{"id":2810,"date":"2022-06-20T01:01:22","date_gmt":"2022-06-20T01:01:22","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=2810"},"modified":"2026-01-18T03:05:41","modified_gmt":"2026-01-18T03:05:41","slug":"7-3-the-inverse-of-a-rational-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/7-3-the-inverse-of-a-rational-function\/","title":{"raw":"7.3: The Inverse of a Rational Function","rendered":"7.3: The Inverse of a Rational 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=\"textbox learning-objectives\">\r\n<h1>Learning Objectives<\/h1>\r\n<ul>\r\n \t<li>Graph the inverse of a rational function<\/li>\r\n \t<li>Find the equation of the inverse function of a one-to-one rational function<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn chapter 3, we discussed that every function has an inverse, but <strong>only a one-to-one function has an inverse function<\/strong>. Some rational functions are one-to-one functions such as [latex]f(x)=\\dfrac{1}{x}[\/latex] or\u00a0[latex]f(x)=\\dfrac{x-1}{x+4}[\/latex]. Therefore, their inverse is a function. Some rational functions are many-to-one functions such as\u00a0[latex]f(x)=\\dfrac{1}{x^2-1}[\/latex] or\u00a0[latex]f(x)=\\dfrac{x^3-4x^2+2}{x^2-1}[\/latex]. Therefore, their inverse is not a function (Figure 1).\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%; text-align: center;\">One-to-one rational functions that have an inverse function<\/th>\r\n<th style=\"width: 50%; text-align: center;\">Many-to-one rational functions that do not have an inverse function<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\"><img class=\"aligncenter wp-image-3301 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092326.109-300x300.png\" alt=\"graph of the function f(x)=1\/x\" width=\"300\" height=\"300\" \/><\/td>\r\n<td style=\"width: 50%; text-align: center;\"><img class=\"aligncenter wp-image-3303 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092927.854-300x300.png\" alt=\"Graph of f(x)=1\/(x^2-1), two separate curve branches in opposite quadrants: one opening left and upward approach a vertical asymptote at x=-1, the other right and upward approaching a vertical asymptote at x=1. A third curve branch opens downward between and approaching the two asymptotes.\" width=\"300\" height=\"300\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\"><img class=\"aligncenter wp-image-3302 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092612.998-300x300.png\" alt=\"Graph of f(x)=(x-1)\/(x+4), two separate curve branches in opposite quadrants: one opening left and upward, the other right and downward, each approaching a vertical asymptote at x=-4 and and a horizontal asymptote at y=1.\" width=\"300\" height=\"300\" \/><\/td>\r\n<td style=\"width: 50%; text-align: center;\"><img class=\"aligncenter wp-image-3307 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T105745.272-300x300.png\" alt=\"Graph of f(x)=(x^3-4x^2+2)\/(x^2-1), two separate curve branches in opposite quadrants: one opening southwest and downward with one side approaching a diagonal line and the other side approaching a vertical line, the other northeast and downward with one side approaching the same diagonal line and the other side approaching another vertical line. A third curve branch opens upward between the two curve branches and approaching the two vertical lines.\" width=\"300\" height=\"300\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\" colspan=\"2\">Figure 1. Graphs of rational functions<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice that three of the graphs in figure 1 have horizontal and vertical asymptotes but the 4th graph has two vertical asymptotes and a <em><strong>slant asymptote<\/strong><\/em>. A slant asymptote is a line of the form [latex]y=mx+b[\/latex] that is neither vertical nor horizontal but that the graph gets closer and closer to as [latex]x[\/latex] approaches positive and negative infinity. Slant asymptotes occur in the graph of a rational function [latex]f(x)=\\dfrac{P(x)}{Q(x)}[\/latex] when the degree of [latex]P(x)[\/latex] is one more than the degree of [latex]Q(x)[\/latex]. For example, in the function\u00a0[latex]f(x)=\\dfrac{x^3-4x^2+2}{x^2-1}[\/latex] (Figure 1), [latex]P(x)=x^3-4x^2+2[\/latex] and has degree 3, while [latex]Q(x)=x^2-1[\/latex] which has degree 2. Since 3 is one more than 2, there is a slant asymptote.\r\n<h2>Graphing the Inverse Function of a Rational Function<\/h2>\r\nWe may graph the inverse of a rational function by creating and using its inverse table. For example, given the function [latex]f(x)=\\dfrac{1}{x}[\/latex], we may graph the function by creating a table of values (Table 1).\r\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 23.301828863241806%;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 19px;\">\r\n<th style=\"width: 12.7637%; height: 10px; text-align: center;\" scope=\"row\">[latex]x[\/latex]<\/th>\r\n<th style=\"width: 12.7637%; height: 10px; text-align: center;\">[latex]y=\\dfrac{1}{x}[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]\u20132[\/latex]<\/th>\r\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-\\dfrac{1}{2}[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]\u20131[\/latex]<\/th>\r\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]\u20131[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-\\dfrac{1}{2}[\/latex]<\/th>\r\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]\u20132[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/th>\r\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]2[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]1[\/latex]<\/th>\r\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]1[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]2[\/latex]<\/th>\r\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]3[\/latex]<\/th>\r\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{3}[\/latex]<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 12.7637%;\" colspan=\"2\">Table 1. Table of values for [latex]f(x)=\\dfrac{1}{x}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left;\">The inverse of the function is found by switching the values of the [latex]x[\/latex] and [latex]y[\/latex] columns so that the inputs become the values of [latex]y[\/latex] and the outputs become the values of [latex]x[\/latex]. The table after switching the values of the [latex]x[\/latex] and [latex]y[\/latex] columns is the inverse table (Table 2).<\/p>\r\n\r\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 25.5274%;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 19px;\">\r\n<th style=\"width: 12.7637%; height: 10px; text-align: center;\" scope=\"row\">[latex]x[\/latex]<\/th>\r\n<th style=\"width: 12.7637%; height: 10px; text-align: center;\">[latex]y=\\dfrac{1}{x}[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-\\dfrac{1}{2}[\/latex]<\/td>\r\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-1[\/latex]<\/td>\r\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-2[\/latex]<\/td>\r\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-\\dfrac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/td>\r\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 19px;\">\r\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{3}[\/latex]<\/td>\r\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 12.7637%;\" colspan=\"2\">Table 2. The inverse table for [latex]f^{-1}(x)=\\dfrac{1}{x}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFigure 2 shows the graph of the inverse of the function [latex]f(x)=\\dfrac{1}{x}[\/latex] drawn from its inverse table. Notice that the graph of the inverse is exactly the same as the graph of the original function [latex]f(x)=\\dfrac{1}{x}[\/latex]. In other words, the function [latex]f(x)=\\dfrac{1}{x}[\/latex] is the inverse of the function itself. The inverse function is a reflection of the original function with respect to the line of symmetry [latex]y=x[\/latex].\r\n\r\n[caption id=\"attachment_2813\" align=\"aligncenter\" width=\"433\"]<img class=\"wp-image-2813\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/20165726\/7-3-Inverse-300x300.png\" alt=\"Graph of f(x)=1\/x with the line of symmetry y=x and several symmetric points on the graph reflected across the line of symmetry.\" width=\"433\" height=\"433\" \/> Figure 2. The inverse of the function [latex]f(x)=\\dfrac{1}{x}[\/latex] is [latex]f^{-1}(x)=\\dfrac{1}{x}[\/latex].[\/caption]<span style=\"color: #373d3f; font-size: 16px; font-weight: 400; orphans: 1;\">Since the graph of the function [latex]f(x)=\\dfrac{1}{x}[\/latex] is symmetric across the line [latex]y=x[\/latex], the inverse function is identical to the original function.<\/span>\r\n<h2>Determining the Inverse Function of an One-to-One Rational Function<\/h2>\r\nTo determine the equation of the inverse function of a one-to-one rational function, we use the same idea of switching the input and output. We start by writing [latex]y=f(x)[\/latex], switch [latex]x[\/latex] and [latex]y[\/latex], and then solve for [latex]y[\/latex].\r\n\r\nFor example, to determine the inverse function of the one-to-one rational function\u00a0[latex]g(x)=\\dfrac{1}{x}[\/latex], we write [latex]y=g(x)[\/latex] then switch [latex]x[\/latex] and [latex]y[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}g(x)&amp;=\\dfrac{1}{x}\\\\\\\\y&amp;=\\dfrac{1}{x}\\\\\\\\x&amp;=\\dfrac{1}{y}\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">At this point, we have the inverse. We now need to solve for [latex]y[\/latex] so we can write the inverse using function notation by replacing [latex]y[\/latex] with [latex]g^{-1}(x)[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&amp;=\\dfrac{1}{y}\\\\\\\\x\\color{blue}{\\cdot y}&amp;=\\dfrac{1}{y}\\color{blue}{\\cdot y}&amp;&amp;\\text{Multiply both sides by }y\\text{ to clear the fractions}\\\\\\\\xy&amp;=1\\\\\\\\y&amp;=\\dfrac{1}{x}\\\\\\\\g^{-1}(x)&amp;=\\dfrac{1}{x}\\end{aligned}[\/latex]<\/p>\r\nTherefore, the equation of the inverse function is\u00a0[latex]g^{-1}(x)=\\dfrac{1}{x}[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nDetermine the\u00a0inverse function of the one-to-one rational function\u00a0[latex]h(x)=\\dfrac{x-1}{x+4}[\/latex].\r\n<h4>Solution<\/h4>\r\nWe start by writing [latex]y=h(x)[\/latex] then switch[latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}y&amp;=\\dfrac{x-1}{x+4}&amp;&amp;\\text{Write }y\\text{ for }h(x)\\\\\\\\x&amp;=\\dfrac{y-1}{y+4}&amp;&amp;\\text{Switch }x\\text{ and }y\\\\\\\\x\\color{blue}{(y+4)}&amp;=\\dfrac{y-1}{y+4}\\color{blue}{\\cdot (y+4)}&amp;&amp;\\text{Multiply both sides by }y+4\\text{ to clear the fractions}\\\\\\\\x(y+4)&amp;=y-1\\\\\\\\xy+4x &amp;=y-1&amp;&amp;\\text{Multiply the left side using the distributive property}\\\\\\\\xy-y &amp;=-4x-1&amp;&amp;\\text{Collect }y\\text{ terms on the left side}\\\\\\\\y(x-1)&amp;=-4x-1&amp;&amp;\\text{Pull }y\\text{ out as a common factor on the left side}\\\\\\\\y&amp;=\\dfrac{-4x-1}{x-1}&amp;&amp;\\text{Divide both sides by }x-1\\end{aligned}[\/latex]<\/p>\r\nNow write the inverse in function notation, [latex]h^{-1}(x)=\\dfrac{-4x-1}{x-1}[\/latex] or by pulling out [latex]-1[\/latex] as a common factor on the numerator,\u00a0[latex]h^{-1}(x)=-\\dfrac{4x+1}{x-1}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nDetermine the\u00a0inverse function of the one-to-one rational function\u00a0[latex]h(x)=\\dfrac{x+5}{x-1}[\/latex].\r\n<h4>Solution<\/h4>\r\nWe start by writing [latex]y=h(x)[\/latex] then switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}y&amp;=\\dfrac{x+5}{x-1}&amp;&amp;\\text{Write }y\\text{ for }h(x)\\\\\\\\x&amp;=\\dfrac{y+5}{y-1}&amp;&amp;\\text{Switch }x\\text{ and }y\\\\\\\\x\\color{blue}{(y-1)}&amp;=\\dfrac{y+5}{y-1}\\color{blue}{\\cdot (y-1)}&amp;&amp;\\text{Multiply both sides by }y-1\\text{ to clear the fractions}\\\\\\\\x(y-1)&amp;=y+5\\\\\\\\xy-x &amp;=y+5&amp;&amp;\\text{Multiply the left side using the distributive property}\\\\\\\\xy-y &amp;=x+5&amp;&amp;\\text{Collect }y\\text{ terms on the left side}\\\\\\\\y(x-1)&amp;=x+5&amp;&amp;\\text{Pull }y\\text{ out as a common factor on the left side}\\\\\\\\y&amp;=\\dfrac{x+5}{x-1}&amp;&amp;\\text{Divide both sides by }x-1\\end{aligned}[\/latex]<\/p>\r\nNow write the inverse in function notation, [latex]h^{-1}(x)=\\dfrac{x+5}{x-1}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nDetermine the\u00a0inverse function of the one-to-one rational function:\r\n<ol>\r\n \t<li>[latex]h(x)=\\dfrac{x+4}{x-6}[\/latex]<\/li>\r\n \t<li>[latex]g(x)=\\dfrac{x+7}{x+4}[\/latex]<\/li>\r\n \t<li>[latex]f(x)=\\dfrac{2x+3}{5x+4}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm547\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm547\"]\r\n<ol>\r\n \t<li>[latex]h^{-1}(x)=\\dfrac{6x+4}{x-1}[\/latex]<\/li>\r\n \t<li>[latex]h^{-1}(x)=\\dfrac{-4x+7}{x-1}[\/latex]<\/li>\r\n \t<li>[latex]h^{-1}(x)=\\dfrac{-4x+3}{5x-2}[\/latex]<\/li>\r\n<\/ol>\r\n[\/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=\"textbox learning-objectives\">\n<h1>Learning Objectives<\/h1>\n<ul>\n<li>Graph the inverse of a rational function<\/li>\n<li>Find the equation of the inverse function of a one-to-one rational function<\/li>\n<\/ul>\n<\/div>\n<p>In chapter 3, we discussed that every function has an inverse, but <strong>only a one-to-one function has an inverse function<\/strong>. Some rational functions are one-to-one functions such as [latex]f(x)=\\dfrac{1}{x}[\/latex] or\u00a0[latex]f(x)=\\dfrac{x-1}{x+4}[\/latex]. Therefore, their inverse is a function. Some rational functions are many-to-one functions such as\u00a0[latex]f(x)=\\dfrac{1}{x^2-1}[\/latex] or\u00a0[latex]f(x)=\\dfrac{x^3-4x^2+2}{x^2-1}[\/latex]. Therefore, their inverse is not a function (Figure 1).<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%; text-align: center;\">One-to-one rational functions that have an inverse function<\/th>\n<th style=\"width: 50%; text-align: center;\">Many-to-one rational functions that do not have an inverse function<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3301 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092326.109-300x300.png\" alt=\"graph of the function f(x)=1\/x\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092326.109-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092326.109-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092326.109-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092326.109-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092326.109-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092326.109-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092326.109.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/td>\n<td style=\"width: 50%; text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3303 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092927.854-300x300.png\" alt=\"Graph of f(x)=1\/(x^2-1), two separate curve branches in opposite quadrants: one opening left and upward approach a vertical asymptote at x=-1, the other right and upward approaching a vertical asymptote at x=1. A third curve branch opens downward between and approaching the two asymptotes.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092927.854-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092927.854-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092927.854-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092927.854-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092927.854-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092927.854-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092927.854.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3302 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092612.998-300x300.png\" alt=\"Graph of f(x)=(x-1)\/(x+4), two separate curve branches in opposite quadrants: one opening left and upward, the other right and downward, each approaching a vertical asymptote at x=-4 and and a horizontal asymptote at y=1.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092612.998-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092612.998-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092612.998-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092612.998-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092612.998-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092612.998-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T092612.998.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/td>\n<td style=\"width: 50%; text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3307 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T105745.272-300x300.png\" alt=\"Graph of f(x)=(x^3-4x^2+2)\/(x^2-1), two separate curve branches in opposite quadrants: one opening southwest and downward with one side approaching a diagonal line and the other side approaching a vertical line, the other northeast and downward with one side approaching the same diagonal line and the other side approaching another vertical line. A third curve branch opens upward between the two curve branches and approaching the two vertical lines.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T105745.272-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T105745.272-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T105745.272-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T105745.272-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T105745.272-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T105745.272-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-15T105745.272.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\" colspan=\"2\">Figure 1. Graphs of rational functions<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice that three of the graphs in figure 1 have horizontal and vertical asymptotes but the 4th graph has two vertical asymptotes and a <em><strong>slant asymptote<\/strong><\/em>. A slant asymptote is a line of the form [latex]y=mx+b[\/latex] that is neither vertical nor horizontal but that the graph gets closer and closer to as [latex]x[\/latex] approaches positive and negative infinity. Slant asymptotes occur in the graph of a rational function [latex]f(x)=\\dfrac{P(x)}{Q(x)}[\/latex] when the degree of [latex]P(x)[\/latex] is one more than the degree of [latex]Q(x)[\/latex]. For example, in the function\u00a0[latex]f(x)=\\dfrac{x^3-4x^2+2}{x^2-1}[\/latex] (Figure 1), [latex]P(x)=x^3-4x^2+2[\/latex] and has degree 3, while [latex]Q(x)=x^2-1[\/latex] which has degree 2. Since 3 is one more than 2, there is a slant asymptote.<\/p>\n<h2>Graphing the Inverse Function of a Rational Function<\/h2>\n<p>We may graph the inverse of a rational function by creating and using its inverse table. For example, given the function [latex]f(x)=\\dfrac{1}{x}[\/latex], we may graph the function by creating a table of values (Table 1).<\/p>\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 23.301828863241806%;\">\n<tbody>\n<tr style=\"height: 19px;\">\n<th style=\"width: 12.7637%; height: 10px; text-align: center;\" scope=\"row\">[latex]x[\/latex]<\/th>\n<th style=\"width: 12.7637%; height: 10px; text-align: center;\">[latex]y=\\dfrac{1}{x}[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 19px;\">\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]\u20132[\/latex]<\/th>\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-\\dfrac{1}{2}[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 19px;\">\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]\u20131[\/latex]<\/th>\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]\u20131[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 19px;\">\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-\\dfrac{1}{2}[\/latex]<\/th>\n<th style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]\u20132[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 19px;\">\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/th>\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]2[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 19px;\">\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]1[\/latex]<\/th>\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]1[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 19px;\">\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]2[\/latex]<\/th>\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 19px;\">\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]3[\/latex]<\/th>\n<th style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{3}[\/latex]<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 12.7637%;\" colspan=\"2\">Table 1. Table of values for [latex]f(x)=\\dfrac{1}{x}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left;\">The inverse of the function is found by switching the values of the [latex]x[\/latex] and [latex]y[\/latex] columns so that the inputs become the values of [latex]y[\/latex] and the outputs become the values of [latex]x[\/latex]. The table after switching the values of the [latex]x[\/latex] and [latex]y[\/latex] columns is the inverse table (Table 2).<\/p>\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 25.5274%;\">\n<tbody>\n<tr style=\"height: 19px;\">\n<th style=\"width: 12.7637%; height: 10px; text-align: center;\" scope=\"row\">[latex]x[\/latex]<\/th>\n<th style=\"width: 12.7637%; height: 10px; text-align: center;\">[latex]y=\\dfrac{1}{x}[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 19px;\">\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-\\dfrac{1}{2}[\/latex]<\/td>\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 19px;\">\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-1[\/latex]<\/td>\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 19px;\">\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-2[\/latex]<\/td>\n<td style=\"width: 12.7637%; height: 34px; text-align: center;\">[latex]-\\dfrac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 19px;\">\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]2[\/latex]<\/td>\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 19px;\">\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]1[\/latex]<\/td>\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 19px;\">\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/td>\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]2[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 19px;\">\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]\\dfrac{1}{3}[\/latex]<\/td>\n<td style=\"width: 12.7637%; height: 19px; text-align: center;\">[latex]3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 12.7637%;\" colspan=\"2\">Table 2. The inverse table for [latex]f^{-1}(x)=\\dfrac{1}{x}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Figure 2 shows the graph of the inverse of the function [latex]f(x)=\\dfrac{1}{x}[\/latex] drawn from its inverse table. Notice that the graph of the inverse is exactly the same as the graph of the original function [latex]f(x)=\\dfrac{1}{x}[\/latex]. In other words, the function [latex]f(x)=\\dfrac{1}{x}[\/latex] is the inverse of the function itself. The inverse function is a reflection of the original function with respect to the line of symmetry [latex]y=x[\/latex].<\/p>\n<div id=\"attachment_2813\" style=\"width: 443px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2813\" class=\"wp-image-2813\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/20165726\/7-3-Inverse-300x300.png\" alt=\"Graph of f(x)=1\/x with the line of symmetry y=x and several symmetric points on the graph reflected across the line of symmetry.\" width=\"433\" height=\"433\" \/><\/p>\n<p id=\"caption-attachment-2813\" class=\"wp-caption-text\">Figure 2. The inverse of the function [latex]f(x)=\\dfrac{1}{x}[\/latex] is [latex]f^{-1}(x)=\\dfrac{1}{x}[\/latex].<\/p>\n<\/div>\n<p><span style=\"color: #373d3f; font-size: 16px; font-weight: 400; orphans: 1;\">Since the graph of the function [latex]f(x)=\\dfrac{1}{x}[\/latex] is symmetric across the line [latex]y=x[\/latex], the inverse function is identical to the original function.<\/span><\/p>\n<h2>Determining the Inverse Function of an One-to-One Rational Function<\/h2>\n<p>To determine the equation of the inverse function of a one-to-one rational function, we use the same idea of switching the input and output. We start by writing [latex]y=f(x)[\/latex], switch [latex]x[\/latex] and [latex]y[\/latex], and then solve for [latex]y[\/latex].<\/p>\n<p>For example, to determine the inverse function of the one-to-one rational function\u00a0[latex]g(x)=\\dfrac{1}{x}[\/latex], we write [latex]y=g(x)[\/latex] then switch [latex]x[\/latex] and [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}g(x)&=\\dfrac{1}{x}\\\\\\\\y&=\\dfrac{1}{x}\\\\\\\\x&=\\dfrac{1}{y}\\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: left;\">At this point, we have the inverse. We now need to solve for [latex]y[\/latex] so we can write the inverse using function notation by replacing [latex]y[\/latex] with [latex]g^{-1}(x)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&=\\dfrac{1}{y}\\\\\\\\x\\color{blue}{\\cdot y}&=\\dfrac{1}{y}\\color{blue}{\\cdot y}&&\\text{Multiply both sides by }y\\text{ to clear the fractions}\\\\\\\\xy&=1\\\\\\\\y&=\\dfrac{1}{x}\\\\\\\\g^{-1}(x)&=\\dfrac{1}{x}\\end{aligned}[\/latex]<\/p>\n<p>Therefore, the equation of the inverse function is\u00a0[latex]g^{-1}(x)=\\dfrac{1}{x}[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Determine the\u00a0inverse function of the one-to-one rational function\u00a0[latex]h(x)=\\dfrac{x-1}{x+4}[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>We start by writing [latex]y=h(x)[\/latex] then switch[latex]x[\/latex] and [latex]y[\/latex] to get the inverse:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}y&=\\dfrac{x-1}{x+4}&&\\text{Write }y\\text{ for }h(x)\\\\\\\\x&=\\dfrac{y-1}{y+4}&&\\text{Switch }x\\text{ and }y\\\\\\\\x\\color{blue}{(y+4)}&=\\dfrac{y-1}{y+4}\\color{blue}{\\cdot (y+4)}&&\\text{Multiply both sides by }y+4\\text{ to clear the fractions}\\\\\\\\x(y+4)&=y-1\\\\\\\\xy+4x &=y-1&&\\text{Multiply the left side using the distributive property}\\\\\\\\xy-y &=-4x-1&&\\text{Collect }y\\text{ terms on the left side}\\\\\\\\y(x-1)&=-4x-1&&\\text{Pull }y\\text{ out as a common factor on the left side}\\\\\\\\y&=\\dfrac{-4x-1}{x-1}&&\\text{Divide both sides by }x-1\\end{aligned}[\/latex]<\/p>\n<p>Now write the inverse in function notation, [latex]h^{-1}(x)=\\dfrac{-4x-1}{x-1}[\/latex] or by pulling out [latex]-1[\/latex] as a common factor on the numerator,\u00a0[latex]h^{-1}(x)=-\\dfrac{4x+1}{x-1}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Determine the\u00a0inverse function of the one-to-one rational function\u00a0[latex]h(x)=\\dfrac{x+5}{x-1}[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>We start by writing [latex]y=h(x)[\/latex] then switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}y&=\\dfrac{x+5}{x-1}&&\\text{Write }y\\text{ for }h(x)\\\\\\\\x&=\\dfrac{y+5}{y-1}&&\\text{Switch }x\\text{ and }y\\\\\\\\x\\color{blue}{(y-1)}&=\\dfrac{y+5}{y-1}\\color{blue}{\\cdot (y-1)}&&\\text{Multiply both sides by }y-1\\text{ to clear the fractions}\\\\\\\\x(y-1)&=y+5\\\\\\\\xy-x &=y+5&&\\text{Multiply the left side using the distributive property}\\\\\\\\xy-y &=x+5&&\\text{Collect }y\\text{ terms on the left side}\\\\\\\\y(x-1)&=x+5&&\\text{Pull }y\\text{ out as a common factor on the left side}\\\\\\\\y&=\\dfrac{x+5}{x-1}&&\\text{Divide both sides by }x-1\\end{aligned}[\/latex]<\/p>\n<p>Now write the inverse in function notation, [latex]h^{-1}(x)=\\dfrac{x+5}{x-1}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Determine the\u00a0inverse function of the one-to-one rational function:<\/p>\n<ol>\n<li>[latex]h(x)=\\dfrac{x+4}{x-6}[\/latex]<\/li>\n<li>[latex]g(x)=\\dfrac{x+7}{x+4}[\/latex]<\/li>\n<li>[latex]f(x)=\\dfrac{2x+3}{5x+4}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm547\">Show Answer<\/span><\/p>\n<div id=\"qhjm547\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]h^{-1}(x)=\\dfrac{6x+4}{x-1}[\/latex]<\/li>\n<li>[latex]h^{-1}(x)=\\dfrac{-4x+7}{x-1}[\/latex]<\/li>\n<li>[latex]h^{-1}(x)=\\dfrac{-4x+3}{5x-2}[\/latex]<\/li>\n<\/ol>\n<\/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-2810\">\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>The Inverse of a Rational Function. <strong>Authored by<\/strong>: Leo Chang and 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><li>All graphs created using desmos graphing calculator. <strong>Authored by<\/strong>: Leo Chang. <strong>Provided by<\/strong>: Utah Valley University. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.desmos.com\/calculator\">http:\/\/www.desmos.com\/calculator<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Examples and Try Its: hjm547. <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>\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":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"The Inverse of a Rational Function\",\"author\":\"Leo Chang and Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All graphs created using desmos graphing calculator\",\"author\":\"Leo Chang\",\"organization\":\"Utah Valley University\",\"url\":\"www.desmos.com\/calculator\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Examples and Try Its: hjm547\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2810","chapter","type-chapter","status-publish","hentry"],"part":2581,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2810","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/users\/422608"}],"version-history":[{"count":27,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2810\/revisions"}],"predecessor-version":[{"id":4829,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2810\/revisions\/4829"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/parts\/2581"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2810\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/media?parent=2810"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2810"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/contributor?post=2810"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/license?post=2810"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}