{"id":920,"date":"2015-11-12T18:37:58","date_gmt":"2015-11-12T18:37:58","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=920"},"modified":"2015-11-12T18:37:58","modified_gmt":"2015-11-12T18:37:58","slug":"solutions-53","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/solutions-53\/","title":{"raw":"Solutions","rendered":"Solutions"},"content":{"raw":"

Solutions for Try Its<\/span><\/h2>\n1.\u00a0[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]\nNo, the functions are not the same.\n\n2.\u00a0A 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.\n\n3.\u00a0[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]\n\n4.\u00a0[latex]g\\left(f\\left(2\\right)\\right)=g\\left(5\\right)=3[\/latex]\n\n5. A. 8; B. 20\n\n6.\u00a0[latex]\\left[-4,0\\right)\\cup \\left(0,\\infty \\right)[\/latex]\n\n7.\u00a0Possible answer:\n

[latex]g\\left(x\\right)=\\sqrt{4+{x}^{2}}[\/latex]<\/p>\n

[latex]h\\left(x\\right)=\\frac{4}{3-x}[\/latex]<\/p>\n

[latex]f=h\\circ g[\/latex]<\/p>\n\u00a0\n\u00a0\n

Solutions to Odd-Numbered Exercises<\/span><\/h2>\n1.\u00a0Find 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.\n\n3.\u00a0Yes. 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].\n\n5.\u00a0[latex]\\left(f+g\\right)\\left(x\\right)=2x+6[\/latex], domain: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]\n\n[latex]\\left(f-g\\right)\\left(x\\right)=2{x}^{2}+2x - 6[\/latex], domain: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]\n\n[latex]\\left(fg\\right)\\left(x\\right)=-{x}^{4}-2{x}^{3}+6{x}^{2}+12x[\/latex], domain: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]\n\n[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]\n\n7.\u00a0[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]\n\n[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]\n\n[latex]\\left(fg\\right)\\left(x\\right)=x+2[\/latex], domain: [latex]\\left(-\\infty ,0\\right)\\cup \\left(0,\\infty \\right)[\/latex]\n\n[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]\n\n9.\u00a0[latex]\\left(f+g\\right)\\left(x\\right)=3{x}^{2}+\\sqrt{x - 5}[\/latex], domain: [latex]\\left[5,\\infty \\right)[\/latex]\n\n[latex]\\left(f-g\\right)\\left(x\\right)=3{x}^{2}-\\sqrt{x - 5}[\/latex], domain: [latex]\\left[5,\\infty \\right)[\/latex]\n\n[latex]\\left(fg\\right)\\left(x\\right)=3{x}^{2}\\sqrt{x - 5}[\/latex], domain: [latex]\\left[5,\\infty \\right)[\/latex]\n\n[latex]\\left(\\frac{f}{g}\\right)\\left(x\\right)=\\frac{3{x}^{2}}{\\sqrt{x - 5}}[\/latex], domain: [latex]\\left(5,\\infty \\right)[\/latex]\n\n11.\u00a0a. 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]\n\n13.\u00a0[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]\n\n15.\u00a0[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]\n\n17.\u00a0[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]\n\n19.\u00a0[latex]f\\left(g\\left(h\\left(x\\right)\\right)\\right)={\\left(\\frac{1}{x+3}\\right)}^{2}+1[\/latex]\n\n21.\u00a0a. [latex]\\left(g\\circ f\\right)\\left(x\\right)=-\\frac{3}{\\sqrt{2 - 4x}}[\/latex]; b. [latex]\\left(-\\infty ,\\frac{1}{2}\\right)[\/latex]\n\n23.\u00a0a. [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]\n\n25.\u00a0[latex]\\left(1,\\infty \\right)[\/latex]\n\n27.\u00a0sample: [latex]\\begin{cases}f\\left(x\\right)={x}^{3}\\\\ g\\left(x\\right)=x - 5\\end{cases}[\/latex]\n\n29.\u00a0sample: [latex]\\begin{cases}f\\left(x\\right)=\\frac{4}{x}\\hfill \\\\ g\\left(x\\right)={\\left(x+2\\right)}^{2}\\hfill \\end{cases}[\/latex]\n\n31.\u00a0sample: [latex]\\begin{cases}f\\left(x\\right)=\\sqrt[3]{x}\\\\ g\\left(x\\right)=\\frac{1}{2x - 3}\\end{cases}[\/latex]\n\n33.\u00a0sample: [latex]\\begin{cases}f\\left(x\\right)=\\sqrt[4]{x}\\\\ g\\left(x\\right)=\\frac{3x - 2}{x+5}\\end{cases}[\/latex]\n\n35. sample:\u00a0[latex]f\\left(x\\right)=\\sqrt{x}[\/latex]\n[latex]g\\left(x\\right)=2x+6[\/latex]\n\n37.sample:\u00a0[latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]\n[latex]g\\left(x\\right)=\\left(x - 1\\right)[\/latex]\n\n39.\u00a0sample: [latex]f\\left(x\\right)={x}^{3}[\/latex]\n[latex]g\\left(x\\right)=\\frac{1}{x - 2}[\/latex]\n\n41.\u00a0sample: [latex]f\\left(x\\right)=\\sqrt{x}[\/latex]\n[latex]g\\left(x\\right)=\\frac{2x - 1}{3x+4}[\/latex]\n\n43. 2\n\n45. 5\n\n47. 4\n\n49. 0\n\n51. 2\n\n53. 1\n\n55. 4\n\n57. 4\n\n59. 9\n\n61. 4\n\n63. 2\n\n65. 3\n\n67. 11\n\n69. 0\n\n71. 7\n\n73.\u00a0[latex]f\\left(g\\left(0\\right)\\right)=27,g\\left(f\\left(0\\right)\\right)=-94[\/latex]\n\n75.\u00a0[latex]f\\left(g\\left(0\\right)\\right)=\\frac{1}{5},g\\left(f\\left(0\\right)\\right)=5[\/latex]\n\n77.\u00a0[latex]18{x}^{2}+60x+51[\/latex]\n\n79.\u00a0[latex]g\\circ g\\left(x\\right)=9x+20[\/latex]\n\n81. 2\n\n83.\u00a0[latex]\\left(-\\infty ,\\infty \\right)[\/latex]\n\n85. False\n\n87.\u00a0[latex]\\left(f\\circ g\\right)\\left(6\\right)=6[\/latex] ; [latex]\\left(g\\circ f\\right)\\left(6\\right)=6[\/latex]\n\n89.\u00a0[latex]\\left(f\\circ g\\right)\\left(11\\right)=11,\\left(g\\circ f\\right)\\left(11\\right)=11[\/latex]\n\n91.\u00a0c. Solve [latex]A\\left(m\\left(t\\right)\\right)=4[\/latex].\n\n93.\u00a0[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\n\n95.\u00a0[latex]A\\left(5\\right)=\\pi {\\left(2\\left(5\\right)+1\\right)}^{2}=121\\pi [\/latex] square units\n\n97.\u00a0a. [latex]N\\left(T\\left(t\\right)\\right)=23{\\left(5t+1.5\\right)}^{2}-56\\left(5t+1.5\\right)+1[\/latex];\nb. 3.38 hours","rendered":"

Solutions for Try Its<\/span><\/h2>\n

1.\u00a0[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]
\nNo, the functions are not the same.<\/p>\n

2.\u00a0A 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.<\/p>\n

3.\u00a0[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]<\/p>\n

4.\u00a0[latex]g\\left(f\\left(2\\right)\\right)=g\\left(5\\right)=3[\/latex]<\/p>\n

5. A. 8; B. 20<\/p>\n

6.\u00a0[latex]\\left[-4,0\\right)\\cup \\left(0,\\infty \\right)[\/latex]<\/p>\n

7.\u00a0Possible answer:<\/p>\n

[latex]g\\left(x\\right)=\\sqrt{4+{x}^{2}}[\/latex]<\/p>\n

[latex]h\\left(x\\right)=\\frac{4}{3-x}[\/latex]<\/p>\n

[latex]f=h\\circ g[\/latex]<\/p>\n

\u00a0
\n\u00a0<\/p>\n

Solutions to Odd-Numbered Exercises<\/span><\/h2>\n

1.\u00a0Find 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.<\/p>\n

3.\u00a0Yes. 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].<\/p>\n

5.\u00a0[latex]\\left(f+g\\right)\\left(x\\right)=2x+6[\/latex], domain: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]<\/p>\n

[latex]\\left(f-g\\right)\\left(x\\right)=2{x}^{2}+2x - 6[\/latex], domain: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]<\/p>\n

[latex]\\left(fg\\right)\\left(x\\right)=-{x}^{4}-2{x}^{3}+6{x}^{2}+12x[\/latex], domain: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]<\/p>\n

[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]<\/p>\n

7.\u00a0[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]<\/p>\n

[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]<\/p>\n

[latex]\\left(fg\\right)\\left(x\\right)=x+2[\/latex], domain: [latex]\\left(-\\infty ,0\\right)\\cup \\left(0,\\infty \\right)[\/latex]<\/p>\n

[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]<\/p>\n

9.\u00a0[latex]\\left(f+g\\right)\\left(x\\right)=3{x}^{2}+\\sqrt{x - 5}[\/latex], domain: [latex]\\left[5,\\infty \\right)[\/latex]<\/p>\n

[latex]\\left(f-g\\right)\\left(x\\right)=3{x}^{2}-\\sqrt{x - 5}[\/latex], domain: [latex]\\left[5,\\infty \\right)[\/latex]<\/p>\n

[latex]\\left(fg\\right)\\left(x\\right)=3{x}^{2}\\sqrt{x - 5}[\/latex], domain: [latex]\\left[5,\\infty \\right)[\/latex]<\/p>\n

[latex]\\left(\\frac{f}{g}\\right)\\left(x\\right)=\\frac{3{x}^{2}}{\\sqrt{x - 5}}[\/latex], domain: [latex]\\left(5,\\infty \\right)[\/latex]<\/p>\n

11.\u00a0a. 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]<\/p>\n

13.\u00a0[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]<\/p>\n

15.\u00a0[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]<\/p>\n

17.\u00a0[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]<\/p>\n

19.\u00a0[latex]f\\left(g\\left(h\\left(x\\right)\\right)\\right)={\\left(\\frac{1}{x+3}\\right)}^{2}+1[\/latex]<\/p>\n

21.\u00a0a. [latex]\\left(g\\circ f\\right)\\left(x\\right)=-\\frac{3}{\\sqrt{2 - 4x}}[\/latex]; b. [latex]\\left(-\\infty ,\\frac{1}{2}\\right)[\/latex]<\/p>\n

23.\u00a0a. [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]<\/p>\n

25.\u00a0[latex]\\left(1,\\infty \\right)[\/latex]<\/p>\n

27.\u00a0sample: [latex]\\begin{cases}f\\left(x\\right)={x}^{3}\\\\ g\\left(x\\right)=x - 5\\end{cases}[\/latex]<\/p>\n

29.\u00a0sample: [latex]\\begin{cases}f\\left(x\\right)=\\frac{4}{x}\\hfill \\\\ g\\left(x\\right)={\\left(x+2\\right)}^{2}\\hfill \\end{cases}[\/latex]<\/p>\n

31.\u00a0sample: [latex]\\begin{cases}f\\left(x\\right)=\\sqrt[3]{x}\\\\ g\\left(x\\right)=\\frac{1}{2x - 3}\\end{cases}[\/latex]<\/p>\n

33.\u00a0sample: [latex]\\begin{cases}f\\left(x\\right)=\\sqrt[4]{x}\\\\ g\\left(x\\right)=\\frac{3x - 2}{x+5}\\end{cases}[\/latex]<\/p>\n

35. sample:\u00a0[latex]f\\left(x\\right)=\\sqrt{x}[\/latex]
\n[latex]g\\left(x\\right)=2x+6[\/latex]<\/p>\n

37.sample:\u00a0[latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]
\n[latex]g\\left(x\\right)=\\left(x - 1\\right)[\/latex]<\/p>\n

39.\u00a0sample: [latex]f\\left(x\\right)={x}^{3}[\/latex]
\n[latex]g\\left(x\\right)=\\frac{1}{x - 2}[\/latex]<\/p>\n

41.\u00a0sample: [latex]f\\left(x\\right)=\\sqrt{x}[\/latex]
\n[latex]g\\left(x\\right)=\\frac{2x - 1}{3x+4}[\/latex]<\/p>\n

43. 2<\/p>\n

45. 5<\/p>\n

47. 4<\/p>\n

49. 0<\/p>\n

51. 2<\/p>\n

53. 1<\/p>\n

55. 4<\/p>\n

57. 4<\/p>\n

59. 9<\/p>\n

61. 4<\/p>\n

63. 2<\/p>\n

65. 3<\/p>\n

67. 11<\/p>\n

69. 0<\/p>\n

71. 7<\/p>\n

73.\u00a0[latex]f\\left(g\\left(0\\right)\\right)=27,g\\left(f\\left(0\\right)\\right)=-94[\/latex]<\/p>\n

75.\u00a0[latex]f\\left(g\\left(0\\right)\\right)=\\frac{1}{5},g\\left(f\\left(0\\right)\\right)=5[\/latex]<\/p>\n

77.\u00a0[latex]18{x}^{2}+60x+51[\/latex]<\/p>\n

79.\u00a0[latex]g\\circ g\\left(x\\right)=9x+20[\/latex]<\/p>\n

81. 2<\/p>\n

83.\u00a0[latex]\\left(-\\infty ,\\infty \\right)[\/latex]<\/p>\n

85. False<\/p>\n

87.\u00a0[latex]\\left(f\\circ g\\right)\\left(6\\right)=6[\/latex] ; [latex]\\left(g\\circ f\\right)\\left(6\\right)=6[\/latex]<\/p>\n

89.\u00a0[latex]\\left(f\\circ g\\right)\\left(11\\right)=11,\\left(g\\circ f\\right)\\left(11\\right)=11[\/latex]<\/p>\n

91.\u00a0c. Solve [latex]A\\left(m\\left(t\\right)\\right)=4[\/latex].<\/p>\n

93.\u00a0[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<\/p>\n

95.\u00a0[latex]A\\left(5\\right)=\\pi {\\left(2\\left(5\\right)+1\\right)}^{2}=121\\pi[\/latex] square units<\/p>\n

97.\u00a0a. [latex]N\\left(T\\left(t\\right)\\right)=23{\\left(5t+1.5\\right)}^{2}-56\\left(5t+1.5\\right)+1[\/latex];
\nb. 3.38 hours<\/p>\n\n\t\t\t

\n\t\t\t

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