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# Solutions

## Solutions for Try Its

1. $\begin{cases}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)=\left(x - 1\right)\left({x}^{2}-1\right)={x}^{3}-{x}^{2}-x+1\\ \left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)=\left(x - 1\right)-\left({x}^{2}-1\right)=x-{x}^{2}\end{cases}$ No, the functions are not the same. 2. A gravitational force is still a force, so $a\left(G\left(r\right)\right)$ makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but $G\left(a\left(F\right)\right)$ does not make sense. 3. $f\left(g\left(1\right)\right)=f\left(3\right)=3$ and $g\left(f\left(4\right)\right)=g\left(1\right)=3$ 4. $g\left(f\left(2\right)\right)=g\left(5\right)=3$ 5. A. 8; B. 20 6. $\left[-4,0\right)\cup \left(0,\infty \right)$ 7. Possible answer:

$g\left(x\right)=\sqrt{4+{x}^{2}}$

$h\left(x\right)=\frac{4}{3-x}$

$f=h\circ g$

## Solutions to Odd-Numbered Exercises

1. Find the numbers that make the function in the denominator $g$ equal to zero, and check for any other domain restrictions on $f$ and $g$, such as an even-indexed root or zeros in the denominator. 3. Yes. Sample answer: Let $f\left(x\right)=x+1\text{ and }g\left(x\right)=x - 1$. Then $f\left(g\left(x\right)\right)=f\left(x - 1\right)=\left(x - 1\right)+1=x$ and $g\left(f\left(x\right)\right)=g\left(x+1\right)=\left(x+1\right)-1=x$. So $f\circ g=g\circ f$. 5. $\left(f+g\right)\left(x\right)=2x+6$, domain: $\left(-\infty ,\infty \right)$ [latex-display]\left(f-g\right)\left(x\right)=2{x}^{2}+2x - 6[/latex], domain: $\left(-\infty ,\infty \right)[/latex-display] [latex-display]\left(fg\right)\left(x\right)=-{x}^{4}-2{x}^{3}+6{x}^{2}+12x$, domain: $\left(-\infty ,\infty \right)[/latex-display] [latex-display]\left(\frac{f}{g}\right)\left(x\right)=\frac{{x}^{2}+2x}{6-{x}^{2}}$, domain: $\left(-\infty ,-\sqrt{6}\right)\cup \left(-\sqrt{6},\sqrt{6}\right)\cup \left(\sqrt{6},\infty \right)[/latex-display] 7. [latex]\left(f+g\right)\left(x\right)=\frac{4{x}^{3}+8{x}^{2}+1}{2x}$, domain: $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ [latex-display]\left(f-g\right)\left(x\right)=\frac{4{x}^{3}+8{x}^{2}-1}{2x}[/latex], domain: $\left(-\infty ,0\right)\cup \left(0,\infty \right)[/latex-display] [latex-display]\left(fg\right)\left(x\right)=x+2$, domain: $\left(-\infty ,0\right)\cup \left(0,\infty \right)[/latex-display] [latex-display]\left(\frac{f}{g}\right)\left(x\right)=4{x}^{3}+8{x}^{2}$, domain: $\left(-\infty ,0\right)\cup \left(0,\infty \right)[/latex-display] 9. [latex]\left(f+g\right)\left(x\right)=3{x}^{2}+\sqrt{x - 5}$, domain: $\left[5,\infty \right)$ [latex-display]\left(f-g\right)\left(x\right)=3{x}^{2}-\sqrt{x - 5}[/latex], domain: $\left[5,\infty \right)[/latex-display] [latex-display]\left(fg\right)\left(x\right)=3{x}^{2}\sqrt{x - 5}$, domain: $\left[5,\infty \right)[/latex-display] [latex-display]\left(\frac{f}{g}\right)\left(x\right)=\frac{3{x}^{2}}{\sqrt{x - 5}}$, domain: $\left(5,\infty \right)[/latex-display] 11. a. 3; b. [latex]f\left(g\left(x\right)\right)=2{\left(3x - 5\right)}^{2}+1$; c. $f\left(g\left(x\right)\right)=6{x}^{2}-2$; d. $\left(g\circ g\right)\left(x\right)=3\left(3x - 5\right)-5=9x - 20$; e. $\left(f\circ f\right)\left(-2\right)=163$ 13. $f\left(g\left(x\right)\right)=\sqrt{{x}^{2}+3}+2,g\left(f\left(x\right)\right)=x+4\sqrt{x}+7$ 15. $f\left(g\left(x\right)\right)=\sqrt[3]{\frac{x+1}{{x}^{3}}}=\frac{\sqrt[3]{x+1}}{x},g\left(f\left(x\right)\right)=\frac{\sqrt[3]{x}+1}{x}$ 17. $\left(f\circ g\right)\left(x\right)=\frac{1}{\frac{2}{x}+4 - 4}=\frac{x}{2},\text{ }\left(g\circ f\right)\left(x\right)=2x - 4$ 19. $f\left(g\left(h\left(x\right)\right)\right)={\left(\frac{1}{x+3}\right)}^{2}+1$ 21. a. $\left(g\circ f\right)\left(x\right)=-\frac{3}{\sqrt{2 - 4x}}$; b. $\left(-\infty ,\frac{1}{2}\right)$ 23. a. $\left(0,2\right)\cup \left(2,\infty \right)$; b. $\left(-\infty ,-2\right)\cup \left(2,\infty \right)$; c. $\left(0,\infty \right)$ 25. $\left(1,\infty \right)$ 27. sample: $\begin{cases}f\left(x\right)={x}^{3}\\ g\left(x\right)=x - 5\end{cases}$ 29. sample: $\begin{cases}f\left(x\right)=\frac{4}{x}\hfill \\ g\left(x\right)={\left(x+2\right)}^{2}\hfill \end{cases}$ 31. sample: $\begin{cases}f\left(x\right)=\sqrt[3]{x}\\ g\left(x\right)=\frac{1}{2x - 3}\end{cases}$ 33. sample: $\begin{cases}f\left(x\right)=\sqrt[4]{x}\\ g\left(x\right)=\frac{3x - 2}{x+5}\end{cases}$ 35. sample: $f\left(x\right)=\sqrt{x}$ [latex-display]g\left(x\right)=2x+6[/latex-display] 37.sample: $f\left(x\right)=\sqrt[3]{x}$ [latex-display]g\left(x\right)=\left(x - 1\right)[/latex-display] 39. sample: $f\left(x\right)={x}^{3}$ [latex-display]g\left(x\right)=\frac{1}{x - 2}[/latex-display] 41. sample: $f\left(x\right)=\sqrt{x}$ [latex-display]g\left(x\right)=\frac{2x - 1}{3x+4}[/latex-display] 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. $f\left(g\left(0\right)\right)=27,g\left(f\left(0\right)\right)=-94$ 75. $f\left(g\left(0\right)\right)=\frac{1}{5},g\left(f\left(0\right)\right)=5$ 77. $18{x}^{2}+60x+51$ 79. $g\circ g\left(x\right)=9x+20$ 81. 2 83. $\left(-\infty ,\infty \right)$ 85. False 87. $\left(f\circ g\right)\left(6\right)=6$ ; $\left(g\circ f\right)\left(6\right)=6$ 89. $\left(f\circ g\right)\left(11\right)=11,\left(g\circ f\right)\left(11\right)=11$ 91. c. Solve $A\left(m\left(t\right)\right)=4$. 93. $A\left(t\right)=\pi {\left(25\sqrt{t+2}\right)}^{2}$ and $A\left(2\right)=\pi {\left(25\sqrt{4}\right)}^{2}=2500\pi$ square inches 95. $A\left(5\right)=\pi {\left(2\left(5\right)+1\right)}^{2}=121\pi$ square units 97. a. $N\left(T\left(t\right)\right)=23{\left(5t+1.5\right)}^{2}-56\left(5t+1.5\right)+1$; b. 3.38 hours