{"id":1372,"date":"2022-04-06T16:38:35","date_gmt":"2022-04-06T16:38:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=1372"},"modified":"2025-12-11T20:25:24","modified_gmt":"2025-12-11T20:25:24","slug":"3-3-2-the-composition-of-a-function-and-its-inverse-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/3-3-2-the-composition-of-a-function-and-its-inverse-function\/","title":{"raw":"3.3.2: The Composition of a Function and Its Inverse Function","rendered":"3.3.2: The Composition of a Function and Its Inverse Function"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"wrapper\">\r\n<div id=\"wrap\">\r\n<div id=\"content\" role=\"main\">\r\n<div id=\"post-365\" class=\"standard post-365 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Objectives<\/h1>\r\n<ul>\r\n \t<li>Determine if two functions are inverses of each other using the composition of functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Composition of a Function and Its Inverse<\/h2>\r\nIn the previous section we discovered that a one-to-one function has an inverse function, and that a one-to-one function and its inverse functions are reflections of one another across the line [latex]y=x[\/latex]. This means that while the green curve in figure 1 is the inverse function of the blue curve, symmetry shows that the blue curve must be the inverse function of the green curve.\r\n\r\n[caption id=\"attachment_1766\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-1766 size-medium\" style=\"text-align: center;\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-73-300x300.png\" alt=\"A function and its inverse are inverses of each other, as described above.\" width=\"300\" height=\"300\" \/> Figure 1. A one-to-one function and its inverse functions are reflections of each other across the line [latex]y=x[\/latex][\/caption]In figure 1, if [latex]y=f(x)[\/latex] is the blue curve, [latex]f(0) = 3[\/latex] implies that [latex]f^{-1}(3) = 0[\/latex]. This is confirmed on the green curve [latex]y=f^{-1}(x)[\/latex]. In other words, if we find the function value at [latex]x=0[\/latex] and get 3, then find the inverse function value at [latex]x=3[\/latex], [latex]f^{-1}(3)[\/latex], we get back to 0. The inverse function <em>undoes<\/em> the original function and we end up back where we started.<\/div>\r\n<div><\/div>\r\n<div class=\"entry-content\">In general, if [latex]f(x)=y[\/latex], then [latex]f^{-1}(y)=x[\/latex]. Consequently, [latex]f^{-1}(f(x)) = f^{-1}(y) = x[\/latex].<span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">Informally, this means that inverse functions \u201cundo\u201d each other. Recall that a\u00a0function must be one-to-one to have an inverse function.<\/span><\/div>\r\n<div class=\"entry-content\">\r\n\r\nLogically, since [latex]f(x)[\/latex] and [latex]f^{-1}(x)[\/latex] are inverse functions of each other, [latex]\\left (f\\circ f^{-1}(x)\\right )=f\\left (f^{-1}(x)\\right )=x[\/latex].\r\n\r\nDiagrammatically,\u00a0<img class=\"aligncenter wp-image-1790 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/f-and-inverse-f-mapping-300x214.png\" alt=\"Since f and f inverse are inverses of each other, the domain and range of one switch.  The domain becomes the range, and the range becomes the domain.\" width=\"300\" height=\"214\" \/>\r\n\r\nGiven a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] and [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] are true.\r\n\r\nFor example, prove that [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.\r\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex]\\left({f}^{}\\circ {f}^{-1}\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>test whether two functions are inverses of each other<\/h3>\r\n<ol>\r\n \t<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\r\n \t<li>If <strong>both<\/strong> statements are true, then [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nIf [latex]f\\left(x\\right)=\\dfrac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\dfrac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n<h4>Solution<\/h4>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} g\\left(f\\left(x\\right)\\right)&amp;=g\\left (\\dfrac{1}{x+2}\\right )\\\\&amp;=\\frac{1}{\\left(\\frac{1}{x+2}\\right)}{-2 }&amp;\\text{The reciprocal of a reciprocal is the number: }\\dfrac{1}{\\frac{1}{a}}=a\\\\[1.5mm]&amp;=({ x }+{ 2 }) -{ 2 }&amp;\\\\[1.5mm]&amp;={ x }&amp; \\end{align}[\/latex]<\/p>\r\nand\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)\\\\&amp;=f\\left (\\dfrac{1}{x}-2\\right )&amp;=\\frac{1}{\\frac{1}{x}-2+2}\\\\[1.5mm] &amp;=\\frac{1}{\\frac{1}{x}} \\\\[1.5mm] &amp;=x \\end{align}[\/latex]<\/p>\r\n&nbsp;\r\n<p style=\"text-align: left;\">Therefore, [latex]g={f}^{-1}\\text{ and }f={g}^{-1}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nIf [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"hjm136\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm136\"]Yes[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nIf you had problems completing the Try It, the problem solution is detailed here.\r\n<iframe src=\"https:\/\/lumenlearning.h5p.com\/content\/1290756028512507138\/embed\" width=\"1088\" height=\"637\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<script src=\"https:\/\/lumenlearning.h5p.com\/js\/h5p-resizer.js\" charset=\"UTF-8\"><\/script>\r\n<div class=\"textbox tryit\">\r\n<h3>TRY IT 2<\/h3>\r\nDetermine if [latex]f(x)=x-1[\/latex] and [latex]g(x)=\\dfrac{1}{x-1}[\/latex] are inverse functions.\r\n\r\n[reveal-answer q=\"hjm507\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm507\"]No, they are not inverse functions.[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nIf [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n<h4>Solution<\/h4>\r\n[latex]f\\left(g\\left(x\\right)\\right)=\\left(\\frac{1}{3}x\\right)^3=\\dfrac{{x}^{3}}{27}\\ne x[\/latex]\r\n\r\nNo, the functions are not inverses.\r\n<h4>Analysis of the Solution<\/h4>\r\nThe correct inverse to [latex]x^3[\/latex] is the cube root [latex]\\sqrt[3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nIf [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}\\text{and}g\\left(x\\right)=\\sqrt[3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"hjm523\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm523\"]Yes[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"wrapper\">\n<div id=\"wrap\">\n<div id=\"content\" role=\"main\">\n<div id=\"post-365\" class=\"standard post-365 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox learning-objectives\">\n<h1>Learning Objectives<\/h1>\n<ul>\n<li>Determine if two functions are inverses of each other using the composition of functions<\/li>\n<\/ul>\n<\/div>\n<h2>The Composition of a Function and Its Inverse<\/h2>\n<p>In the previous section we discovered that a one-to-one function has an inverse function, and that a one-to-one function and its inverse functions are reflections of one another across the line [latex]y=x[\/latex]. This means that while the green curve in figure 1 is the inverse function of the blue curve, symmetry shows that the blue curve must be the inverse function of the green curve.<\/p>\n<div id=\"attachment_1766\" style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1766\" class=\"wp-image-1766 size-medium\" style=\"text-align: center;\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-73-300x300.png\" alt=\"A function and its inverse are inverses of each other, as described above.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-73-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-73-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-73-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-73-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-73-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-73-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-73.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p id=\"caption-attachment-1766\" class=\"wp-caption-text\">Figure 1. A one-to-one function and its inverse functions are reflections of each other across the line [latex]y=x[\/latex]<\/p>\n<\/div>\n<p>In figure 1, if [latex]y=f(x)[\/latex] is the blue curve, [latex]f(0) = 3[\/latex] implies that [latex]f^{-1}(3) = 0[\/latex]. This is confirmed on the green curve [latex]y=f^{-1}(x)[\/latex]. In other words, if we find the function value at [latex]x=0[\/latex] and get 3, then find the inverse function value at [latex]x=3[\/latex], [latex]f^{-1}(3)[\/latex], we get back to 0. The inverse function <em>undoes<\/em> the original function and we end up back where we started.<\/div>\n<div><\/div>\n<div class=\"entry-content\">In general, if [latex]f(x)=y[\/latex], then [latex]f^{-1}(y)=x[\/latex]. Consequently, [latex]f^{-1}(f(x)) = f^{-1}(y) = x[\/latex].<span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">Informally, this means that inverse functions \u201cundo\u201d each other. Recall that a\u00a0function must be one-to-one to have an inverse function.<\/span><\/div>\n<div class=\"entry-content\">\n<p>Logically, since [latex]f(x)[\/latex] and [latex]f^{-1}(x)[\/latex] are inverse functions of each other, [latex]\\left (f\\circ f^{-1}(x)\\right )=f\\left (f^{-1}(x)\\right )=x[\/latex].<\/p>\n<p>Diagrammatically,\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1790 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/f-and-inverse-f-mapping-300x214.png\" alt=\"Since f and f inverse are inverses of each other, the domain and range of one switch.  The domain becomes the range, and the range becomes the domain.\" width=\"300\" height=\"214\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/f-and-inverse-f-mapping-300x214.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/f-and-inverse-f-mapping-768x547.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/f-and-inverse-f-mapping-65x46.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/f-and-inverse-f-mapping-225x160.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/f-and-inverse-f-mapping-350x249.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/f-and-inverse-f-mapping.png 780w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Given a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] and [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] are true.<\/p>\n<p>For example, prove that [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.<\/p>\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex]\\left({f}^{}\\circ {f}^{-1}\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>test whether two functions are inverses of each other<\/h3>\n<ol>\n<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\n<li>If <strong>both<\/strong> statements are true, then [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>If [latex]f\\left(x\\right)=\\dfrac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\dfrac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p style=\"text-align: center;\">[latex]\\begin{align} g\\left(f\\left(x\\right)\\right)&=g\\left (\\dfrac{1}{x+2}\\right )\\\\&=\\frac{1}{\\left(\\frac{1}{x+2}\\right)}{-2 }&\\text{The reciprocal of a reciprocal is the number: }\\dfrac{1}{\\frac{1}{a}}=a\\\\[1.5mm]&=({ x }+{ 2 }) -{ 2 }&\\\\[1.5mm]&={ x }& \\end{align}[\/latex]<\/p>\n<p>and<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)\\\\&=f\\left (\\dfrac{1}{x}-2\\right )&=\\frac{1}{\\frac{1}{x}-2+2}\\\\[1.5mm] &=\\frac{1}{\\frac{1}{x}} \\\\[1.5mm] &=x \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left;\">Therefore, [latex]g={f}^{-1}\\text{ and }f={g}^{-1}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>If [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm136\">Show Answer<\/span><\/p>\n<div id=\"qhjm136\" class=\"hidden-answer\" style=\"display: none\">Yes<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>If you had problems completing the Try It, the problem solution is detailed here.<br \/>\n<iframe loading=\"lazy\" src=\"https:\/\/lumenlearning.h5p.com\/content\/1290756028512507138\/embed\" width=\"1088\" height=\"637\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\n<script src=\"https:\/\/lumenlearning.h5p.com\/js\/h5p-resizer.js\" charset=\"UTF-8\"><\/script><\/p>\n<div class=\"textbox tryit\">\n<h3>TRY IT 2<\/h3>\n<p>Determine if [latex]f(x)=x-1[\/latex] and [latex]g(x)=\\dfrac{1}{x-1}[\/latex] are inverse functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm507\">Show Answer<\/span><\/p>\n<div id=\"qhjm507\" class=\"hidden-answer\" style=\"display: none\">No, they are not inverse functions.<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>If [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>[latex]f\\left(g\\left(x\\right)\\right)=\\left(\\frac{1}{3}x\\right)^3=\\dfrac{{x}^{3}}{27}\\ne x[\/latex]<\/p>\n<p>No, the functions are not inverses.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The correct inverse to [latex]x^3[\/latex] is the cube root [latex]\\sqrt[3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}\\text{and}g\\left(x\\right)=\\sqrt[3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm523\">Show Answer<\/span><\/p>\n<div id=\"qhjm523\" class=\"hidden-answer\" style=\"display: none\">Yes<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1372\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Authored by<\/strong>: Hazel McKenna and Leo Chang. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Authored by<\/strong>: Hazel McKenna and Leo Chang. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All Try Its and Examples. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":422608,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"Hazel McKenna and Leo Chang\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"Hazel McKenna and Leo Chang\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All Try Its and Examples\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley 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