Solutions for Try Its
1. [latex]\begin{cases}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)=\left(x - 1\right)\left({x}^{2}-1\right)={x}^{3}-{x}^{2}-x+1\\ \left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)=\left(x - 1\right)-\left({x}^{2}-1\right)=x-{x}^{2}\end{cases}[/latex]
No, the functions are not the same.
2. A gravitational force is still a force, so [latex]a\left(G\left(r\right)\right)[/latex] makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but [latex]G\left(a\left(F\right)\right)[/latex] does not make sense.
3. [latex]f\left(g\left(1\right)\right)=f\left(3\right)=3[/latex] and [latex]g\left(f\left(4\right)\right)=g\left(1\right)=3[/latex]
4. [latex]g\left(f\left(2\right)\right)=g\left(5\right)=3[/latex]
5. A. 8; B. 20
6. [latex]\left[-4,0\right)\cup \left(0,\infty \right)[/latex]
7. Possible answer:
[latex]g\left(x\right)=\sqrt{4+{x}^{2}}[/latex]
[latex]h\left(x\right)=\frac{4}{3-x}[/latex]
[latex]f=h\circ g[/latex]
Solutions to Odd-Numbered Exercises
1. Find the numbers that make the function in the denominator [latex]g[/latex] equal to zero, and check for any other domain restrictions on [latex]f[/latex] and [latex]g[/latex], such as an even-indexed root or zeros in the denominator.
3. Yes. Sample answer: Let [latex]f\left(x\right)=x+1\text{ and }g\left(x\right)=x - 1[/latex]. Then [latex]f\left(g\left(x\right)\right)=f\left(x - 1\right)=\left(x - 1\right)+1=x[/latex] and [latex]g\left(f\left(x\right)\right)=g\left(x+1\right)=\left(x+1\right)-1=x[/latex]. So [latex]f\circ g=g\circ f[/latex].
5. [latex]\left(f+g\right)\left(x\right)=2x+6[/latex], domain: [latex]\left(-\infty ,\infty \right)[/latex]
[latex]\left(f-g\right)\left(x\right)=2{x}^{2}+2x - 6[/latex], domain: [latex]\left(-\infty ,\infty \right)[/latex]
[latex]\left(fg\right)\left(x\right)=-{x}^{4}-2{x}^{3}+6{x}^{2}+12x[/latex], domain: [latex]\left(-\infty ,\infty \right)[/latex]
[latex]\left(\frac{f}{g}\right)\left(x\right)=\frac{{x}^{2}+2x}{6-{x}^{2}}[/latex], domain: [latex]\left(-\infty ,-\sqrt{6}\right)\cup \left(-\sqrt{6},\sqrt{6}\right)\cup \left(\sqrt{6},\infty \right)[/latex]
7. [latex]\left(f+g\right)\left(x\right)=\frac{4{x}^{3}+8{x}^{2}+1}{2x}[/latex], domain: [latex]\left(-\infty ,0\right)\cup \left(0,\infty \right)[/latex]
[latex]\left(f-g\right)\left(x\right)=\frac{4{x}^{3}+8{x}^{2}-1}{2x}[/latex], domain: [latex]\left(-\infty ,0\right)\cup \left(0,\infty \right)[/latex]
[latex]\left(fg\right)\left(x\right)=x+2[/latex], domain: [latex]\left(-\infty ,0\right)\cup \left(0,\infty \right)[/latex]
[latex]\left(\frac{f}{g}\right)\left(x\right)=4{x}^{3}+8{x}^{2}[/latex], domain: [latex]\left(-\infty ,0\right)\cup \left(0,\infty \right)[/latex]
9. [latex]\left(f+g\right)\left(x\right)=3{x}^{2}+\sqrt{x - 5}[/latex], domain: [latex]\left[5,\infty \right)[/latex]
[latex]\left(f-g\right)\left(x\right)=3{x}^{2}-\sqrt{x - 5}[/latex], domain: [latex]\left[5,\infty \right)[/latex]
[latex]\left(fg\right)\left(x\right)=3{x}^{2}\sqrt{x - 5}[/latex], domain: [latex]\left[5,\infty \right)[/latex]
[latex]\left(\frac{f}{g}\right)\left(x\right)=\frac{3{x}^{2}}{\sqrt{x - 5}}[/latex], domain: [latex]\left(5,\infty \right)[/latex]
11. a. 3; b. [latex]f\left(g\left(x\right)\right)=2{\left(3x - 5\right)}^{2}+1[/latex]; c. [latex]f\left(g\left(x\right)\right)=6{x}^{2}-2[/latex]; d. [latex]\left(g\circ g\right)\left(x\right)=3\left(3x - 5\right)-5=9x - 20[/latex]; e. [latex]\left(f\circ f\right)\left(-2\right)=163[/latex]
13. [latex]f\left(g\left(x\right)\right)=\sqrt{{x}^{2}+3}+2,g\left(f\left(x\right)\right)=x+4\sqrt{x}+7[/latex]
15. [latex]f\left(g\left(x\right)\right)=\sqrt[3]{\frac{x+1}{{x}^{3}}}=\frac{\sqrt[3]{x+1}}{x},g\left(f\left(x\right)\right)=\frac{\sqrt[3]{x}+1}{x}[/latex]
17. [latex]\left(f\circ g\right)\left(x\right)=\frac{1}{\frac{2}{x}+4 - 4}=\frac{x}{2},\text{ }\left(g\circ f\right)\left(x\right)=2x - 4[/latex]
19. [latex]f\left(g\left(h\left(x\right)\right)\right)={\left(\frac{1}{x+3}\right)}^{2}+1[/latex]
21. a. [latex]\left(g\circ f\right)\left(x\right)=-\frac{3}{\sqrt{2 - 4x}}[/latex]; b. [latex]\left(-\infty ,\frac{1}{2}\right)[/latex]
23. a. [latex]\left(0,2\right)\cup \left(2,\infty \right)[/latex]; b. [latex]\left(-\infty ,-2\right)\cup \left(2,\infty \right)[/latex]; c. [latex]\left(0,\infty \right)[/latex]
25. [latex]\left(1,\infty \right)[/latex]
27. sample: [latex]\begin{cases}f\left(x\right)={x}^{3}\\ g\left(x\right)=x - 5\end{cases}[/latex]
29. sample: [latex]\begin{cases}f\left(x\right)=\frac{4}{x}\hfill \\ g\left(x\right)={\left(x+2\right)}^{2}\hfill \end{cases}[/latex]
31. sample: [latex]\begin{cases}f\left(x\right)=\sqrt[3]{x}\\ g\left(x\right)=\frac{1}{2x - 3}\end{cases}[/latex]
33. sample: [latex]\begin{cases}f\left(x\right)=\sqrt[4]{x}\\ g\left(x\right)=\frac{3x - 2}{x+5}\end{cases}[/latex]
35. sample: [latex]f\left(x\right)=\sqrt{x}[/latex]
[latex]g\left(x\right)=2x+6[/latex]
37.sample: [latex]f\left(x\right)=\sqrt[3]{x}[/latex]
[latex]g\left(x\right)=\left(x - 1\right)[/latex]
39. sample: [latex]f\left(x\right)={x}^{3}[/latex]
[latex]g\left(x\right)=\frac{1}{x - 2}[/latex]
41. sample: [latex]f\left(x\right)=\sqrt{x}[/latex]
[latex]g\left(x\right)=\frac{2x - 1}{3x+4}[/latex]
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. [latex]f\left(g\left(0\right)\right)=27,g\left(f\left(0\right)\right)=-94[/latex]
75. [latex]f\left(g\left(0\right)\right)=\frac{1}{5},g\left(f\left(0\right)\right)=5[/latex]
77. [latex]18{x}^{2}+60x+51[/latex]
79. [latex]g\circ g\left(x\right)=9x+20[/latex]
81. 2
83. [latex]\left(-\infty ,\infty \right)[/latex]
85. False
87. [latex]\left(f\circ g\right)\left(6\right)=6[/latex] ; [latex]\left(g\circ f\right)\left(6\right)=6[/latex]
89. [latex]\left(f\circ g\right)\left(11\right)=11,\left(g\circ f\right)\left(11\right)=11[/latex]
91. c. Solve [latex]A\left(m\left(t\right)\right)=4[/latex].
93. [latex]A\left(t\right)=\pi {\left(25\sqrt{t+2}\right)}^{2}[/latex] and [latex]A\left(2\right)=\pi {\left(25\sqrt{4}\right)}^{2}=2500\pi[/latex] square inches
95. [latex]A\left(5\right)=\pi {\left(2\left(5\right)+1\right)}^{2}=121\pi[/latex] square units
97. a. [latex]N\left(T\left(t\right)\right)=23{\left(5t+1.5\right)}^{2}-56\left(5t+1.5\right)+1[/latex];
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.