Solutions for Try Its
1. {(fg)(x)=f(x)g(x)=(x−1)(x2−1)=x3−x2−x+1(f−g)(x)=f(x)−g(x)=(x−1)−(x2−1)=x−x2{(fg)(x)=f(x)g(x)=(x−1)(x2−1)=x3−x2−x+1(f−g)(x)=f(x)−g(x)=(x−1)−(x2−1)=x−x2
No, the functions are not the same.
2. A gravitational force is still a force, so a(G(r))a(G(r)) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but G(a(F))G(a(F)) does not make sense.
3. f(g(1))=f(3)=3f(g(1))=f(3)=3 and g(f(4))=g(1)=3g(f(4))=g(1)=3
4. g(f(2))=g(5)=3g(f(2))=g(5)=3
5. A. 8; B. 20
6. [−4,0)∪(0,∞)[−4,0)∪(0,∞)
7. Possible answer:
g(x)=√4+x2g(x)=√4+x2
h(x)=43−xh(x)=43−x
f=h∘gf=h∘g
Solutions to Odd-Numbered Exercises
1. Find the numbers that make the function in the denominator gg equal to zero, and check for any other domain restrictions on ff and gg, such as an even-indexed root or zeros in the denominator.
3. Yes. Sample answer: Let f(x)=x+1 and g(x)=x−1f(x)=x+1 and g(x)=x−1. Then f(g(x))=f(x−1)=(x−1)+1=xf(g(x))=f(x−1)=(x−1)+1=x and g(f(x))=g(x+1)=(x+1)−1=xg(f(x))=g(x+1)=(x+1)−1=x. So f∘g=g∘ff∘g=g∘f.
5. (f+g)(x)=2x+6(f+g)(x)=2x+6, domain: (−∞,∞)(−∞,∞)
(f−g)(x)=2x2+2x−6(f−g)(x)=2x2+2x−6, domain: (−∞,∞)(−∞,∞)
(fg)(x)=−x4−2x3+6x2+12x(fg)(x)=−x4−2x3+6x2+12x, domain: (−∞,∞)(−∞,∞)
(fg)(x)=x2+2x6−x2(fg)(x)=x2+2x6−x2, domain: (−∞,−√6)∪(−√6,√6)∪(√6,∞)(−∞,−√6)∪(−√6,√6)∪(√6,∞)
7. (f+g)(x)=4x3+8x2+12x(f+g)(x)=4x3+8x2+12x, domain: (−∞,0)∪(0,∞)(−∞,0)∪(0,∞)
(f−g)(x)=4x3+8x2−12x(f−g)(x)=4x3+8x2−12x, domain: (−∞,0)∪(0,∞)(−∞,0)∪(0,∞)
(fg)(x)=x+2(fg)(x)=x+2, domain: (−∞,0)∪(0,∞)(−∞,0)∪(0,∞)
(fg)(x)=4x3+8x2(fg)(x)=4x3+8x2, domain: (−∞,0)∪(0,∞)(−∞,0)∪(0,∞)
9. (f+g)(x)=3x2+√x−5(f+g)(x)=3x2+√x−5, domain: [5,∞)[5,∞)
(f−g)(x)=3x2−√x−5(f−g)(x)=3x2−√x−5, domain: [5,∞)[5,∞)
(fg)(x)=3x2√x−5(fg)(x)=3x2√x−5, domain: [5,∞)[5,∞)
(fg)(x)=3x2√x−5(fg)(x)=3x2√x−5, domain: (5,∞)(5,∞)
11. a. 3; b. f(g(x))=2(3x−5)2+1f(g(x))=2(3x−5)2+1; c. f(g(x))=6x2−2f(g(x))=6x2−2; d. (g∘g)(x)=3(3x−5)−5=9x−20(g∘g)(x)=3(3x−5)−5=9x−20; e. (f∘f)(−2)=163(f∘f)(−2)=163
13. f(g(x))=√x2+3+2,g(f(x))=x+4√x+7f(g(x))=√x2+3+2,g(f(x))=x+4√x+7
15. f(g(x))=3√x+1x3=3√x+1x,g(f(x))=3√x+1xf(g(x))=3√x+1x3=3√x+1x,g(f(x))=3√x+1x
17. (f∘g)(x)=12x+4−4=x2, (g∘f)(x)=2x−4(f∘g)(x)=12x+4−4=x2, (g∘f)(x)=2x−4
19. f(g(h(x)))=(1x+3)2+1f(g(h(x)))=(1x+3)2+1
21. a. (g∘f)(x)=−3√2−4x(g∘f)(x)=−3√2−4x; b. (−∞,12)(−∞,12)
23. a. (0,2)∪(2,∞)(0,2)∪(2,∞); b. (−∞,−2)∪(2,∞)(−∞,−2)∪(2,∞); c. (0,∞)(0,∞)
25. (1,∞)(1,∞)
27. sample: {f(x)=x3g(x)=x−5{f(x)=x3g(x)=x−5
29. sample: {f(x)=4xg(x)=(x+2)2{f(x)=4xg(x)=(x+2)2
31. sample: {f(x)=3√xg(x)=12x−3{f(x)=3√xg(x)=12x−3
33. sample: {f(x)=4√xg(x)=3x−2x+5{f(x)=4√xg(x)=3x−2x+5
35. sample: f(x)=√xf(x)=√x
g(x)=2x+6g(x)=2x+6
37.sample: f(x)=3√xf(x)=3√x
g(x)=(x−1)g(x)=(x−1)
39. sample: f(x)=x3f(x)=x3
g(x)=1x−2g(x)=1x−2
41. sample: f(x)=√xf(x)=√x
g(x)=2x−13x+4g(x)=2x−13x+4
43. 2
45. 5
47. 4
49. 0
51. 2
53. 1
55. 4
57. 4
59. 9
61. 4
63. 2
65. 3
67. 11
69. 0
71. 7
73. f(g(0))=27,g(f(0))=−94f(g(0))=27,g(f(0))=−94
75. f(g(0))=15,g(f(0))=5f(g(0))=15,g(f(0))=5
77. 18x2+60x+5118x2+60x+51
79. g∘g(x)=9x+20g∘g(x)=9x+20
81. 2
83. (−∞,∞)(−∞,∞)
85. False
87. (f∘g)(6)=6(f∘g)(6)=6 ; (g∘f)(6)=6(g∘f)(6)=6
89. (f∘g)(11)=11,(g∘f)(11)=11(f∘g)(11)=11,(g∘f)(11)=11
91. c. Solve A(m(t))=4A(m(t))=4.
93. A(t)=π(25√t+2)2A(t)=π(25√t+2)2 and A(2)=π(25√4)2=2500πA(2)=π(25√4)2=2500π square inches
95. A(5)=π(2(5)+1)2=121πA(5)=π(2(5)+1)2=121π square units
97. a. N(T(t))=23(5t+1.5)2−56(5t+1.5)+1N(T(t))=23(5t+1.5)2−56(5t+1.5)+1;
b. 3.38 hours
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.