{"id":11013,"date":"2015-07-14T17:40:14","date_gmt":"2015-07-14T17:40:14","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&#038;p=11013"},"modified":"2021-08-23T06:32:59","modified_gmt":"2021-08-23T06:32:59","slug":"inverse-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/inverse-functions\/","title":{"raw":"Section 5.2: One-to-One Functions; Inverse Functions","rendered":"Section 5.2: One-to-One Functions; Inverse Functions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>LEARNING OBJECTIVES<\/h3>\r\nBy the end of this lesson, you will be able to:\r\n<ul>\r\n \t<li>Determine whether a function is one-to-one.<\/li>\r\n \t<li>Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.<\/li>\r\n \t<li>Find or evaluate the inverse of a function.<\/li>\r\n \t<li>Use the graph of a one-to-one function to graph its inverse function on the same axes.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Determining Whether a Function is One-to-One<\/span><\/h2>\r\n<p id=\"fs-id1165135245630\">Some functions have only one input value for each output value, as well as having only one output for each input. We call these functions <strong>one-to-one functions<\/strong>. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in.<\/p>\r\n\r\n<table id=\"Table_01_01_13\" summary=\"Two columns and five rows. The first column is labeled, \"><colgroup> <col \/> <col \/> <\/colgroup>\r\n<thead>\r\n<tr>\r\n<th>Letter grade<\/th>\r\n<th>Grade point average<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>A<\/td>\r\n<td>4.0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>B<\/td>\r\n<td>3.0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C<\/td>\r\n<td>2.0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>D<\/td>\r\n<td>1.0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165137561844\">This grading system represents a one-to-one function, because each letter input yields one particular grade point average output and each grade point average corresponds to one input letter.<\/p>\r\nTo visualize this concept, let\u2019s look again at the two simple functions sketched in (a)and (b) of Figure 10.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010531\/CNX_Precalc_Figure_01_01_0012.jpg\" alt=\"Three relations that demonstrate what constitute a function.\" width=\"975\" height=\"243\" \/> <b>Figure 10<\/b>[\/caption]\r\n\r\nThe function in part (a) shows a relationship that is not a one-to-one function because inputs [latex]q[\/latex] and [latex]r[\/latex] both give output [latex]n[\/latex]. The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output.\r\n<div class=\"textbox\">\r\n<h3>A General Note: One-to-One Function<\/h3>\r\nA one-to-one function is a function in which each output value corresponds to exactly one input value.\r\n\r\n<\/div>\r\n<div id=\"Example_01_07_01\" class=\"example\">\r\n<div id=\"fs-id1165137656641\" class=\"exercise\">\r\n<div id=\"fs-id1165137922642\" class=\"problem textbox shaded\">\r\n<h3>Example 1: determining if a set of ordered pairs is a one-to-one function<\/h3>\r\n<p id=\"fs-id1165137659325\">For each set of ordered pairs, determine if it represents a function and, if so, if the function is one-to-one.<\/p>\r\na)\u00a0[latex] \\{(-3,27),(-2,8),(-1,1),(0,0),(1,1),(2,8),(3,27)\\}[\/latex]\r\n\r\nb)\u00a0[latex] \\{(0,0),(1,1),(4,2),(9,3),(16,4)\\}[\/latex]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137768306\" class=\"solution textbox shaded\">\r\n<h3>Solution<\/h3>\r\n<p id=\"fs-id1165137737081\">a) Each <em>x<\/em>-value is matched with only one <em>y<\/em>-value. So this relation is a function.\u00a0 But each <em>y<\/em>-value is not paired with only one <em>x<\/em>-value, [latex](-3,27)[\/latex] and [latex](3,27)[\/latex], for example. So this function is not one-to-one.<\/p>\r\nb) Each <em>x<\/em>-value is matched with only one <em>y<\/em>-value. So this relation is a function. Since each <em>y<\/em>-value is paired with only one <em>x<\/em>-value, this function is one-to-one.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 1<\/h3>\r\n&nbsp;\r\n<p id=\"fs-id1165137659325\">For each set of ordered pairs, determine if it represents a function and, if so, if the function is one-to-one.<\/p>\r\na)\u00a0[latex] \\{(-3,-6),(-2,-4),(-1,-2),(0,0),(1,2),(2,4),(3,6)\\}[\/latex]\r\n\r\nb)\u00a0[latex] \\{(-4,-8),(-2,4),(-1,2),(0,0),(1,2),(2,4),(4,8)\\}[\/latex]\r\n\r\n[reveal-answer q=\"380740\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380740\"]\r\n<div id=\"q676921\" class=\"hidden-answer\">\r\n<p id=\"fs-id1165137737081\">a) Each <em>x<\/em>-value is matched with only one <em>y<\/em>-value. So this relation is a function.\u00a0Since each <em>y<\/em>-value is paired with only one <em>x<\/em>-value, this function is one-to-one.<\/p>\r\nb) Each <em>x<\/em>-value is matched with only one <em>y<\/em>-value. So this relation is a function. But each <em>y<\/em>-value is not paired with only one <em>x<\/em>-value, [latex](-2,4)[\/latex] and [latex](2,4)[\/latex], for example. So this function is not one-to-one.\r\n\r\n<\/div>\r\n<div id=\"q676921\">\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nTo help us determine whether a function is one-to-one, we use the horizontal line test. We use a horizontal line and check that each horizontal line intersects the graph in only one point. The horizontal line is representing a y-value and we check that it intersects the graph in only one x-value. If every horizontal line intersects the graph of a function in at most one point, it is a one-to-one function. This is the <strong>horizontal line test.<\/strong>\r\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Using the Horizontal Line Test<\/span><\/h2>\r\n<p id=\"fs-id1165137871503\">Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the <strong>horizontal line test<\/strong>. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function.<\/p>\r\n\r\n<div id=\"fs-id1165137445319\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137635406\" style=\"text-align: center;\"><strong>How To: Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function.<\/strong><\/h3>\r\n<ol id=\"fs-id1165137611853\">\r\n \t<li>Inspect the graph to see if any horizontal line drawn would intersect the curve more than once.<\/li>\r\n \t<li>If there is any such line, determine that the function is not one-to-one.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Applying the Horizontal Line Test<\/h3>\r\nConsider the functions (a), (b), and (c) shown in\u00a0the graphs in Figure 1.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005018\/CNX_Precalc_Figure_01_01_013abc.jpg\" alt=\"Graph of a polynomial, line, circle\" width=\"975\" height=\"418\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\nAre either of the functions one-to-one?\r\n\r\n[reveal-answer q=\"549926\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"549926\"]\r\n<p id=\"fs-id1165135185190\">The function in (a) is not one-to-one. The horizontal line shown in Figure 2\u00a0intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)<\/p>\r\n\r\n<figure id=\"Figure_01_01_010\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005019\/CNX_Precalc_Figure_01_01_010.jpg\" alt=\"\" width=\"487\" height=\"445\" \/> <b>Figure 2<\/b>[\/caption]<\/figure>\r\n<p id=\"fs-id1165135151243\">The function in (b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.<\/p>\r\nThe graph in (c) in not one-to-one.\u00a0 Any horizontal line will intersect a circle at two points.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Inverse Function<\/h3>\r\nSuppose [latex]y=f\\left(x\\right)[\/latex] is a one-to-one function. An <strong>inverse function<\/strong>\u00a0is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.\u00a0 The symbol\u00a0[latex]{f}^{-1}\\left(x\\right)[\/latex] is used to represent the inverse function of [latex]f[\/latex].\u00a0 In other words, if\u00a0[latex]y=f\\left(x\\right)[\/latex] is a one-to-one function, then [latex]f[\/latex] has an inverse function\u00a0[latex]{f}^{-1}\\left(x\\right)[\/latex] and [latex]x={f}^{-1}\\left(y\\right)[\/latex] .\r\n\r\n<\/div>\r\n<p id=\"fs-id1165135528385\">\u00a0The raised [latex]-1[\/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[\/latex] . In other words, [latex]{f}^{-1}\\left(x\\right)[\/latex] does <em>not<\/em> mean [latex]\\frac{1}{f\\left(x\\right)}[\/latex] because [latex]\\frac{1}{f\\left(x\\right)}[\/latex] is the reciprocal of [latex]f[\/latex] and not the inverse.<\/p>\r\n\r\n<div id=\"fs-id1165137933105\" class=\"note textbox\">\r\n<h3 class=\"title\">Verifying an inverse function<\/h3>\r\n[latex]f[\/latex] and [latex]g[\/latex] are inverses of each other if:\r\n\r\n[latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] where [latex]x[\/latex]\u00a0 is in the domain of [latex]f[\/latex]\r\n\r\n[latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] where\u00a0[latex]x[\/latex]\u00a0 is in the domain of [latex]g[\/latex]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137655153\">Given a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether either [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] or [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)<\/p>\r\n<p id=\"fs-id1165135397975\">For example, [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.<\/p>\r\n\r\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\left(g\\circ f\\right)\\left(x\\right)=g\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/div>\r\n<p id=\"fs-id1165137767233\">and<\/p>\r\n\r\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\left({f}^{}\\circ g\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/div>\r\n<div><\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137438777\">A few coordinate pairs from the graph of the function [latex]y=4x[\/latex] are (\u22122, \u22128), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function [latex]y=\\frac{1}{4}x[\/latex] are (\u22128, \u22122), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.<\/p>\r\n\r\n<div id=\"fs-id1165137656641\" class=\"exercise\">\r\n<div id=\"fs-id1165137922642\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Identifying an Inverse Function for a Given Input-Output Pair<\/h3>\r\n<p id=\"fs-id1165137659325\">If for a particular one-to-one function [latex]f\\left(2\\right)=4[\/latex] and [latex]f\\left(5\\right)=12[\/latex], what are the corresponding input and output values for the inverse function?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137768306\" class=\"solution textbox shaded\">\r\n<h3>Solution<\/h3>\r\n<p id=\"fs-id1165137737081\">The inverse function reverses the input and output quantities, so if<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(2\\right)=4,\\text{ then }{f}^{-1}\\left(4\\right)=2;\\\\ f\\left(5\\right)=12,{\\text{ then f}}^{-1}\\left(12\\right)=5\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137659464\">Alternatively, if we want to name the inverse function [latex]g[\/latex], then [latex]g\\left(4\\right)=2[\/latex] and [latex]g\\left(12\\right)=5[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135245520\" class=\"commentary\">\r\n<h3>Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165135508518\">Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.<\/p>\r\n\r\n<table id=\"Table_01_07_01\" summary=\"For (x,f(x)) we have the values (2, 4) and (5, 12); for (x, g(x)), we have the values (4, 2) and (12, 5).\">\r\n<thead>\r\n<tr>\r\n<th>[latex]\\left(x,f\\left(x\\right)\\right)[\/latex]<\/th>\r\n<th>[latex]\\left(x,g\\left(x\\right)\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\left(2,4\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(4,2\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\left(5,12\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(12,5\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 2<\/h3>\r\n<p id=\"fs-id1165137659089\">Given that [latex]{h}^{-1}\\left(6\\right)=2[\/latex], what are the corresponding input and output values of the original function [latex]h?[\/latex]<\/p>\r\n[reveal-answer q=\"380743\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380743\"]\r\n\r\nYou would reverse the 6 and 2 and rewrite as: [latex]{h}\\left(2\\right)=6[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/TSztRfzmk0M\r\n<div id=\"fs-id1165134357354\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135434077\">How To: Given two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex], test whether the functions are inverses of each other.<\/h3>\r\n<ol id=\"fs-id1165137452358\">\r\n \t<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] or [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\r\n \t<li>If either statement is true, then both are true, and [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then both are false, and [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_07_02\" class=\"example\">\r\n<div id=\"fs-id1165137557051\" class=\"exercise\">\r\n<div id=\"fs-id1165137679032\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Testing Inverse Relationships Algebraically<\/h3>\r\n<p id=\"fs-id1165135519417\">If [latex]f\\left(x\\right)=\\frac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\frac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137627632\" class=\"solution\">\r\n<div id=\"fs-id1165137675509\" class=\"equation unnumbered textbox shaded\">\r\n<h3>Solution<\/h3>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} g\\left(f\\left(x\\right)\\right)=\\frac{1}{\\left(\\frac{1}{x+2}\\right)}{-2 }\\hfill\\\\={ x }+{ 2 } -{ 2 }\\hfill\\\\={ x }\\hfill\\end{align}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165135678636\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\text{So }g={f}^{-1}\\text{ and }f={g}^{-1}\\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165135436648\">This is enough to answer yes to the question, but we can also verify the other formula.<\/p>\r\n&nbsp;\r\n<div id=\"fs-id1165137784350\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)=\\frac{1}{\\frac{1}{x}-2+2}\\\\ =\\frac{1}{\\frac{1}{x}}\\hfill \\\\ =x\\hfill \\end{align}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137733685\" class=\"commentary\">\r\n<h3>Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165135389000\">Notice the inverse operations are in reverse order of the operations from the original function.<\/p>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 3<\/h3>\r\n<p id=\"fs-id1165135160550\">If [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\r\n[reveal-answer q=\"380741\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380741\"]\r\n\r\nYes [latex]g={f}^{-1}[\/latex] since:\r\n<div id=\"q676921\" class=\"hidden-answer\">\r\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(g(x))=f(x^{3}-4)=\\sqrt[3]{x^3-4+4}=\\sqrt[3]{x^{3}}=x\\\\g(f(x))=g(\\sqrt[3]{x+4})=(\\sqrt[3]{x+4})^{3}-4=x+4-4=x[\/latex]<\/div>\r\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\r\n<p id=\"fs-id1165137737081\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_01_07_03\" class=\"example\">\r\n<div id=\"fs-id1165135259560\" class=\"exercise\">\r\n<div id=\"fs-id1165134042918\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Determining Inverse Relationships for Power Functions<\/h3>\r\n<p id=\"fs-id1165137441834\">If [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137442603\" class=\"solution\">\r\n<div id=\"fs-id1165137591632\" class=\"equation unnumbered textbox shaded\">\r\n<h3>Solution<\/h3>\r\n<p style=\"text-align: center;\">[latex]f\\left(g\\left(x\\right)\\right)=\\frac{{x}^{3}}{27}\\ne x[\/latex]<\/p>\r\n<p id=\"fs-id1165137694053\">No, the functions are not inverses.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135317479\" class=\"commentary\">\r\n<h3>Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165134192978\">The correct inverse to the cube is, of course, the cube root [latex]\\sqrt[3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.<\/p>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 4<\/h3>\r\n<p id=\"fs-id1165137573532\">If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\r\n[reveal-answer q=\"380750\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380750\"]\r\n\r\nYes [latex]g={f}^{-1}[\/latex] since:\r\n<div id=\"q676921\" class=\"hidden-answer\">\r\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(g(x))=f(\\sqrt[3]{x}+1)=(\\sqrt[3]{x}+1-1)^{3}=(\\sqrt[3]{x})^{3}=x\\\\g(f(x))=g((x-1)^{3})=\\sqrt[3]{(x-1)^{3}}+1=x-1+1=x[\/latex]<\/div>\r\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\r\n<p id=\"fs-id1165137737081\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>Finding Domain and Range of Inverse Functions<\/h2>\r\nThe outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010624\/CNX_Precalc_Figure_01_07_0032.jpg\" alt=\"Domain and range of a function and its inverse.\" width=\"487\" height=\"143\" \/> <b>Figure 3.<\/b> Domain and range of a function and its inverse[\/caption]\r\n<p id=\"fs-id1165135557891\">When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] is [latex]{f}^{-1}\\left(x\\right)={x}^{2}[\/latex], because a square \"undoes\" a square root; but the square is only the inverse of the square root on the domain [latex]\\left[0,\\infty \\right)[\/latex], since that is the range of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex].<\/p>\r\n<p id=\"fs-id1165137730185\">We can look at this problem from the other side, starting with the square (toolkit quadratic) function [latex]f\\left(x\\right)={x}^{2}[\/latex]. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and \u20133. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the \"inverse\" is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.<\/p>\r\n<p id=\"fs-id1165137823552\">In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).<\/p>\r\n<p id=\"fs-id1165132037000\">If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}[\/latex] on [latex]\\left[1,\\infty \\right)[\/latex], then the inverse function is [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+1[\/latex].<\/p>\r\n\r\n<ul id=\"fs-id1165137851227\">\r\n \t<li>The domain of [latex]f[\/latex] = range of [latex]{f}^{-1}[\/latex] = [latex]\\left[1,\\infty \\right)[\/latex].<\/li>\r\n \t<li>The domain of [latex]{f}^{-1}[\/latex] = range of [latex]f[\/latex] = [latex]\\left[0,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1165137733804\" class=\"note precalculus qa textbox\">\r\n<p id=\"fs-id1165137723526\"><strong>Q &amp; A<\/strong><\/p>\r\n<strong>Is it possible for a function to have more than one inverse?<\/strong>\r\n<p id=\"fs-id1165137456608\"><em>No. If two supposedly different functions, say, [latex]g[\/latex] and [latex]h[\/latex], both meet the definition of being inverses of another function [latex]f[\/latex], then you can prove that [latex]g=h[\/latex]. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137704938\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Domain and Range of Inverse Functions<\/h3>\r\n<p id=\"fs-id1165135319550\">The range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165137673886\">The domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135308785\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137605040\"><strong>How To: Given a function, find the domain and range of its inverse.\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id1165137530434\">\r\n \t<li>If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.<\/li>\r\n \t<li>If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_07_05\" class=\"example\">\r\n<div id=\"fs-id1165137667922\" class=\"exercise\">\r\n<div id=\"fs-id1165135511321\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Finding the Inverses of Toolkit Functions<\/h3>\r\n<p id=\"fs-id1165137448020\">Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed below. We restrict the domain in such a fashion that the function assumes all <em>y<\/em>-values exactly once.<\/p>\r\n\r\n<table id=\"Table_01_07_02\" summary=\"A list of the toolkit function. The constant function is f(x) = c where c is the constant; the identity function is f(x) = x; the absolute function is f(x)=|x|; the quadratic function is f(x) = x^2; the cubic function is f(x)=x^3; the reciprocal function is f(x)=1\/x; the reciprocal squared function is f(x)=1\/x^2; the square root function is f(x)=sqrt(x); the cube root function is f(x) = x^(1\/3).\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>Constant<\/td>\r\n<td>Identity<\/td>\r\n<td>Quadratic<\/td>\r\n<td>Cubic<\/td>\r\n<td>Reciprocal<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(x\\right)=c[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=x[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)={x}^{2}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)={x}^{3}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Reciprocal squared<\/td>\r\n<td>Cube root<\/td>\r\n<td>Square root<\/td>\r\n<td>Absolute value<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=\\sqrt{x}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=|x|[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id1165137767030\" class=\"solution textbox shaded\">\r\n<h3>Solution<\/h3>\r\n<p id=\"fs-id1165132988445\">The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.<\/p>\r\n<p id=\"fs-id1165134080947\">The absolute value function can be restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], where it is equal to the identity function.<\/p>\r\n<p id=\"fs-id1165137642849\">The reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137901280\" class=\"commentary\">\r\n<h3>Analysis of the Solution<\/h3>\r\nWe can see that these functions (if unrestricted) are not one-to-one by looking at their graphs.\u00a0They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010624\/CNX_Precalc_Figure_01_07_004ab2.jpg\" alt=\"Graph of an absolute function.\" width=\"975\" height=\"404\" \/> <b>Figure 4.<\/b> (a) Absolute value (b) Reciprocal squared[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 5<\/h3>\r\n<p id=\"fs-id1165137507853\">The domain of function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of function [latex]f[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex]. Find the domain and range of the inverse function.<\/p>\r\n[reveal-answer q=\"380755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380755\"]\r\n<div id=\"q676921\" class=\"hidden-answer\">\r\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center;\">The domain of function [latex]{f}^{-1}[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex] and the range of function [latex]{f}^{-1}[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex].<\/div>\r\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\r\n<p id=\"fs-id1165137737081\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>Finding and Evaluating Inverse Functions<\/h2>\r\n<p id=\"fs-id1165137761017\">Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.<\/p>\r\n\r\n<section id=\"fs-id1165135466392\">\r\n<h2 style=\"text-align: center;\"><\/h2>\r\n<h2 style=\"text-align: center;\"><\/h2>\r\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Inverting Tabular Functions<\/span><\/h2>\r\n<p id=\"fs-id1165135190714\">Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.<\/p>\r\n<p id=\"fs-id1165137422578\">Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.<\/p>\r\n\r\n<div id=\"Example_01_07_06\" class=\"example\">\r\n<div id=\"fs-id1165135544995\" class=\"exercise\">\r\n<div id=\"fs-id1165137698262\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Interpreting the Inverse of a Tabular Function<\/h3>\r\n<p id=\"fs-id1165135435474\">A function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes. Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].<\/p>\r\n\r\n<table id=\"Table_01_07_03\" summary=\"Two rows and five columns. The first row is labeled \"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\r\n<td>30<\/td>\r\n<td>50<\/td>\r\n<td>70<\/td>\r\n<td>90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\r\n<td>20<\/td>\r\n<td>40<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id1165137407569\" class=\"solution textbox shaded\">\r\n<h3>Solution<\/h3>\r\n<p id=\"fs-id1165137640334\">The inverse function takes an output of [latex]f[\/latex] and returns an input for [latex]f[\/latex]. So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.<\/p>\r\n<p id=\"fs-id1165135181841\">Alternatively, recall that the definition of the inverse was that if [latex]f\\left(a\\right)=b[\/latex], then [latex]{f}^{-1}\\left(b\\right)=a[\/latex]. By this definition, if we are given [latex]{f}^{-1}\\left(70\\right)=a[\/latex], then we are looking for a value [latex]a[\/latex] so that [latex]f\\left(a\\right)=70[\/latex]. In this case, we are looking for a [latex]t[\/latex] so that [latex]f\\left(t\\right)=70[\/latex], which is when [latex]t=90[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 6<\/h3>\r\n<p id=\"fs-id1165134108483\">Using the table below,\u00a0find and interpret (a) [latex]\\text{ }f\\left(60\\right)[\/latex], and (b) [latex]\\text{ }{f}^{-1}\\left(60\\right)[\/latex].<\/p>\r\n\r\n<table id=\"Table_01_07_04\" summary=\"Two rows and five columns. The first row is labeled \"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>[latex]t\\text{ (minutes)}[\/latex]<\/td>\r\n<td>30<\/td>\r\n<td>50<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<td>90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex]<\/td>\r\n<td>20<\/td>\r\n<td>40<\/td>\r\n<td>50<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"q676921\" class=\"hidden-answer\">[reveal-answer q=\"380760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380760\"][latex]\\text{ }f\\left(60\\right)=50[\/latex], in 60 minutes, 50 miles are traveled.\r\n[latex]\\text{ }{f}^{-1}\\left(60\\right)=70[\/latex], to travel 60 miles, it will take 70 minutes.[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<section id=\"fs-id1165137418615\">\r\n<h2 style=\"text-align: center;\"><\/h2>\r\n<h2 style=\"text-align: center;\"><\/h2>\r\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/span><\/h2>\r\n<p id=\"fs-id1165137400045\">We saw in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/csn-precalculus\/chapter\/determine-whether-a-relation-represents-a-function\/\" target=\"_blank\" rel=\"noopener\">Functions and Function Notation<\/a> that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the <em>vertical<\/em> extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the <em>horizontal<\/em> extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function\u2019s graph.<\/p>\r\n\r\n<div id=\"fs-id1165133045388\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135333128\">How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\r\n<ol id=\"fs-id1165137464840\">\r\n \t<li>Find the desired input on the <em>y<\/em>-axis of the given graph.<\/li>\r\n \t<li>Read the inverse function\u2019s output from the <em>x<\/em>-axis of the given graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_07_07\" class=\"example\">\r\n<div id=\"fs-id1165135434803\" class=\"exercise\">\r\n<div id=\"fs-id1165135434805\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/h3>\r\nA function [latex]g\\left(x\\right)[\/latex] is given in Figure 5. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137468840\" class=\"solution textbox shaded\">\r\n<h3>Solution<\/h3>\r\n<p id=\"fs-id1165137468842\">To evaluate [latex]g\\left(3\\right)[\/latex], we find 3 on the <em>x<\/em>-axis and find the corresponding output value on the <em>y<\/em>-axis. The point [latex]\\left(3,1\\right)[\/latex] tells us that [latex]g\\left(3\\right)=1[\/latex].<\/p>\r\nTo evaluate [latex]{g}^{-1}\\left(3\\right)[\/latex], recall that by definition [latex]{g}^{-1}\\left(3\\right)[\/latex] means the value of <em>x<\/em> for which [latex]g\\left(x\\right)=3[\/latex]. By looking for the output value 3 on the vertical axis, we find the point [latex]\\left(5,3\\right)[\/latex] on the graph, which means [latex]g\\left(5\\right)=3[\/latex], so by definition, [latex]{g}^{-1}\\left(3\\right)=5[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 7<\/h3>\r\n<p id=\"fs-id1165137812560\">Using the graph in Example 6, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].<\/p>\r\n\r\n<div id=\"q676921\" class=\"hidden-answer\">[reveal-answer q=\"380765\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380765\"]\r\na.)\u00a0 1 \u00a0b.)\u00a0 5.5\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137605437\">\r\n<h2 style=\"text-align: center;\"><\/h2>\r\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Finding Inverses of Functions Represented by Formulas<\/span><\/h2>\r\n<p id=\"fs-id1165137433184\">Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula\u2014 for example, [latex]y[\/latex] as a function of [latex]x\\text{-\\hspace{0.17em}}[\/latex] we can often find the inverse function by solving to obtain [latex]x[\/latex] as a function of [latex]y[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165137652548\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135195849\">How To: Given a one-to-one function represented by a formula, find the inverse.<\/h3>\r\n<ol id=\"fs-id1165135443898\">\r\n \t<li>Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex] .<\/li>\r\n \t<li>Switch [latex]x[\/latex] and [latex]y[\/latex].<\/li>\r\n \t<li>Solve for [latex]y[\/latex]<\/li>\r\n \t<li>Replace [latex]y[\/latex] with [latex]{f}^{-1}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_07_09\" class=\"example\">\r\n<div id=\"fs-id1165134065146\" class=\"exercise\">\r\n<div id=\"fs-id1165137409366\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Solving to Find an Inverse Function<\/h3>\r\n<p id=\"fs-id1165137891504\">Find the inverse of the function [latex]f\\left(x\\right)=\\frac{2}{x - 3}+4[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137482074\" class=\"solution textbox shaded\">\r\n<h3 class=\"equation unnumbered\">Solution<\/h3>\r\n<div id=\"fs-id1165135189953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}y=\\frac{2}{x - 3}+4\\hspace{0.5in} &amp; \\text{Set up an equation}.\\\\ x=\\frac{2}{y - 3}+4 \\hspace{0.5in} &amp; \\text{Switch x and y} \\\\ x(y-3)=2+4(y-3)\\hspace{0.5in} &amp; \\text{Multiply both sides by }y - 3 \\\\xy-3x=2+4y-12\\hspace{0.5in}&amp; \\text{Simplify} \\\\xy-4y=3x-10\\hspace{0.5in}&amp;\\text{Get all y's on one side} \\\\y(x-4)=3x-10 \\hspace{0.5in} &amp; \\text{Factor out a y} \\\\ y = \\frac{3x-10}{x-4}\\hspace{0.5in} &amp; \\text{Divide by x-4}\\\\{f}^{-1}\\left(x\\right)=\\frac{3x-10}{x-4} \\hspace{0.5in} &amp; \\text{This is the inverse}\\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137678168\"><\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137864156\" class=\"commentary\">\r\n<h3>Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165135394231\">The domain and range of [latex]f[\/latex] exclude the values 3 and 4, respectively. [latex]f[\/latex] and [latex]{f}^{-1}[\/latex] are equal at two points but are not the same function, as we can see by creating\u00a0the table below.<\/p>\r\n\r\n<table id=\"Table_01_07_05\" summary=\"The values of f(x) are: f(1)=3, f(2)=2, and f(5)=5. So f^(-1)(y)=y.\">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>5<\/td>\r\n<td>[latex]{f}^{-1}\\left(y\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\r\n<td>3<\/td>\r\n<td>2<\/td>\r\n<td>5<\/td>\r\n<td>[latex]y[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_01_07_10\" class=\"example\">\r\n<div id=\"fs-id1165137603677\" class=\"exercise\">\r\n<div id=\"fs-id1165137547656\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Solving to Find an Inverse with Radicals<\/h3>\r\n<p id=\"fs-id1165137841687\">Find the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt{x - 4}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135193684\" class=\"solution textbox shaded\">\r\n<h3 class=\"equation unnumbered\">Solution<\/h3>\r\n<div id=\"fs-id1165137828173\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}y=2+\\sqrt{x - 4}\\hfill \\\\x=2+\\sqrt{y - 4}\\hfill \\\\ {\\left(x - 2\\right)}^{2}=y - 4\\hfill \\\\ y={\\left(x - 2\\right)}^{2}+4\\hfill \\\\ {f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4\\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137900392\">The domain of [latex]f[\/latex] is [latex]\\left[4,\\infty \\right)[\/latex]. Notice that the range of [latex]f[\/latex] is [latex]\\left[2,\\infty \\right)[\/latex], so this means that the domain of the inverse function [latex]{f}^{-1}[\/latex] is also [latex]\\left[2,\\infty \\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137667328\" class=\"commentary\">\r\n<h3>Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165135546050\">The formula we found for [latex]{f}^{-1}\\left(x\\right)[\/latex] looks like it would be valid for all real [latex]x[\/latex]. However, [latex]{f}^{-1}[\/latex] itself must have an inverse (namely, [latex]f[\/latex] ) so we have to restrict the domain of [latex]{f}^{-1}[\/latex] to [latex]\\left[2,\\infty \\right)[\/latex] in order to make [latex]{f}^{-1}[\/latex] a one-to-one function. This domain of [latex]{f}^{-1}[\/latex] is exactly the range of [latex]f[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It\u00a08<\/h3>\r\n<p id=\"fs-id1165137756074\">What is the inverse of the function [latex]f\\left(x\\right)=2-\\sqrt{x}?[\/latex] State the domains of both the function and the inverse function.<\/p>\r\n\r\n<div id=\"q676921\" class=\"hidden-answer\">[reveal-answer q=\"380770\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380770\"]\r\n[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2};\\text{domain of }f:\\left[0,\\infty \\right);\\text{domain of }{f}^{-1}:\\left(-\\infty ,2\\right][\/latex]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\nNow that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as in Figure 7.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/> <b>Figure 7.<\/b> Quadratic function with domain restricted to [0, \u221e).[\/caption]\r\n<p id=\"fs-id1165137419977\"><strong>Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.<\/p>\r\n<p id=\"fs-id1165137656093\">We already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]<\/p>\r\nWe notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown in Figure 8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0092.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"251\" \/> <b>Figure 8.<\/b> Square and square-root functions on the non-negative domain[\/caption]\r\n<p id=\"fs-id1165137393212\">This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.<\/p>\r\n\r\n<div id=\"Example_01_07_11\" class=\"example\">\r\n<div id=\"fs-id1165134430460\" class=\"exercise\">\r\n<div id=\"fs-id1165134430463\" class=\"problem textbox shaded\">\r\n<h3>Example 10: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\r\nGiven the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137407658\" class=\"solution textbox shaded\">\r\n<h3>Solution<\/h3>\r\n<p id=\"fs-id1165137407660\">This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\r\nIf we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in Figure 10.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0112.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"363\" \/> <b>Figure 10.<\/b> The function and its inverse, showing reflection about the identity line[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137911739\" class=\"solution\">\r\n<div id=\"fs-id1165137627081\" class=\"note precalculus qa textbox\">\r\n<p id=\"fs-id1165134388228\"><strong>Q &amp; A <\/strong><\/p>\r\n<strong>Is there any function that is equal to its own inverse?<\/strong>\r\n<p id=\"fs-id1165137602656\"><em>Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em><\/p>\r\n\r\n<div id=\"fs-id1165135205827\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/div>\r\n<p id=\"fs-id1165137897050\"><em>Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em><\/p>\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>A function is one-to-one if each output value corresponds to only one input value.<\/li>\r\n \t<li>The graph of a one-to-one function passes the horizontal line test.<\/li>\r\n \t<li>If [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex], then<\/li>\r\n \t<li>[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex].<\/li>\r\n \t<li>Each of the toolkit functions has an inverse.<\/li>\r\n \t<li>For a function to have an inverse, it must be one-to-one (pass the horizontal line test).<\/li>\r\n \t<li>A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.<\/li>\r\n \t<li>For a tabular function, exchange the input and output rows to obtain the inverse.<\/li>\r\n \t<li>The inverse of a function can be determined at specific points on its graph.<\/li>\r\n \t<li>To find the inverse of a formula, solve the equation [latex]y=f\\left(x\\right)[\/latex] for [latex]x[\/latex] as a function of\u00a0[latex]y[\/latex]. Then exchange the labels [latex]x[\/latex] and [latex]y[\/latex].<\/li>\r\n \t<li>The graph of an inverse function is the reflection of the graph of the original function across the line [latex]y=x[\/latex].<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137932588\" class=\"definition\">\r\n \t<dt><strong>horizontal line test<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134149777\">a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more than once<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137441703\" class=\"definition\">\r\n \t<dt>inverse function<\/dt>\r\n \t<dd id=\"fs-id1165137441708\">for any one-to-one function [latex]f\\left(x\\right)[\/latex], the inverse is a function [latex]{f}^{-1}\\left(x\\right)[\/latex] such that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]; this also implies that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135511364\" class=\"definition\">\r\n \t<dt><strong>one-to-one function<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135511369\">a function for which each value of the output is associated with a unique input value<\/dd>\r\n<\/dl>\r\n<h2><\/h2>\r\n<h2 style=\"text-align: center;\">Section 5.2 Homework Exercises<\/h2>\r\n1. How can you determine if a relation is a one-to-one function?\r\n\r\n2. Why does the horizontal line test tell us whether the graph of a function is one-to-one?\r\n\r\n3. Why do we restrict the domain of the function [latex]f\\left(x\\right)={x}^{2}[\/latex] to find the function\u2019s inverse?\r\n\r\n4. Can a function be its own inverse? Explain.\r\n\r\n5. Are one-to-one functions either always increasing or always decreasing? Why or why not?\r\n\r\n6. How do you find the inverse of a function algebraically?\r\n\r\n7. Show that the function [latex]f\\left(x\\right)=a-x[\/latex] is its own inverse for all real numbers [latex]a[\/latex].\r\n\r\nFor the following exercises, find [latex]{f}^{-1}\\left(x\\right)[\/latex] for each function.\r\n\r\n8. [latex]f\\left(x\\right)=x+3[\/latex]\r\n\r\n9.\u00a0[latex]f\\left(x\\right)=x+5[\/latex]\r\n\r\n10. [latex]f\\left(x\\right)=2-x[\/latex]\r\n\r\n11.\u00a0[latex]f\\left(x\\right)=3-x[\/latex]\r\n\r\n12.\u00a0[latex]f\\left(x\\right)=\\frac{x}{x+2}[\/latex]\r\n\r\n13.\u00a0[latex]f\\left(x\\right)=\\frac{2x+3}{5x+4}[\/latex]\r\n\r\nFor the following exercises, find a domain on which each function [latex]f[\/latex] is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of [latex]f[\/latex] restricted to that domain.\r\n\r\n14. [latex]f\\left(x\\right)={\\left(x+7\\right)}^{2}[\/latex]\r\n\r\n15.\u00a0[latex]f\\left(x\\right)={\\left(x - 6\\right)}^{2}[\/latex]\r\n\r\n16. [latex]f\\left(x\\right)={x}^{2}-5[\/latex]\r\n\r\n17.\u00a0Given [latex]f\\left(x\\right)=\\frac{x}{2}+x[\/latex] and [latex]g\\left(x\\right)=\\frac{2x}{1-x}[\/latex]\r\n\r\na. Find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex]\r\n\r\nb. What does the answer tell us about the relationship between [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)?[\/latex]\r\n\r\nFor the following exercises, use function composition to verify that [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] are inverse functions.\r\n\r\n18. [latex]f\\left(x\\right)=\\sqrt[3]{x - 1}[\/latex] and [latex]g\\left(x\\right)={x}^{3}+1[\/latex]\r\n\r\n19.\u00a0[latex]f\\left(x\\right)=-3x+5[\/latex] and [latex]g\\left(x\\right)=\\frac{x - 5}{-3}[\/latex]\r\n\r\nFor the following exercises, determine if the given graph is a one-to-one function.\r\n\r\n20. Graph of a circle.\r\n\r\n21. Graph of a parabola.\r\n\r\n22. Graph of a rotated cubic function.\r\n\r\n23. Graph of half of 1\/x.\r\n\r\n24. Graph of a one-to-one function.\r\n\r\nFor the following exercises, determine whether the graph represents a one-to-one function.\r\n\r\n25.\r\n<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010629\/CNX_Precalc_Figure_01_07_2012.jpg\" alt=\"Graph of an upright parabola with vertex at (10, -10), passing through (0,0) and (25,0)\" width=\"487\" height=\"253\" \/>\r\n\r\n26.\r\n<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010629\/CNX_Precalc_Figure_01_07_2022.jpg\" alt=\"Graph of a step-function, with y = -5 for {x|-10&lt;=x,0} and y = 0 for {x|0&lt;=x&lt;10}\" width=\"487\" height=\"376\" \/>\r\n\r\nFor the following exercises, use the graph of [latex]f[\/latex] shown below.\r\n<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010629\/CNX_Precalc_Figure_01_07_2032.jpg\" alt=\"Graph of the line y = (-3\/2)x + 3\" width=\"487\" height=\"368\" \/>\r\n27. Find [latex]f\\left(0\\right)[\/latex].\r\n\r\n28. Solve [latex]f\\left(x\\right)=0[\/latex].\r\n\r\n29. Find [latex]{f}^{-1}\\left(0\\right)[\/latex].\r\n\r\n30.\u00a0Solve [latex]{f}^{-1}\\left(x\\right)=0[\/latex].\r\n\r\nFor the following exercises, use the graph of the one-to-one function shown below.\r\n<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010629\/CNX_Precalc_Figure_01_07_2042.jpg\" alt=\"Graph of a square root function for {x|x&gt;=2}\" width=\"487\" height=\"254\" \/>\r\n31. Sketch the graph of [latex]{f}^{-1}[\/latex].\r\n\r\n32. Find [latex]f\\left(6\\right)\\text{ and }{f}^{-1}\\left(2\\right)[\/latex].\r\n\r\n33. If the complete graph of [latex]f[\/latex] is shown, find the domain of [latex]f[\/latex].\r\n\r\n34. If the complete graph of [latex]f[\/latex] is shown, find the range of [latex]f[\/latex].\r\n\r\nFor the following exercises, evaluate or solve, assuming that the function [latex]f[\/latex] is one-to-one.\r\n\r\n35. If [latex]f\\left(6\\right)=7[\/latex], find [latex]{f}^{-1}\\left(7\\right)[\/latex].\r\n\r\n36.\u00a0If [latex]f\\left(3\\right)=2[\/latex], find [latex]{f}^{-1}\\left(2\\right)[\/latex].\r\n\r\n37. If [latex]{f}^{-1}\\left(-4\\right)=-8[\/latex], find [latex]f\\left(-8\\right)[\/latex].\r\n\r\n38.\u00a0If [latex]{f}^{-1}\\left(-2\\right)=-1[\/latex], find [latex]f\\left(-1\\right)[\/latex].\r\n\r\nFor the following exercises, use the values listed in the table below\u00a0to evaluate or solve.\r\n<table id=\"Table_01_07_06\" summary=\"Two column and ten rows. The first column is labeled, \">\r\n<tbody>\r\n<tr>\r\n<td style=\"text-align: center;\"><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td style=\"text-align: center;\"><strong> [latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">0<\/td>\r\n<td style=\"text-align: center;\">8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">1<\/td>\r\n<td style=\"text-align: center;\">0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">2<\/td>\r\n<td style=\"text-align: center;\">7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">3<\/td>\r\n<td style=\"text-align: center;\">4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">4<\/td>\r\n<td style=\"text-align: center;\">2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">5<\/td>\r\n<td style=\"text-align: center;\">6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">6<\/td>\r\n<td style=\"text-align: center;\">5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">7<\/td>\r\n<td style=\"text-align: center;\">3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">8<\/td>\r\n<td style=\"text-align: center;\">9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;\">9<\/td>\r\n<td style=\"text-align: center;\">1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n39. Find [latex]f\\left(1\\right)[\/latex].\r\n\r\n40.\u00a0Solve [latex]f\\left(x\\right)=3[\/latex].\r\n\r\n41. Find [latex]{f}^{-1}\\left(0\\right)[\/latex].\r\n\r\n42.\u00a0Solve [latex]{f}^{-1}\\left(x\\right)=7[\/latex].\r\n\r\n43. Use the tabular representation of [latex]f[\/latex] to create a table for [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\r\n<td>3<\/td>\r\n<td>6<\/td>\r\n<td>9<\/td>\r\n<td>13<\/td>\r\n<td>14<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\r\n<td>1<\/td>\r\n<td>4<\/td>\r\n<td>7<\/td>\r\n<td>12<\/td>\r\n<td>16<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFor the following exercises, find the inverse function. Then, graph the function and its inverse.\r\n\r\n44. [latex]f\\left(x\\right)=\\frac{3}{x - 2}[\/latex]\r\n\r\n45. [latex]f\\left(x\\right)={x}^{3}-1[\/latex]\r\n\r\n46. Find the inverse function of [latex]f\\left(x\\right)=\\frac{1}{x - 1}[\/latex]. Use a graphing utility to find its domain and range. Write the domain and range in interval notation.\r\n\r\n47.\u00a0To convert from [latex]x[\/latex] degrees Celsius to [latex]y[\/latex] degrees Fahrenheit, we use the formula [latex]f\\left(x\\right)=\\frac{9}{5}x+32[\/latex]. Find the inverse function, if it exists, and explain its meaning.\r\n\r\n48.\u00a0The circumference [latex]C[\/latex] of a circle is a function of its radius given by [latex]C\\left(r\\right)=2\\pi r[\/latex]. Express the radius of a circle as a function of its circumference. Call this function [latex]r\\left(C\\right)[\/latex]. Find [latex]r\\left(36\\pi \\right)[\/latex] and interpret its meaning.\r\n\r\n49. A car travels at a constant speed of 50 miles per hour. The distance the car travels in miles is a function of time, [latex]t[\/latex], in hours given by [latex]d\\left(t\\right)=50t[\/latex]. Find the inverse function by expressing the time of travel in terms of the distance traveled. Call this function [latex]t\\left(d\\right)[\/latex]. Find [latex]t\\left(180\\right)[\/latex] and interpret its meaning.\r\n\r\n<\/div>\r\n<\/section>\r\n<h2><\/h2>\r\n<h2 style=\"text-align: center;\">Section 1.8 Solutions to Odd-Numbered Exercises<\/h2>\r\n1.\u00a0Each output of a function must have exactly one input for the function to be one-to-one.\r\n\r\n3. Without any domain restriction,\u00a0[latex]f(x)=x^{2}[\/latex] does not have an inverse function as it fails the horizontal line test. But if we restrict the domain to be [latex]x&gt;0[\/latex], then we find that it passes the horizontal line test and therefore has an inverse function.\r\n\r\n5. One-to-one functions are always increasing or decreasing. This ensures there is only one input for every output.\r\n\r\n7. [latex]f(x)=a-x\\\\y=a-x\\\\x=a-y\\\\x+y=a\\\\y=a-x\\\\f^{-1}=a-x[\/latex]\r\n\r\n9. [latex]f^{-1}(x)=x-5[\/latex]\r\n\r\n11. [latex]f^{-1}(x)=3-x[\/latex]\r\n\r\n13. [latex]f^{-1}(x)=\\frac{3-4x}{5x-2}[\/latex]\r\n\r\n15. Domain of [latex]f(x)[\/latex] is [latex]\\left[6,\\infty \\right)[\/latex] and\u00a0 [latex]f^{-1}=\\sqrt{x}+6[\/latex]\r\n\r\n17. a. [latex]f(g(x))=\\frac{3x}{1-x}[\/latex] and [latex]g(f(x))=\\frac{6x}{2-3x}[\/latex]\r\nb. [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are not inverses.\r\n\r\n19. [latex]f(g(x))=x[\/latex] and [latex]g(f(x))=x[\/latex]\r\n\r\n21. Not one-to-one\r\n\r\n23. One-to-one\r\n\r\n25. Not one-to-one\r\n\r\n27. [latex]f(0)=3[\/latex]\r\n\r\n29. [latex]f^{-1}(0)=2[\/latex]\r\n\r\n31.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010630\/CNX_Precalc_Figure_01_07_2052.jpg\" alt=\"Graph of a square root function and its inverse.\" \/>\r\n\r\n33. [latex]\\left[2,\\infty \\right)[\/latex]\r\n\r\n35.[latex]f^{-1}(7)=6[\/latex]\r\n\r\n37. [latex]f(-8)=-4[\/latex]\r\n\r\n39. [latex]f(1)=0[\/latex]\r\n\r\n41. [latex]f^{-1}(0)=1[\/latex]\r\n\r\n43.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]x[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>4<\/td>\r\n<td>7<\/td>\r\n<td>12<\/td>\r\n<td>16<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]{f}^{-1}\\left(x\\right)[\/latex]<\/td>\r\n<td>3<\/td>\r\n<td>6<\/td>\r\n<td>9<\/td>\r\n<td>13<\/td>\r\n<td>14<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n45. [latex]{f}^{-1}\\left(x\\right)={\\left(1+x\\right)}^{1\/3}[\/latex]\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005116\/CNX_Precalc_Figure_01_07_207.jpg\" alt=\"Graph of a cubic function and its inverse.\" \/>\r\n\r\n47. [latex]{f}^{-1}\\left(x\\right)=\\frac{5}{9}\\left(x - 32\\right)[\/latex]. Given the Fahrenheit temperature, [latex]x[\/latex], this formula allows you to calculate the Celsius temperature.\r\n\r\n49. [latex]t\\left(d\\right)=\\frac{d}{50}[\/latex], [latex]t\\left(180\\right)=\\frac{180}{50}[\/latex]. The time for the car to travel 180 miles is 3.6 hours.\r\n\r\n<section id=\"fs-id1165137725994\"><\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>LEARNING OBJECTIVES<\/h3>\n<p>By the end of this lesson, you will be able to:<\/p>\n<ul>\n<li>Determine whether a function is one-to-one.<\/li>\n<li>Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.<\/li>\n<li>Find or evaluate the inverse of a function.<\/li>\n<li>Use the graph of a one-to-one function to graph its inverse function on the same axes.<\/li>\n<\/ul>\n<\/div>\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Determining Whether a Function is One-to-One<\/span><\/h2>\n<p id=\"fs-id1165135245630\">Some functions have only one input value for each output value, as well as having only one output for each input. We call these functions <strong>one-to-one functions<\/strong>. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in.<\/p>\n<table id=\"Table_01_01_13\" summary=\"Two columns and five rows. The first column is labeled,\">\n<colgroup>\n<col \/>\n<col \/> <\/colgroup>\n<thead>\n<tr>\n<th>Letter grade<\/th>\n<th>Grade point average<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>A<\/td>\n<td>4.0<\/td>\n<\/tr>\n<tr>\n<td>B<\/td>\n<td>3.0<\/td>\n<\/tr>\n<tr>\n<td>C<\/td>\n<td>2.0<\/td>\n<\/tr>\n<tr>\n<td>D<\/td>\n<td>1.0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137561844\">This grading system represents a one-to-one function, because each letter input yields one particular grade point average output and each grade point average corresponds to one input letter.<\/p>\n<p>To visualize this concept, let\u2019s look again at the two simple functions sketched in (a)and (b) of Figure 10.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010531\/CNX_Precalc_Figure_01_01_0012.jpg\" alt=\"Three relations that demonstrate what constitute a function.\" width=\"975\" height=\"243\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<p>The function in part (a) shows a relationship that is not a one-to-one function because inputs [latex]q[\/latex] and [latex]r[\/latex] both give output [latex]n[\/latex]. The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: One-to-One Function<\/h3>\n<p>A one-to-one function is a function in which each output value corresponds to exactly one input value.<\/p>\n<\/div>\n<div id=\"Example_01_07_01\" class=\"example\">\n<div id=\"fs-id1165137656641\" class=\"exercise\">\n<div id=\"fs-id1165137922642\" class=\"problem textbox shaded\">\n<h3>Example 1: determining if a set of ordered pairs is a one-to-one function<\/h3>\n<p id=\"fs-id1165137659325\">For each set of ordered pairs, determine if it represents a function and, if so, if the function is one-to-one.<\/p>\n<p>a)\u00a0[latex]\\{(-3,27),(-2,8),(-1,1),(0,0),(1,1),(2,8),(3,27)\\}[\/latex]<\/p>\n<p>b)\u00a0[latex]\\{(0,0),(1,1),(4,2),(9,3),(16,4)\\}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137768306\" class=\"solution textbox shaded\">\n<h3>Solution<\/h3>\n<p id=\"fs-id1165137737081\">a) Each <em>x<\/em>-value is matched with only one <em>y<\/em>-value. So this relation is a function.\u00a0 But each <em>y<\/em>-value is not paired with only one <em>x<\/em>-value, [latex](-3,27)[\/latex] and [latex](3,27)[\/latex], for example. So this function is not one-to-one.<\/p>\n<p>b) Each <em>x<\/em>-value is matched with only one <em>y<\/em>-value. So this relation is a function. Since each <em>y<\/em>-value is paired with only one <em>x<\/em>-value, this function is one-to-one.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 1<\/h3>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137659325\">For each set of ordered pairs, determine if it represents a function and, if so, if the function is one-to-one.<\/p>\n<p>a)\u00a0[latex]\\{(-3,-6),(-2,-4),(-1,-2),(0,0),(1,2),(2,4),(3,6)\\}[\/latex]<\/p>\n<p>b)\u00a0[latex]\\{(-4,-8),(-2,4),(-1,2),(0,0),(1,2),(2,4),(4,8)\\}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380740\">Show Solution<\/span><\/p>\n<div id=\"q380740\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"q676921\" class=\"hidden-answer\">\n<p id=\"fs-id1165137737081\">a) Each <em>x<\/em>-value is matched with only one <em>y<\/em>-value. So this relation is a function.\u00a0Since each <em>y<\/em>-value is paired with only one <em>x<\/em>-value, this function is one-to-one.<\/p>\n<p>b) Each <em>x<\/em>-value is matched with only one <em>y<\/em>-value. So this relation is a function. But each <em>y<\/em>-value is not paired with only one <em>x<\/em>-value, [latex](-2,4)[\/latex] and [latex](2,4)[\/latex], for example. So this function is not one-to-one.<\/p>\n<\/div>\n<div id=\"q676921\">\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>To help us determine whether a function is one-to-one, we use the horizontal line test. We use a horizontal line and check that each horizontal line intersects the graph in only one point. The horizontal line is representing a y-value and we check that it intersects the graph in only one x-value. If every horizontal line intersects the graph of a function in at most one point, it is a one-to-one function. This is the <strong>horizontal line test.<\/strong><\/p>\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Using the Horizontal Line Test<\/span><\/h2>\n<p id=\"fs-id1165137871503\">Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the <strong>horizontal line test<\/strong>. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function.<\/p>\n<div id=\"fs-id1165137445319\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137635406\" style=\"text-align: center;\"><strong>How To: Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function.<\/strong><\/h3>\n<ol id=\"fs-id1165137611853\">\n<li>Inspect the graph to see if any horizontal line drawn would intersect the curve more than once.<\/li>\n<li>If there is any such line, determine that the function is not one-to-one.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Applying the Horizontal Line Test<\/h3>\n<p>Consider the functions (a), (b), and (c) shown in\u00a0the graphs in Figure 1.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005018\/CNX_Precalc_Figure_01_01_013abc.jpg\" alt=\"Graph of a polynomial, line, circle\" width=\"975\" height=\"418\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p>Are either of the functions one-to-one?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q549926\">Show Solution<\/span><\/p>\n<div id=\"q549926\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135185190\">The function in (a) is not one-to-one. The horizontal line shown in Figure 2\u00a0intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)<\/p>\n<figure id=\"Figure_01_01_010\" class=\"small\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005019\/CNX_Precalc_Figure_01_01_010.jpg\" alt=\"\" width=\"487\" height=\"445\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165135151243\">The function in (b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.<\/p>\n<p>The graph in (c) in not one-to-one.\u00a0 Any horizontal line will intersect a circle at two points.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Inverse Function<\/h3>\n<p>Suppose [latex]y=f\\left(x\\right)[\/latex] is a one-to-one function. An <strong>inverse function<\/strong>\u00a0is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.\u00a0 The symbol\u00a0[latex]{f}^{-1}\\left(x\\right)[\/latex] is used to represent the inverse function of [latex]f[\/latex].\u00a0 In other words, if\u00a0[latex]y=f\\left(x\\right)[\/latex] is a one-to-one function, then [latex]f[\/latex] has an inverse function\u00a0[latex]{f}^{-1}\\left(x\\right)[\/latex] and [latex]x={f}^{-1}\\left(y\\right)[\/latex] .<\/p>\n<\/div>\n<p id=\"fs-id1165135528385\">\u00a0The raised [latex]-1[\/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[\/latex] . In other words, [latex]{f}^{-1}\\left(x\\right)[\/latex] does <em>not<\/em> mean [latex]\\frac{1}{f\\left(x\\right)}[\/latex] because [latex]\\frac{1}{f\\left(x\\right)}[\/latex] is the reciprocal of [latex]f[\/latex] and not the inverse.<\/p>\n<div id=\"fs-id1165137933105\" class=\"note textbox\">\n<h3 class=\"title\">Verifying an inverse function<\/h3>\n<p>[latex]f[\/latex] and [latex]g[\/latex] are inverses of each other if:<\/p>\n<p>[latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] where [latex]x[\/latex]\u00a0 is in the domain of [latex]f[\/latex]<\/p>\n<p>[latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] where\u00a0[latex]x[\/latex]\u00a0 is in the domain of [latex]g[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1165137655153\">Given a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether either [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] or [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)<\/p>\n<p id=\"fs-id1165135397975\">For example, [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.<\/p>\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\left(g\\circ f\\right)\\left(x\\right)=g\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/div>\n<p id=\"fs-id1165137767233\">and<\/p>\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\left({f}^{}\\circ g\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/div>\n<div><\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137438777\">A few coordinate pairs from the graph of the function [latex]y=4x[\/latex] are (\u22122, \u22128), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function [latex]y=\\frac{1}{4}x[\/latex] are (\u22128, \u22122), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.<\/p>\n<div id=\"fs-id1165137656641\" class=\"exercise\">\n<div id=\"fs-id1165137922642\" class=\"problem textbox shaded\">\n<h3>Example 3: Identifying an Inverse Function for a Given Input-Output Pair<\/h3>\n<p id=\"fs-id1165137659325\">If for a particular one-to-one function [latex]f\\left(2\\right)=4[\/latex] and [latex]f\\left(5\\right)=12[\/latex], what are the corresponding input and output values for the inverse function?<\/p>\n<\/div>\n<div id=\"fs-id1165137768306\" class=\"solution textbox shaded\">\n<h3>Solution<\/h3>\n<p id=\"fs-id1165137737081\">The inverse function reverses the input and output quantities, so if<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(2\\right)=4,\\text{ then }{f}^{-1}\\left(4\\right)=2;\\\\ f\\left(5\\right)=12,{\\text{ then f}}^{-1}\\left(12\\right)=5\\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137659464\">Alternatively, if we want to name the inverse function [latex]g[\/latex], then [latex]g\\left(4\\right)=2[\/latex] and [latex]g\\left(12\\right)=5[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165135245520\" class=\"commentary\">\n<h3>Analysis of the Solution<\/h3>\n<p id=\"fs-id1165135508518\">Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.<\/p>\n<table id=\"Table_01_07_01\" summary=\"For (x,f(x)) we have the values (2, 4) and (5, 12); for (x, g(x)), we have the values (4, 2) and (12, 5).\">\n<thead>\n<tr>\n<th>[latex]\\left(x,f\\left(x\\right)\\right)[\/latex]<\/th>\n<th>[latex]\\left(x,g\\left(x\\right)\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\left(2,4\\right)[\/latex]<\/td>\n<td>[latex]\\left(4,2\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\left(5,12\\right)[\/latex]<\/td>\n<td>[latex]\\left(12,5\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\n<p id=\"fs-id1165137659089\">Given that [latex]{h}^{-1}\\left(6\\right)=2[\/latex], what are the corresponding input and output values of the original function [latex]h?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380743\">Show Solution<\/span><\/p>\n<div id=\"q380743\" class=\"hidden-answer\" style=\"display: none\">\n<p>You would reverse the 6 and 2 and rewrite as: [latex]{h}\\left(2\\right)=6[\/latex]\n<\/p><\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Find an Inverse Function From a Table\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TSztRfzmk0M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"fs-id1165134357354\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135434077\">How To: Given two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex], test whether the functions are inverses of each other.<\/h3>\n<ol id=\"fs-id1165137452358\">\n<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] or [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\n<li>If either statement is true, then both are true, and [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then both are false, and [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_07_02\" class=\"example\">\n<div id=\"fs-id1165137557051\" class=\"exercise\">\n<div id=\"fs-id1165137679032\" class=\"problem textbox shaded\">\n<h3>Example 4: Testing Inverse Relationships Algebraically<\/h3>\n<p id=\"fs-id1165135519417\">If [latex]f\\left(x\\right)=\\frac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\frac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137627632\" class=\"solution\">\n<div id=\"fs-id1165137675509\" class=\"equation unnumbered textbox shaded\">\n<h3>Solution<\/h3>\n<p style=\"text-align: center;\">[latex]\\begin{align} g\\left(f\\left(x\\right)\\right)=\\frac{1}{\\left(\\frac{1}{x+2}\\right)}{-2 }\\hfill\\\\={ x }+{ 2 } -{ 2 }\\hfill\\\\={ x }\\hfill\\end{align}[\/latex]<\/p>\n<div id=\"fs-id1165135678636\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\text{So }g={f}^{-1}\\text{ and }f={g}^{-1}\\end{align}[\/latex]<\/div>\n<div><\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165135436648\">This is enough to answer yes to the question, but we can also verify the other formula.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1165137784350\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)=\\frac{1}{\\frac{1}{x}-2+2}\\\\ =\\frac{1}{\\frac{1}{x}}\\hfill \\\\ =x\\hfill \\end{align}[\/latex]<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137733685\" class=\"commentary\">\n<h3>Analysis of the Solution<\/h3>\n<p id=\"fs-id1165135389000\">Notice the inverse operations are in reverse order of the operations from the original function.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 3<\/h3>\n<p id=\"fs-id1165135160550\">If [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380741\">Show Solution<\/span><\/p>\n<div id=\"q380741\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes [latex]g={f}^{-1}[\/latex] since:<\/p>\n<div id=\"q676921\" class=\"hidden-answer\">\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(g(x))=f(x^{3}-4)=\\sqrt[3]{x^3-4+4}=\\sqrt[3]{x^{3}}=x\\\\g(f(x))=g(\\sqrt[3]{x+4})=(\\sqrt[3]{x+4})^{3}-4=x+4-4=x[\/latex]<\/div>\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<p id=\"fs-id1165137737081\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_01_07_03\" class=\"example\">\n<div id=\"fs-id1165135259560\" class=\"exercise\">\n<div id=\"fs-id1165134042918\" class=\"problem textbox shaded\">\n<h3>Example 4: Determining Inverse Relationships for Power Functions<\/h3>\n<p id=\"fs-id1165137441834\">If [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137442603\" class=\"solution\">\n<div id=\"fs-id1165137591632\" class=\"equation unnumbered textbox shaded\">\n<h3>Solution<\/h3>\n<p style=\"text-align: center;\">[latex]f\\left(g\\left(x\\right)\\right)=\\frac{{x}^{3}}{27}\\ne x[\/latex]<\/p>\n<p id=\"fs-id1165137694053\">No, the functions are not inverses.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135317479\" class=\"commentary\">\n<h3>Analysis of the Solution<\/h3>\n<p id=\"fs-id1165134192978\">The correct inverse to the cube is, of course, the cube root [latex]\\sqrt[3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 4<\/h3>\n<p id=\"fs-id1165137573532\">If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380750\">Show Solution<\/span><\/p>\n<div id=\"q380750\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes [latex]g={f}^{-1}[\/latex] since:<\/p>\n<div id=\"q676921\" class=\"hidden-answer\">\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(g(x))=f(\\sqrt[3]{x}+1)=(\\sqrt[3]{x}+1-1)^{3}=(\\sqrt[3]{x})^{3}=x\\\\g(f(x))=g((x-1)^{3})=\\sqrt[3]{(x-1)^{3}}+1=x-1+1=x[\/latex]<\/div>\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<p id=\"fs-id1165137737081\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Domain and Range of Inverse Functions<\/h2>\n<p>The outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010624\/CNX_Precalc_Figure_01_07_0032.jpg\" alt=\"Domain and range of a function and its inverse.\" width=\"487\" height=\"143\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3.<\/b> Domain and range of a function and its inverse<\/p>\n<\/div>\n<p id=\"fs-id1165135557891\">When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] is [latex]{f}^{-1}\\left(x\\right)={x}^{2}[\/latex], because a square &#8220;undoes&#8221; a square root; but the square is only the inverse of the square root on the domain [latex]\\left[0,\\infty \\right)[\/latex], since that is the range of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex].<\/p>\n<p id=\"fs-id1165137730185\">We can look at this problem from the other side, starting with the square (toolkit quadratic) function [latex]f\\left(x\\right)={x}^{2}[\/latex]. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and \u20133. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the &#8220;inverse&#8221; is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.<\/p>\n<p id=\"fs-id1165137823552\">In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).<\/p>\n<p id=\"fs-id1165132037000\">If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}[\/latex] on [latex]\\left[1,\\infty \\right)[\/latex], then the inverse function is [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+1[\/latex].<\/p>\n<ul id=\"fs-id1165137851227\">\n<li>The domain of [latex]f[\/latex] = range of [latex]{f}^{-1}[\/latex] = [latex]\\left[1,\\infty \\right)[\/latex].<\/li>\n<li>The domain of [latex]{f}^{-1}[\/latex] = range of [latex]f[\/latex] = [latex]\\left[0,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<div id=\"fs-id1165137733804\" class=\"note precalculus qa textbox\">\n<p id=\"fs-id1165137723526\"><strong>Q &amp; A<\/strong><\/p>\n<p><strong>Is it possible for a function to have more than one inverse?<\/strong><\/p>\n<p id=\"fs-id1165137456608\"><em>No. If two supposedly different functions, say, [latex]g[\/latex] and [latex]h[\/latex], both meet the definition of being inverses of another function [latex]f[\/latex], then you can prove that [latex]g=h[\/latex]. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.<\/em><\/p>\n<\/div>\n<div id=\"fs-id1165137704938\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Domain and Range of Inverse Functions<\/h3>\n<p id=\"fs-id1165135319550\">The range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<p id=\"fs-id1165137673886\">The domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165135308785\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137605040\"><strong>How To: Given a function, find the domain and range of its inverse.<br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id1165137530434\">\n<li>If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.<\/li>\n<li>If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_07_05\" class=\"example\">\n<div id=\"fs-id1165137667922\" class=\"exercise\">\n<div id=\"fs-id1165135511321\" class=\"problem textbox shaded\">\n<h3>Example 4: Finding the Inverses of Toolkit Functions<\/h3>\n<p id=\"fs-id1165137448020\">Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed below. We restrict the domain in such a fashion that the function assumes all <em>y<\/em>-values exactly once.<\/p>\n<table id=\"Table_01_07_02\" summary=\"A list of the toolkit function. The constant function is f(x) = c where c is the constant; the identity function is f(x) = x; the absolute function is f(x)=|x|; the quadratic function is f(x) = x^2; the cubic function is f(x)=x^3; the reciprocal function is f(x)=1\/x; the reciprocal squared function is f(x)=1\/x^2; the square root function is f(x)=sqrt(x); the cube root function is f(x) = x^(1\/3).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>Constant<\/td>\n<td>Identity<\/td>\n<td>Quadratic<\/td>\n<td>Cubic<\/td>\n<td>Reciprocal<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(x\\right)=c[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=x[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)={x}^{2}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)={x}^{3}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Reciprocal squared<\/td>\n<td>Cube root<\/td>\n<td>Square root<\/td>\n<td>Absolute value<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=\\sqrt{x}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=|x|[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137767030\" class=\"solution textbox shaded\">\n<h3>Solution<\/h3>\n<p id=\"fs-id1165132988445\">The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.<\/p>\n<p id=\"fs-id1165134080947\">The absolute value function can be restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], where it is equal to the identity function.<\/p>\n<p id=\"fs-id1165137642849\">The reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165137901280\" class=\"commentary\">\n<h3>Analysis of the Solution<\/h3>\n<p>We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs.\u00a0They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010624\/CNX_Precalc_Figure_01_07_004ab2.jpg\" alt=\"Graph of an absolute function.\" width=\"975\" height=\"404\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4.<\/b> (a) Absolute value (b) Reciprocal squared<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 5<\/h3>\n<p id=\"fs-id1165137507853\">The domain of function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of function [latex]f[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex]. Find the domain and range of the inverse function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380755\">Show Solution<\/span><\/p>\n<div id=\"q380755\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"q676921\" class=\"hidden-answer\">\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center;\">The domain of function [latex]{f}^{-1}[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex] and the range of function [latex]{f}^{-1}[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex].<\/div>\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<p id=\"fs-id1165137737081\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding and Evaluating Inverse Functions<\/h2>\n<p id=\"fs-id1165137761017\">Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.<\/p>\n<section id=\"fs-id1165135466392\">\n<h2 style=\"text-align: center;\"><\/h2>\n<h2 style=\"text-align: center;\"><\/h2>\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Inverting Tabular Functions<\/span><\/h2>\n<p id=\"fs-id1165135190714\">Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.<\/p>\n<p id=\"fs-id1165137422578\">Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.<\/p>\n<div id=\"Example_01_07_06\" class=\"example\">\n<div id=\"fs-id1165135544995\" class=\"exercise\">\n<div id=\"fs-id1165137698262\" class=\"problem textbox shaded\">\n<h3>Example 5: Interpreting the Inverse of a Tabular Function<\/h3>\n<p id=\"fs-id1165135435474\">A function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes. Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].<\/p>\n<table id=\"Table_01_07_03\" summary=\"Two rows and five columns. The first row is labeled\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\n<td>30<\/td>\n<td>50<\/td>\n<td>70<\/td>\n<td>90<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137407569\" class=\"solution textbox shaded\">\n<h3>Solution<\/h3>\n<p id=\"fs-id1165137640334\">The inverse function takes an output of [latex]f[\/latex] and returns an input for [latex]f[\/latex]. So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.<\/p>\n<p id=\"fs-id1165135181841\">Alternatively, recall that the definition of the inverse was that if [latex]f\\left(a\\right)=b[\/latex], then [latex]{f}^{-1}\\left(b\\right)=a[\/latex]. By this definition, if we are given [latex]{f}^{-1}\\left(70\\right)=a[\/latex], then we are looking for a value [latex]a[\/latex] so that [latex]f\\left(a\\right)=70[\/latex]. In this case, we are looking for a [latex]t[\/latex] so that [latex]f\\left(t\\right)=70[\/latex], which is when [latex]t=90[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 6<\/h3>\n<p id=\"fs-id1165134108483\">Using the table below,\u00a0find and interpret (a) [latex]\\text{ }f\\left(60\\right)[\/latex], and (b) [latex]\\text{ }{f}^{-1}\\left(60\\right)[\/latex].<\/p>\n<table id=\"Table_01_07_04\" summary=\"Two rows and five columns. The first row is labeled\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>[latex]t\\text{ (minutes)}[\/latex]<\/td>\n<td>30<\/td>\n<td>50<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<td>90<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex]<\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>50<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"q676921\" class=\"hidden-answer\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380760\">Show Solution<\/span><\/p>\n<div id=\"q380760\" class=\"hidden-answer\" style=\"display: none\">[latex]\\text{ }f\\left(60\\right)=50[\/latex], in 60 minutes, 50 miles are traveled.<br \/>\n[latex]\\text{ }{f}^{-1}\\left(60\\right)=70[\/latex], to travel 60 miles, it will take 70 minutes.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<section id=\"fs-id1165137418615\">\n<h2 style=\"text-align: center;\"><\/h2>\n<h2 style=\"text-align: center;\"><\/h2>\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/span><\/h2>\n<p id=\"fs-id1165137400045\">We saw in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/csn-precalculus\/chapter\/determine-whether-a-relation-represents-a-function\/\" target=\"_blank\" rel=\"noopener\">Functions and Function Notation<\/a> that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the <em>vertical<\/em> extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the <em>horizontal<\/em> extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function\u2019s graph.<\/p>\n<div id=\"fs-id1165133045388\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135333128\">How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\n<ol id=\"fs-id1165137464840\">\n<li>Find the desired input on the <em>y<\/em>-axis of the given graph.<\/li>\n<li>Read the inverse function\u2019s output from the <em>x<\/em>-axis of the given graph.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_07_07\" class=\"example\">\n<div id=\"fs-id1165135434803\" class=\"exercise\">\n<div id=\"fs-id1165135434805\" class=\"problem textbox shaded\">\n<h3>Example 6: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/h3>\n<p>A function [latex]g\\left(x\\right)[\/latex] is given in Figure 5. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137468840\" class=\"solution textbox shaded\">\n<h3>Solution<\/h3>\n<p id=\"fs-id1165137468842\">To evaluate [latex]g\\left(3\\right)[\/latex], we find 3 on the <em>x<\/em>-axis and find the corresponding output value on the <em>y<\/em>-axis. The point [latex]\\left(3,1\\right)[\/latex] tells us that [latex]g\\left(3\\right)=1[\/latex].<\/p>\n<p>To evaluate [latex]{g}^{-1}\\left(3\\right)[\/latex], recall that by definition [latex]{g}^{-1}\\left(3\\right)[\/latex] means the value of <em>x<\/em> for which [latex]g\\left(x\\right)=3[\/latex]. By looking for the output value 3 on the vertical axis, we find the point [latex]\\left(5,3\\right)[\/latex] on the graph, which means [latex]g\\left(5\\right)=3[\/latex], so by definition, [latex]{g}^{-1}\\left(3\\right)=5[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 7<\/h3>\n<p id=\"fs-id1165137812560\">Using the graph in Example 6, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].<\/p>\n<div id=\"q676921\" class=\"hidden-answer\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380765\">Show Solution<\/span><\/p>\n<div id=\"q380765\" class=\"hidden-answer\" style=\"display: none\">\na.)\u00a0 1 \u00a0b.)\u00a0 5.5\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137605437\">\n<h2 style=\"text-align: center;\"><\/h2>\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Finding Inverses of Functions Represented by Formulas<\/span><\/h2>\n<p id=\"fs-id1165137433184\">Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula\u2014 for example, [latex]y[\/latex] as a function of [latex]x\\text{-\\hspace{0.17em}}[\/latex] we can often find the inverse function by solving to obtain [latex]x[\/latex] as a function of [latex]y[\/latex].<\/p>\n<div id=\"fs-id1165137652548\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135195849\">How To: Given a one-to-one function represented by a formula, find the inverse.<\/h3>\n<ol id=\"fs-id1165135443898\">\n<li>Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex] .<\/li>\n<li>Switch [latex]x[\/latex] and [latex]y[\/latex].<\/li>\n<li>Solve for [latex]y[\/latex]<\/li>\n<li>Replace [latex]y[\/latex] with [latex]{f}^{-1}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_07_09\" class=\"example\">\n<div id=\"fs-id1165134065146\" class=\"exercise\">\n<div id=\"fs-id1165137409366\" class=\"problem textbox shaded\">\n<h3>Example 8: Solving to Find an Inverse Function<\/h3>\n<p id=\"fs-id1165137891504\">Find the inverse of the function [latex]f\\left(x\\right)=\\frac{2}{x - 3}+4[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165137482074\" class=\"solution textbox shaded\">\n<h3 class=\"equation unnumbered\">Solution<\/h3>\n<div id=\"fs-id1165135189953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}y=\\frac{2}{x - 3}+4\\hspace{0.5in} & \\text{Set up an equation}.\\\\ x=\\frac{2}{y - 3}+4 \\hspace{0.5in} & \\text{Switch x and y} \\\\ x(y-3)=2+4(y-3)\\hspace{0.5in} & \\text{Multiply both sides by }y - 3 \\\\xy-3x=2+4y-12\\hspace{0.5in}& \\text{Simplify} \\\\xy-4y=3x-10\\hspace{0.5in}&\\text{Get all y's on one side} \\\\y(x-4)=3x-10 \\hspace{0.5in} & \\text{Factor out a y} \\\\ y = \\frac{3x-10}{x-4}\\hspace{0.5in} & \\text{Divide by x-4}\\\\{f}^{-1}\\left(x\\right)=\\frac{3x-10}{x-4} \\hspace{0.5in} & \\text{This is the inverse}\\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137678168\">\n<\/div>\n<div id=\"fs-id1165137864156\" class=\"commentary\">\n<h3>Analysis of the Solution<\/h3>\n<p id=\"fs-id1165135394231\">The domain and range of [latex]f[\/latex] exclude the values 3 and 4, respectively. [latex]f[\/latex] and [latex]{f}^{-1}[\/latex] are equal at two points but are not the same function, as we can see by creating\u00a0the table below.<\/p>\n<table id=\"Table_01_07_05\" summary=\"The values of f(x) are: f(1)=3, f(2)=2, and f(5)=5. So f^(-1)(y)=y.\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>5<\/td>\n<td>[latex]{f}^{-1}\\left(y\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\n<td>3<\/td>\n<td>2<\/td>\n<td>5<\/td>\n<td>[latex]y[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_01_07_10\" class=\"example\">\n<div id=\"fs-id1165137603677\" class=\"exercise\">\n<div id=\"fs-id1165137547656\" class=\"problem textbox shaded\">\n<h3>Example 9: Solving to Find an Inverse with Radicals<\/h3>\n<p id=\"fs-id1165137841687\">Find the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt{x - 4}[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165135193684\" class=\"solution textbox shaded\">\n<h3 class=\"equation unnumbered\">Solution<\/h3>\n<div id=\"fs-id1165137828173\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}y=2+\\sqrt{x - 4}\\hfill \\\\x=2+\\sqrt{y - 4}\\hfill \\\\ {\\left(x - 2\\right)}^{2}=y - 4\\hfill \\\\ y={\\left(x - 2\\right)}^{2}+4\\hfill \\\\ {f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4\\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137900392\">The domain of [latex]f[\/latex] is [latex]\\left[4,\\infty \\right)[\/latex]. Notice that the range of [latex]f[\/latex] is [latex]\\left[2,\\infty \\right)[\/latex], so this means that the domain of the inverse function [latex]{f}^{-1}[\/latex] is also [latex]\\left[2,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165137667328\" class=\"commentary\">\n<h3>Analysis of the Solution<\/h3>\n<p id=\"fs-id1165135546050\">The formula we found for [latex]{f}^{-1}\\left(x\\right)[\/latex] looks like it would be valid for all real [latex]x[\/latex]. However, [latex]{f}^{-1}[\/latex] itself must have an inverse (namely, [latex]f[\/latex] ) so we have to restrict the domain of [latex]{f}^{-1}[\/latex] to [latex]\\left[2,\\infty \\right)[\/latex] in order to make [latex]{f}^{-1}[\/latex] a one-to-one function. This domain of [latex]{f}^{-1}[\/latex] is exactly the range of [latex]f[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It\u00a08<\/h3>\n<p id=\"fs-id1165137756074\">What is the inverse of the function [latex]f\\left(x\\right)=2-\\sqrt{x}?[\/latex] State the domains of both the function and the inverse function.<\/p>\n<div id=\"q676921\" class=\"hidden-answer\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380770\">Show Solution<\/span><\/p>\n<div id=\"q380770\" class=\"hidden-answer\" style=\"display: none\">\n[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2};\\text{domain of }f:\\left[0,\\infty \\right);\\text{domain of }{f}^{-1}:\\left(-\\infty ,2\\right][\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as in Figure 7.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7.<\/b> Quadratic function with domain restricted to [0, \u221e).<\/p>\n<\/div>\n<p id=\"fs-id1165137419977\"><strong>Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.<\/p>\n<p id=\"fs-id1165137656093\">We already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]<\/p>\n<p>We notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown in Figure 8.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0092.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"251\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8.<\/b> Square and square-root functions on the non-negative domain<\/p>\n<\/div>\n<p id=\"fs-id1165137393212\">This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.<\/p>\n<div id=\"Example_01_07_11\" class=\"example\">\n<div id=\"fs-id1165134430460\" class=\"exercise\">\n<div id=\"fs-id1165134430463\" class=\"problem textbox shaded\">\n<h3>Example 10: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\n<p>Given the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137407658\" class=\"solution textbox shaded\">\n<h3>Solution<\/h3>\n<p id=\"fs-id1165137407660\">This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<p>If we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in Figure 10.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0112.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10.<\/b> The function and its inverse, showing reflection about the identity line<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137911739\" class=\"solution\">\n<div id=\"fs-id1165137627081\" class=\"note precalculus qa textbox\">\n<p id=\"fs-id1165134388228\"><strong>Q &amp; A <\/strong><\/p>\n<p><strong>Is there any function that is equal to its own inverse?<\/strong><\/p>\n<p id=\"fs-id1165137602656\"><em>Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em><\/p>\n<div id=\"fs-id1165135205827\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/div>\n<p id=\"fs-id1165137897050\"><em>Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em><\/p>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>A function is one-to-one if each output value corresponds to only one input value.<\/li>\n<li>The graph of a one-to-one function passes the horizontal line test.<\/li>\n<li>If [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex], then<\/li>\n<li>[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex].<\/li>\n<li>Each of the toolkit functions has an inverse.<\/li>\n<li>For a function to have an inverse, it must be one-to-one (pass the horizontal line test).<\/li>\n<li>A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.<\/li>\n<li>For a tabular function, exchange the input and output rows to obtain the inverse.<\/li>\n<li>The inverse of a function can be determined at specific points on its graph.<\/li>\n<li>To find the inverse of a formula, solve the equation [latex]y=f\\left(x\\right)[\/latex] for [latex]x[\/latex] as a function of\u00a0[latex]y[\/latex]. Then exchange the labels [latex]x[\/latex] and [latex]y[\/latex].<\/li>\n<li>The graph of an inverse function is the reflection of the graph of the original function across the line [latex]y=x[\/latex].<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137932588\" class=\"definition\">\n<dt><strong>horizontal line test<\/strong><\/dt>\n<dd id=\"fs-id1165134149777\">a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more than once<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137441703\" class=\"definition\">\n<dt>inverse function<\/dt>\n<dd id=\"fs-id1165137441708\">for any one-to-one function [latex]f\\left(x\\right)[\/latex], the inverse is a function [latex]{f}^{-1}\\left(x\\right)[\/latex] such that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]; this also implies that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135511364\" class=\"definition\">\n<dt><strong>one-to-one function<\/strong><\/dt>\n<dd id=\"fs-id1165135511369\">a function for which each value of the output is associated with a unique input value<\/dd>\n<\/dl>\n<h2><\/h2>\n<h2 style=\"text-align: center;\">Section 5.2 Homework Exercises<\/h2>\n<p>1. How can you determine if a relation is a one-to-one function?<\/p>\n<p>2. Why does the horizontal line test tell us whether the graph of a function is one-to-one?<\/p>\n<p>3. Why do we restrict the domain of the function [latex]f\\left(x\\right)={x}^{2}[\/latex] to find the function\u2019s inverse?<\/p>\n<p>4. Can a function be its own inverse? Explain.<\/p>\n<p>5. Are one-to-one functions either always increasing or always decreasing? Why or why not?<\/p>\n<p>6. How do you find the inverse of a function algebraically?<\/p>\n<p>7. Show that the function [latex]f\\left(x\\right)=a-x[\/latex] is its own inverse for all real numbers [latex]a[\/latex].<\/p>\n<p>For the following exercises, find [latex]{f}^{-1}\\left(x\\right)[\/latex] for each function.<\/p>\n<p>8. [latex]f\\left(x\\right)=x+3[\/latex]<\/p>\n<p>9.\u00a0[latex]f\\left(x\\right)=x+5[\/latex]<\/p>\n<p>10. [latex]f\\left(x\\right)=2-x[\/latex]<\/p>\n<p>11.\u00a0[latex]f\\left(x\\right)=3-x[\/latex]<\/p>\n<p>12.\u00a0[latex]f\\left(x\\right)=\\frac{x}{x+2}[\/latex]<\/p>\n<p>13.\u00a0[latex]f\\left(x\\right)=\\frac{2x+3}{5x+4}[\/latex]<\/p>\n<p>For the following exercises, find a domain on which each function [latex]f[\/latex] is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of [latex]f[\/latex] restricted to that domain.<\/p>\n<p>14. [latex]f\\left(x\\right)={\\left(x+7\\right)}^{2}[\/latex]<\/p>\n<p>15.\u00a0[latex]f\\left(x\\right)={\\left(x - 6\\right)}^{2}[\/latex]<\/p>\n<p>16. [latex]f\\left(x\\right)={x}^{2}-5[\/latex]<\/p>\n<p>17.\u00a0Given [latex]f\\left(x\\right)=\\frac{x}{2}+x[\/latex] and [latex]g\\left(x\\right)=\\frac{2x}{1-x}[\/latex]<\/p>\n<p>a. Find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex]<\/p>\n<p>b. What does the answer tell us about the relationship between [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)?[\/latex]<\/p>\n<p>For the following exercises, use function composition to verify that [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] are inverse functions.<\/p>\n<p>18. [latex]f\\left(x\\right)=\\sqrt[3]{x - 1}[\/latex] and [latex]g\\left(x\\right)={x}^{3}+1[\/latex]<\/p>\n<p>19.\u00a0[latex]f\\left(x\\right)=-3x+5[\/latex] and [latex]g\\left(x\\right)=\\frac{x - 5}{-3}[\/latex]<\/p>\n<p>For the following exercises, determine if the given graph is a one-to-one function.<\/p>\n<p>20. Graph of a circle.<\/p>\n<p>21. Graph of a parabola.<\/p>\n<p>22. Graph of a rotated cubic function.<\/p>\n<p>23. Graph of half of 1\/x.<\/p>\n<p>24. Graph of a one-to-one function.<\/p>\n<p>For the following exercises, determine whether the graph represents a one-to-one function.<\/p>\n<p>25.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010629\/CNX_Precalc_Figure_01_07_2012.jpg\" alt=\"Graph of an upright parabola with vertex at (10, -10), passing through (0,0) and (25,0)\" width=\"487\" height=\"253\" \/><\/p>\n<p>26.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010629\/CNX_Precalc_Figure_01_07_2022.jpg\" alt=\"Graph of a step-function, with y = -5 for {x|-10&lt;=x,0} and y = 0 for {x|0&lt;=x&lt;10}\" width=\"487\" height=\"376\" \/><\/p>\n<p>For the following exercises, use the graph of [latex]f[\/latex] shown below.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010629\/CNX_Precalc_Figure_01_07_2032.jpg\" alt=\"Graph of the line y = (-3\/2)x + 3\" width=\"487\" height=\"368\" \/><br \/>\n27. Find [latex]f\\left(0\\right)[\/latex].<\/p>\n<p>28. Solve [latex]f\\left(x\\right)=0[\/latex].<\/p>\n<p>29. Find [latex]{f}^{-1}\\left(0\\right)[\/latex].<\/p>\n<p>30.\u00a0Solve [latex]{f}^{-1}\\left(x\\right)=0[\/latex].<\/p>\n<p>For the following exercises, use the graph of the one-to-one function shown below.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010629\/CNX_Precalc_Figure_01_07_2042.jpg\" alt=\"Graph of a square root function for {x|x&gt;=2}\" width=\"487\" height=\"254\" \/><br \/>\n31. Sketch the graph of [latex]{f}^{-1}[\/latex].<\/p>\n<p>32. Find [latex]f\\left(6\\right)\\text{ and }{f}^{-1}\\left(2\\right)[\/latex].<\/p>\n<p>33. If the complete graph of [latex]f[\/latex] is shown, find the domain of [latex]f[\/latex].<\/p>\n<p>34. If the complete graph of [latex]f[\/latex] is shown, find the range of [latex]f[\/latex].<\/p>\n<p>For the following exercises, evaluate or solve, assuming that the function [latex]f[\/latex] is one-to-one.<\/p>\n<p>35. If [latex]f\\left(6\\right)=7[\/latex], find [latex]{f}^{-1}\\left(7\\right)[\/latex].<\/p>\n<p>36.\u00a0If [latex]f\\left(3\\right)=2[\/latex], find [latex]{f}^{-1}\\left(2\\right)[\/latex].<\/p>\n<p>37. If [latex]{f}^{-1}\\left(-4\\right)=-8[\/latex], find [latex]f\\left(-8\\right)[\/latex].<\/p>\n<p>38.\u00a0If [latex]{f}^{-1}\\left(-2\\right)=-1[\/latex], find [latex]f\\left(-1\\right)[\/latex].<\/p>\n<p>For the following exercises, use the values listed in the table below\u00a0to evaluate or solve.<\/p>\n<table id=\"Table_01_07_06\" summary=\"Two column and ten rows. The first column is labeled,\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\"><strong> [latex]x[\/latex] <\/strong><\/td>\n<td style=\"text-align: center;\"><strong> [latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">0<\/td>\n<td style=\"text-align: center;\">8<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">1<\/td>\n<td style=\"text-align: center;\">0<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\">7<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">5<\/td>\n<td style=\"text-align: center;\">6<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">6<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">7<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">8<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">9<\/td>\n<td style=\"text-align: center;\">1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>39. Find [latex]f\\left(1\\right)[\/latex].<\/p>\n<p>40.\u00a0Solve [latex]f\\left(x\\right)=3[\/latex].<\/p>\n<p>41. Find [latex]{f}^{-1}\\left(0\\right)[\/latex].<\/p>\n<p>42.\u00a0Solve [latex]{f}^{-1}\\left(x\\right)=7[\/latex].<\/p>\n<p>43. Use the tabular representation of [latex]f[\/latex] to create a table for [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<table>\n<tbody>\n<tr>\n<td><strong> [latex]x[\/latex] <\/strong><\/td>\n<td>3<\/td>\n<td>6<\/td>\n<td>9<\/td>\n<td>13<\/td>\n<td>14<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\n<td>1<\/td>\n<td>4<\/td>\n<td>7<\/td>\n<td>12<\/td>\n<td>16<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>For the following exercises, find the inverse function. Then, graph the function and its inverse.<\/p>\n<p>44. [latex]f\\left(x\\right)=\\frac{3}{x - 2}[\/latex]<\/p>\n<p>45. [latex]f\\left(x\\right)={x}^{3}-1[\/latex]<\/p>\n<p>46. Find the inverse function of [latex]f\\left(x\\right)=\\frac{1}{x - 1}[\/latex]. Use a graphing utility to find its domain and range. Write the domain and range in interval notation.<\/p>\n<p>47.\u00a0To convert from [latex]x[\/latex] degrees Celsius to [latex]y[\/latex] degrees Fahrenheit, we use the formula [latex]f\\left(x\\right)=\\frac{9}{5}x+32[\/latex]. Find the inverse function, if it exists, and explain its meaning.<\/p>\n<p>48.\u00a0The circumference [latex]C[\/latex] of a circle is a function of its radius given by [latex]C\\left(r\\right)=2\\pi r[\/latex]. Express the radius of a circle as a function of its circumference. Call this function [latex]r\\left(C\\right)[\/latex]. Find [latex]r\\left(36\\pi \\right)[\/latex] and interpret its meaning.<\/p>\n<p>49. A car travels at a constant speed of 50 miles per hour. The distance the car travels in miles is a function of time, [latex]t[\/latex], in hours given by [latex]d\\left(t\\right)=50t[\/latex]. Find the inverse function by expressing the time of travel in terms of the distance traveled. Call this function [latex]t\\left(d\\right)[\/latex]. Find [latex]t\\left(180\\right)[\/latex] and interpret its meaning.<\/p>\n<\/div>\n<\/section>\n<h2><\/h2>\n<h2 style=\"text-align: center;\">Section 1.8 Solutions to Odd-Numbered Exercises<\/h2>\n<p>1.\u00a0Each output of a function must have exactly one input for the function to be one-to-one.<\/p>\n<p>3. Without any domain restriction,\u00a0[latex]f(x)=x^{2}[\/latex] does not have an inverse function as it fails the horizontal line test. But if we restrict the domain to be [latex]x>0[\/latex], then we find that it passes the horizontal line test and therefore has an inverse function.<\/p>\n<p>5. One-to-one functions are always increasing or decreasing. This ensures there is only one input for every output.<\/p>\n<p>7. [latex]f(x)=a-x\\\\y=a-x\\\\x=a-y\\\\x+y=a\\\\y=a-x\\\\f^{-1}=a-x[\/latex]<\/p>\n<p>9. [latex]f^{-1}(x)=x-5[\/latex]<\/p>\n<p>11. [latex]f^{-1}(x)=3-x[\/latex]<\/p>\n<p>13. [latex]f^{-1}(x)=\\frac{3-4x}{5x-2}[\/latex]<\/p>\n<p>15. Domain of [latex]f(x)[\/latex] is [latex]\\left[6,\\infty \\right)[\/latex] and\u00a0 [latex]f^{-1}=\\sqrt{x}+6[\/latex]<\/p>\n<p>17. a. [latex]f(g(x))=\\frac{3x}{1-x}[\/latex] and [latex]g(f(x))=\\frac{6x}{2-3x}[\/latex]<br \/>\nb. [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are not inverses.<\/p>\n<p>19. [latex]f(g(x))=x[\/latex] and [latex]g(f(x))=x[\/latex]<\/p>\n<p>21. Not one-to-one<\/p>\n<p>23. One-to-one<\/p>\n<p>25. Not one-to-one<\/p>\n<p>27. [latex]f(0)=3[\/latex]<\/p>\n<p>29. [latex]f^{-1}(0)=2[\/latex]<\/p>\n<p>31.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010630\/CNX_Precalc_Figure_01_07_2052.jpg\" alt=\"Graph of a square root function and its inverse.\" \/><\/p>\n<p>33. [latex]\\left[2,\\infty \\right)[\/latex]<\/p>\n<p>35.[latex]f^{-1}(7)=6[\/latex]<\/p>\n<p>37. [latex]f(-8)=-4[\/latex]<\/p>\n<p>39. [latex]f(1)=0[\/latex]<\/p>\n<p>41. [latex]f^{-1}(0)=1[\/latex]<\/p>\n<p>43.<\/p>\n<table>\n<tbody>\n<tr>\n<td>[latex]x[\/latex]<\/td>\n<td>1<\/td>\n<td>4<\/td>\n<td>7<\/td>\n<td>12<\/td>\n<td>16<\/td>\n<\/tr>\n<tr>\n<td>[latex]{f}^{-1}\\left(x\\right)[\/latex]<\/td>\n<td>3<\/td>\n<td>6<\/td>\n<td>9<\/td>\n<td>13<\/td>\n<td>14<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>45. [latex]{f}^{-1}\\left(x\\right)={\\left(1+x\\right)}^{1\/3}[\/latex]<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005116\/CNX_Precalc_Figure_01_07_207.jpg\" alt=\"Graph of a cubic function and its inverse.\" \/><\/p>\n<p>47. [latex]{f}^{-1}\\left(x\\right)=\\frac{5}{9}\\left(x - 32\\right)[\/latex]. Given the Fahrenheit temperature, [latex]x[\/latex], this formula allows you to calculate the Celsius temperature.<\/p>\n<p>49. [latex]t\\left(d\\right)=\\frac{d}{50}[\/latex], [latex]t\\left(180\\right)=\\frac{180}{50}[\/latex]. The time for the car to travel 180 miles is 3.6 hours.<\/p>\n<section id=\"fs-id1165137725994\"><\/section>\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-11013\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/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":276,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-11013","chapter","type-chapter","status-publish","hentry"],"part":13696,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/11013","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":152,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/11013\/revisions"}],"predecessor-version":[{"id":17931,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/11013\/revisions\/17931"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/parts\/13696"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/11013\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=11013"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=11013"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=11013"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=11013"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}