{"id":1370,"date":"2022-04-06T16:37:52","date_gmt":"2022-04-06T16:37:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=1370"},"modified":"2025-12-11T20:16:53","modified_gmt":"2025-12-11T20:16:53","slug":"3-3-1-inverse-and-inverse-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/3-3-1-inverse-and-inverse-function\/","title":{"raw":"3.3.1: The Inverse of a Polynomial Function","rendered":"3.3.1: The Inverse of a Polynomial 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>Explain the inverse of a function<\/li>\r\n \t<li>Evaluate the inverse of polynomial function<\/li>\r\n \t<li>Graph the inverse of a polynomial function\r\n<ul>\r\n \t<li>by switching domain and range<\/li>\r\n \t<li>using the line of symmetry [latex]y=x[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Determine whether the inverse of a polynomial function is a function\r\n<ul>\r\n \t<li>by mapping<\/li>\r\n \t<li>by determining if the original function is one-to-one<\/li>\r\n \t<li>by the vertical line test<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Find the equation of an inverse function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Inverse of a Function<\/h2>\r\nWe explored the concept of inverse functions in section 1.3 where we defined the inverse of a function as the switch between the domain and range of the function (or mapping). It is the exchange of the input and output of a function. We discovered that while all functions have inverses, only one-to-one-functions have inverses that are also functions. The notation for the inverse of a one-to-one function [latex]f[\/latex] is [latex]f^{-1}[\/latex], which is read \"the inverse of [latex]f[\/latex]. Recall that t<span style=\"font-size: 1em;\">he symbol \"[latex]-1[\/latex]\" is not an exponent; it means \"the inverse\". Therefore, the notation [latex]f^{-1}[\/latex] means the inverse of the function [latex]f[\/latex].\u00a0<\/span>\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">For example, if a function is defined as the set of ordered pairs<\/span>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]f:\\;\\{(1, 2), (3, 8), (4, 2)\\}[\/latex], <\/span><\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">it has an inverse <\/span>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]\\{(2, 1), (8, 3), (2, 4)\\}[\/latex].<\/span><\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">However, this inverse is NOT a function because the input 2 has two outputs, 1 and 4. This is because the original function [latex]f[\/latex] was NOT a one-to-one function (the function value 2 is the output of two domain values 1 and 4). <\/span>\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">On the other hand, if a one-to-one function is defined as the set of ordered pairs <\/span>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]g:\\;\\{(1, 2), (3, 8), (4, 9)\\}[\/latex], <\/span><\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">then its inverse function is\u00a0<\/span>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]g^{-1}:\\; \\{(2, 1), (8, 3), (9, 4)\\}[\/latex],<\/span><\/p>\r\nwhich is a function. Consequently, we can use the\u00a0<span style=\"font-size: 1rem; text-align: initial;\">inverse function notation [latex]g^{-1}[\/latex], which we could not use with the inverse of [latex]f[\/latex].<\/span>\r\n<div class=\"textbox shaded\">\r\n<h3>Inverse of a function<\/h3>\r\nEvery function has an inverse, but <strong>only a one-to-one function has an inverse function<\/strong>.\r\n\r\n<\/div>\r\nTo evaluate the inverse of a one-to-one function, we need to switch the input and output of the function. The output of the original function becomes the input of its inverse function,\u00a0and the input of the original function becomes the output of its inverse function. To illustrate,\r\n<p style=\"text-align: center;\">[latex]f(1)=2[\/latex], means [latex]f^{-1}(2)=1[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]f(3)=8[\/latex], means [latex]f^{-1}(8)=3[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]f(4)=9[\/latex], means [latex]f^{-1}(9)=4[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nFor the function [latex]f(x)=3x+5[\/latex], find:\r\n<ol>\r\n \t<li>\u00a0[latex]f^{-1}(17)[\/latex]<\/li>\r\n \t<li>\u00a0[latex]f^{-1}(-1)[\/latex]<\/li>\r\n \t<li>\u00a0[latex]f^{-1}(0)[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\n1. 17 is the input of the inverse function so this means that 17 was the output of the original function. To find\u00a0[latex]f^{-1}(17)[\/latex] we need to find the associated [latex]x[\/latex]-value when [latex]f(x)=17[\/latex].\r\n\r\n[latex]\\begin{aligned}f(x)&amp;=3x+5\\\\17&amp;=3x+5\\\\12&amp;=3x\\\\4&amp;=x\\end{aligned}[\/latex]\r\n\r\nSo, since [latex]f(4)=17[\/latex],\u00a0[latex]f^{-1}(17)=4[\/latex].\r\n\r\n2. \u20131\u00a0is the input of the inverse function so this means that \u20131 was the output of the original function. To find\u00a0[latex]f^{-1}(\u20131)[\/latex] we need to find the associated [latex]x[\/latex]-value when [latex]f(x)=-1[\/latex].\r\n\r\n[latex]\\begin{aligned}f(x)&amp;=3x+5\\\\-1&amp;=3x+5\\\\-6&amp;=3x\\\\-2&amp;=x\\end{aligned}[\/latex]\r\n\r\nSo, since [latex]f(-2)=-1[\/latex],\u00a0[latex]f^{-1}(-1)=-2[\/latex].\r\n\r\n3. 0\u00a0is the input of the inverse function so this means that 0 was the output of the original function. To find\u00a0[latex]f^{-1}(0)[\/latex] we need to find the associated [latex]x[\/latex]-value when [latex]f(x)=0[\/latex].\r\n\r\n[latex]\\begin{aligned}f(x)&amp;=3x+5\\\\0&amp;=3x+5\\\\-5&amp;=3x\\\\-\\dfrac{5}{3}&amp;=x\\end{aligned}[\/latex]\r\n\r\nSo, since [latex]f(-\\dfrac{5}{3})=0[\/latex],\u00a0[latex]f^{-1}(0)=-\\dfrac{5}{3}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nFor the function [latex]f(x)=-2x+7[\/latex], find:\r\n<ol>\r\n \t<li>\u00a0[latex]f^{-1}(5)[\/latex]<\/li>\r\n \t<li>\u00a0[latex]f^{-1}(-3)[\/latex]<\/li>\r\n \t<li>\u00a0[latex]f^{-1}(0)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm473\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm473\"]\r\n<ol>\r\n \t<li>[latex]f^{-1}(5)=1[\/latex]<\/li>\r\n \t<li>[latex]f^{-1}(-3)=5[\/latex]<\/li>\r\n \t<li>[latex]f^{-1}(0)=\\dfrac{7}{2}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Graphing the Inverse of a Polynomial Function<\/h2>\r\nIf we are given the graph of a polynomial function on the coordinate plane we can graph its inverse by switching the [latex]x[\/latex] and [latex]y[\/latex] coordinates for every point on the graph. Figure 1 shows that if we switch the coordinates of the points on the blue graph (\u20132, \u20134), (0, 4) and (2, 12) to (\u20134, \u20132), (4, 0), and (12, 2) we get the green graph, which is the inverse of the blue graph.\r\n\r\n[caption id=\"attachment_1764\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-1764 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-72-300x300.png\" alt=\"function and its inverse demonstrating that each point on one gives you a point on the other by switching the x and y coordinates.\" width=\"300\" height=\"300\" \/> Figure 1. Graphing an inverse function[\/caption]\r\n\r\nAnother way to graph the inverse is to reflect the graph across the line [latex]y = x[\/latex]. Figure 1 also shows that the blue curve and its inverse the green curve are reflections of one another across the line of symmetry [latex]y = x[\/latex]. The green curve is the inverse of the blue curve, and the blue curve is the inverse of the green curve.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nGiven the graph of the function [latex]y=f(x)[\/latex], graph the inverse function [latex]y=f^{-1}(x)[\/latex] using symmetry.\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-1756 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-64-300x300.png\" alt=\"graph of a line f of x equals 5x -2.\" width=\"300\" height=\"300\" \/><\/p>\r\n\r\n<h4>Solution<\/h4>\r\nFirst graph the line [latex]y=x[\/latex] so we can reflect the graph across this line. Choose a couple of points on the graph and switch their coordinates to help graph the reflection. (0, \u20134) and (2, 6) lie on the graph of the [latex]y=f(x)[\/latex], so (\u20134, 0) and (6, 2) must lie on the graph of the inverse function [latex]y=f^{-1}(x)[\/latex].\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-1757 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-65-300x300.png\" alt=\"graph of a line f of x equals 5x -2, and it's inverse.  F goes through (0,-4) and (2,6), and f inverse goes through (-4,0) and (6,2).\" width=\"300\" height=\"300\" \/><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nGiven the graph of the function [latex]y=f(x)[\/latex], graph the inverse function [latex]y=f^{-1}(x)[\/latex] using symmetry.\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-1760 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-69-300x300.png\" alt=\"Graph of the right side of a parabola starting at (0,-4) and passing through (2,0).\" width=\"300\" height=\"300\" \/><\/p>\r\n[reveal-answer q=\"hjm277\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm277\"]\r\n\r\n<img class=\"aligncenter wp-image-1761 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-68-300x300.png\" alt=\"Graph of the function above and its inverse, the top half of a sideways parabola starting at (-4,0) and passing through (0,2).\" width=\"300\" height=\"300\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIt is important to note that the inverse of a function is not necessarily a function. For example, given the function [latex]f[\/latex]: {(1, 2), (4, 2), (3, 6)}, its inverse is not a function. Its inverse is the set {(2, 1), (2, 4), (6, 3)}. This set is not a function because it is a one-to-many mapping (i.e., 2 is mapped to both 1 and 4). In other words, as long as there is one case of one input with many outputs in a mapping, it is not a function.\r\n\r\nFigure 2 shows a parabola that opens up and its inverse, which opens right. The inverse is not a function because it does not pass the vertical line test. A vertical line test on the inverse graph shows that there are two intersection points with any vertical line [latex]x = c,\\;c&gt;0[\/latex] and the graph of the inverse, which opens right, has one input with two outputs (a one-to-many mapping). Therefore, the inverse of the parabola is not a function.\r\n\r\n[caption id=\"attachment_1763\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-1763 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-71-300x300.png\" alt=\"Inverse that is not a function. The complete parabolas from above and a vertical line showing the inverse graph is not a function.\" width=\"300\" height=\"300\" \/> Figure 2. A parabola and its inverse[\/caption]\r\n\r\nOf course, we can tell from the graph of the original function whether or not the inverse will be a function. If the original graph is NOT one-to-one, the inverse is NOT a function. As stated previously, all functions have inverses, but only one-to-one functions have inverse functions.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nGraph the inverse of the function. Determine if the inverse is a function.\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%;\"><strong>A one to one function<\/strong><\/th>\r\n<th style=\"width: 50%;\"><strong>Not a one to one function<\/strong><\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">1.\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-1765 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-74-300x300.png\" alt=\"graph of a cube function Passing through the points (-5,-9.5), (0,3), and (4,9.5).\" width=\"300\" height=\"300\" \/><\/p>\r\n<\/td>\r\n<td style=\"width: 50%;\">2.\r\n\r\n<img class=\"aligncenter wp-image-1767 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-76-300x300.png\" alt=\"graph of a polynomial function passing through (-2,11), (-1,-4), (1,4), and (2,11).\" width=\"300\" height=\"300\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Solution<\/h4>\r\nChoose a few points on the original graph, switch the coordinates, and graph the new points. These new points will lie on the inverse. Graph the line [latex]y=x[\/latex] and use the newly plotted points to help draw the reflection of the original graph across the line [latex]y=x[\/latex].\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%; vertical-align: top; text-align: center;\"><strong>A cubic and it's inverse<\/strong><\/th>\r\n<th style=\"width: 50%; text-align: center;\"><strong>A 4th degree and it's inverse<\/strong><\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\">1.\r\n\r\n<img class=\"aligncenter 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=\"The cubic function above with it's inverse cube root function, passing through the points (-9.5,-5), (3,0), and (9.5,4).\" width=\"300\" height=\"300\" \/>\r\n\r\nThe original graph is one-to-one (it passes the horizontal line test), so the inverse is a function. In addition, the inverse is a function since any vertical line passes through only 1 point.<\/td>\r\n<td style=\"width: 50%;\">2.\r\n\r\n<img class=\"aligncenter wp-image-1768 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-75-300x300.png\" alt=\"The 4th degree polynomial above and it's inverse, passing through (-5,0), (-4,1), (-4,-1), (11,2), and (11,-2).\" width=\"300\" height=\"300\" \/>\r\n\r\nThe original graph is NOT one-to-one (it fails the horizontal line test), so the inverse is NOT a function.Also, the inverse is not a function as there are vertical lines that pass through the graph at 2 points.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nGraph the inverse of the function. Determine if the inverse is a function.\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;\"><strong>A linear function<\/strong><\/th>\r\n<th style=\"width: 50%; text-align: center;\"><strong>A quadratic function<\/strong><\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">1.\r\n\r\n<img class=\"aligncenter wp-image-1772 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-80-300x300.png\" alt=\"A linear function passing through (0, -1) and (2,5).\" width=\"300\" height=\"300\" \/><\/td>\r\n<td style=\"width: 50%;\">2.\r\n\r\n<img class=\"aligncenter wp-image-1775 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-84-300x300.png\" alt=\"An inverted quadratic function passing through (-2, -10), (0,8), and (2, -10).\" width=\"300\" height=\"300\" \/>\r\n\r\n&nbsp;<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"hjm172\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm172\"]\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%; vertical-align: top; text-align: center;\">The linear function and its inverse function<\/th>\r\n<th style=\"width: 50%; text-align: center;\">The quadratic function and its inverse graph<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; vertical-align: top;\">1.\r\n\r\n<img class=\"aligncenter wp-image-1770 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-791-300x300.png\" alt=\"The linear function described above and its inverse passing through (-4,1), (-1,0), (2,1), and (5,2).\" width=\"300\" height=\"300\" \/>\r\n\r\nThe inverse is a function.<\/td>\r\n<td style=\"width: 50%;\">2.\r\n\r\n<img class=\"aligncenter wp-image-1776 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-85-300x300.png\" alt=\"The quadratic function above and its inverse passing through (-8,4), (-8,-4), and (8,0).\" width=\"300\" height=\"300\" \/>\r\n\r\nThe inverse is not a function.\r\n\r\n&nbsp;<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding the Inverse Algebraically<\/h2>\r\nTo find the equation of the inverse of a polynomial function we use the same idea of switching the input and output. We start by writing [latex]y=f(x)[\/latex] then switch the [latex]x[\/latex] and [latex]y[\/latex] variables. For the function\u00a0[latex]f(x)=x^3+2[\/latex], we start by writing [latex]y=x^3+2[\/latex]. To find the inverse, we switch the input and the output i.e., the [latex]x[\/latex] and [latex]y[\/latex] variables. This results in the equation [latex]x = y^3+2[\/latex], which is the equation of the inverse of the function[latex]f(x)=x^3+2[\/latex]. Solving for [latex]y[\/latex] we get:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&amp;=y^3+2\\\\x-2&amp;=y^3\\\\ \\sqrt[3]{x-2}&amp;=y\\end{aligned}[\/latex]<\/p>\r\n<span style=\"font-size: 1em;\">Since [latex]f(x)=x^3+2[\/latex] is a one-to-one function, its inverse is also a function.<\/span><span style=\"font-size: 1em;\">\u00a0<\/span>The next step is to write the inverse as a function. Because it is the inverse of the one-to-one function\u00a0[latex]f[\/latex], we use the notation\u00a0[latex]f^{-1}[\/latex] to represent it as the inverse function of [latex]f[\/latex]:\r\n<p style=\"text-align: center;\">[latex]f^{-1}(x) = \\sqrt[3]{x - 2}[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 4<\/h3>\r\nFind the inverse function of the polynomial function [latex]f(x)=5x-4[\/latex].\r\n<h4>Solution<\/h4>\r\nStart by writing [latex]y[\/latex] for [latex]f(x)[\/latex]:\r\n<p style=\"text-align: center;\">[latex]y=5x-4[\/latex]<\/p>\r\nSwitch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse equation:\r\n<p style=\"text-align: center;\">[latex]x=5y-4[\/latex]<\/p>\r\nSolve for [latex]y[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&amp;=5y-4\\\\x+4&amp;=5y\\\\\\dfrac{x+4}{5}&amp;=y\\\\ \\dfrac{1}{5}x+\\dfrac{4}{5}&amp;=y\\end{aligned}[\/latex]<\/p>\r\nSince [latex]f(x)[\/latex] is a linear function (with a non-zero slope), it is one-to-one. Write the inverse function using function notation:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}y&amp;=\\dfrac{1}{5}x+\\dfrac{4}{5}\\\\f^{-1}(x)&amp;=\\dfrac{1}{5}x+\\dfrac{4}{5}\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 5<\/h3>\r\nFind the inverse function of the polynomial function [latex]g(x)=x^2+6x-1[\/latex].\r\n<h4>Solution<\/h4>\r\nStart by writing [latex]y[\/latex] for [latex]g(x)[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}g(x)=x^2+6x-1\\\\y=x^2+6x-1\\end{aligned}[\/latex]<\/p>\r\nSwitch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse equation:\r\n<p style=\"text-align: center;\">[latex]x=y^2+6y-1[\/latex]<\/p>\r\nAt this point we should solve for [latex]y[\/latex], but we don't know, yet, how to do that. Consequently, we leave the inverse equation as\u00a0[latex]x=y^2+6y-1[\/latex].\r\n\r\n<\/div>\r\nNote: Using Desmos to look at the graph of\u00a0[latex]g(x)=x^2+6x-1[\/latex], (figure 3), we find that [latex]g(x)[\/latex] is not a one-to-one function since it does not pass the horizontal line test. Consequently, its inverse is not a function--it does not pass the vertical line test.\r\n\r\n[caption id=\"attachment_1782\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-1782 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-86-300x300.png\" alt=\"quadratic function and its inverse graph.  Since the function is not one to one, its inverse is not a function.\" width=\"300\" height=\"300\" \/> Figure 3. [latex]g(x)=x^2+6x-1[\/latex] and its inverse[\/caption]\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nFind the inverse function of the polynomial functions:\r\n<ol>\r\n \t<li>\u00a0[latex]f(x)=x^3-4[\/latex]<\/li>\r\n \t<li>\u00a0[latex]g(x)=2x-6[\/latex]<\/li>\r\n \t<li>\u00a0[latex]h(x)=x^2-4x[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm215\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm215\"]\r\n<ol>\r\n \t<li>[latex]f^{-1}(x)=\\sqrt[3]{x+4}[\/latex]<\/li>\r\n \t<li>[latex]g^{-1}(x)=\\dfrac{1}{2}x+3[\/latex]<\/li>\r\n \t<li>The inverse equation is [latex]x=y^2-4y[\/latex], which is not a function in terms of [latex]x[\/latex]; [latex]h(x)=x^2-4x[\/latex] is not a one-to-one function so its inverse is not a function.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Inverse Function of the Function [latex]f(x)=x^n+c[\/latex]<\/h2>\r\nAny odd polynomial function of the form [latex]f(x)=ax^n+c[\/latex], where [latex]a[\/latex] is a real number and [latex]n[\/latex] is an odd, positive integer, is a one-to-one function. Consequently, all odd polynomial functions of the form\u00a0[latex]f(x)=ax^n+c[\/latex] have an inverse function. To illustrate, let's find that inverse function for the function [latex]2x^3+4[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=2x^3+4&amp;\\\\ y&amp;=2x^3+4&amp;\\text{Write }y\\text{ for }f(x)\\\\ x&amp;=2y^3+4&amp;\\text{Switch }x\\text{ and }y\\\\ x-4&amp;=2y^3&amp;\\text{Solve for }y\\text{: subtract }4\\text{ from both sides}\\\\ \\dfrac{1}{2}(x-4)&amp;=y^3&amp;\\text{Divide both sides by }2\\\\ \\sqrt[3]{\\dfrac{1}{2}(x-4)}&amp;=y&amp;\\text{Take the 3th root of both sides}\\\\ f^{-1}(x)&amp;=\\sqrt[3]{\\dfrac{1}{2}(x-4)}&amp;\\text{Write function notation}\\end{aligned}[\/latex]<\/p>\r\nTherefore, the general form of the inverse function of\u00a0[latex]f(x)=ax^n+c[\/latex] when [latex]n[\/latex] is odd is [latex]f^{-1}(x) = \\sqrt[n]{\\frac{1}{a}(x - c)}[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 6<\/h3>\r\nFind the inverse function:\r\n<ol>\r\n \t<li>\u00a0[latex]f(x)=4x^5+7[\/latex]<\/li>\r\n \t<li>\u00a0[latex]g(x)=-3x^7+4[\/latex]<\/li>\r\n \t<li>\u00a0[latex]h(x)=2x^8+5[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\nYou may use the method described above, which is exchanging [latex]x[\/latex] and\u00a0[latex]y[\/latex] and then solve for [latex]y[\/latex].\r\n\r\nYou may also use a formula. The inverse function of\u00a0[latex]f(x)=ax^n+c[\/latex] when [latex]n[\/latex] is odd is [latex]f^{-1}(x) = \\sqrt[n]{\\frac{1}{a}(x - c)}[\/latex].\r\n\r\nIdentify [latex]a[\/latex], [latex]c[\/latex] and [latex]n[\/latex] then use the general form.\r\n<ol>\r\n \t<li>\u00a0[latex]a=4,\\;c=7,\\;n=5[\/latex]\u00a0 so\u00a0 [latex]f^{-1}(x)=\\sqrt[5]{\\dfrac{1}{4}(x - 7)}[\/latex]<\/li>\r\n \t<li>\u00a0[latex]a=-3,\\;c=4,\\;n=7[\/latex]\u00a0 so\u00a0 [latex]f^{-1}(x)=\\sqrt[7]{-\\dfrac{1}{3}(x-4)}[\/latex]<\/li>\r\n \t<li>\u00a0[latex]a=2,\\;c=5,\\;n=8[\/latex]. Since [latex]n=8[\/latex] is even, not odd, [latex]h(x)[\/latex] does not have an inverse function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 5<\/h3>\r\nFind the inverse function:\r\n<ol>\r\n \t<li>\u00a0[latex]f(x)=9x^3+5[\/latex]<\/li>\r\n \t<li>\u00a0[latex]g(x)=-6x^8+4[\/latex]<\/li>\r\n \t<li>\u00a0[latex]h(x)=2x^9-3[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm483\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm483\"]\r\n<ol>\r\n \t<li>\u00a0[latex]f^{-1}(x)=\\sqrt[3]{\\dfrac{1}{9}(x - 5)}[\/latex]<\/li>\r\n \t<li>\u00a0[latex]g(x)[\/latex] does not have an inverse function.<\/li>\r\n \t<li>\u00a0[latex]h^{-1}(x)=\\sqrt[9]{\\dfrac{1}{2}(x + 3)}[\/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>Explain the inverse of a function<\/li>\n<li>Evaluate the inverse of polynomial function<\/li>\n<li>Graph the inverse of a polynomial function\n<ul>\n<li>by switching domain and range<\/li>\n<li>using the line of symmetry [latex]y=x[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>Determine whether the inverse of a polynomial function is a function\n<ul>\n<li>by mapping<\/li>\n<li>by determining if the original function is one-to-one<\/li>\n<li>by the vertical line test<\/li>\n<\/ul>\n<\/li>\n<li>Find the equation of an inverse function<\/li>\n<\/ul>\n<\/div>\n<h2>The Inverse of a Function<\/h2>\n<p>We explored the concept of inverse functions in section 1.3 where we defined the inverse of a function as the switch between the domain and range of the function (or mapping). It is the exchange of the input and output of a function. We discovered that while all functions have inverses, only one-to-one-functions have inverses that are also functions. The notation for the inverse of a one-to-one function [latex]f[\/latex] is [latex]f^{-1}[\/latex], which is read &#8220;the inverse of [latex]f[\/latex]. Recall that t<span style=\"font-size: 1em;\">he symbol &#8220;[latex]-1[\/latex]&#8221; is not an exponent; it means &#8220;the inverse&#8221;. Therefore, the notation [latex]f^{-1}[\/latex] means the inverse of the function [latex]f[\/latex].\u00a0<\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">For example, if a function is defined as the set of ordered pairs<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]f:\\;\\{(1, 2), (3, 8), (4, 2)\\}[\/latex], <\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">it has an inverse <\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]\\{(2, 1), (8, 3), (2, 4)\\}[\/latex].<\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">However, this inverse is NOT a function because the input 2 has two outputs, 1 and 4. This is because the original function [latex]f[\/latex] was NOT a one-to-one function (the function value 2 is the output of two domain values 1 and 4). <\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">On the other hand, if a one-to-one function is defined as the set of ordered pairs <\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]g:\\;\\{(1, 2), (3, 8), (4, 9)\\}[\/latex], <\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">then its inverse function is\u00a0<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]g^{-1}:\\; \\{(2, 1), (8, 3), (9, 4)\\}[\/latex],<\/span><\/p>\n<p>which is a function. Consequently, we can use the\u00a0<span style=\"font-size: 1rem; text-align: initial;\">inverse function notation [latex]g^{-1}[\/latex], which we could not use with the inverse of [latex]f[\/latex].<\/span><\/p>\n<div class=\"textbox shaded\">\n<h3>Inverse of a function<\/h3>\n<p>Every function has an inverse, but <strong>only a one-to-one function has an inverse function<\/strong>.<\/p>\n<\/div>\n<p>To evaluate the inverse of a one-to-one function, we need to switch the input and output of the function. The output of the original function becomes the input of its inverse function,\u00a0and the input of the original function becomes the output of its inverse function. To illustrate,<\/p>\n<p style=\"text-align: center;\">[latex]f(1)=2[\/latex], means [latex]f^{-1}(2)=1[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]f(3)=8[\/latex], means [latex]f^{-1}(8)=3[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]f(4)=9[\/latex], means [latex]f^{-1}(9)=4[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>For the function [latex]f(x)=3x+5[\/latex], find:<\/p>\n<ol>\n<li>\u00a0[latex]f^{-1}(17)[\/latex]<\/li>\n<li>\u00a0[latex]f^{-1}(-1)[\/latex]<\/li>\n<li>\u00a0[latex]f^{-1}(0)[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>1. 17 is the input of the inverse function so this means that 17 was the output of the original function. To find\u00a0[latex]f^{-1}(17)[\/latex] we need to find the associated [latex]x[\/latex]-value when [latex]f(x)=17[\/latex].<\/p>\n<p>[latex]\\begin{aligned}f(x)&=3x+5\\\\17&=3x+5\\\\12&=3x\\\\4&=x\\end{aligned}[\/latex]<\/p>\n<p>So, since [latex]f(4)=17[\/latex],\u00a0[latex]f^{-1}(17)=4[\/latex].<\/p>\n<p>2. \u20131\u00a0is the input of the inverse function so this means that \u20131 was the output of the original function. To find\u00a0[latex]f^{-1}(\u20131)[\/latex] we need to find the associated [latex]x[\/latex]-value when [latex]f(x)=-1[\/latex].<\/p>\n<p>[latex]\\begin{aligned}f(x)&=3x+5\\\\-1&=3x+5\\\\-6&=3x\\\\-2&=x\\end{aligned}[\/latex]<\/p>\n<p>So, since [latex]f(-2)=-1[\/latex],\u00a0[latex]f^{-1}(-1)=-2[\/latex].<\/p>\n<p>3. 0\u00a0is the input of the inverse function so this means that 0 was the output of the original function. To find\u00a0[latex]f^{-1}(0)[\/latex] we need to find the associated [latex]x[\/latex]-value when [latex]f(x)=0[\/latex].<\/p>\n<p>[latex]\\begin{aligned}f(x)&=3x+5\\\\0&=3x+5\\\\-5&=3x\\\\-\\dfrac{5}{3}&=x\\end{aligned}[\/latex]<\/p>\n<p>So, since [latex]f(-\\dfrac{5}{3})=0[\/latex],\u00a0[latex]f^{-1}(0)=-\\dfrac{5}{3}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>For the function [latex]f(x)=-2x+7[\/latex], find:<\/p>\n<ol>\n<li>\u00a0[latex]f^{-1}(5)[\/latex]<\/li>\n<li>\u00a0[latex]f^{-1}(-3)[\/latex]<\/li>\n<li>\u00a0[latex]f^{-1}(0)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm473\">Show Answer<\/span><\/p>\n<div id=\"qhjm473\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]f^{-1}(5)=1[\/latex]<\/li>\n<li>[latex]f^{-1}(-3)=5[\/latex]<\/li>\n<li>[latex]f^{-1}(0)=\\dfrac{7}{2}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Graphing the Inverse of a Polynomial Function<\/h2>\n<p>If we are given the graph of a polynomial function on the coordinate plane we can graph its inverse by switching the [latex]x[\/latex] and [latex]y[\/latex] coordinates for every point on the graph. Figure 1 shows that if we switch the coordinates of the points on the blue graph (\u20132, \u20134), (0, 4) and (2, 12) to (\u20134, \u20132), (4, 0), and (12, 2) we get the green graph, which is the inverse of the blue graph.<\/p>\n<div id=\"attachment_1764\" style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1764\" class=\"wp-image-1764 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-72-300x300.png\" alt=\"function and its inverse demonstrating that each point on one gives you a point on the other by switching the x and y coordinates.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-72-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-72-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-72-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-72-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-72-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-72-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-72.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p id=\"caption-attachment-1764\" class=\"wp-caption-text\">Figure 1. Graphing an inverse function<\/p>\n<\/div>\n<p>Another way to graph the inverse is to reflect the graph across the line [latex]y = x[\/latex]. Figure 1 also shows that the blue curve and its inverse the green curve are reflections of one another across the line of symmetry [latex]y = x[\/latex]. The green curve is the inverse of the blue curve, and the blue curve is the inverse of the green curve.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Given the graph of the function [latex]y=f(x)[\/latex], graph the inverse function [latex]y=f^{-1}(x)[\/latex] using symmetry.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1756 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-64-300x300.png\" alt=\"graph of a line f of x equals 5x -2.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-64-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-64-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-64-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-64-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-64-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-64-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-64.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<h4>Solution<\/h4>\n<p>First graph the line [latex]y=x[\/latex] so we can reflect the graph across this line. Choose a couple of points on the graph and switch their coordinates to help graph the reflection. (0, \u20134) and (2, 6) lie on the graph of the [latex]y=f(x)[\/latex], so (\u20134, 0) and (6, 2) must lie on the graph of the inverse function [latex]y=f^{-1}(x)[\/latex].<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1757 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-65-300x300.png\" alt=\"graph of a line f of x equals 5x -2, and it's inverse.  F goes through (0,-4) and (2,6), and f inverse goes through (-4,0) and (6,2).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-65-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-65-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-65-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-65-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-65-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-65-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-65.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Given the graph of the function [latex]y=f(x)[\/latex], graph the inverse function [latex]y=f^{-1}(x)[\/latex] using symmetry.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1760 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-69-300x300.png\" alt=\"Graph of the right side of a parabola starting at (0,-4) and passing through (2,0).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-69-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-69-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-69-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-69-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-69-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-69-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-69.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm277\">Show Answer<\/span><\/p>\n<div id=\"qhjm277\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1761 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-68-300x300.png\" alt=\"Graph of the function above and its inverse, the top half of a sideways parabola starting at (-4,0) and passing through (0,2).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-68-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-68-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-68-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-68-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-68-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-68-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-68.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>It is important to note that the inverse of a function is not necessarily a function. For example, given the function [latex]f[\/latex]: {(1, 2), (4, 2), (3, 6)}, its inverse is not a function. Its inverse is the set {(2, 1), (2, 4), (6, 3)}. This set is not a function because it is a one-to-many mapping (i.e., 2 is mapped to both 1 and 4). In other words, as long as there is one case of one input with many outputs in a mapping, it is not a function.<\/p>\n<p>Figure 2 shows a parabola that opens up and its inverse, which opens right. The inverse is not a function because it does not pass the vertical line test. A vertical line test on the inverse graph shows that there are two intersection points with any vertical line [latex]x = c,\\;c>0[\/latex] and the graph of the inverse, which opens right, has one input with two outputs (a one-to-many mapping). Therefore, the inverse of the parabola is not a function.<\/p>\n<div id=\"attachment_1763\" style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1763\" class=\"wp-image-1763 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-71-300x300.png\" alt=\"Inverse that is not a function. The complete parabolas from above and a vertical line showing the inverse graph is not a function.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-71-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-71-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-71-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-71-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-71-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-71-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-71.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p id=\"caption-attachment-1763\" class=\"wp-caption-text\">Figure 2. A parabola and its inverse<\/p>\n<\/div>\n<p>Of course, we can tell from the graph of the original function whether or not the inverse will be a function. If the original graph is NOT one-to-one, the inverse is NOT a function. As stated previously, all functions have inverses, but only one-to-one functions have inverse functions.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Graph the inverse of the function. Determine if the inverse is a function.<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%;\"><strong>A one to one function<\/strong><\/th>\n<th style=\"width: 50%;\"><strong>Not a one to one function<\/strong><\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">1.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1765 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-74-300x300.png\" alt=\"graph of a cube function Passing through the points (-5,-9.5), (0,3), and (4,9.5).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-74-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-74-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-74-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-74-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-74-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-74-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-74.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/td>\n<td style=\"width: 50%;\">2.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1767 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-76-300x300.png\" alt=\"graph of a polynomial function passing through (-2,11), (-1,-4), (1,4), and (2,11).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-76-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-76-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-76-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-76-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-76-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-76-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-76.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Solution<\/h4>\n<p>Choose a few points on the original graph, switch the coordinates, and graph the new points. These new points will lie on the inverse. Graph the line [latex]y=x[\/latex] and use the newly plotted points to help draw the reflection of the original graph across the line [latex]y=x[\/latex].<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%; vertical-align: top; text-align: center;\"><strong>A cubic and it&#8217;s inverse<\/strong><\/th>\n<th style=\"width: 50%; text-align: center;\"><strong>A 4th degree and it&#8217;s inverse<\/strong><\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\">1.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter 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=\"The cubic function above with it's inverse cube root function, passing through the points (-9.5,-5), (3,0), and (9.5,4).\" 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>The original graph is one-to-one (it passes the horizontal line test), so the inverse is a function. In addition, the inverse is a function since any vertical line passes through only 1 point.<\/td>\n<td style=\"width: 50%;\">2.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1768 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-75-300x300.png\" alt=\"The 4th degree polynomial above and it's inverse, passing through (-5,0), (-4,1), (-4,-1), (11,2), and (11,-2).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-75-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-75-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-75-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-75-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-75-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-75-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-75.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>The original graph is NOT one-to-one (it fails the horizontal line test), so the inverse is NOT a function.Also, the inverse is not a function as there are vertical lines that pass through the graph at 2 points.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Graph the inverse of the function. Determine if the inverse is a function.<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%; text-align: center;\"><strong>A linear function<\/strong><\/th>\n<th style=\"width: 50%; text-align: center;\"><strong>A quadratic function<\/strong><\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">1.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1772 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-80-300x300.png\" alt=\"A linear function passing through (0, -1) and (2,5).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-80-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-80-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-80-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-80-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-80-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-80-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-80.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/td>\n<td style=\"width: 50%;\">2.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1775 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-84-300x300.png\" alt=\"An inverted quadratic function passing through (-2, -10), (0,8), and (2, -10).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-84-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-84-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-84-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-84-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-84-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-84-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-84.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>&nbsp;<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm172\">Show Answer<\/span><\/p>\n<div id=\"qhjm172\" class=\"hidden-answer\" style=\"display: none\">\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%; vertical-align: top; text-align: center;\">The linear function and its inverse function<\/th>\n<th style=\"width: 50%; text-align: center;\">The quadratic function and its inverse graph<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%; vertical-align: top;\">1.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1770 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-791-300x300.png\" alt=\"The linear function described above and its inverse passing through (-4,1), (-1,0), (2,1), and (5,2).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-791-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-791-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-791-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-791-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-791-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-791-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-791.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>The inverse is a function.<\/td>\n<td style=\"width: 50%;\">2.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1776 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-85-300x300.png\" alt=\"The quadratic function above and its inverse passing through (-8,4), (-8,-4), and (8,0).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-85-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-85-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-85-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-85-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-85-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-85-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-85.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>The inverse is not a function.<\/p>\n<p>&nbsp;<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding the Inverse Algebraically<\/h2>\n<p>To find the equation of the inverse of a polynomial function we use the same idea of switching the input and output. We start by writing [latex]y=f(x)[\/latex] then switch the [latex]x[\/latex] and [latex]y[\/latex] variables. For the function\u00a0[latex]f(x)=x^3+2[\/latex], we start by writing [latex]y=x^3+2[\/latex]. To find the inverse, we switch the input and the output i.e., the [latex]x[\/latex] and [latex]y[\/latex] variables. This results in the equation [latex]x = y^3+2[\/latex], which is the equation of the inverse of the function[latex]f(x)=x^3+2[\/latex]. Solving for [latex]y[\/latex] we get:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&=y^3+2\\\\x-2&=y^3\\\\ \\sqrt[3]{x-2}&=y\\end{aligned}[\/latex]<\/p>\n<p><span style=\"font-size: 1em;\">Since [latex]f(x)=x^3+2[\/latex] is a one-to-one function, its inverse is also a function.<\/span><span style=\"font-size: 1em;\">\u00a0<\/span>The next step is to write the inverse as a function. Because it is the inverse of the one-to-one function\u00a0[latex]f[\/latex], we use the notation\u00a0[latex]f^{-1}[\/latex] to represent it as the inverse function of [latex]f[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]f^{-1}(x) = \\sqrt[3]{x - 2}[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4<\/h3>\n<p>Find the inverse function of the polynomial function [latex]f(x)=5x-4[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Start by writing [latex]y[\/latex] for [latex]f(x)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]y=5x-4[\/latex]<\/p>\n<p>Switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse equation:<\/p>\n<p style=\"text-align: center;\">[latex]x=5y-4[\/latex]<\/p>\n<p>Solve for [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&=5y-4\\\\x+4&=5y\\\\\\dfrac{x+4}{5}&=y\\\\ \\dfrac{1}{5}x+\\dfrac{4}{5}&=y\\end{aligned}[\/latex]<\/p>\n<p>Since [latex]f(x)[\/latex] is a linear function (with a non-zero slope), it is one-to-one. Write the inverse function using function notation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}y&=\\dfrac{1}{5}x+\\dfrac{4}{5}\\\\f^{-1}(x)&=\\dfrac{1}{5}x+\\dfrac{4}{5}\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 5<\/h3>\n<p>Find the inverse function of the polynomial function [latex]g(x)=x^2+6x-1[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Start by writing [latex]y[\/latex] for [latex]g(x)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}g(x)=x^2+6x-1\\\\y=x^2+6x-1\\end{aligned}[\/latex]<\/p>\n<p>Switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse equation:<\/p>\n<p style=\"text-align: center;\">[latex]x=y^2+6y-1[\/latex]<\/p>\n<p>At this point we should solve for [latex]y[\/latex], but we don&#8217;t know, yet, how to do that. Consequently, we leave the inverse equation as\u00a0[latex]x=y^2+6y-1[\/latex].<\/p>\n<\/div>\n<p>Note: Using Desmos to look at the graph of\u00a0[latex]g(x)=x^2+6x-1[\/latex], (figure 3), we find that [latex]g(x)[\/latex] is not a one-to-one function since it does not pass the horizontal line test. Consequently, its inverse is not a function&#8211;it does not pass the vertical line test.<\/p>\n<div id=\"attachment_1782\" style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1782\" class=\"wp-image-1782 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-86-300x300.png\" alt=\"quadratic function and its inverse graph.  Since the function is not one to one, its inverse is not a function.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-86-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-86-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-86-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-86-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-86-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-86-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/04\/desmos-graph-86.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p id=\"caption-attachment-1782\" class=\"wp-caption-text\">Figure 3. [latex]g(x)=x^2+6x-1[\/latex] and its inverse<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Find the inverse function of the polynomial functions:<\/p>\n<ol>\n<li>\u00a0[latex]f(x)=x^3-4[\/latex]<\/li>\n<li>\u00a0[latex]g(x)=2x-6[\/latex]<\/li>\n<li>\u00a0[latex]h(x)=x^2-4x[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm215\">Show Answer<\/span><\/p>\n<div id=\"qhjm215\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]f^{-1}(x)=\\sqrt[3]{x+4}[\/latex]<\/li>\n<li>[latex]g^{-1}(x)=\\dfrac{1}{2}x+3[\/latex]<\/li>\n<li>The inverse equation is [latex]x=y^2-4y[\/latex], which is not a function in terms of [latex]x[\/latex]; [latex]h(x)=x^2-4x[\/latex] is not a one-to-one function so its inverse is not a function.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Inverse Function of the Function [latex]f(x)=x^n+c[\/latex]<\/h2>\n<p>Any odd polynomial function of the form [latex]f(x)=ax^n+c[\/latex], where [latex]a[\/latex] is a real number and [latex]n[\/latex] is an odd, positive integer, is a one-to-one function. Consequently, all odd polynomial functions of the form\u00a0[latex]f(x)=ax^n+c[\/latex] have an inverse function. To illustrate, let&#8217;s find that inverse function for the function [latex]2x^3+4[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=2x^3+4&\\\\ y&=2x^3+4&\\text{Write }y\\text{ for }f(x)\\\\ x&=2y^3+4&\\text{Switch }x\\text{ and }y\\\\ x-4&=2y^3&\\text{Solve for }y\\text{: subtract }4\\text{ from both sides}\\\\ \\dfrac{1}{2}(x-4)&=y^3&\\text{Divide both sides by }2\\\\ \\sqrt[3]{\\dfrac{1}{2}(x-4)}&=y&\\text{Take the 3th root of both sides}\\\\ f^{-1}(x)&=\\sqrt[3]{\\dfrac{1}{2}(x-4)}&\\text{Write function notation}\\end{aligned}[\/latex]<\/p>\n<p>Therefore, the general form of the inverse function of\u00a0[latex]f(x)=ax^n+c[\/latex] when [latex]n[\/latex] is odd is [latex]f^{-1}(x) = \\sqrt[n]{\\frac{1}{a}(x - c)}[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 6<\/h3>\n<p>Find the inverse function:<\/p>\n<ol>\n<li>\u00a0[latex]f(x)=4x^5+7[\/latex]<\/li>\n<li>\u00a0[latex]g(x)=-3x^7+4[\/latex]<\/li>\n<li>\u00a0[latex]h(x)=2x^8+5[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>You may use the method described above, which is exchanging [latex]x[\/latex] and\u00a0[latex]y[\/latex] and then solve for [latex]y[\/latex].<\/p>\n<p>You may also use a formula. The inverse function of\u00a0[latex]f(x)=ax^n+c[\/latex] when [latex]n[\/latex] is odd is [latex]f^{-1}(x) = \\sqrt[n]{\\frac{1}{a}(x - c)}[\/latex].<\/p>\n<p>Identify [latex]a[\/latex], [latex]c[\/latex] and [latex]n[\/latex] then use the general form.<\/p>\n<ol>\n<li>\u00a0[latex]a=4,\\;c=7,\\;n=5[\/latex]\u00a0 so\u00a0 [latex]f^{-1}(x)=\\sqrt[5]{\\dfrac{1}{4}(x - 7)}[\/latex]<\/li>\n<li>\u00a0[latex]a=-3,\\;c=4,\\;n=7[\/latex]\u00a0 so\u00a0 [latex]f^{-1}(x)=\\sqrt[7]{-\\dfrac{1}{3}(x-4)}[\/latex]<\/li>\n<li>\u00a0[latex]a=2,\\;c=5,\\;n=8[\/latex]. Since [latex]n=8[\/latex] is even, not odd, [latex]h(x)[\/latex] does not have an inverse function.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 5<\/h3>\n<p>Find the inverse function:<\/p>\n<ol>\n<li>\u00a0[latex]f(x)=9x^3+5[\/latex]<\/li>\n<li>\u00a0[latex]g(x)=-6x^8+4[\/latex]<\/li>\n<li>\u00a0[latex]h(x)=2x^9-3[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm483\">Show Answer<\/span><\/p>\n<div id=\"qhjm483\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>\u00a0[latex]f^{-1}(x)=\\sqrt[3]{\\dfrac{1}{9}(x - 5)}[\/latex]<\/li>\n<li>\u00a0[latex]g(x)[\/latex] does not have an inverse function.<\/li>\n<li>\u00a0[latex]h^{-1}(x)=\\sqrt[9]{\\dfrac{1}{2}(x + 3)}[\/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-1370\">\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>Inverse and Inverse 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>: 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>All Examples and Try Its. <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":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Inverse and Inverse 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\":\"Hazel McKenna and Leo Chang\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All Examples and Try Its\",\"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-1370","chapter","type-chapter","status-publish","hentry"],"part":1176,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1370","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":57,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1370\/revisions"}],"predecessor-version":[{"id":4742,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1370\/revisions\/4742"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/parts\/1176"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1370\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/media?parent=1370"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1370"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/contributor?post=1370"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/license?post=1370"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}