Solutions
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-display]\left(f-g\right)\left(x\right)=2{x}^{2}+2x - 6[/latex], domain: [latex]\left(-\infty ,\infty \right)[/latex-display] [latex-display]\left(fg\right)\left(x\right)=-{x}^{4}-2{x}^{3}+6{x}^{2}+12x[/latex], domain: [latex]\left(-\infty ,\infty \right)[/latex-display] [latex-display]\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-display] 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-display]\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-display] [latex-display]\left(fg\right)\left(x\right)=x+2[/latex], domain: [latex]\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}[/latex], domain: [latex]\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}[/latex], domain: [latex]\left[5,\infty \right)[/latex] [latex-display]\left(f-g\right)\left(x\right)=3{x}^{2}-\sqrt{x - 5}[/latex], domain: [latex]\left[5,\infty \right)[/latex-display] [latex-display]\left(fg\right)\left(x\right)=3{x}^{2}\sqrt{x - 5}[/latex], domain: [latex]\left[5,\infty \right)[/latex-display] [latex-display]\left(\frac{f}{g}\right)\left(x\right)=\frac{3{x}^{2}}{\sqrt{x - 5}}[/latex], domain: [latex]\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[/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-display]g\left(x\right)=2x+6[/latex-display] 37.sample: [latex]f\left(x\right)=\sqrt[3]{x}[/latex] [latex-display]g\left(x\right)=\left(x - 1\right)[/latex-display] 39. sample: [latex]f\left(x\right)={x}^{3}[/latex] [latex-display]g\left(x\right)=\frac{1}{x - 2}[/latex-display] 41. sample: [latex]f\left(x\right)=\sqrt{x}[/latex] [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. [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 hoursLicenses & Attributions
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- Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution. License terms: Download For Free at : http://cnx.org/contents/[email protected]..
