{"id":1049,"date":"2016-10-21T00:35:29","date_gmt":"2016-10-21T00:35:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1049"},"modified":"2020-11-17T22:50:38","modified_gmt":"2020-11-17T22:50:38","slug":"introduction-inverse-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-collegealgebra\/chapter\/introduction-inverse-functions\/","title":{"raw":"Inverse Functions","rendered":"Inverse Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul class=\"ul1\">\r\n \t<li class=\"li2\"><span class=\"s1\">Verify inverse functions.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Find or evaluate the inverse of a function.<\/span><\/li>\r\n \t<li class=\"li3\"><span class=\"s8\">Use the graph of a one-to-one function to graph its inverse function on the same axes.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165135358875\">A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.<\/p>\r\nIf some physical machines can run in two directions, we might ask whether some of the function \"machines\" we have been studying can also run backwards. Figure 1\u00a0provides a visual representation of this question. In this section, we will consider the reverse nature of functions.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205509\/CNX_Precalc_Figure_01_07_0012.jpg\" alt=\"Diagram of a function and what would be its inverse.\" width=\"731\" height=\"305\" \/> <b>Figure 1.<\/b> Can a function \"machine\" operate in reverse?[\/caption]\r\n<h2>Characteristics of Inverse Functions<\/h2>\r\nSuppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the <strong>Celsius<\/strong> scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees <strong>Fahrenheit<\/strong> to degrees Celsius. She finds the formula [latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex] and substitutes 75 for [latex]F[\/latex] to calculate [latex]\\frac{5}{9}\\left(75 - 32\\right)\\approx {24}^{ \\circ} {C}[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205512\/CNX_Precalc_Figure_01_07_0022.jpg\" alt=\"A forecast of Monday\u2019s through Thursday\u2019s weather.\" width=\"731\" height=\"226\" \/>\r\n\r\nKnowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week\u2019s weather forecast\u00a0for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.\r\n\r\nAt first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for [latex]F[\/latex] after substituting a value for [latex]C[\/latex]. For example, to convert 26 degrees Celsius, she could write\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;26=\\frac{5}{9}\\left(F - 32\\right) \\\\[1.5mm] &amp;26\\cdot \\frac{9}{5}=F - 32 \\\\[1.5mm] &amp;F=26\\cdot \\frac{9}{5}+32\\approx 79 \\end{align}[\/latex]<\/p>\r\nAfter considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.\r\n\r\nThe formula for which Betty is searching corresponds to the idea of an <strong>inverse function<\/strong>, which is 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.\r\n\r\nGiven a function [latex]f\\left(x\\right)[\/latex], we represent its inverse as [latex]{f}^{-1}\\left(x\\right)[\/latex], read as \"[latex]f[\/latex] inverse of [latex]x[\/latex].\" The 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.\r\n\r\nThe \"exponent-like\" notation comes from an analogy between function composition and multiplication: just as [latex]{a}^{-1}a=1[\/latex] (1 is the identity element for multiplication) for any nonzero number [latex]a[\/latex], so [latex]{f}^{-1}\\circ f[\/latex] equals the identity function, that is,\r\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(y\\right)=x[\/latex]<\/p>\r\nThis holds for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. Informally, this means that inverse functions \"undo\" each other. However, just as zero does not have a <strong>reciprocal<\/strong>, some functions do not have inverses.\r\n\r\nGiven a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether 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.)\r\n\r\nFor example, [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.\r\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex]\\left({f}^{}\\circ {f}^{-1}\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/p>\r\nA 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.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Inverse Function<\/h3>\r\nFor any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex]. This can also be written as [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. It also follows 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] if [latex]{f}^{-1}[\/latex] is the inverse of [latex]f[\/latex].\r\n\r\nThe notation [latex]{f}^{-1}[\/latex] is read \"[latex]f[\/latex] inverse.\" Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x[\/latex]\".\r\n\r\nKeep in mind that [latex]{f}^{-1}\\left(x\\right)\\ne \\frac{1}{f\\left(x\\right)}[\/latex] and not all functions have inverses.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying an Inverse Function for a Given Input-Output Pair<\/h3>\r\nIf 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?\r\n\r\n[reveal-answer q=\"348517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348517\"]\r\n\r\nThe inverse function reverses the input and output quantities, so if\r\n<p style=\"text-align: center;\">[latex]f\\left(2\\right)=4[\/latex], then [latex]{f}^{-1}\\left(4\\right)=2[\/latex]<\/p>\r\n<p style=\"text-align: left;\">and if<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(5\\right)=12[\/latex], then [latex]{f}^{-1}\\left(12\\right)=5[\/latex]<\/p>\r\nAlternatively, 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].\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.\r\n<table 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[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven that [latex]{h}^{-1}\\left(6\\right)=2[\/latex], what are the corresponding input and output values of the original function [latex]h?[\/latex]\r\n\r\n[reveal-answer q=\"664604\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"664604\"]\r\n\r\n[latex]h(2)=6[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/TSztRfzmk0M\r\n<div class=\"textbox\">\r\n<h3>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>\r\n \t<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\r\n \t<li>If both statements are true, then [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Testing Inverse Relationships Algebraically<\/h3>\r\nIf [latex]f\\left(x\\right)=\\dfrac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\dfrac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"421291\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421291\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align} g\\left(f\\left(x\\right)\\right)&amp;=\\frac{1}{\\left(\\frac{1}{x+2}\\right)}{-2 }\\\\[1.5mm]&amp;={ x }+{ 2 } -{ 2 }\\\\[1.5mm]&amp;={ x } \\end{align}[\/latex]<\/p>\r\nso\r\n<p style=\"text-align: center;\">[latex]g={f}^{-1}\\text{ and }f={g}^{-1}[\/latex]<\/p>\r\nThis is enough to answer yes to the question, but we can also verify the other formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)&amp;=\\frac{1}{\\frac{1}{x}-2+2}\\\\[1.5mm] &amp;=\\frac{1}{\\frac{1}{x}} \\\\[1.5mm] &amp;=x \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice the inverse operations are in reverse order of the operations from the original function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIf [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"226656\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"226656\"]\r\n\r\nYes\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Determining Inverse Relationships for Power Functions<\/h3>\r\nIf [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"637419\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"637419\"]\r\n\r\n[latex]f\\left(g\\left(x\\right)\\right)=\\left(\\frac{1}{3}x\\right)^3=\\dfrac{{x}^{3}}{27}\\ne x[\/latex]\r\n\r\nNo, the functions are not inverses.\r\n<h4>Analysis of the Solution<\/h4>\r\nThe correct inverse to [latex]x^3[\/latex] is the cube root [latex]\\sqrt[3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIf [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}\\text{and}g\\left(x\\right)=\\sqrt[3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"384680\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384680\"]\r\n\r\nYes\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Determine the Domain and Range of an Inverse Function<\/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\/wp-content\/uploads\/sites\/896\/2016\/10\/18205514\/CNX_Precalc_Figure_01_07_0032.jpg\" alt=\"Domain and range of a function and its inverse.\" width=\"487\" height=\"143\" \/> Domain and range of a function and its inverse[\/caption]\r\n\r\nWhen 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].\r\n\r\nWe 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.\r\n\r\nIn 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).\r\n\r\nIf [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].\r\n<ul>\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 class=\"textbox\">\r\n\r\n<strong>Q &amp; A<\/strong>\r\n\r\n<strong>Is it possible for a function to have more than one inverse?<\/strong>\r\n\r\n<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>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Domain and Range of Inverse Functions<\/h3>\r\nThe range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n\r\nThe domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3><strong>How To: Given a function, find the domain and range of its inverse.\r\n<\/strong><\/h3>\r\n<ol>\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 class=\"textbox exercises\">\r\n<h3>Example: Finding the Inverses of Toolkit Functions<\/h3>\r\nIdentify 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.\r\n<table 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[reveal-answer q=\"229708\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"229708\"]\r\n\r\nThe 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.\r\n\r\nThe absolute value function can be restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], where it is equal to the identity function.\r\n\r\nThe reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\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\/wp-content\/uploads\/sites\/896\/2016\/10\/18205517\/CNX_Precalc_Figure_01_07_004ab2.jpg\" alt=\"Graph of an absolute function.\" width=\"975\" height=\"404\" \/> (a) Absolute value (b) Reciprocal squared[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe domain of the function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of the function [latex]f[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex]. Find the domain and range of the inverse function.\r\n\r\n[reveal-answer q=\"146498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"146498\"]\r\n\r\nThe domain of the function [latex]{f}^{-1}[\/latex] is [latex]\\left(-\\infty \\text{,}-2\\right)[\/latex] and the range of the function [latex]{f}^{-1}[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Define and Graph an Inverse<\/h2>\r\nOnce 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.\r\n<h2>Inverting Tabular Functions<\/h2>\r\nSuppose 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.\r\n\r\nEach 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.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Interpreting the Inverse of a Tabular Function<\/h3>\r\nA 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].\r\n<table 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[reveal-answer q=\"61261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"61261\"]\r\n\r\nThe 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.\r\n\r\nAlternatively, 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].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the table below,\u00a0find and interpret (a) [latex]\\text{ }f\\left(60\\right)[\/latex], and (b) [latex]\\text{ }{f}^{-1}\\left(60\\right)[\/latex].\r\n<table 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[reveal-answer q=\"401025\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"401025\"]\r\n\r\na.\u00a0[latex]f\\left(60\\right)=50[\/latex]. In 60 minutes, 50 miles are traveled.\r\n\r\nb. [latex]{f}^{-1}\\left(60\\right)=70[\/latex]. To travel 60 miles, it will take 70 minutes.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/h2>\r\nThe 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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\r\n<ol>\r\n \t<li>Find the desired input of the inverse function on the [latex]y[\/latex]-axis of the given graph.<\/li>\r\n \t<li>Read the inverse function\u2019s output from the [latex]x[\/latex]-axis of the given graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/h3>\r\nA function [latex]g\\left(x\\right)[\/latex] is given\u00a0below. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205520\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/>\r\n\r\n[reveal-answer q=\"334632\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"334632\"]\r\n\r\nTo evaluate [latex]g\\left(3\\right)[\/latex], we find 3 on the <em>x<\/em>-axis and find the corresponding output value on the [latex]y[\/latex]-axis. The point [latex]\\left(3,1\\right)[\/latex] tells us that [latex]g\\left(3\\right)=1[\/latex].\r\n\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<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205521\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the graph in the previous example, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"350455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"350455\"]\r\n\r\na. 3; b. 5.6\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Inverses of Functions Represented by Formulas<\/h2>\r\nSometimes 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\u2014for example, [latex]y[\/latex] as a function of [latex]x-[\/latex] we can often find the inverse function by solving to obtain [latex]x[\/latex] as a function of [latex]y[\/latex].\r\n<div class=\"textbox\">\r\n<h3>How To: Given a function represented by a formula, find the inverse.<\/h3>\r\n<ol>\r\n \t<li>Verify that\u00a0[latex]f[\/latex] is a one-to-one function.<\/li>\r\n \t<li>Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex].<\/li>\r\n \t<li>Interchange [latex]x[\/latex]\u00a0and [latex]y[\/latex].<\/li>\r\n \t<li>Solve for [latex]y[\/latex], and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Inverting the Fahrenheit-to-Celsius Function<\/h3>\r\nFind a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.\r\n<p style=\"text-align: center;\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"625400\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"625400\"]\r\n<p style=\"text-align: center;\">[latex]{ C }=\\frac{5}{9}\\left(F - 32\\right)[\/latex]\r\n[latex]C\\cdot \\frac{9}{5}=F - 32[\/latex]\r\n[latex]F=\\frac{9}{5}C+32[\/latex]<\/p>\r\nBy solving in general, we have uncovered the inverse function. If\r\n<p style=\"text-align: center;\">[latex]C=h\\left(F\\right)=\\frac{5}{9}\\left(F - 32\\right)[\/latex],<\/p>\r\nthen\r\n<p style=\"text-align: center;\">[latex]F={h}^{-1}\\left(C\\right)=\\frac{9}{5}C+32[\/latex].<\/p>\r\nIn this case, we introduced a function [latex]h[\/latex] to represent the conversion because the input and output variables are descriptive, and writing [latex]{C}^{-1}[\/latex] could get confusing.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve for [latex]x[\/latex] in terms of [latex]y[\/latex] given [latex]y=\\frac{1}{3}\\left(x - 5\\right)[\/latex]\r\n\r\n[reveal-answer q=\"875458\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"875458\"]\r\n\r\n[latex]x=3y+5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving to Find an Inverse Function<\/h3>\r\nFind the inverse of the function [latex]f\\left(x\\right)=\\dfrac{2}{x - 3}+4[\/latex].\r\n\r\n[reveal-answer q=\"903464\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"903464\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=\\frac{2}{x - 3}+4 &amp;&amp; \\text{Change }f(x)\\text{ to }y. \\\\[1.5mm]&amp;x=\\frac{2}{y - 3}+4 &amp;&amp; \\text{Switch }x\\text{ and }y. \\\\[1.5mm] &amp;y - 4=\\frac{2}{x - 3} &amp;&amp; \\text{Subtract 4 from both sides}. \\\\[1.5mm] &amp;y - 3=\\frac{2}{x - 4} &amp;&amp; \\text{Multiply both sides by }y - 3\\text{ and divide by }x - 4. \\\\[1.5mm] &amp;y=\\frac{2}{x - 4}+3 &amp;&amp; \\text{Add 3 to both sides}.\\\\[-3mm]&amp;\\end{align}[\/latex]<\/p>\r\nSo [latex]{f}^{-1}\\left(x\\right)=\\dfrac{2}{x - 4}+3[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe 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.\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[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving to Find an Inverse with Radicals<\/h3>\r\nFind the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt{x - 4}[\/latex].\r\n\r\n[reveal-answer q=\"444147\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"444147\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=2+\\sqrt{x - 4}\\\\[1.5mm]&amp;x=2+\\sqrt{y - 4}\\\\[1.5mm] &amp;{\\left(x - 2\\right)}^{2}=y - 4 \\\\[1.5mm] &amp;y={\\left(x- 2\\right)}^{2}+4 \\end{align}[\/latex]<\/p>\r\nSo [latex]{f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4[\/latex].\r\n\r\nThe 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].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe 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].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWhat 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.\r\nUse an online graphing tool to graph the function, its inverse, and [latex]f(x) = x[\/latex] to check whether you are correct.\r\n\r\n[reveal-answer q=\"327817\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"327817\"]\r\n\r\n[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2}[\/latex]; domain of\u00a0 [latex]f:\\left[0,\\infty \\right)[\/latex]; domain of [latex]{ f}^{-1}:\\left(-\\infty ,2\\right][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Graph a Function's Inverse<\/h2>\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 below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205523\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/> Quadratic function with domain restricted to [0, \u221e).[\/caption]<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.\r\n\r\nWe 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]\r\n\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 below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205525\/CNX_Precalc_Figure_01_07_0092.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"251\" \/> Square and square-root functions on the non-negative domain[\/caption]\r\n\r\nThis 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.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: 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<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205527\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/>\r\n[reveal-answer q=\"694066\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"694066\"]\r\n\r\nThis 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].\r\n\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 the graph below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205529\/CNX_Precalc_Figure_01_07_0112.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"363\" \/> The function and its inverse, showing reflection about the identity line[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n\r\n<strong>Q &amp; A<\/strong>\r\n\r\n<strong>Is there any function that is equal to its own inverse?<\/strong>\r\n\r\n<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>\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/p>\r\n<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>\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>If [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex], then [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, except [latex]y=c[\/latex] has an inverse. Some need a restricted domain.<\/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 function [latex]y=f\\left(x\\right)[\/latex], switch the variables [latex]x[\/latex] and [latex]y[\/latex]. Then solve for [latex]y[\/latex] as a function of [latex]x[\/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-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<section id=\"fs-id1165137660004\"><\/section>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul class=\"ul1\">\n<li class=\"li2\"><span class=\"s1\">Verify inverse functions.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Find or evaluate the inverse of a function.<\/span><\/li>\n<li class=\"li3\"><span class=\"s8\">Use the graph of a one-to-one function to graph its inverse function on the same axes.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165135358875\">A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.<\/p>\n<p>If some physical machines can run in two directions, we might ask whether some of the function &#8220;machines&#8221; we have been studying can also run backwards. Figure 1\u00a0provides a visual representation of this question. In this section, we will consider the reverse nature of functions.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205509\/CNX_Precalc_Figure_01_07_0012.jpg\" alt=\"Diagram of a function and what would be its inverse.\" width=\"731\" height=\"305\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> Can a function &#8220;machine&#8221; operate in reverse?<\/p>\n<\/div>\n<h2>Characteristics of Inverse Functions<\/h2>\n<p>Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the <strong>Celsius<\/strong> scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees <strong>Fahrenheit<\/strong> to degrees Celsius. She finds the formula [latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex] and substitutes 75 for [latex]F[\/latex] to calculate [latex]\\frac{5}{9}\\left(75 - 32\\right)\\approx {24}^{ \\circ} {C}[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205512\/CNX_Precalc_Figure_01_07_0022.jpg\" alt=\"A forecast of Monday\u2019s through Thursday\u2019s weather.\" width=\"731\" height=\"226\" \/><\/p>\n<p>Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week\u2019s weather forecast\u00a0for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.<\/p>\n<p>At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for [latex]F[\/latex] after substituting a value for [latex]C[\/latex]. For example, to convert 26 degrees Celsius, she could write<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&26=\\frac{5}{9}\\left(F - 32\\right) \\\\[1.5mm] &26\\cdot \\frac{9}{5}=F - 32 \\\\[1.5mm] &F=26\\cdot \\frac{9}{5}+32\\approx 79 \\end{align}[\/latex]<\/p>\n<p>After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.<\/p>\n<p>The formula for which Betty is searching corresponds to the idea of an <strong>inverse function<\/strong>, which is 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.<\/p>\n<p>Given a function [latex]f\\left(x\\right)[\/latex], we represent its inverse as [latex]{f}^{-1}\\left(x\\right)[\/latex], read as &#8220;[latex]f[\/latex] inverse of [latex]x[\/latex].&#8221; The 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<p>The &#8220;exponent-like&#8221; notation comes from an analogy between function composition and multiplication: just as [latex]{a}^{-1}a=1[\/latex] (1 is the identity element for multiplication) for any nonzero number [latex]a[\/latex], so [latex]{f}^{-1}\\circ f[\/latex] equals the identity function, that is,<\/p>\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(y\\right)=x[\/latex]<\/p>\n<p>This holds for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. Informally, this means that inverse functions &#8220;undo&#8221; each other. However, just as zero does not have a <strong>reciprocal<\/strong>, some functions do not have inverses.<\/p>\n<p>Given a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether 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>For example, [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.<\/p>\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex]\\left({f}^{}\\circ {f}^{-1}\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/p>\n<p>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 class=\"textbox\">\n<h3>A General Note: Inverse Function<\/h3>\n<p>For any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex]. This can also be written as [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. It also follows 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] if [latex]{f}^{-1}[\/latex] is the inverse of [latex]f[\/latex].<\/p>\n<p>The notation [latex]{f}^{-1}[\/latex] is read &#8220;[latex]f[\/latex] inverse.&#8221; Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x[\/latex]&#8220;.<\/p>\n<p>Keep in mind that [latex]{f}^{-1}\\left(x\\right)\\ne \\frac{1}{f\\left(x\\right)}[\/latex] and not all functions have inverses.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying an Inverse Function for a Given Input-Output Pair<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348517\">Show Solution<\/span><\/p>\n<div id=\"q348517\" class=\"hidden-answer\" style=\"display: none\">\n<p>The inverse function reverses the input and output quantities, so if<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(2\\right)=4[\/latex], then [latex]{f}^{-1}\\left(4\\right)=2[\/latex]<\/p>\n<p style=\"text-align: left;\">and if<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(5\\right)=12[\/latex], then [latex]{f}^{-1}\\left(12\\right)=5[\/latex]<\/p>\n<p>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<h4>Analysis of the Solution<\/h4>\n<p>Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.<\/p>\n<table 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=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q664604\">Show Solution<\/span><\/p>\n<div id=\"q664604\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]h(2)=6[\/latex]<\/p>\n<\/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 class=\"textbox\">\n<h3>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>\n<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\n<li>If both statements are true, then [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Testing Inverse Relationships Algebraically<\/h3>\n<p>If [latex]f\\left(x\\right)=\\dfrac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\dfrac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q421291\">Show Solution<\/span><\/p>\n<div id=\"q421291\" class=\"hidden-answer\" style=\"display: none\">\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 }\\\\[1.5mm]&={ x }+{ 2 } -{ 2 }\\\\[1.5mm]&={ x } \\end{align}[\/latex]<\/p>\n<p>so<\/p>\n<p style=\"text-align: center;\">[latex]g={f}^{-1}\\text{ and }f={g}^{-1}[\/latex]<\/p>\n<p>This is enough to answer yes to the question, but we can also verify the other formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)&=\\frac{1}{\\frac{1}{x}-2+2}\\\\[1.5mm] &=\\frac{1}{\\frac{1}{x}} \\\\[1.5mm] &=x \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice the inverse operations are in reverse order of the operations from the original function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>If [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q226656\">Show Solution<\/span><\/p>\n<div id=\"q226656\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determining Inverse Relationships for Power Functions<\/h3>\n<p>If [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q637419\">Show Solution<\/span><\/p>\n<div id=\"q637419\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(g\\left(x\\right)\\right)=\\left(\\frac{1}{3}x\\right)^3=\\dfrac{{x}^{3}}{27}\\ne x[\/latex]<\/p>\n<p>No, the functions are not inverses.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The correct inverse to [latex]x^3[\/latex] is the cube root [latex]\\sqrt[3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}\\text{and}g\\left(x\\right)=\\sqrt[3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q384680\">Show Solution<\/span><\/p>\n<div id=\"q384680\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Determine the Domain and Range of an Inverse Function<\/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\/wp-content\/uploads\/sites\/896\/2016\/10\/18205514\/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\">Domain and range of a function and its inverse<\/p>\n<\/div>\n<p>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>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>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>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>\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 class=\"textbox\">\n<p><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><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 class=\"textbox\">\n<h3>A General Note: Domain and Range of Inverse Functions<\/h3>\n<p>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>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 class=\"textbox\">\n<h3><strong>How To: Given a function, find the domain and range of its inverse.<br \/>\n<\/strong><\/h3>\n<ol>\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 class=\"textbox exercises\">\n<h3>Example: Finding the Inverses of Toolkit Functions<\/h3>\n<p>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 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q229708\">Show Solution<\/span><\/p>\n<div id=\"q229708\" class=\"hidden-answer\" style=\"display: none\">\n<p>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>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>The reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\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\/wp-content\/uploads\/sites\/896\/2016\/10\/18205517\/CNX_Precalc_Figure_01_07_004ab2.jpg\" alt=\"Graph of an absolute function.\" width=\"975\" height=\"404\" \/><\/p>\n<p class=\"wp-caption-text\">(a) Absolute value (b) Reciprocal squared<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The domain of the function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of the 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=\"q146498\">Show Solution<\/span><\/p>\n<div id=\"q146498\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain of the function [latex]{f}^{-1}[\/latex] is [latex]\\left(-\\infty \\text{,}-2\\right)[\/latex] and the range of the function [latex]{f}^{-1}[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Define and Graph an Inverse<\/h2>\n<p>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<h2>Inverting Tabular Functions<\/h2>\n<p>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>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 class=\"textbox exercises\">\n<h3>Example: Interpreting the Inverse of a Tabular Function<\/h3>\n<p>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 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q61261\">Show Solution<\/span><\/p>\n<div id=\"q61261\" class=\"hidden-answer\" style=\"display: none\">\n<p>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>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=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q401025\">Show Solution<\/span><\/p>\n<div id=\"q401025\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex]f\\left(60\\right)=50[\/latex]. In 60 minutes, 50 miles are traveled.<\/p>\n<p>b. [latex]{f}^{-1}\\left(60\\right)=70[\/latex]. To travel 60 miles, it will take 70 minutes.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/h2>\n<p>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 class=\"textbox\">\n<h3>How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\n<ol>\n<li>Find the desired input of the inverse function on the [latex]y[\/latex]-axis of the given graph.<\/li>\n<li>Read the inverse function\u2019s output from the [latex]x[\/latex]-axis of the given graph.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: 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\u00a0below. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205520\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q334632\">Show Solution<\/span><\/p>\n<div id=\"q334632\" class=\"hidden-answer\" style=\"display: none\">\n<p>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 [latex]y[\/latex]-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<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205521\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the graph in the previous example, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q350455\">Show Solution<\/span><\/p>\n<div id=\"q350455\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. 3; b. 5.6<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Inverses of Functions Represented by Formulas<\/h2>\n<p>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\u2014for example, [latex]y[\/latex] as a function of [latex]x-[\/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 class=\"textbox\">\n<h3>How To: Given a function represented by a formula, find the inverse.<\/h3>\n<ol>\n<li>Verify that\u00a0[latex]f[\/latex] is a one-to-one function.<\/li>\n<li>Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex].<\/li>\n<li>Interchange [latex]x[\/latex]\u00a0and [latex]y[\/latex].<\/li>\n<li>Solve for [latex]y[\/latex], and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Inverting the Fahrenheit-to-Celsius Function<\/h3>\n<p>Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.<\/p>\n<p style=\"text-align: center;\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q625400\">Show Solution<\/span><\/p>\n<div id=\"q625400\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]{ C }=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<br \/>\n[latex]C\\cdot \\frac{9}{5}=F - 32[\/latex]<br \/>\n[latex]F=\\frac{9}{5}C+32[\/latex]<\/p>\n<p>By solving in general, we have uncovered the inverse function. If<\/p>\n<p style=\"text-align: center;\">[latex]C=h\\left(F\\right)=\\frac{5}{9}\\left(F - 32\\right)[\/latex],<\/p>\n<p>then<\/p>\n<p style=\"text-align: center;\">[latex]F={h}^{-1}\\left(C\\right)=\\frac{9}{5}C+32[\/latex].<\/p>\n<p>In this case, we introduced a function [latex]h[\/latex] to represent the conversion because the input and output variables are descriptive, and writing [latex]{C}^{-1}[\/latex] could get confusing.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve for [latex]x[\/latex] in terms of [latex]y[\/latex] given [latex]y=\\frac{1}{3}\\left(x - 5\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q875458\">Show Solution<\/span><\/p>\n<div id=\"q875458\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=3y+5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving to Find an Inverse Function<\/h3>\n<p>Find the inverse of the function [latex]f\\left(x\\right)=\\dfrac{2}{x - 3}+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q903464\">Show Solution<\/span><\/p>\n<div id=\"q903464\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=\\frac{2}{x - 3}+4 && \\text{Change }f(x)\\text{ to }y. \\\\[1.5mm]&x=\\frac{2}{y - 3}+4 && \\text{Switch }x\\text{ and }y. \\\\[1.5mm] &y - 4=\\frac{2}{x - 3} && \\text{Subtract 4 from both sides}. \\\\[1.5mm] &y - 3=\\frac{2}{x - 4} && \\text{Multiply both sides by }y - 3\\text{ and divide by }x - 4. \\\\[1.5mm] &y=\\frac{2}{x - 4}+3 && \\text{Add 3 to both sides}.\\\\[-3mm]&\\end{align}[\/latex]<\/p>\n<p>So [latex]{f}^{-1}\\left(x\\right)=\\dfrac{2}{x - 4}+3[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>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 class=\"textbox exercises\">\n<h3>Example: Solving to Find an Inverse with Radicals<\/h3>\n<p>Find the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt{x - 4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q444147\">Show Solution<\/span><\/p>\n<div id=\"q444147\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=2+\\sqrt{x - 4}\\\\[1.5mm]&x=2+\\sqrt{y - 4}\\\\[1.5mm] &{\\left(x - 2\\right)}^{2}=y - 4 \\\\[1.5mm] &y={\\left(x- 2\\right)}^{2}+4 \\end{align}[\/latex]<\/p>\n<p>So [latex]{f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4[\/latex].<\/p>\n<p>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<h4>Analysis of the Solution<\/h4>\n<p>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=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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.<br \/>\nUse an online graphing tool to graph the function, its inverse, and [latex]f(x) = x[\/latex] to check whether you are correct.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q327817\">Show Solution<\/span><\/p>\n<div id=\"q327817\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2}[\/latex]; domain of\u00a0 [latex]f:\\left[0,\\infty \\right)[\/latex]; domain of [latex]{ f}^{-1}:\\left(-\\infty ,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Graph a Function&#8217;s Inverse<\/h2>\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 below.<\/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\/wp-content\/uploads\/sites\/896\/2016\/10\/18205523\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\">Quadratic function with domain restricted to [0, \u221e).<\/p>\n<\/div>\n<p><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>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 below.<\/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\/wp-content\/uploads\/sites\/896\/2016\/10\/18205525\/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\">Square and square-root functions on the non-negative domain<\/p>\n<\/div>\n<p>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 class=\"textbox exercises\">\n<h3>Example: 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<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205527\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q694066\">Show Solution<\/span><\/p>\n<div id=\"q694066\" class=\"hidden-answer\" style=\"display: none\">\n<p>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 the graph below.<\/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\/wp-content\/uploads\/sites\/896\/2016\/10\/18205529\/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\">The function and its inverse, showing reflection about the identity line<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<p><strong>Q &amp; A<\/strong><\/p>\n<p><strong>Is there any function that is equal to its own inverse?<\/strong><\/p>\n<p><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<p style=\"text-align: center;\">[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/p>\n<p><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>If [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex], then [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, except [latex]y=c[\/latex] has an inverse. Some need a restricted domain.<\/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 function [latex]y=f\\left(x\\right)[\/latex], switch the variables [latex]x[\/latex] and [latex]y[\/latex]. Then solve for [latex]y[\/latex] as a function of [latex]x[\/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-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<section id=\"fs-id1165137660004\"><\/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-1049\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Ex: Find an Inverse Function From a Table. <strong>Authored by<\/strong>: Mathispower4u. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/TSztRfzmk0M\">https:\/\/youtu.be\/TSztRfzmk0M<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/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":21,"menu_order":34,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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