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Section 8.4 Inverse Functions 815. Version: Fall 2007. 8.4 Exercises. In Exercises 1-12, use the graph to de- termine whether the function is one-to-.
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Section 8.4 Inverse Functions 815
In Exercises 1 - 12 , use the graph to de-
termine whether the function is one-to-
one.
Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/ 1
816 Chapter 8 Exponential and Logarithmic Functions
In Exercises 13 - 28 , evaluate the com-
position g(f (x)) and simplify your an-
swer.
13. g(x) =
x
, f (x) = − 2 x 2
14. f (x) = −
x
, g(x) = − 4 x 2
15. g(x) = 2
x, f (x) = − x − 3
16. f (x) = 3x 2 − 3 x − 5 , g(x) =
x
17. g(x) = 3
x, f (x) = 4x + 1
18. f (x) = − 3 x − 5 , g(x) = − x − 2 19. g(x) = − 5 x 2 + 3x − 4 , f (x) =
x
20. g(x) = 3x + 3, f (x) = 4x 2 − 2 x − 2 21. g(x) = 6
x, f (x) = − 4 x + 4
22. g(x) = 5x − 3 , f (x) = − 2 x − 4 23. g(x) = 3
x, f (x) = − 2 x + 1
24. g(x) =
x
, f (x) = − 5 x 2 − 5 x − 4
25. f (x) =
x
, g(x) = − x + 1
26. f (x) = 4x 2 + 3x − 4 , g(x) =
x
818 Chapter 8 Exponential and Logarithmic Functions
In Exercises 37 - 68 , find the formula for
the inverse function f − 1 (x).
37. f (x) = 5x 3 − 5 38. f (x) = 4x 7 − 3 39. f (x) = −
9 x − 3
7 x + 6
40. f (x) = 6x − 4 41. f (x) = 7x − 9 42. f (x) = 7x + 4 43. f (x) = 3x 5 − 9 44. f (x) = 6x + 7 45. f (x) =
4 x + 2
4 x + 3
46. f (x) = 5x 7 + 4 47. f (x) =
4 x − 1
2 x + 2
48. f (x) =
7
8 x − 3
49. f (x) =
3
− 6 x − 4
50. f (x) =
8 x − 7
3 x − 6
51. f (x) = 7
− 3 x − 5
52. f (x) =
9
8 x + 2
53. f (x) =
3
6 x + 7
54. f (x) =
3 x + 7
2 x + 8
55. f (x) = − 5 x + 2 56. f (x) = 6x + 8 57. f (x) = 9x 9 + 5 58. f (x) = 4x 5 − 9 59. f (x) =
9 x − 3
9 x + 7
60. f (x) = 3
9 x − 7
61. f (x) = x 4 , x ≤ 0 62. f (x) = x 4 , x ≥ 0 63. f (x) = x 2 − 1 , x ≤ 0 64. f (x) = x 2 + 2, x ≥ 0 65. f (x) = x 4 + 3, x ≤ 0 66. f (x) = x 4 − 5 , x ≥ 0 67. f (x) = (x − 1) 2 , x ≤ 1 68. f (x) = (x + 2) 2 , x ≥ − 2
Section 8.4 Inverse Functions
1. The graph fails the horizontal line test. For example, the horizontal line y = 1 cuts
the graph in more than one place. Therefore, the function not is one-to-one.
3. The graph fails the horizontal line test. For example, the horizontal line y = 0 cuts
the graph in more than one place. Therefore, the function not is one-to-one.
5. The graph fails the horizontal line test. For example, the horizontal line y = 4 cuts
the graph in more than one place. Therefore, the function not is one-to-one.
7. The graph meets the horizontal line test. Every horizontal line intersects the graph
no more than once. Therefore, the function is one-to-one.
9. The graph meets the horizontal line test. Every horizontal line intersects the graph
no more than once. Therefore, the function is one-to-one.
Section 8.4 Inverse Functions
29. If you reflect the given graph across the line y = x (pictured in black), you obtain
the inverse, shown in red.
31. If you reflect the given graph across the line y = x (pictured in black), you obtain
the inverse, shown in red.
33. If you reflect the given graph across the line y = x (pictured in black), you obtain
the inverse, shown in red.
35. If you reflect the given graph across the line y = x (pictured in black), you obtain
the inverse, shown in red.
Chapter 8 Exponential and Logarithmic Functions
37. Start with the equation y = 5x 3 − 5.
Interchange x and y: x = 5y 3 − 5.
Then solve for y: y =
3
x +
39. Start with the equation y = −
9 x − 3
7 x + 6
Interchange x and y: x = −
9 y − 3
7 y + 6
Then solve for y:
x(7y + 6) = − 9 y + 3 = ⇒ (7x + 9)y = − 6 x + 3 = ⇒ y = −
6 x − 3
7 x + 9
41. Start with the equation y = 7x − 9.
Interchange x and y: x = 7y − 9.
Then solve for y: y = x + 7
43. Start with the equation y = 3x 5 − 9.
Interchange x and y: x = 3y 5 − 9.
Then solve for y: y =
5
x +
45. Start with the equation y =
4 x + 2
4 x + 3
Interchange x and y: x =
4 y + 2
4 y + 3
Then solve for y:
x(4y + 3) = 4y + 2 = ⇒ (4x − 4)y = − 3 x + 2 = ⇒ y = −
3 x − 2
4 x − 4
47. Start with the equation y =
4 x − 1
2 x + 2
Interchange x and y: x =
4 y − 1
2 y + 2
Then solve for y:
x(2y + 2) = 4y − 1 = ⇒ (2x − 4)y = − 2 x − 1 = ⇒ y = −
2 x + 1
2 x − 4
49. Start with the equation y =
3
− 6 x − 4.
Interchange x and y: x = 3
− 6 y − 4.
Then solve for y: y = −
x^3 +
Chapter 8 Exponential and Logarithmic Functions
67. Start with the equation y = (x − 1) 2 with the domain condition x ≤ 1.
Interchange x and y: x = (y − 1) 2 , y ≤ 1.
Solve for y: y = ±
x + 1, y ≤ 1.
The condition y ≤ 1 then implies that y = −
x + 1.
