Solutions to Try Its
1. f−1(f(x))=f−1(x+53)=3(x+53)−5=(x−5)+5=xf−1(f(x))=f−1(x+53)=3(x+53)−5=(x−5)+5=x and f(f−1(x))=f(3x−5)=(3x−5)+53=3x3=xf(f−1(x))=f(3x−5)=(3x−5)+53=3x3=x
2. f−1(x)=x3−4f−1(x)=x3−4
3. f−1(x)=√x−1f−1(x)=√x−1
4. f−1(x)=x2−32,x≥0f−1(x)=x2−32,x≥0
5. f−1(x)=2x+3x−1f−1(x)=2x+3x−1
Solutions to Odd-Numbered Exercises
1. It can be too difficult or impossible to solve for x in terms of y.
3. We will need a restriction on the domain of the answer.
5. f−1(x)=√x+4f−1(x)=√x+4
7. f−1(x)=√x+3−1f−1(x)=√x+3−1
9. f−1(x)=−√x−53f−1(x)=−√x−53
11. f(x)=√9−xf(x)=√9−x
13. f−1(x)=3√x−5f−1(x)=3√x−5
15. f−1(x)=3√4−xf−1(x)=3√4−x
17. f−1(x)=x2−12,[0,∞)f−1(x)=x2−12,[0,∞)
19. f−1(x)=(x−9)2+44,[9,∞)f−1(x)=(x−9)2+44,[9,∞)
21. f−1(x)=(x−92)3f−1(x)=(x−92)3
23. f−1(x)=2−8xxf−1(x)=2−8xx
25. f−1(x)=7x−31−xf−1(x)=7x−31−x
27. f−1(x)=5x−44x+3f−1(x)=5x−44x+3
29. f−1(x)=√x+1−1f−1(x)=√x+1−1
31. f−1(x)=√x+6+3f−1(x)=√x+6+3
33. f−1(x)=√4−xf−1(x)=√4−x

35. f−1(x)=√x+4f−1(x)=√x+4

37. f−1(x)=3√1−xf−1(x)=3√1−x

39. f−1(x)=√x+8+3f−1(x)=√x+8+3

41. f−1(x)=√1xf−1(x)=√1x

43. [−2,1)∪[3,∞)[−2,1)∪[3,∞)

45. [−4,2)∪[5,∞)[−4,2)∪[5,∞)

47. (−2,0);(4,2);(22,3)(−2,0);(4,2);(22,3)

49. (−4,0);(0,1);(10,2)(−4,0);(0,1);(10,2)

51. (−3,−1);(1,0);(7,1)(−3,−1);(1,0);(7,1)

53. f−1(x)=√x+b24−b2f−1(x)=√x+b24−b2
55. f−1(x)=x3−baf−1(x)=x3−ba
57. t(h)=√200−h4.9t(h)=√200−h4.9, 5.53 seconds
59. r(V)=3√3V4πr(V)=3√3V4π, 3.63 feet
61. n(C)=100C−25.6−Cn(C)=100C−25.6−C, 250 mL
63. r(V)=√V6πr(V)=√V6π, 3.99 meters
65. r(V)=√V4πr(V)=√V4π, 1.99 inches
Candela Citations
- Precalculus. Authored by: Jay Abramson, et al.. Provided by: OpenStax. Located at: http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175. License: CC BY: Attribution. License Terms: Download For Free at : http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.