . The graph passes the vertical line test. Thus, the equation represents a function. 1 . The graph passes the vertical line x test. Thus, the equation represents a function. 33. Graph y  66 Copyright © 2020 Pearson Education, Inc. Section 2.1: Functions 3x  1 . The graph passes the vertical x2 line test. Thus, the equation represents a function. f. 40. Graph y  f  x  1  3  x  1  2  x  1  4 2    3 x2  2 x  1  2x  2  4  3×2  6 x  3  2 x  2  4  3×2  8 x  1 41. g. f  2 x   3  2 x   2  2 x   4  12 x 2  4 x  4 h. f  x  h  3 x  h  2  x  h  4 2 2   3 x 2  2 xh  h 2  2 x  2h  4 Solve for y: y  2 x  3 or y  (2 x  3)  3x 2  6 xh  3h 2  2 x  2h  4 For x  1, y  5 or y  5 . Thus, 1,5  and 1, 5  are on the graph. This is not a function, 44. since a distinct x-value corresponds to two different y-values. 42. x 2  4 y 2  1 Solve for y: x 2  4 y 2  1 2 2 4 y  x 1 x2  1 y2  4 f  x    2 x2  x  1 a. f  0    2  0   0  1  1 b. f 1   2 1  1  1   2 c. f  1   2  1   1  1   4 d. f   x    2   x     x   1   2×2  x 1 e.  f  x    2 x 2  x  1  2 x 2  x  1 f. f  x  1   2  x  1   x  1  1 2 2 2 2      2 x2  2 x  1  x  1 1  x2  1 2 1 1  For x  2, y   . Thus,  2,  and 2 2  1    2,   are on the graph. This is not a 2  function, since a distinct x-value corresponds to two different y-values.   2 x2  4x  2  x   2 x 2  3x  2 g. f  2 x    2  2 x    2 x   1   8×2  2 x  1 h. f  x  h    2( x  h) 2   x  h   1 2     2 x 2  2 xh  h 2  x  h  1   2 x 2  4 xh  2h 2  x  h  1 f  x   3x 2  2 x  4 a. f  0  3 0  2  0  4   4 b. f 1  3 1  2 1  4  3  2  4  1 2 2 c. f  1  3  1  2  1  4  3  2  4  3 d. f   x   3   x   2   x   4  3x 2  2 x  4 e.  f  x    3x 2  2 x  4  3x 2  2 x  4 45. f  x  a. 2 2   2 y 43.  y  2x  3  b. c. x 2 x 1 0 0 0 1 1 1 1 f 1  2  1 1 2 1 1 1 f  1    2  1  1 1  1 2 f  0  0 2  x x d. f x  e. x  x   f  x    2  2  1 x x 1   67 Copyright © 2020 Pearson Education, Inc. x 1 2  2 x 1 Chapter 2: Functions and Their Graphs f. x 1 f  x  1   x  1  1 x 1  x2  2 x  1  1 x 1  g. h. 46. x2  2 x  2 2x 2x f 2x   2 2  2x 1 4x  1 f  x  h  f  x  xh  x  h 2  1  48. xh x 2  2 xh  h 2  1 x2  1 x4 02  1 1 1   04 4 4 a. f  0  b. 12  1 0 f 1   0 1 4 5 c. f  1   12  1 1  4 f x   x  4  x  4 e.  f  x     x  4   x  4 f. f  x  1  x  1  4 g. f 2x  2x  4  2 x  4 h. f  x  h  x  h  4  0 0 3 f  x   x2  x a. f  0   02  0  0  0 b. f 1  12  1  2 c. f  1   12   1  1  1  0  0 d. f x    x 2    x   e.  f  x    x  x   x  x f. f  x  1   x  12   x  1 d. x2  1 f x   x  4 x  4 e.  x2  1   x2  1  f  x       x  4  x4  f.  x  1  1  x  1  4 f  x  1  49. 4×2  1 2x  4 g. f 2x  h.  x  h 2  1 x 2  2 xh  h 2  1 f  x  h   xh4  x  h  4 2x  4 f  x  x  4 a. f 0  0  4  0  4  4 b. f 1  1  4  1  4  5 c. f  1   1  4  1  4  5 g. f 2x   2 x 2  2 x  h. f  x  h  4 x2  2 x  x  h 2   x  h   x 2  2 xh  h 2  x  h x2  2 x  1  1 x2  2 x  x5 x5  2  x 2  3x  2 2  2 x 2  1 2 x2  x  x2  2 x  1  x  1   x 2  1  47. d. 2 f  x  2x 1 3x  5 2  0  1 0 1 1  05 5 a. f  0  b. f 1  c. f  1  d. f x  e.  2x  1   2x 1  f  x      3x  5  3x  5 68 Copyright © 2020 Pearson Education, Inc. 3 0  5 2 1  1 3 1  5   2 1 3 3   35 2 2 2  1  1 3  1  5 2x 1 3 x  5   2  1 1 1   3  5  8 8   2x  1 2x 1  3 x  5 3x  5 Section 2.1: Functions f. f  x  1  g. f 2x  h. 50. 2  2x   1 3 2x  5   2x  2 1 2x  3  3x  3  5 3x  2 4x 1 6x  5 2  x  h 1 3 x  h  5 2 x  2h  1 3 x  3h  5  0  2  1 2 1 c. f  1  1  1  2  f   x  1  2  1 1  1  2  2 1 3  4 4 1 8  9 9 1  1  0 1 57. F ( x)  x3  x x3  x  0 f  x  1  1  1  x  1  2  2x  2 2  1 2  1 1  x  3 4  x  1 2 x  0, x 2  1 Domain:  x x  0 58. G ( x)  x  4x x  4x  0 2 x( x 2  4)  0 x  0, x 2  4 x  0, x  2 1  x  h  2 2 f ( x)  x 2  2 Domain:  x x is any real number x4 3 3 1 f ( x)  5 x  4 f ( x)  x2 x( x 2  1)  0 Domain:  x x is any real number 53. x 2  4  x  2 Domain:  x x   2, x  2   x  2 2 1 2x 2 x 4 x 4  0 1 f  x  h  1 x 2 2 f. h. 52. 56. h( x)  e. f 2x  1  x2  1 Domain:  x x is any real number x  16 x  16  0   1 1   f  x    1  1 2   x  2    x  2 2   g. x2 x 2  16  x  4 Domain:  x x   4, x  4 1 f 1  1  f ( x)  2  x  2 2 f 0  1  54. 55. g ( x)  1 b. d. 51. 3  x  1  5 f  x  h  f  x  1 a. 2  x  1  1 Domain:  x x   2, x  0, x  2 59. h( x )  3 x  12 3x  12  0 3x  12 x4 Domain:  x x  4 x 1 2 x2  8 Domain:  x x is any real number 69 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs Also 3t  21  0 60. G ( x)  1  x 1 x  0  x  1 x 1 Domain:  x x  1 61. p( x)  3t  21  0 3t  21 t7 Domain: t t  4, t  7 x 2x  3 1 z 3 z2 z 3 0 66. h( z )  2x  3 1  0 2x  3  1 z  3 Also z  2  0 z2 Domain:  z z  3, z  2 2 x  3  1 or 2 x  3  1 2 x  4 2 x  2 x  2 x  1 Domain:  x x   2, x  1 67. 62. p( x)  x 1 3x  1  4 Domain:  x x is any real number . 68. g (t )  t 2  3 t 2  7t 3x  1  4  0 Domain: t t is any real number . 3x  1  4 3 x  1  4 or 3 x  1  4 69. M (t )  5 3 x  3 3x  5 5 x  1 x 3  5 Domain:  x x  1, x   3  63. t 1 2 t  5t  14 t 2  5t  14  0 (t  2)(t  7)  0 t  2  0 or t  7  0 t  2 t 7 Domain: t t  2, x  7 x f ( x)  f ( x)  3 5 x  4 x4 x4  0 x4 Domain:  x x  4 70. N ( p )  5 2 p  98 2 2 p  98  0 2( p 2  49)  0 x  x2 64. q( x )  p 2 2( p  7)( p  7)  0 p  7  0 or p  7  0 x  2  0 x  2 p  7 Domain:  p p  7, x  7 x  2 Domain:  x x   2 65. P (t )  p7 71. t4 3t  21 f ( x)  3x  4 a. g ( x)  2 x  3 ( f  g )( x)  3 x  4  2 x  3  5 x  1 Domain:  x x is any real number . t4 0 t4 70 Copyright © 2020 Pearson Education, Inc. Section 2.1: Functions b. ( f  g )( x)  (3 x  4)  (2 x  3)  3x  4  2 x  3  x7 Domain:  x x is any real number . c. e. ( f  g )(3)  5(3)  1  15  1  14 f. ( f  g )(4)  4  3  1 g. ( f  g )(2)  6(2) 2  2  2  6(4)  2  2  24  2  2  20 ( f  g )( x)  (3x  4)(2 x  3)  6 x 2  9 x  8 x  12 h.  6 x 2  x  12 Domain:  x x is any real number . d. 73. f  3x  4   ( x)  2x  3 g b. ( f  g )(3)  5(3)  1  15  1  16 f. ( f  g )(4)  4  7  11 g. ( f  g )(2)  6(2) 2  2  12  24  2  12  10 h.  f  3(1)  4 3  4 7    7   (1)  2(1)  3 2  3 1 g a. c. Domain:  x x is any real number . c. d. g ( x)  3 x  2 ( f  g )( x)  2 x  1  3 x  2  5 x  1 ( f  g )( x)  (2 x  1)  (3 x  2)  2 x  1  3x  2  x  3 Domain:  x x is any real number .  6 x2  x  2 Domain:  x x is any real number .  f  2x 1   ( x)  g x2 3   3x  2  0 2 3 2  Domain:  x x   . 3  ( f  g )(3)  2(3) 2  3  1  2(9)  3  1  18  3  1  20 f. ( f  g )(4)   2(4) 2  4  1  2(16)  4  1  32  4  1  29 g. ( f  g )(2)  2(2)3  2(2) 2  2(8)  2(4)  16  8  8 h. 74.  f  x 1   ( x)  2 g 2x   Domain:  x x  0 . e. ( f  g )( x)  (2 x  1)(3 x  2) 3x  2  x  ( f  g )( x)  ( x  1)(2 x 2 )  2 x3  2 x 2 Domain:  x x is any real number .  6 x 2  4 x  3x  2 d. ( f  g )( x)  ( x  1)  (2 x 2 )   2 x2  x  1 Domain:  x x is any real number . b. ( f  g )( x)  x  1  2 x 2  2 x 2  x  1  x  1  2×2 e. f ( x)  2 x  1 g ( x)  2 x 2 Domain:  x x is any real number . 3 2  3 Domain:  x x   . 2  72. f ( x)  x  1 a. 2x  3  0  2x  3  x   f  2(1)  1 2  1 3   3   (1)  g 3(1)  2 32 1    f  11 0 0   0   (1)  2 g 2(1) 2 2(1)   f ( x)  2 x 2  3 71 Copyright © 2020 Pearson Education, Inc. g ( x)  4 x3  1 Chapter 2: Functions and Their Graphs a. ( f  g )( x)  2 x 2  3  4 x3  1 3 b.  4x  2x  4 Domain:  x x is any real number . b.    2 c.  3 d.   4 x3  2 x 2  2 Domain:  x x is any real number .   3x  5  x   f  2 x2  3   ( x)  3 4x 1 g 3 4x  1  0 e. 4 x3  1 f. g. h. ( f  g )(3)  4(3)3  2(3) 2  4 76. a. ( f  g )(4)   4(4)3  2(4) 2  2 b.  8(32)  12(8)  2(4)  3 c.  256  96  8  3  363 ( f  g )( x)  x  x ( f  g )( x)  x  x  x x Domain:  x x is any real number . 2 a. ( f  g )( x)  x  x Domain:  x x is any real number . ( f  g )(2)  8(2)5  12(2)3  2(2) 2  3  f  2(1)  3 2(1)  3 2  3 5    1   (1)  g 4(1)3  1 4(1)  1 4  1 5   f ( x)  x g ( x)  x Domain:  x x is any real number .  256  32  2  222 75.  f  1 1 1 1      (1)  g 3(1)  5 3  5  2 2   f ( x)  x  4(64)  2(16)  2 h. ( f  g )(2)  3(2) 2  5 2  6 2 5 2  2  108  18  4  130 g. ( f  g )(4)  4  3(4)  5  2  12  5  5  4(27)  2(9)  4 f. ( f  g )(3)  3  3(3)  5  3 95  3  4 3 1 1 2 x3    x  3    4 4 2 3   2 Domain:  x x   . 2   e.  f  x   ( x)  3x  5 g x  0 and 3 x  5  0 5 3  5 Domain:  x x  0 and x   . 3   ( f  g )( x)  2 x 2  3 4 x3  1  8 x5  12 x3  2 x 2  3 Domain:  x x is any real number . d. ( f  g )( x)  x (3x  5)  3x x  5 x Domain:  x x  0 . ( f  g )( x)  2 x  3  4 x  1  2 x 2  3  4 x3  1 c. ( f  g )( x)  x  (3 x  5)  x  3 x  5 Domain:  x x  0 . 2 d. g ( x)  3 x  5 ( f  g )( x)  x  3 x  5 Domain:  x x  0 . x  f    ( x)  x g Domain:  x x  0 . e. ( f  g )(3)  3  3  3  3  6 f. ( f  g )(4)  4  4  4  4  0 g. ( f  g )(2)  2 2  2  2  4 72 Copyright © 2020 Pearson Education, Inc. Section 2.1: Functions h. 77. 1 1  f   1   (1)  1 1 g f ( x)  1  a. 1 x g ( x)  ( f  g )( x)  1  c. ( f  g )( x)  1  1 x x4 Domain:  x 1  x  4 . 1 1 2   1 x x x d. 1 1  1 x x Domain:  x x  0 . c. Domain:  x 1  x  4 . e. 78. ( f  g )(4)  1 g. ( f  g )(2)  h.  f    (1)  1  1  2 g f ( x)  x  1 a. f. ( f  g )(4)  4  1  4  4  3  0  3 0  3 g. ( f  g )(2)  (2)2  5(2)  4  4  10  4  2 2 5  3 3 f. ( f  g )(3)  3  1  4  3  2  1  2 1 1 x 1  f  x  x  x 1  x  x 1 d.   ( x)  1 1 x 1 g x x Domain:  x x  0 . 1 ( f  g )(3)  1   f  x 1 x 1    ( x)  4 x 4 x g x  1  0 and 4  x  0 x  1 and  x  4 x4  11 1 1 ( f  g )( x)  1     2  xx x x Domain:  x x  0 . e. h. 1 1 1 1 3     2 (2) 2 2 4 4 79.  f  1 1 0   0 0   (1)  g 4 1 3    f ( x)  a. 2x  3 3x  2 ( f  g )( x)  x  1  4  x g ( x)  3x  2  0 3x  2  x  2 3 Domain: x x  2 . 3  x4 4x 3x  2 2x  3 4x  3x  2 3x  2 2x  3  4x 6x  3   3x  2 3x  2 ( f  g )( x)  g ( x)  4  x x  1  0 and 4  x  0 x  1 and  x  4 Domain:  x 1  x  4 . b.  x  1 4  x    x2  5x  4 x  1  0 and 4  x  0 x  1 and  x  4 Domain:  x x  0 . b. ( f  g )( x)  ( f  g )( x)  x  1  4  x x  1  0 and 4  x  0 x  1 and  x  4 x4 Domain:  x 1  x  4 . 73 Copyright © 2020 Pearson Education, Inc.  Chapter 2: Functions and Their Graphs b. 2x  3 4x  3x  2 3x  2 2x  3  4x  2x  3   3x  2 3x  2 ( f  g )( x)  a. x 1  0 2 3  2 Domain:  x x   . 3  3x  2  x  b. c. 2 3  2 Domain:  x x   . 3  2 2 x 1  x x x0 ( f  g )( x)  x  1  x 1  0 and x  1 Domain:  x x  1, and x  0 . 2x  3  f  3x  2 2 x  3  3x  2  2 x  3   ( x)  4 x  3x  2 4 x 4x g 3x  2 3x  2  0 and x  0 d.  f    ( x)  g x 1  0 x 1 x x 1  2 2 x and x  0 x  1 Domain:  x x  1, and x  0 . 3x  2 2 3  2  Domain:  x x  and x  0  . 3   e. ( f  g )(3)  3  1  2 2 2 8  4   2  3 3 3 3 f. ( f  g )(4)  4  1  2 1  5 4 2 ( f  g )(3)  6(3)  3 18  3 21   3 3(3)  2 9  2 7 g. ( f  g )(2)  f. ( f  g )(4)   2(4)  3  8  3  5 1    3(4)  2 12  2 10 2 h.  f  1 11 2    (1)  2 2 g g. ( f  g )(2)  e.  h. 80. and x  1 Domain:  x x  1, and x  0 . 2  2 x  3   4 x  8 x  12 x ( f  g )( x)      3 x  2   3 x  2  (3x  2) 2 3x  2  0 x 2 x x0 ( f  g )( x)  x  1  x 1  0 3x  2  x  d. and x  1 Domain:  x x  1, and x  0 . 3x  2  0 c. 2 x x0 ( f  g )( x)  x  1  8(2) 2  12(2)  3(2)  2 2 8(4)  24  6  2 2  32  24  4 2 81.  56 7  16 2 g ( x)  ( f  g )( x)  6  1 x  3 x  1  g ( x) 2 7 5  x  g ( x) 2 7 g ( x)  5  x 2 6 f  2(1)  3 2  3 5     (1)  4(1) 4 4 g f ( x)  x  1 f ( x)  3x  1 2 2 1 2 3   3 2 2 2 x 74 Copyright © 2020 Pearson Education, Inc. 1 x 2 Section 2.1: Functions 82. f ( x)  1 x f ( x  h)  f ( x ) h 3( x  h) 2  2  (3 x 2  2)  h 2 3 x  6 xh  3h 2  2  3 x 2  2  h 6 xh  3h 2  h  6 x  3h  f  x 1   ( x)  2 g x x   1  x x 2  x g ( x) 1 1 x2  x g ( x)  x   x 1 x x 1 x2  x 1 x( x  1) x  1    x x 1 x 1 x 1 83. 87. f ( x  h)  f ( x ) h ( x  h) 2  ( x  h)  4  ( x 2  x  4)  h 2 2 x  2 xh  h  x  h  4  x 2  x  4  h 2 xh  h 2  h  h  2x  h 1 f ( x)  4 x  3 f ( x  h)  f ( x) 4( x  h)  3  (4 x  3)  h h 4 x  4h  3  4 x  3  h 4h  4 h 84. f ( x)  3x  1 f ( x  h)  f ( x) 3( x  h)  1  (3x  1)  h h 3x  3h  1  3x  1  h 3h   3 h 85. f ( x)  x 2  4 f ( x  h)  f ( x ) h ( x  h) 2  4  ( x 2  4)  h 2 x  2 xh  h 2  4  x 2  4  h 2 2 xh  h  h  2x  h 86. f ( x)  x 2  x  4 88. f  x   3x 2  2 x  6 f  x  h  f  x h 3  x  h  2  2  x  h   6    3 x 2  2 x  6       h    3 x 2  2 xh  h 2  2 x  2h  6  3x 2  2 x  6 h 3x  6 xh  3h  2h  3 x 2 6 xh  3h 2  2h   h h  6 x  3h  2 2 f ( x)  3 x 2  2 75 Copyright © 2020 Pearson Education, Inc. 2 Chapter 2: Functions and Their Graphs 89. f ( x)  5 4x  3 91. 2 2 2 x  6 x  2hx  6h  2 x  6 x  2 xh  x  h  3 x  3  h    1 6h      x  h  3 x  3   h   92. f ( x)  2x x3 2( x  h) 2x  f ( x  h)  f ( x ) x  h  3 x  3  h h 2( x  h)( x  3)  2 x  x  3  h   x  h  3 x  3  h 5 5  f ( x  h)  f ( x) 4( x  h)  3 4 x  3  h h 5(4 x  3)  5  4( x  h)  3  4( x  h)  3 4 x  3  h  20 x  15  20 x  15  20h   1         4( x  h)  3 4 x  3   h    1  20h      4( ) 3 4 3 x h x       h   20   4( x  h)  3 4 x  3 90. f ( x)  1 x3 f ( x)  6   3 x  3 x h  5x x4 5( x  h) 5x  f ( x  h)  f ( x ) x  h  4 x  4  h h 5( x  h)( x  4)  5 x  x  4  h   x  h  4 x  4  h 1 1  f ( x  h)  f ( x ) x  h  3 x  3  h h x  3   x  3  h  x  h  3 x  3  h  x  3 x  3 h  1         x  h  3 x  3   h    1  h      x h x 3 3       h   1   x  h  3 x  3 5 x  20 x  5hx  20 h  5 x  20 x  5 xh 2  2  x  h  4 x  4 h    1 20h      x  h  4 x  4   h   76 Copyright © 2020 Pearson Education, Inc. 20   x h 4 x  4  Section 2.1: Functions 93. f  x  x2 95. f  x  h  f  x     94. h f ( x  h)  f ( x )  x  h   h h  2 2 h x  x 2  2 xh  h 2 x  x  h 2   x  h  2  x  2  2 h 1 2   h2 xh      h  x 2  x  h 2 1 xh2 x2  1  h  2 x  h     h  x 2  x  h 2 f ( x)  x  1 f  x  h  f  x  h x  h 1  x 1 h x  h 1  x 1 x  h 1  x 1   h x  h 1  x 1 x  h  1  ( x  1) h   h x  h 1  x 1 h x  h 1  x 1   x2  x  h    x  h  2  x  2  1 x2 x2   x  h  h h  2 xh2  x2 h 2 xh  x2 xh2 x2  h xh2 x2 xh2 x2  1 x2 1 h  f  x    96. f  x  2 x  h x  x  h 2 2   2x  h x2  x  h  1 x 1 f ( x  h)  f ( x )  x  h   1  h h 2  1 x 1  2  x2  1   x  h   1 1 2 ( x  1)( x  h   1) 2 2 x  h 1  x 1 2 2 1     2 h  x  1  x 2  2 xh  h 2  1 ( x  1)( x  h   1) 2 2  h 2 xh  h 2 1    h  ( x 2  1)( x  h 2  1) h  2 x  h  1   2  h  ( x  1)( x  h 2  1)   77 Copyright © 2020 Pearson Education, Inc. 2 x  h ( x  1)( x  h   1) 2 2   2x  h ( x  1)( x  h   1) 2 2 Chapter 2: Functions and Their Graphs 97. f ( x)  4  x 2 99. f  x  h  f  x 0  x2  2 x  8 0  ( x  4)( x  2) x  4  0 or x  2  0 x4 or x  2 h 4  ( x  h) 2  4  x 2  h 4  ( x  h) 2  4  x 2  h h     98. The solution set is:  2, 4 4  ( x  h) 2  4  x 2  4  ( x  h) 2  4  x 2 100. 4  ( x  h) 2  (4  x 2 )   4  ( x  h)  4  x  2 2 4  ( x 2  2 xh  h 2 )  (4  x 2 ) h  x  h  1  x  1 2 xh  h 2 h  x  h  1  x  1 2 x  h 4  ( x  h) 2  4  x 2 4  ( x  h) 2  4  x 2 101. 1 x2  xh2 x2  xh2 102. x2  xh2 h x2 xh2 x2  xh2 x  2  ( x  h  2)  h( x  2) x  h  2  ( x  h  2) x  2 x2 xh2  h( x  2) x  h  2  ( x  h  2) x  2 h  h( x  2) x  h  2  ( x  h  2) x  2 1  ( x  2) x  h  2  ( x  h  2) x  2 f ( x)  3x 2  Bx  4 and f (1)  12 : f (1)  3(1) 2  B (1)  4 12  3  B  4 B5   f ( x)  2 x3  Ax 2  4 x  5 and f (2)  5 f (2)  2(2)3  A(2) 2  4(2)  5 5  16  4 A  8  5 5  4 A  19 14  4 A 14 7 A  4 2 x2 f ( x  h)  f ( x )  h 1 1  xh2 x2  h h x2 xh2 7 5 3  x 16 6 4 7 3 5    x 16 4 6 5 7 12 x  6 16 16 5 5 x 6 16 5 6 3 x   16 5 8  3 The solution set is:   8 (2 x  h) f  x  11  x 2  2 x  3  103. 3x  8 and f (0)  2 2x  A 3(0)  8 f (0)  2(0)  A 8 2 A  2A  8 A  4 f ( x)  78 Copyright © 2020 Pearson Education, Inc. Section 2.1: Functions 104. 2x  B 1 and f (2)  3x  4 2 2(2)  B f (2)  3(2)  4 1 4B  2 10 5  4B b. f ( x)  H  x   15 : 15  20  4.9 x 2 5   4.9 x 2 x 2  1.0204 x  1.01 seconds H  x   10 : B  1 10  20  4.9 x 2 10   4.9 x 2 105. Let x represent the length of the rectangle. x Then, represents the width of the rectangle 2 since the length is twice the width. The function x x2 1 2 for the area is: A( x )  x    x 2 2 2 106. Let x represent the length of one of the two equal sides. The function for the area is: 1 1 A( x )   x  x  x 2 2 2 x 2  2.0408 x  1.43 seconds H  x  5 : 5  20  4.9 x 2 15   4.9 x 2 x 2  3.0612 x  1.75 seconds c. 107. Let x represent the number of hours worked. The function for the gross salary is: G ( x)  16 x 108. Let x represent the number of items sold. The function for the gross salary is: G ( x)  10 x  100 109. a. H 1  20  4.9 1 H  x  0 0  20  4.9 x 2  20   4.9 x 2 x 2  4.0816 x  2.02 seconds 110. a. H 1  20  13 1  20  13  7 meters 2 H 1.1  20  13 1.1  20  13 1.21 2 2  20  15.73  4.27 meters  20  4.9  15.1 meters H 1.1  20  4.9 1.1 2 H 1.2   20  13 1.2   20  13 1.44  2  20  4.9 1.21  20  18.72  1.28 meters  20  5.929  14.071 meters H 1.2   20  4.9 1.2  2 b. H  x   15 15  20  13 x 2  20  4.9 1.44  5  13 x 2  20  7.056  12.944 meters x 2  0.3846 x  0.62 seconds H  x   10 10  20  13 x 2 10  13 x 2 x 2  0.7692 x  0.88 seconds 79 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs H  x  5 2 c. 5  20  13x 2 2 8 5 8 5 2 2 A   4  1      3 3 3 3 9 3 3     15   13 x 2  2 x  1.1538 x  1.07 seconds c. L  x L 113. R  x      x   P  x P H  x  0 0  20  13x 2  20  13x 2 114. T  x   V  P  x   V  x   P  x  x 2  1.5385 115. H  x    P  I  x   P  x   I  x  x  1.24 seconds 116. N  x    I  T  x   I  x   T  x  x 36, 000 111. C  x   100   x 10 a. 117. a.  0.05 x 3  0.8 x 2  155 x  500 b. 450 36, 000 C  450   100   10 450  100  45  80 c. 600 36, 000  10 600  100  60  60 118. a. b. 400 36, 000  10 400  100  40  90 P is the dependent variable; a is the independent variable P (20)  0.027(20) 2  6.530(20)  363.804  244.004 In 2015 there are 244.004 million people who are 20 years of age or older. c. 2 P (0)  0.027(0) 2  6.530(0)  363.804  363.804 In 2015 there are 363.804 million people. 2 1 4 8 4 2 2 1 1 A   4  1      3 3 9 3 3 3 3 119. a. R (v)  2.2v; B (v)  0.05v 2  0.4 v  15 D (v )  R (v )  B (v ) 8 2  1.26 ft 2 9  2.2v  0.05v 2  0.4 v  15  0.05v 2  2.6v  15 2 b. When 15 hundred smartphones are sold, the profit is $1836.25.  10.8  130.6  363.804 C  400   100   2  $1836.25  $230 a. 3 P (15)  0.05(15)  0.8(15)  155(15)  500  168.75  180  2325  500 C  600   100  112. A  x   4 x 1  x   1.2 x 2  220 x  0.05 x 3  2 x 2  65 x  500  $220 d.    1.2 x 2  220 x  0.05 x 3  2 x 2  65 x  500  $225 c. P ( x)  R( x)  C ( x)  500 36, 000  10 500  100  50  72 C  500   100   $222 b. 8 5  1.99 ft 2 9 1 3 3 1 1 A   4  1    2  2 2 4 2 2 2 b.  3  1.73 ft 2 2 D (60)  0.05(60)  2.6(60)  15  180  156  15  321 80 Copyright © 2020 Pearson Education, Inc. Section 2.1: Functions c. 120. a. The car will need 321 feet to stop once the impediment is observed. c. F  a  b   5  a  b   2  5a  5b  2 h  x  2x Since 5a  5b  2  5a  2  5b  2  F  a   F  b  , h  a  b   2  a  b   2a  2b F  x   5 x  2 does not have the property.  h  a   h b  h  x   2 x has the property. b. F  x   5x  2 d. g  x   x2 G  x  1 x G a  b  g  a  b    a  b   a 2  2ab  b 2 2 Since a 2  2ab  b 2  a 2  b 2  g  a   g  b  , G  x  1 1 1    G  a   G b ab a b 1 does not have the property. x g ( x)  x 2 does not have the property. 121. f ( x  h)  f ( x ) 3 x  h  3 x   h h 1 1   x  h 3  x 3 h 1 1  2 1 1 2 2 1 1 2  x  h  3  x 3 ( x  h) 3  x 3 ( x  h) 3  x 3  h ( x  h) 3  x 3 ( x  h) 3  x 3 h  2 1 1 2 ( x  h) 3  x 3 ( x  h) 3  x 3   xhx h  ( x  h) 3  x 3 ( x  h) 3  x 3    1 2 2 1 1 1 1 2  h h ( x  h) 3  x 3 ( x  h) 3  x 3    2 1 1 2 2 ( x  h) 3  x 3 ( x  h) 3  x 3 122.  x4  2 f   3x  2  5x  4  x4  1. Solve 5x  4 x4 1 5x  4 x  4  5x  4 123. We need x2  1  0 . Since x 2  1  0 for all 7  3x  1 real numbers x, we need 7  3 x  1  0 . 7  3x  1  0 3x  1  7 7  3 x  1  7 x2 Therefore, f 1  3(2)  2  10 2 2  x  8 3 8 8   The domain of f is  x | 2  x   , or  2,  3 3   in interval notation. 81 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs 124. No. The domain of f is  x x is any real number , but the domain of g is 129. Let x represent the amount of the 7% fat hamburger added. % fat tot. amt. amt. of fat 20% 12  0.2012 7% x  0.07  x  15% 12  x  0.1512  x   x x  1 . 125. 3x  x3 ( your age)  0.2012   0.07 x    0.1512  x  126. Answers will vary. 2.4  0.07 x  1.8  0.15 x 127. ( x  12)2  y 2  16 x-intercept (y=0): ( x  12) 2  02  16 0.6  .08 x x  7.5 7.5 lbs. of the 7% fat hamburger must be added, producing 19.5 lbs. of the 15% fat hamburger. ( x  12) 2  16 ( x  12)  4 x 3  9 x  2 x 2  18 130. x  12  4 x 3  2 x 2  9 x  18  0 x  16, x  8 ( 16, 0), ( 8, 0) y-intercept (x=0): (0  12) 2  y 2  16 ( x 3  2 x 2 )  (9 x  18)  0 x 2 ( x  2)  9( x  2)  0 ( x 2  9)( x  2)  0 (12) 2  y 2  16 ( x  3)( x  3)( x  2)  0 ( x  3)  0 or ( x  3)  0 or ( x  2)  0 2 y  16  144  128 There are no real solutions so there are no yintercepts. Symmetry: ( x  12) 2  (  y ) 2  16 x  3, x  3, x  2 The solution set is:  3, 3, 2 131. ( x  12) 2  y 2  16 This shows x-axis symmetry. a  bx  ac  d a  ac  d  bx a(1  c)  d  bx 128. y  3 x 2  8 x a y  3( 1) 2  8 1 There is no solution so (-1,-5) is NOT a solution. y  3×2  8 x 132. y  3(4) 2  8 4 d  bx 1 c r  kd 2 0.4  k (0.6) 2 10 k 9 Thus, 10 r  (1.5) 2 9  2.5 kg  m 2  48  16  32 So (4,32) is a solution. y  3×2  8 x y  3(9) 2  8 9  243  24  219  171 So (9,171) is NOT a solution. 82 Copyright © 2020 Pearson Education, Inc. Section 2.2: The Graph of a Function 3. vertical 133. 3x  10 y  12 10 y  3x  12 3 6 y  x 10 5 f  5   3 5. f  x   ax 2  4 a  1  4  2  a  2 2 3 . The slope of a 10 10 perpendicular line would be  . 3 6. False. The graph must pass the vertical line test in order to be the graph of a function. (4 x 2  7)  3  (3 x  5)  8 x 7. False; e.g. y  The slope of the line is 134. 4. (4 x 2  7) 2 12 x 2  21  (24 x 2  40 x) (4 x 2  7) 2 2 2 12 x  21  24 x  40 x 2 (4 x  7) 2  8. True   9. c 2 12 x  40 x  21  (4 x 2  7) 2 12 x 2  40 x  21 10. a 11. a. Section 2.2 b. c. f (3) is positive since f (3)  3.7. d. f (4) is negative since f (4)   1. e. f ( x)  0 when x  3, x  6, and x  10. f. f ( x)  0 when  3  x  6, and 10  x  11. g. The domain of f is  x  6  x  11 or   6, 11 . x 2  4  0   16 2 x 2  16 h. x  4   4, 0  ,  4, 0  y-intercepts:  0   4 y 2  16 2 4 y 2  16 y2  4 y  2   0, 2  ,  0, 2  2. False; f (6)  0 since (6, 0) is on the graph. f (11)  1 since (11, 1) is on the graph. 2 1. x  4 y  16 x-intercepts: f (0)  3 since (0,3) is on the graph. f ( 6)  3 since ( 6, 3) is on the graph. (4 x 2  7) 2 135. Add the powers of x to obtain a degree of 7. 2 1 . x x  2y  2 2  2 y  2 0  2y 0 y   3, 4 . i. The x-intercepts are 3 , 6, and 10. j. The y-intercept is 3. k. The line y  l. The line x  5 intersects the graph 1 time. m. f ( x)  3 when x  0 and x  4. n. f ( x)   2 when x  5 and x  8. 12. a. The point  2, 0  is on the graph. The range of f is  y  3  y  4 or 1 intersects the graph 3 times. 2 f (0)  0 since (0, 0) is on the graph. f (6)  0 since ( 6, 0) is on the graph. 83 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs b. f (2)  2 since (2,  2) is on the graph. c. f (2)  1 since (2, 1) is on the graph. Symmetry about y-axis. 16. Function c. f (3) is negative since f (3)  1. d. f (1) is positive since f (1)  1.0. e. f ( x)  0 when x  0, x  4, and x  6. b. Intercepts:  , 0  ,  , 0  , (0, 0) f. f ( x)  0 when 0  x  4. c. g. The domain of f is  x  4  x  6 or a. Range:  y  1  y  1 h. The range of f is  y  2  y  3 or  2, 3 . The x-intercepts are 0, 4, and 6. j. The y-intercept is 0. a. b. Intercepts: (0, 0) l. The line x  1 intersects the graph 1 time. m. f ( x)  3 when x  5. n. f ( x)   2 when x  2. c. a. b. Intercepts: (2, 0)(2, 0)(0, 2)(0, 2) c. Domain:  x x  1 or x  1 ; a. Domain:  x 0  x  3 ; Range:  y y 0 and shifted down k units if k < 0,   x   2     1      exp  . Then   2 2     stretch/compress vertically by a factor of so the range of g is  k ,   . 97. The domain of g ( x)  1  to get   x   2     1 1    exp    f ( x)     2  2     1   multiply all the y -coordinates by   .   1. Stretch/compress horizontally by a factor or  (stretch if   1 ) 2. Shift horizontally  units (left if   0 and right if   0 ). 3.Stretch/compress vertically by a factor of (compress if   1 ) 95. The graph of y   x is the graph of y  x but reflected about the y-axis. Therefore, our region is simply rotated about the y-axis and does not change shape. Instead of the region being bounded on the right by x  4 , it is bounded on the left by x  4 . Thus, the area of 16 the second region would also be square 3 units. g ( x)  f ( x)  k is the graph of f shifted up k and right if   0 ) to get f ( x)  93. The graph of y  4 f ( x) is a vertical stretch of the graph of f by a factor of 4, while the graph of y  f (4 x) is a horizontal compression of the 1  x is  0,   . The graph of g ( x  k ) is the graph of g shifted k units to the right, so the domaine of g is  k ,   . 98. 3x  5 y  30 5 y  3x  30 3 y  x6 5 3 The slope is and the y-intercept is -6. 5 13.1 13.1   8.4214 . The 7 2 total distance is 26.2 mile. Thus the average 26.2  3.11 mph . speed is 8.4214 99. The total time run is 138 Copyright © 2020 Pearson Education, Inc. Section 2.6: Mathematical Models: Building Functions 100. g  1.75m g  1.75(9)  15.75 gal 101. y 2  x  4 x-intercepts: (0) 2  x  4 0 x4 x  4 105. y-intercepts: y2  0  4 y2  4 y  2 The intercepts are  4, 0 ,  0, 2 and  0, 2 . Test x-axis symmetry: Let y   y f ( x  h)  f ( x )  h 3( x  h) 2  2( x  h)  1  (3x 2  2 x  1)  h 3( x 2  2 xh  h 2 )  2 x  2h  1  3 x 2  2 x  1  h 3x 2  6 xh  3h 2  2 x  2h  1  3x 2  2 x  1 h 2 6 xh  h  2h h(6 x  h  2)   6x  h  2 h h z 3  216   z    6  3   y 2  x  4 106. y 2  x  4 same 3  ( z  6)( z 2  6 z  36) Test y-axis symmetry: Let x   x y 2   x  4 different Test origin symmetry: Let x   x and y   y .   y 2   x  4 Section 2.6 y 2   x  4 different 1. a. Therefore, the graph will have x-axis symmetry. d  x 2  y 2 . Since P is a point on the 102. The denominator must not be zero. x 2  5 x  14  0 ( x  7)( x  2)  0 x  7, x  2 graph of y  x 2  8 , we have: d ( x)  x 2  ( x 2  8) 2  x 4  15 x 2  64 So the domain is:  x | x  7, x  2 103. 16t 2  96t  200  88 16t 2  96t  112  0 16(t 2  6t  7)  0 16(t  7)(t  1)  0 t  7, t  1 Since t represents time the only answer that is reasonable is 7 seconds. 104. 3 The distance d from P to the origin is b. d (0)  04  15(0) 2  64  64  8 c. d (1)  (1) 4  15(1) 2  64  1  15  64  50  5 2  7.07  d.    16 x5 y 6 z  3 8  2 x3 x 2 y 6 z  2 xy 2 3 2 x 2 z e. d is smallest when x  2.74 or when x  2.74 . 139 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs 2. a. The distance d from P to (0, –1) is 2 4. a. 2 The distance d from P to the origin is d  x 2  y 2 . Since P is a point on the d  x  ( y  1) . Since P is a point on the graph of y  x 2  8 , we have: graph of y  d ( x)  x 2  ( x 2  8  1) 2 2  2  x  x 7 4 2 2 b. d (0)  04  13(0) 2  49  49  7 c. d (1)  (1)4  13(1) 2  49  37  6.08 d. 2 1 1 d ( x)  x     x 2  2 x x     x  13x  49 2 1 , we have: x b. 8 10 5 –5 0 c. –4 d is smallest when x  1 or x  1 . 4 0 e. d is smallest when x  2.55 or when x  2.55 . d. d (1)  3. a. the graph of y  x , we have: d ( x)  ( x  1) 2   x   x  x 1 2 2 where x  0 . b. 1  2 5. By definition, a triangle has area 1 A  b h, b  base, h  height. From the figure, 2 we know that b  x and h  y. Expressing the area of the triangle as a function of x , we have: 1 1 1 A( x )  xy  x x3  x 4 . 2 2 2   2 0 2 0 d is smallest when x  12 . 6. By definition, a triangle has area 1 A  b h, b=base, h  height. Because one 2 vertex of the triangle is at the origin and the other is on the x-axis, we know that b  x and h  y. Expressing the area of the triangle as a function of x , we have: 1 1 9 1 A( x )  xy  x 9  x 2  x  x3 . 2 2 2 2  7. a. d.  12  1 The distance d from P to the point (1, 0) is d  ( x  1) 2  y 2 . Since P is a point on c. 12  1  2; d (1)  1 12 1 3 d ( x)   1  2 2 2   A( x )  xy  x 16  x 2  b. Domain:  x 0  x  4 140 Copyright © 2020 Pearson Education, Inc. Section 2.6: Mathematical Models: Building Functions c. The area is largest when x  2.31 . e. The largest area is A(1.41)  2 1.41 4  1.412  4 square 30 units. The largest perimeter is p (1.79)  4 1.79   2 4  1.792  8.94 4 0 units. 0 9. a. In Quadrant I, x 2  y 2  4  y  4  x 2 A( x)  (2 x)(2 y )  4 x 4  x 2 d. The largest area is  b. p ( x)  2(2 x)  2(2 y )  4 x  4 4  x 2 c. Graphing the area equation: 10  A(2.31)  2.31 16  2.312  24.63 square units. 8. a. A( x)  2 xy  2 x 4  x 2 0 b. p( x)  2(2 x)  2( y )  4 x  2 4  x 2 c. Graphing the area equation: 2 0 4 The area is largest when x  1.41 . d. Graphing the perimeter equation: 0 12 2 0 0 2 0 The area is largest when x  1.41 . d. Graphing the perimeter equation: 10 The perimeter is largest when x  1.41 . 10. a. 0 2 b. 0 11. a. A(r )  (2r )(2r )  4r 2 p (r )  4(2r )  8r C  circumference, A  total area, r  radius, x  side of square C  2r  10  4 x  r  52 x The perimeter is largest when x  1.79 . 141 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs Total Area  area square + area circle  x 2   r 2  A( x )  x   52 x 2  2 c. 8 25  20 x  4 x 2 x   2 b. Since the lengths must be positive, we have: 10  4 x  0 and x  0  4 x  10 and x  0 x  2.5 and x  0 Domain:  x 0  x  2.5 c. The area is smallest when x  2.08 meters. 0 3.33 0 The total area is smallest when x  1.40 meters. 8 13. a. 0 Since the wire of length x is bent into a circle, the circumference is x . Therefore, C ( x)  x . b. Since C  x  2 r , r  2.5 0 x . 2 2 x2  x  . A( x )   r 2      4  2  14. a. 12. a. C  circumference, A  total area, r  radius, x  side of equilateral triangle 10  3x C  2r  10  3x  r  2 The height of the equilateral triangle is Total Area  area triangle  area circle 3 2  10  3 x  x    4  2  b. Since P  x  4s, s  1 x , we have 4 2 3 x. 2 1  3   x  x    r2 2  2  A( x)  Since the wire of length x is bent into a square, the perimeter is x . Therefore, p( x)  x . 2 3 2 100  60 x  9 x 2  x  4 4 1 1  A( x )  s 2   x   x 2 . 16 4  15. a. A  area, r  radius; diameter  2r A(r )  (2r )(r )  2r 2 b. p  perimeter p(r )  2(2r )  2r  6r 16. C  circumference, r  radius; x  length of a side of the triangle b. Since the lengths must be positive, we have: 10  3x  0 and x  0  3x  10 and x  0 10 x and x  0 3  10  Domain:  x 0  x   3  Since ABC is equilateral, EM  142 Copyright © 2020 Pearson Education, Inc. 3x . 2 Section 2.6: Mathematical Models: Building Functions d 2  3  40t 3x 3x  OE  r 2 2 Therefore, OM  2   x   3x r In OAM , r 2      2  2  2 d1  2  30t d 2 x 3  x 2  3 rx  r 2 4 4 3 rx  x 2 r2  b. The distance is smallest at t  0.07 hours. x 3 Therefore, the circumference of the circle is  x  2 3 C ( x )  2 r  2  x  3  3 r 20. r  radius of cylinder, h  height of cylinder, V  volume of cylinder 17. Area of the equilateral triangle 1 3 3 2 A  x x x 2 2 4 2 h2 h2 h r 2     R2  r 2   R2  r 2  R2  4 4 2 2 V  r h x2 . 3 Area inside the circle, but outside the triangle: 3 2 A( x )   r 2  x 4 3 2  3 2 x2   x     x 3 4 3 4  From problem 16, we have r 2    h2  h2  V (h)    R 2   h  h  R 2   4  4    21. r  radius of cylinder, h  height of cylinder, V  volume of cylinder H H h  R r Hr  R  H  h  By similar triangles: 18. d 2  d12  d 2 2 d 2   30t    40t  2 Hr  RH  Rh 2 Rh  RH  Hr d  t   900 t  1600 t  2500 t  50 t 2 2 2 RH  Hr H  R  r   R R  H  R  r    H  R  r  r2 V (r )   r 2 h   r 2   R R   h d2 =40t d1=30t d 22. a. 19. a. d 2  d12  d 2 2 d 2   2  30t    3  40t  2 d t   2  2  30t 2   3  40t 2 The total cost of installing the cable along the road is 500x . If cable is installed x miles along the road, there are 5  x miles between the road to the house and where the cable ends along the road.  4  120t  900t 2  9  240t  1600t 2  2500t 2  360t  13 143 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs House 23. a. time on land is given by d 2 Town P d  (5  x) 2  22 T ( x)  C ( x)  500 x  700 x 2  10 x  29 Domain:  x 0  x  5  C (1)  500 1  700 1  10 1  29 x 12  x d1 12  x x2  4    5 3 5 3 b. Domain:  x 0  x  12  2  500  700 20  $3630.50 c. C (3)  500  3  700 32  10  3  29  T (4)    1500  700 8  $3479.90 d. T (8)    e. 12–x d1  x 2  22  x 2  4 The total time for the trip is:  25  10 x  x 2  4  x 2  10 x  29 The total cost of installing the cable is: d. 12  x . 5 d1 2 x 5x c. d1 . The 3 Island Box b. The time on the boat is given by 3000 24. a.  Using MINIMUM, the graph indicates that x  2.96 miles results in the least cost. 12  4 42  4  5 3 8 20   3.09 hours 5 3 12  8 82  4  5 3 4 68   3.55 hours 5 3 Let A  amount of material , x  length of the base , h  height , and V  volume . 10 V  x 2 h  10  h  2 x Total Area A   Area base    4   Area side   x 2  4 xh  10   x2  4 x  2  x  40  x2  x 40 2 A x  x  x 40  1  40  41 ft 2 1 b. A 1  12  c. A  2   22  144 Copyright © 2020 Pearson Education, Inc. 40  4  20  24 ft 2 2 Section 2.6: Mathematical Models: Building Functions d. y1  x 2  26. Consider the diagrams shown below. 40 x 100 10 0 0     The amount of material is least when x  2.71 ft. e. The largest area is A  2.71  2.712  25. 40  22.1 ft 2 2.71 a. length = 24  2x ; width = 24  2x ; height = x V ( x)  x(24  2 x)(24  2 x)  x(24  2 x) 2 b. V (3)  3(24  2(3)) 2  3(18) 2 There is a pair of similar triangles in the diagram. This allows us to write r 4 r 1 1    r h h 16 h 4 4 Substituting into the volume formula for the conical portion of water gives  3(324)  972 in 3 . c. V (10)  10(24  2(10))2  10(4) 2 2 1 1 1   3 V  h   r 2h    h  h  h . 3 3 4  48  10(16)  160 in 3 . d. y1  x(24  2 x )2 27. a. 1100 0 0 12 Use MAXIMUM.  The total cost is the sum of the shipment cost, storage cost, and product cost. Since each shipment will contain x units, there are 600/x shipments per year, each costing $15.  600   9000  So the shipment cost is 15   = .  x   x  The storage cost for the year is given as 1.60 x. The product costs is 600(4.85)  2910. So, the total cost is C ( x)     The volume is largest when x  4 inches. e. The largest volume is V (4)  4(24  2(4)) 2  1024 in 3 145 Copyright © 2020 Pearson Education, Inc. 9000  1.60 x  2910. x Chapter 2: Functions and Their Graphs b. 31. 10 14  4 x 10 x  4(14) 10 x  56 x  5.6 32. x  u  1 u 1 u 1 y  u 11 u The retailer should order 75 drives per order for a minimum yearly cost of $3150. 28. 33. x5 2 1 x 3  3x 3  2 x  3  5  2 2x  3  3  2 x  3  3 or 2 x  3  3 2 x  0 or 2x  6 x  0 or 2 3x 3 3x 3 x  5  3x 2 3x 3 4x  5 2 3x  2  4 Convert this to miles-per-hour. 5 5 1 5 sec  min  hr  hr. 60 3600 720 66 66 ft  mi 5280 66 distance 5280  1  9 mph time 720 Since the truck is traveling 55 mph, the Fusion must travel 55 + 9 = 64 mph. y2  y1 6  ( 2) 8    4 1 3 x2  x1 2 3x 34.  3x  2  4 x3 29. In order for the 16-foot long Ford Fusion to pass the 50-foot truck, the Ford Fusion must travel the length of the truck and the length of itself in the time frame of 5 seconds. Thus the Fusion must travel an additional 66 feet in 5 seconds. 30. m   2 3x 3 The solution set is  0,3 . speed= x5 No solution since a square root cannot be negative. 35. Since the graph is symmetric is symmetric about the origin then (3, -2) is symmetric to (-3, 2). 36. v 2.6t d2 vd 2  2.6t E P E P 2  vd 2  E  2.6t   P   v2 d 4 E  2 P 6.76t 2 4 Pv d E 6.76t 2 6.76t 2 E P 2 4 v d 146 Copyright © 2020 Pearson Education, Inc. Chapter 2 Review Exercises 3x 2  7 x  4 x  2 37. 2 3x  11x  2  0 b 2  4ac  (11) 2  4(3)(2)  121  24  97 d.  3x  3 x  f ( x)    2   x  1  x2  1 e. f ( x  2)  3( x  2) ( x  2) 2  1 3 x  2 3x  6  2 x  4x  4 1 x  4x  3  f. Chapter 2 Review Exercises 1. a. Domain {8, 16, 20, 24} 7. Range {$6.30, $12.32, $13.99} b. {(8,$6.30), (16,$13.99), (20,$12.32), (24,$13.99)} c. d. f (2 x)  2 3(2 x ) 6x  2 2 (2 x)  1 4 x  1 f ( x)  x 2  4 a. f (2)  22  4  4  4  0  0 b. f (2)  c. f ( x)  ( x) 2  4  x 2  4 d.  f ( x)   x 2  4 e. f ( x  2)  ( x  2) 2  4  2 2  4  44  0  0  x2  4 x  4  4  x2  4 x f. f (2 x)  (2 x) 2  4  4 x 2  4    4 x2  1  2 x2  1 2. This relation represents a function. Domain = {–1, 2, 4}; Range = {0, 3}. 8. 3. Domain {2,4}; Range {-1,1,2} Not a function x2  4 x2 22  4 4  4 0   0 4 4 22 a. f (2)  b. f (2)   2 2  4 4  4 0   0 4 4  2 2 c. f ( x)  ( x) 2  4 x 2  4  x2 ( x) 2 a. 3(2) 6 6 f (2)    2 2 4 1 3  (2)  1 d.  x2  4  4  x2 x2  4  f ( x)    2     x2 x2  x  b. 3(2) 6 6 f (2)     2 (2) 2  1 4  1 3 e. f ( x  2)  c. 3( x) 3 x f ( x)   2 2 ( x)  1 x  1 4. not a function; domain [-1, 3]; range [-2, 2] 5. function; domain: all real numbers; range  3,   6. f ( x)  f ( x)  3x x 1 2  147 Copyright © 2020 Pearson Education, Inc. ( x  2) 2  4 x 2  4 x  4  4  ( x  2) 2 ( x  2) 2 x2  4 x x  x  4  ( x  2) 2 ( x  2) 2 Chapter 2: Functions and Their Graphs f. f (2 x)   9. (2 x) 2  4 4 x 2  4  (2 x) 2 4 x2  14.   x 1 4 x2  1 2 4 x2 x2 x  8 Domain:  x x  8 x x 9 The denominator cannot be zero: x2  9  0 f ( x)  2 15. g ( x)  3 x  1  2  x  3x  1  2 x  3 Domain:  x x is any real number x  3 or 3 Domain:  x x  3, x  3 ( f  g )( x)  f  x   g ( x)  2  x   3 x  1 f ( x)  2  x The radicand must be non-negative: 2 x  0  2  x  3x  1  4 x  1 Domain:  x x is any real number x2 Domain:  x x  2 or  , 2 ( f  g )( x)  f ( x)  g  x    2  x  3 x  1 x 11. g ( x)  x The denominator cannot be zero: x0  6 x  2  3x 2  x  3x 2  5 x  2 Domain:  x x is any real number Domain:  x x  0 f  x 2  x  f   g  ( x)  g x  3x  1     3x  1  0 x 12. f ( x)  2 x  2x  3 The denominator cannot be zero: x2  2 x  3  0 1 3  1 Domain:  x x    3  3 x  1  x    x  3 x  1  0 x  3 or 1 Domain: x x  3, x  1 13. f ( x)  2  x ( f  g )( x)  f  x   g ( x) ( x  3)( x  3)  0 10. x x8 The radicand must be non-negative and not zero: x8 0 f ( x)  16. f ( x)  3x 2  x  1 g ( x)  3x ( f  g )( x)  f  x   g ( x) x 1 x2  4 The denominator cannot be zero: x2  4  0 f ( x)   3×2  x  1  3x  3×2  4 x  1 Domain:  x x is any real number  x  2 x  2  0 ( f  g )( x)  f  x   g ( x) x  2 or 2 Also, the radicand must be non-negative: x 1  0  3x 2  x  1  3x  3x 2  2 x  1 Domain:  x x is any real number x  1 Domain:  1, 2   2,   148 Copyright © 2020 Pearson Education, Inc. Chapter 2 Review Exercises ( f  g )( x)  f ( x)  g  x   18.  f ( x)  2 x 2  x  1 f  x  h  f  x h  3x 2  x  1  3x   9 x3  3x 2  3 x Domain:  x x is any real number 2  f  x  3x 2  x  1  f   g  ( x)  g x  3x     3x  0  x  0  x 1 1 g ( x)  x 1 x    f g x f x g x ( )( )   ( ) f ( x)    x  1 1 x  x  1  1 x  1   x 1 x x  x  1 2  h 2 2 x  2 xh  h 2   x  h  1  2x  x 1 19. a. Range: 2 x  x  x 1 x  2x 1  x  x  1 x  x  1  y  3  y  3  ;  3, 3 b. Intercept:  0, 0  c. f  2   1 ( f  g )( x)  f  x   g ( x) d. f  x   3 when x = –4 e. f ( x)  0 when 0  x  3  x  1 1 x  x  1  1 x  1   x 1 x x  x  1 2 2 Domain:  x  4  x  3  ;  4, 3 Domain:  x x  0, x  1   h 2 x 2  4 xh  2h 2  x  h  1  2 x 2  x  1  h 4 xh  2h 2  h h  4 x  2h  1   h h  4 x  2h  1 Domain:  x x  0 17.  2  x  h    x  h   1  2 x 2  x  1  x | 0  x  3 2 x  x  x 1 x 1  x  x  1 x  x  1 f. To graph y  f  x  3 , shift the graph of f horizontally 3 units to the right. Domain:  x x  0, x  1 x 1  x 1 1  ( f  g )( x)  f ( x)  g  x     x   x  1 x  x  1    Domain:  x x  0, x  1 x 1 f  x  x  1  x  1   x  x( x  1)  f   g  ( x)  g x  1   x  1   1   x  1        x Domain:  x x  0, x  1 149 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs g. 4  x2 1  x4 4  ( x) 2 4  x 2 g ( x)    g ( x) 1  ( x) 4 1  x 4 g is even. 1  To graph y  f  x  , stretch the graph of 2  f horizontally by a factor of 2. 22. g ( x)  23. G ( x)  1  x  x3 G (  x )  1  (  x )  (  x )3  1  x  x 3  G ( x) or G ( x ) G is neither even nor odd. h. To graph y   f  x  , reflect the graph of f x 1  x2 x x f ( x)     f ( x) 1  ( x) 2 1  x 2 f is odd. 24. f ( x)  25. f  x   2 x3  5 x  1 on the interval  3,3 vertically about the y-axis. Use MAXIMUM and MINIMUM on the graph of y1  2 x3  5 x  1 . 20 20. a. 3 Domain:  , 4 Range:  ,3 20 Decreasing:  2, 2 e. The graph has no symmetry. f. The function is neither. g. x-intercepts:  3, 0  ,  0, 0  ,  3, 0  ; Use MAXIMUM and MINIMUM on the graph of y1  2 x 4  5 x3  2 x  1 . 20 2 f ( x)  x3  4 x 20 3 2 10 20 3 f ( x)  ( x)  4( x)   x  4 x  20 f  x   2 x 4  5 x3  2 x  1 on the interval  2,3 26. y-intercept: (0,0) 3 3 f is decreasing on:  0.91, 0.91 . Local minimum is 1 at x  2 ; Local maximum is 1 at x  2 d. No absolute minimum; Absolute maximum is 3 at x  4 21. 3 3 local maximum value: 4.04 when x  0.91 local minimum value: 2.04 when x  0.91 f is increasing on:  3, 0.91 and  0.91,3 ; b. Increasing:  , 2 and  2, 4 ; c. 20  3 10   x3  4 x   f ( x) 2 f is odd. 3 10 local maximum: 1.53 when x  0.41 150 Copyright © 2020 Pearson Education, Inc. Chapter 2 Review Exercises local minimal values: 0.54 when x  0.34 , 3.56 when x  1.80 f is increasing on:  0.34, 0.41 and 1.80, 3 ; 33. f ( x)  x 34. f ( x)  x f is decreasing on:  2, 0.34 and  0.41, 1.80 . 27. f ( x)  8 x 2  x a. b. c. 28. f (2)  f (1) 8(2) 2  2  [8(1) 2  1]  2 1 1  32  2  (7)  23 f (1)  f (0) 8(1) 2  1  [8(0) 2  0]  1 0 1  8  1  0  7 f (4)  f (2) 8(4) 2  4  [8(2) 2  2]  42 2 128  4  (30) 94    47 2 2 f ( x)  2  5 x f (3)  f (2)  2  5  3    2  5  2    3 2 32  2  15    2  10   1  13   8   5 29. f ( x)  3x  4 x 2 35. F ( x )  x  4 . Using the graph of y  x , 2  2  f (3)  f (2) 3  3  4  3   3  2   4  2    3 2 3 2  9  36    6  16   1  27  10  17 vertically shift the graph downward 4 units. 30. Refer to question 29 for the slope. y  10  17( x  2) y  10  17 x  34 y  17 x  24 Intercepts: (–4,0), (4,0), (0,–4) Domain:  x x is any real number 31. The graph does not pass the Vertical Line Test and is therefore not a function. Range:  y y   4 or  4,   32. The graph passes the Vertical Line Test and is therefore a function. 36. g ( x)   2 x . Reflect the graph of y  x about the x-axis and vertically stretch the graph by a factor of 2. 151 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs 39. h( x )  ( x  1) 2  2 . Using the graph of y  x 2 , horizontally shift the graph to the right 1 unit and vertically shift the graph up 2 units. Intercepts: (0, 0) Domain:  x x is any real number Range:  y y  0 or  , 0 Intercepts: (0, 3) Domain:  x x is any real number Range:  y y  2 or  2,   37. h( x)  x  1 . Using the graph of y  x , horizontally shift the graph to the right 1 unit. 40. g ( x)   2( x  2)3  8 Using the graph of y  x3 , horizontally shift the graph to the left 2 units, vertically stretch the graph by a factor of 2, reflect about the x-axis, and vertically shift the graph down 8 units. y x   2  3 4, 0  Intercept: (1, 0) Domain:  x x  1 or 1,   5  Range:  y y  0 or  0,   38.  f ( x)  1  x  ( x  1) . Reflect the graph of  y  x about the y-axis and horizontally shift the graph to the right 1 unit.   Intercepts: (0,–24),  2  3 4, 0   3.6, 0  Domain:  x x is any real number Range:  y y is any real number 41. 3x f ( x)   x 1 a. if  2  x  1 if x  1 Domain:  x x  2  or  2,   b. Intercept:  0, 0  Intercepts: (1, 0), (0, 1) Domain:  x x  1 or  , 1 Range:  y y  0 or  0,   152 Copyright © 2020 Pearson Education, Inc. Chapter 2 Review Exercises c. Graph: 44. a. x 2 h  10  h  10 x2 A( x )  2 x 2  4 x h  10   2 x2  4 x  2  x  40  2 x2  x d. Range:  y | y  6  or  6,   42. x  f ( x)  1 3x  a. A(1)  2 12  c. A(2)  2  22  40  8  20  28 ft 2 2 d. Graphing: if  4  x  0 50 if x  0 if x  0 Domain:  x x   4 or  4,   5 0 b. Intercept: (0, 1) c. 40  2  40  42 ft 2 1 b. 0 Graph: The area is smallest when x  2.15 feet. 45. a. Consider the following diagram: P(x,y) y y  10  x d. Range:  y y   4, y  0 43. 2 x Ax  5 and f (1)  4 6x  2 A(1)  5 4 6(1)  2 A5 4 4 A  5  16 f ( x)  The area of the rectangle is A  xy . Thus, the area function for the rectangle is: A( x )  x(10  x 2 ) A  11 b. The maximum area is roughly: A(1.83)  (1.83)3  10(1.83)  12.17 square units The maximum value occurs at the vertex: 153 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs can never equal 0. This means that x  2 . Domain:  x | x  2 Chapter 2 Test 1. a.  2,5 ,  4, 6  ,  6, 7  , 8,8 g  1  This relation is a function because there are no ordered pairs that have the same first element and different second elements. Domain: 2, 4, 6,8 x4 x 2  5 x  36 The function tells us to divide x  4 by 4. h  x   Range: 5, 6, 7,8 b. 1,3 ,  4, 2  ,  3,5 , 1, 7  x 2  5 x  36 . Since division by 0 is not defined, we need to exclude any values which make the denominator 0. x 2  5 x  36  0 This relation is not a function because there are two ordered pairs that have the same first element but different second elements.  x  9  x  4   0 Domain: 3,1, 4 x  9 or x  4 Domain:  x | x  9, x  4 Range: 2,3,5, 7 c. This relation is not a function because the graph fails the vertical line test. (note: there is a common factor of x  4 but we must determine the domain prior to simplifying) Domain:  1,   h  1  Range:  x x is any real number d. This relation is a function because it passes the vertical line test. Domain:  x x is any real number 5. a. Range:  y | y  2 or  2,   2. f  x   4  5x The function tells us to take the square root of 4  5x . Only nonnegative numbers have real square roots so we need 4  5 x  0 . 4  5x  0 4  5x  4  0  4 5 x  4 5 x 4  5 5 4 x 5  4 4  Domain:  x x   or  ,  5 5      1  4 5 1    1  5  1  36 40 8 2 To find the domain, note that all the points on the graph will have an x-coordinate between 5 and 5, inclusive. To find the range, note that all the points on the graph will have a y-coordinate between 3 and 3, inclusive. Domain:  x | 5  x  5 or  5, 5 Range:  y | 3  y  3 or  3, 3 b. The intercepts are  0, 2  ,  2, 0  , and  2, 0  . x-intercepts: 2, 2 y-intercept: 2 c. f 1 is the value of the function when x  1 . According to the graph, f 1  3 . d. Since  5, 3 and  3, 3 are the only points on the graph for which y  f  x   3 , we have f  x   3 when x  5 and x  3 . f  1  4  5  1  4  5  9  3 3. g  x    1  2 1  1  1  2 1 e. x2 x2 To solve f  x   0 , we want to find xvalues such that the graph is below the xaxis. The graph is below the x-axis for values in the domain that are less than 2 and greater than 2. Therefore, the solution set is  x | 5  x  2 or 2  x  5 . In The function tells us to divide x  2 by x  2 . Division by 0 is undefined, so the denominator 154 Copyright © 2020 Pearson Education, Inc. Chapter 2 Chapter Test keep the part for which x  1 . interval notation we would write the solution set as  5, 2    2,5 . 6. f  x    x 4  2 x3  4 x 2  2 We set Xmin = 5 and Xmax = 5. The standard Ymin and Ymax will not be good enough to see the whole picture so some adjustment must be made. b. To find the intercepts, notice that the only piece that hits either axis is y  x  4 . y  x4 y  x4 y  04 0  x4 y  4 4x The intercepts are  0, 4  and  4, 0  . c. To find g  5  we first note that x  5 so we must use the first “piece” because 5  1 . g  5   2  5   1  10  1  9 d. To find g  2  we first note that x  2 so we We see that the graph has a local maximum value of 0.86 (rounded to two places) when x  0.85 and another local maximum value of 15.55 when x  2.35 . There is a local minimum value of 2 when x  0 . Thus, we have Local maxima: f  0.85   0.86 must use the second “piece” because 2  1 . g  2   2  4  2 8. a. The average rate of change from 3 to 4 is given by f  4   f  3 43 f  2.35   15.55 Local minima: f  0   2  3  4   3  4   4    3  3  3  3  4   2 The function is increasing on the intervals  5, 0.85 and 0, 2.35 and decreasing on the 43 40  22 18    18 43 1 intervals  0.85, 0 and  2.35,5 . 7. a. 2 x  1 2 x  1 f  x   x  1  x4 To graph the function, we graph each “piece”. First we graph the line y  2 x  1 but only keep the part for which x  1 . Then we plot the line y  x  4 but only b. y  40  18( x  4) y  40  18 x  72 y  18 x  32 9. a.   ( f  g )( x)  2 x 2  1   3 x  2 2  2 x  1  3x  2  2 x 2  3x  3 b.   ( f  g )( x)  2 x 2  1  3 x  2 3  6 x  4 x 2  3x  2 155 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs c. y f  x  h  f  x        2  x  2 xh  h   1   2 x  1  2  x  h   1  2 x2  1 2 2 2   2  x   2 x 2  4 xh  2h 2  1  2 x 2  1   4 xh  2h 2 10. a. y  2 x  1 3 3 The basic function is y  x so we start with the graph of this function. y The last step is to shift this graph up 3 units to obtain the graph of y  2  x  1  3 . 3 y  x3  y        x  x    Next we shift this graph 1 unit to the left to y  2 x  1  3 3 obtain the graph of y   x  1 . 3 y   x  1 3 y b. The basic function is y  x so we start with the graph of this function.  y y x     x    x   Next we reflect this graph about the x-axis to obtain the graph of y    x  1 . 3 Next we shift this graph 4 units to the left to obtain the graph of y  x  4 . y   y    y  x4  x    y    x  1 3 Next we stretch this graph vertically by a factor of 2 to obtain the graph of  x Next we shift this graph up 2 units to obtain y  2  x  1 . 3 156 Copyright © 2020 Pearson Education, Inc. Chapter 2 Cumulative Review b. If the rink is 90 feet wide, then we have x  90 . the graph of y  x  4  2 . y y  x4 2 V  90     2 The volume of ice is roughly 3460.29 ft 3 .   902 10  90    90     3460.29 3 3 24 x 13. f   x   (  x ) 2  7   x 2  7 same The function is even. 11. a. f ( x  h)  f ( x) ( x  h) 2  3( x  h)  ( x 2  3 x)  h h 2 2 x  2 xh  h  3 x  3h  x 2  3x  h 2 xh  h 2  3h  h h(2 x  h  3)   2x  h  3 h 12. a. Let x = width of the rink in feet. Then the length of the rectangular portion is given by 2 x  20 . The radius of the semicircular x portions is half the width, or r  . 2 To find the volume, we first find the area of the surface and multiply by the thickness of the ice. The two semicircles can be combined to form a complete circle, so the area is given by A  l  w   r2  x   2 x  20  x      2 2  2 x  20 x  1. 3 x  8  10 3x  8  8  10  8 3 x  18 3x 18  3 3 x6 The solution set is 6 . 2. 3x 2  x  0 x  3 x  1  0 x  0 or 3x  1  0 3x  1 1 3  1 The solution set is 0,  .  3 x 2 3.  x2 x2  8x  9  0  x  9  x  1  0 4 We have expressed our measures in feet so we need to convert the thickness to feet as well. 1 ft 2 1 2 in   ft  ft 12 in 12 6 Now we multiply this by the area to obtain the volume. That is, 1  x2  V  x    2 x 2  20 x  6 4  V  x  Chapter 2 Cumulative Review x  9  0 or x  1  0 x9 x  1 The solution set is 1,9 . x 2 10 x  x 2   3 3 24 157 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs 4. 6 x2  5x  1  0 4  Interval notation:  ,   3   3x  1 2 x  1  0 3x  1  0 or 2 x  1  0 3x  1 2x  1 1 2 1 1  The solution set is  ,  . 3 2  x 5. 1 3 x 8. 3  2 x  5  3 2  2x  8 1 x  4 Solution set:  x |1  x  4 2x  3  4 2 x  3  4 or 2 x  3  4 2 x  7 2x  1 6. 7 2 Interval notation: 1, 4  1 2  7 1 The solution set is  ,  .  2 2 x x 9.  2x  3   2 2 4x 1  7 4 x  1  7 or 4 x  1  7 4 x  8 4x  6 3 x  2 x 2 3  Solution set:  x | x  2 or x   2  3  Interval notation:  , 2    ,   2  2x  3  2 2 2x  3  4 2x  1 x 2x  5  3 1 2 Check: ? 1 2   3  2 2 ? 1 3  2 10. a. ? 4 2 22 T 1  The solution set is   . 2  x2  x1    y2  y1  2 2   3   2    5   3   3  2    5  3 2 2 2  52   2   25  4 2 7. 2  3x  6 3x  4 x d  29 4 3 4  Solution set:  x | x    3  158 Copyright © 2020 Pearson Education, Inc. 2 Chapter 2 Cumulative Review b. c.  x  x y  y2  M  1 2 , 1  2   2  2  3 3   5    ,   2  2   1    , 4  2  m 12. x  y 2  x, y  2 2 x   2   4  4, 2  2 1 x   1  1 1, 1 0 x  02  0  0, 0  1 x  12  1 1,1 2 2 x2 4  4, 2  y y2  y1 5   3 2 2    x2  x1 5 5 3   2  x  y2 11. 3 x  2 y  12 x-intercept: 3x  2  0   12 3 x  12 x4 The point  4, 0  is on the graph. y-intercept: 3  0   2 y  12 13. x 2   y  3  16 2 2 y  12 y  6 This is the equation of a circle with radius r  16  4 and center at  0,3 . Starting at the The point  0, 6  is on the graph. center we can obtain some points on the graph by moving 4 units up, down, left, and right. The corresponding points are  0, 7  ,  0, 1 ,  4,3 , and  4,3 , respectively. 159 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs The graph of the equation has y-axis symmetry. 14. y  x x 0 1 4  x, y  y  0  0  0, 0  y  1  1 1,1 y  4  2  4, 2  16. First we find the slope: 84 4 1 m   8 2 6   2  y x Next we use the slope and the given point  6,8  in the point-slope form of the equation of a line: y  y1  m  x  x1  1  x  6 2 1 y 8  x 3 2 1 y  x5 2 y 8  17. 15. 3 x 2  4 y  12 x-intercepts: 3x 2  4  0   12 f  x    x  2  3 2 Starting with the graph of y  x 2 , shift the graph 2 units to the left  y   x  2   and down 3   2 2 units  y   x  2   3 .   3 x 2  12 x2  4 x  2 y-intercept: 3  0   4 y  12 2 4 y  12 y  3 The intercepts are  2, 0  ,  2, 0  , and  0, 3 . Check x-axis symmetry: 3x 2  4   y   12 3x 2  4 y  12 different Check y-axis symmetry: 3   x   4 y  12 2 3 x 2  4 y  12 same Check origin symmetry: 3   x   4   y   12 2 3x 2  4 y  12 different 160 Copyright © 2020 Pearson Education, Inc. Chapter 2 Projects 18. f  x  Project II 1 x 1 x y x 1 1 y   1 1 1 y  1 1 1 1 y 2 2 1. Silver: C  x   20  0.16  x  200   0.16 x  12  x, y  20  C ( x)   0.16 x  12  1, 1 0  x  200 x  200 Gold: C  x   50  0.08  x  1000   0.08 x  30 1,1 50.00 0  x  1000  C ( x)   0.08 x  30 x  1000   1  2,   2 Platinum: C  x   100  0.04  x  3000   0.04 x  20 C ( x)  100.00 0  x  3000   0.04 x 20 x  3000  Cost (dollars) C(x) 300 2  x if x  2 19. f  x    if x  2  x Graph the line y  2  x for x  2 . Two points Silver Gold 200 Platinum 100 0 1000 2000 3000 4000 x K-Bytes on the graph are  0, 2  and  2, 0  . 3. Let y = #K-bytes of service over the plan minimum. Graph the line y  x for x  2 . There is a hole in the graph at x  2 . Silver: 20  0.16 y  50 0.16 y  30 y  187.5 Silver is the best up to 187.5  200  387.5 K-bytes of service. Gold: 50  0.08 y  100 0.08 y  50 y  625 Gold is the best from 387.5 K-bytes to 625  1000  1625 K-bytes of service. Platinum: Platinum will be the best if more than 1625 K-bytes is needed. 4. Answers will vary. Chapter 2 Projects Project I – Internet-based Project – Answers will vary 161 Copyright © 2020 Pearson Education, Inc. Chapter 2: Functions and Their Graphs Project III 6. C(4.5) = 100(4.5) + 140 4  (5  4.5) 2 1.  $738.62 The cost for the Steven’s cable would be $738.62. Possible route 1 Driveway 2 miles 7. 5000(738.62) = $3,693,100 State legislated 5000(695.96) = $3,479,800 cheapest cost It will cost the company $213,300 more. Cable box 5 miles Possible route 2 Highway House 2. $140/mile L 4  (5  x )2 Project IV 2 miles 1. A   r 2 Cable box 2. r  2.2t 5 miles $10 0/mile C ( x)  100 x  140 L 3. r  2.2  2   4.4 ft C ( x)  100 x  140 4  (5  x) 3. 2 r  2.2  2.5   5.5 ft x C  x 4. A   (4.4) 2  60.82 ft 2 0 100  0   140 4  25  $753.92 A   (5.5)2  95.03 ft 2 1 100 1  140 4  16  $726.10 5. A   (2.2t ) 2  4.84 t 2 2 100  2   140 4  9  $704.78 6. A  4.84 (2) 2  60.82 ft 2 3 100  3  140 4  4  $695.98 A  4.84 (2.5) 2  95.03 ft 2 4 100  4   140 4  1  $713.05 5 100  5   140 4  0  $780.00 The choice where the cable goes 3 miles down the road then cutting up to the house seems to yield the lowest cost. 4. Since all of the costs are less than $800, there would be a profit made with any of the plans.  A(2.5)  A(2) 95.03  60.82   68.42 ft/hr 2.5  2 0.5 8. A(3.5)  A(3) 186.27  136.85   98.84 ft/hr 3.5  3 0.5 9. The average rate of change is increasing. 10. 150 yds = 450 ft r  2.2t 450 t  204.5 hours 2.2 C(x ) dollars   7. 11. 6 miles = 31680 ft Therefore, we need a radius of 15,840 ft. 15,840 t  7200 hours 2.2  x miles Using the MINIMUM function on a graphing calculator, the minimum occurs at x  2.96 . C(x) dollars     x miles The minimum cost occurs when the cable runs for 2.96 mile along the road. 162 Copyright © 2020 Pearson Education, Inc.