{"id":235,"date":"2026-01-30T22:59:57","date_gmt":"2026-01-30T22:59:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/working-with-functions-get-stronger-key-precalculus-practice-page-answer-keys\/"},"modified":"2026-08-07T16:44:47","modified_gmt":"2026-08-07T16:44:47","slug":"working-with-functions-get-stronger-key-precalculus-practice-page-answer-keys","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/working-with-functions-get-stronger-key-precalculus-practice-page-answer-keys\/","title":{"raw":"Working with Functions: Get Stronger Key -- Precalculus Practice Page Answer Keys","rendered":"Working with Functions: Get Stronger Key — Precalculus Practice Page Answer Keys"},"content":{"raw":"

<\/h1>\r\n
\r\n

Composition of Functions Solutions<\/h2>\r\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.\r\n\r\n2.\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].\r\n\r\n3.\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]\r\n\r\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]\r\n\r\n[latex]\\left(fg\\right)\\left(x\\right)=x+2[\/latex], domain: [latex]\\left(-\\infty ,0\\right)\\cup \\left(0,\\infty \\right)[\/latex]\r\n\r\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]\r\n\r\n4.\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]\r\n\r\n5.\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]\r\n\r\n6.\u00a0[latex]\\left(1,\\infty \\right)[\/latex]\r\n\r\n7.\u00a0sample: [latex]\\begin{cases}f\\left(x\\right)={x}^{3}\\\\ g\\left(x\\right)=x - 5\\end{cases}[\/latex]\r\n\r\n8. sample:\u00a0[latex]f\\left(x\\right)=\\sqrt{x}[\/latex]\r\n[latex]g\\left(x\\right)=2x+6[\/latex]\r\n\r\n9. 2\r\n\r\n10. 5\r\n\r\n11. 9\r\n\r\n12. 4\r\n\r\n13. 2\r\n\r\n14.\u00a0[latex]f\\left(g\\left(0\\right)\\right)=27,g\\left(f\\left(0\\right)\\right)=-94[\/latex]\r\n\r\n15.\u00a0c. Solve [latex]A\\left(m\\left(t\\right)\\right)=4[\/latex].\r\n

Transformation of Functions Solutions<\/h2>\r\n1.\u00a0A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.\r\n\r\n2.\u00a0A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.\r\n\r\n3.\u00a0[latex]g\\left(x\\right)=|x - 1|-3[\/latex]\r\n\r\n4.\u00a0The graph of [latex]f\\left(x+43\\right)[\/latex] is a horizontal shift to the left 43 units of the graph of [latex]f[\/latex].\r\n\r\n5.\u00a0The graph of [latex]f\\left(x\\right)+8[\/latex] is a vertical shift up 8 units of the graph of [latex]f[\/latex].\r\n\r\n6. The graph of [latex]f\\left(x+4\\right)-1[\/latex] is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of [latex]f[\/latex].\r\n\r\n7.\u00a0decreasing on [latex]\\left(-\\infty ,-3\\right)[\/latex] and increasing on [latex]\\left(-3,\\infty \\right)[\/latex]\r\n\r\n8.\u00a0decreasing on [latex]\\left(0,\\infty \\right)[\/latex]\r\n\r\n9.\r\n\"Graph\r\n\r\n10.\r\n\"Graph\r\n\r\n11.\u00a0[latex]g\\left(x\\right)=f\\left(x - 1\\right),h\\left(x\\right)=f\\left(x\\right)+1[\/latex]\r\n\r\n12.\u00a0[latex]f\\left(x\\right)=|x - 3|-2[\/latex]\r\n\r\n13.\u00a0[latex]f\\left(x\\right)=\\sqrt{x+3}-1[\/latex]\r\n\r\n14.\u00a0[latex]f\\left(x\\right)=|x+3|-2[\/latex]\r\n\r\n15. even\r\n\r\n16. odd\r\n\r\n17.\u00a0The graph of [latex]g[\/latex] is a vertical reflection (across the [latex]x[\/latex] -axis) of the graph of [latex]f[\/latex].\r\n\r\n18.\u00a0The graph of [latex]g[\/latex] is a horizontal compression by a factor of [latex]\\frac{1}{5}[\/latex] of the graph of [latex]f[\/latex].\r\n\r\n19.\u00a0The graph of [latex]g[\/latex] is a horizontal stretch by a factor of 3 of the graph of [latex]f[\/latex].\r\n\r\n20.\u00a0The graph of [latex]g[\/latex] is a horizontal reflection across the [latex]y[\/latex] -axis and a vertical stretch by a factor of 3 of the graph of [latex]f[\/latex].\r\n\r\n21.\u00a0[latex]g\\left(x\\right)=\\frac{1}{3{\\left(x+2\\right)}^{2}}-3[\/latex]\r\n\r\n22.\u00a0[latex]g\\left(x\\right)=\\frac{1}{2}{\\left(x - 5\\right)}^{2}+1[\/latex]\r\n\r\n23.\u00a0The graph of the function [latex]f\\left(x\\right)={x}^{2}[\/latex] is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.\r\n\"Graph<\/span><\/span>\r\n\r\n24.\u00a0The graph of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] is shifted right 4 units and then reflected across the vertical line [latex]x=4[\/latex].\r\n\"Graph<\/span>\r\n

Inverse Functions Solutions<\/h2>\r\n1.\u00a0Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that [latex]y[\/latex] -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no [latex]y[\/latex] -values repeat and the function is one-to-one.\r\n\r\n2.\u00a0Yes. For example, [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] is its own inverse.\r\n\r\n3.\u00a0Given a function [latex]y=f\\left(x\\right)[\/latex], solve for [latex]x[\/latex] in terms of [latex]y[\/latex]. Interchange the [latex]x[\/latex] and [latex]y[\/latex]. Solve the new equation for [latex]y[\/latex]. The expression for [latex]y[\/latex] is the inverse, [latex]y={f}^{-1}\\left(x\\right)[\/latex].\r\n\r\n4.\u00a0[latex]{f}^{-1}\\left(x\\right)=x - 3[\/latex]\r\n\r\n5.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\frac{-2x}{x - 1}[\/latex]\r\n\r\n6.\u00a0domain of [latex]f\\left(x\\right):\\left[-7,\\infty \\right);{f}^{-1}\\left(x\\right)=\\sqrt[\\leftroot{0}\\uproot{0}]{x}-7[\/latex]\r\n\r\n7.\u00a0domain of [latex]f\\left(x\\right):\\left[0,\\infty \\right);{f}^{-1}\\left(x\\right)=\\sqrt[\\leftroot{0}\\uproot{0}]{x+5}[\/latex]\r\n\r\n8.\u00a0[latex]f\\left(g\\left(x\\right)\\right)=x,g\\left(f\\left(x\\right)\\right)=x[\/latex]\r\n\r\n9.\u00a0[latex]3[\/latex]\r\n\r\n10.\u00a0[latex]2[\/latex]\r\n\r\n11.\r\n\"Graph\r\n\r\n12.\u00a0[latex]6[\/latex]\r\n\r\n13.\u00a0[latex]-4[\/latex]\r\n\r\n14.\u00a0[latex]0[\/latex]\r\n\r\n15.\u00a0[latex]1[\/latex]\r\n\r\n16.\r\n\r\n\r\n\r\n\r\n
[latex]x[\/latex]<\/td>\r\n1<\/td>\r\n4<\/td>\r\n7<\/td>\r\n12<\/td>\r\n16<\/td>\r\n<\/tr>\r\n
[latex]{f}^{-1}\\left(x\\right)[\/latex]<\/td>\r\n3<\/td>\r\n6<\/td>\r\n9<\/td>\r\n13<\/td>\r\n14<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n17. [latex]\\begin{aligned} y &= \\frac{9}{5}x + 32 \\ y - 32 &= \\frac{9}{5}x \\text{ (subtract 32 from both sides)} \\ \\frac{5}{9}(y - 32) &= x \\text{ (multiply both sides by } \\frac{5}{9}) \\ f^{-1}(x) &= \\frac{5}{9}(x - 32) \\text{ (swap variables)} \\end{aligned}[\/latex]\r\n\r\nThe inverse function converts temperature from degrees Fahrenheit to degrees Celsius.\r\n\r\n<\/div>","rendered":"

<\/h1>\n
\n

Composition of Functions Solutions<\/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

2.\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

3.\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

4.\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

5.\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

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

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

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

9. 2<\/p>\n

10. 5<\/p>\n

11. 9<\/p>\n

12. 4<\/p>\n

13. 2<\/p>\n

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

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

Transformation of Functions Solutions<\/h2>\n

1.\u00a0A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.<\/p>\n

2.\u00a0A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.<\/p>\n

3.\u00a0[latex]g\\left(x\\right)=|x - 1|-3[\/latex]<\/p>\n

4.\u00a0The graph of [latex]f\\left(x+43\\right)[\/latex] is a horizontal shift to the left 43 units of the graph of [latex]f[\/latex].<\/p>\n

5.\u00a0The graph of [latex]f\\left(x\\right)+8[\/latex] is a vertical shift up 8 units of the graph of [latex]f[\/latex].<\/p>\n

6. The graph of [latex]f\\left(x+4\\right)-1[\/latex] is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of [latex]f[\/latex].<\/p>\n

7.\u00a0decreasing on [latex]\\left(-\\infty ,-3\\right)[\/latex] and increasing on [latex]\\left(-3,\\infty \\right)[\/latex]<\/p>\n

8.\u00a0decreasing on [latex]\\left(0,\\infty \\right)[\/latex]<\/p>\n

9.
\n\"Graph<\/p>\n

10.
\n\"Graph<\/p>\n

11.\u00a0[latex]g\\left(x\\right)=f\\left(x - 1\\right),h\\left(x\\right)=f\\left(x\\right)+1[\/latex]<\/p>\n

12.\u00a0[latex]f\\left(x\\right)=|x - 3|-2[\/latex]<\/p>\n

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

14.\u00a0[latex]f\\left(x\\right)=|x+3|-2[\/latex]<\/p>\n

15. even<\/p>\n

16. odd<\/p>\n

17.\u00a0The graph of [latex]g[\/latex] is a vertical reflection (across the [latex]x[\/latex] -axis) of the graph of [latex]f[\/latex].<\/p>\n

18.\u00a0The graph of [latex]g[\/latex] is a horizontal compression by a factor of [latex]\\frac{1}{5}[\/latex] of the graph of [latex]f[\/latex].<\/p>\n

19.\u00a0The graph of [latex]g[\/latex] is a horizontal stretch by a factor of 3 of the graph of [latex]f[\/latex].<\/p>\n

20.\u00a0The graph of [latex]g[\/latex] is a horizontal reflection across the [latex]y[\/latex] -axis and a vertical stretch by a factor of 3 of the graph of [latex]f[\/latex].<\/p>\n

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

22.\u00a0[latex]g\\left(x\\right)=\\frac{1}{2}{\\left(x - 5\\right)}^{2}+1[\/latex]<\/p>\n

23.\u00a0The graph of the function [latex]f\\left(x\\right)={x}^{2}[\/latex] is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.
\n\"Graph<\/span><\/span><\/p>\n

24.\u00a0The graph of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] is shifted right 4 units and then reflected across the vertical line [latex]x=4[\/latex].
\n\"Graph<\/span><\/p>\n

Inverse Functions Solutions<\/h2>\n

1.\u00a0Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that [latex]y[\/latex] -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no [latex]y[\/latex] -values repeat and the function is one-to-one.<\/p>\n

2.\u00a0Yes. For example, [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] is its own inverse.<\/p>\n

3.\u00a0Given a function [latex]y=f\\left(x\\right)[\/latex], solve for [latex]x[\/latex] in terms of [latex]y[\/latex]. Interchange the [latex]x[\/latex] and [latex]y[\/latex]. Solve the new equation for [latex]y[\/latex]. The expression for [latex]y[\/latex] is the inverse, [latex]y={f}^{-1}\\left(x\\right)[\/latex].<\/p>\n

4.\u00a0[latex]{f}^{-1}\\left(x\\right)=x - 3[\/latex]<\/p>\n

5.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\frac{-2x}{x - 1}[\/latex]<\/p>\n

6.\u00a0domain of [latex]f\\left(x\\right):\\left[-7,\\infty \\right);{f}^{-1}\\left(x\\right)=\\sqrt[\\leftroot{0}\\uproot{0}]{x}-7[\/latex]<\/p>\n

7.\u00a0domain of [latex]f\\left(x\\right):\\left[0,\\infty \\right);{f}^{-1}\\left(x\\right)=\\sqrt[\\leftroot{0}\\uproot{0}]{x+5}[\/latex]<\/p>\n

8.\u00a0[latex]f\\left(g\\left(x\\right)\\right)=x,g\\left(f\\left(x\\right)\\right)=x[\/latex]<\/p>\n

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

10.\u00a0[latex]2[\/latex]<\/p>\n

11.
\n\"Graph<\/p>\n

12.\u00a0[latex]6[\/latex]<\/p>\n

13.\u00a0[latex]-4[\/latex]<\/p>\n

14.\u00a0[latex]0[\/latex]<\/p>\n

15.\u00a0[latex]1[\/latex]<\/p>\n

16.<\/p>\n\n\n\n\n
[latex]x[\/latex]<\/td>\n1<\/td>\n4<\/td>\n7<\/td>\n12<\/td>\n16<\/td>\n<\/tr>\n
[latex]{f}^{-1}\\left(x\\right)[\/latex]<\/td>\n3<\/td>\n6<\/td>\n9<\/td>\n13<\/td>\n14<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

17. [latex]\\begin{aligned} y &= \\frac{9}{5}x + 32 \\ y - 32 &= \\frac{9}{5}x \\text{ (subtract 32 from both sides)} \\ \\frac{5}{9}(y - 32) &= x \\text{ (multiply both sides by } \\frac{5}{9}) \\ f^{-1}(x) &= \\frac{5}{9}(x - 32) \\text{ (swap variables)} \\end{aligned}[\/latex]<\/p>\n

The inverse function converts temperature from degrees Fahrenheit to degrees Celsius.<\/p>\n<\/div>\n","protected":false},"author":13,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":224,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/235"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/235\/revisions"}],"predecessor-version":[{"id":390,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/235\/revisions\/390"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/224"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/235\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=235"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=235"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=235"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=235"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}