{"id":138,"date":"2025-04-09T17:16:10","date_gmt":"2025-04-09T17:16:10","guid":{"rendered":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/chapter\/3-7-inverse-functions-college-algebra-2e-openstax\/"},"modified":"2025-08-19T19:30:42","modified_gmt":"2025-08-19T19:30:42","slug":"3-7-inverse-functions","status":"publish","type":"chapter","link":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/chapter\/3-7-inverse-functions\/","title":{"raw":"3.7 Inverse Functions","rendered":"3.7 Inverse Functions"},"content":{"raw":"<div id=\"main-content\" class=\"MainContent__ContentStyles-sc-6yy1if-0 NnXKu\" tabindex=\"-1\" data-dynamic-style=\"true\">\r\n<div id=\"page_f592aad0-19d8-42d6-94b9-086bdd84c2b5\" class=\"chapter-content-module\" data-type=\"page\" data-book-content=\"true\">\r\n<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Learning Objectives<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Verify inverse functions.<\/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<\/div>\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\n<p id=\"fs-id1165135701544\">If some physical machines can run in two directions, we might ask whether some of the function \u201cmachines\u201d we have been studying can also run backwards. Figure 1 provides a visual representation of this question. In this section, we will consider the reverse nature of functions.<\/p>\r\n&nbsp;\r\n\r\n[caption id=\"attachment_626\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-626\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-1-300x125.jpeg\" alt=\"\" width=\"300\" height=\"125\" \/> Figure 1. Can a function \u201cmachine\u201d operate in reverse?[\/caption]\r\n\r\n<section id=\"fs-id1165137725994\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Verifying That Two Functions Are Inverse Functions<\/h2>\r\n<p id=\"fs-id1165135705795\">Betty is traveling to Milan for a fashion show and wants to know what the temperature will be. She is not familiar with the <span id=\"term-00007\" class=\"no-emphasis\" data-type=\"term\">Celsius<\/span> scale. To get an idea of how temperature measurements are related, Betty wants to convert 75 degrees <span id=\"term-00008\" class=\"no-emphasis\" data-type=\"term\">Fahrenheit<\/span> to degrees Celsius using the formula<\/p>\r\n<p style=\"text-align: center;\">[latex] C=\\frac{5}{9}(F-32) [\/latex]<\/p>\r\n<p id=\"fs-id1165135433486\">and substitutes 75 for\u00a0[latex] F [\/latex] to calculate<\/p>\r\n<p style=\"text-align: center;\">[latex] \\frac{5}{9}(75-32)\\approx24^\\circ C [\/latex]<\/p>\r\n<p id=\"fs-id1165137409312\">Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, Betty gets the week\u2019s weather forecast from Figure 2 for Milan, and wants to convert all of the temperatures to degrees Fahrenheit.<\/p>\r\n&nbsp;\r\n\r\n[caption id=\"attachment_627\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-627\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-2-300x93.jpeg\" alt=\"\" width=\"300\" height=\"93\" \/> Figure 2[\/caption]\r\n<p id=\"fs-id1165137724415\">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\u00a0[latex] F [\/latex] after substituting a value for\u00a0[latex] C. [\/latex] For example, to convert 26 degrees Celsius, she could write<\/p>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rcl}26 &amp;=&amp; \\frac{5}{9}(F-32) \\\\26 \\cdot \\frac{9}{5} &amp;=&amp; F-32 \\\\F &amp;=&amp; 26 \\cdot \\frac{9}{5} + 32 \\approx 79\\end{array} [\/latex]<\/p>\r\n<p id=\"fs-id1165137540705\">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>\r\n<p id=\"fs-id1165137827441\">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>\r\n<p id=\"fs-id1165135528385\">Given a function\u00a0[latex] f(x), [\/latex] we represent its inverse as\u00a0[latex] f^{-1}(x), [\/latex] read as\u00a0\"[latex] f [\/latex] inverse of\u00a0[latex] x. [\/latex]\" The raised\u00a0[latex] -1 [\/latex] is part of the notation. It is not an exponent; it does not imply a power of\u00a0[latex] -1. [\/latex] \u00a0In other words, [latex] f^{-1}(x) [\/latex] does <em data-effect=\"italics\">not<\/em> mean\u00a0[latex] \\frac{1}{f(x)} [\/latex] because\u00a0[latex] \\frac{1}{f(x)} [\/latex] is the reciprocal of\u00a0[latex] f [\/latex] and not the inverse.<\/p>\r\n<p id=\"fs-id1165137724926\">The \u201cexponent-like\u201d notation comes from an analogy between function composition and multiplication: just as\u00a0[latex]- a^{-1}a=1 [\/latex] (1 is the identity element for multiplication) for any nonzero number\u00a0[latex] a [\/latex] so\u00a0[latex] f^{-1}\\circ f [\/latex] equals the identity function, that is,<\/p>\r\n<p style=\"text-align: center;\">[latex] (f^{-1}\\circ f)(x)=f^{-1}(f(x))=f^{-1}(y)=x [\/latex]<\/p>\r\nThis holds for all\u00a0[latex] [\/latex] in the domain of\u00a0[latex] [\/latex] Informally, this means that inverse functions \u201cundo\u201d each other. However, just as zero does not have a <span id=\"term-00009\" class=\"no-emphasis\" data-type=\"term\">reciprocal<\/span>, some functions do not have inverses.\r\n<p id=\"fs-id1165137655153\">Given a function\u00a0[latex] f(x), [\/latex] we can verify whether some other function\u00a0[latex] g(x) [\/latex] is the inverse of\u00a0[latex] f(x) [\/latex] by checking if both\u00a0[latex] g(f(x))=x [\/latex] and\u00a0[latex] f(g(x))=x [\/latex] are true.<\/p>\r\n<p id=\"fs-id1165135397975\">For example,\u00a0[latex] y=4x [\/latex] and\u00a0[latex] y=\\frac{1}{4}x [\/latex] are inverse functions.<\/p>\r\n<p style=\"text-align: center;\">[latex] f^{-1}\\circ f)(x)=f^{-1}(4x)=\\frac{1}{4}(4x)=x [\/latex]<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex] (f\\circ f^{-1})(x)=f(\\frac{1}{4}x)=4(\\frac{1}{4}x)=x [\/latex]<\/p>\r\n<p id=\"fs-id1165137438777\">A few coordinate pairs from the graph of the function\u00a0[latex] y=4x [\/latex] are\u00a0[latex] (-2, -8), (0, 0), [\/latex] and\u00a0[latex] (2, 8). [\/latex] A few coordinate pairs from the graph of the function\u00a0[latex] y=\\frac{1}{4}x [\/latex] are\u00a0[latex] (-8, -2), (0, 0), [\/latex] and\u00a0[latex] (8, 2). [\/latex] 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 class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Inverse Function<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFor any one-to-one function\u00a0[latex] f(x)=y, [\/latex] a function\u00a0 is an <strong>inverse function<\/strong> of\u00a0[latex] f [\/latex] if\u00a0[latex] f^{-1}(y)=x. [\/latex] This can also be written as\u00a0[latex] f^{-1}(f(x))=x [\/latex] for all\u00a0[latex] x [\/latex] in the domain of\u00a0[latex] f. [\/latex] It also follows that\u00a0[latex] f(f^{-1}(x))=x [\/latex] for all\u00a0[latex] x [\/latex] in the domain of\u00a0[latex] f^{-1} [\/latex] if\u00a0[latex] f^{-1} [\/latex] is the inverse of [latex] f. [\/latex]\r\n\r\nThe notation\u00a0[latex] f^{-1} [\/latex] is read \u201c[latex] f [\/latex] inverse.\" Like any other function, we can use any variable name as the input for\u00a0[latex] f^{-1}, [\/latex] so we will often write\u00a0[latex] f^{-1}(x), [\/latex] which we read as \"[latex] f [\/latex] inverse of [latex] x. [\/latex]\" Keep in mind that\r\n<p style=\"text-align: center;\">[latex] f^{-1}(x)\\not=\\frac{1}{f(x)} [\/latex]<\/p>\r\nand not all functions have inverses.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 1: Identifying an Inverse Function for a Given Input-Output Pair<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIf for a particular one-to-one function\u00a0[latex] f(2)=4 [\/latex] and\u00a0[latex] f(5)=12, [\/latex] what are the corresponding input and output values for the inverse function?\r\n\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>The inverse function reverses the input and output quantities, so if\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rcl}f(2) &amp;=&amp; 4, \\, \\text{then } f^{-1}(4) = 2; \\\\f(5) &amp;=&amp; 12, \\, \\text{then } f^{-1}(12) = 5.\\end{array} [\/latex]<\/p>\r\nAlternatively, if we want to name the inverse function\u00a0[latex] g, [\/latex] then\u00a0[latex] g(4)=2 [\/latex] and [latex] g(12)=5. [\/latex]\r\n<h3>Analysis<\/h3>\r\nNotice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table 1.\r\n<table class=\"grid\" border=\"0\"><caption>Table 1<\/caption>\r\n<tbody>\r\n<tr>\r\n<td>[latex] (x, f(x)) [\/latex]<\/td>\r\n<td>[latex] (x, g(x)) [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] (2, 4) [\/latex]<\/td>\r\n<td>[latex] (4, 2) [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] (5, 12) [\/latex]<\/td>\r\n<td>[latex] (12, 5) [\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/details><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nGiven that\u00a0[latex] h^{-1}(6)=2, [\/latex] what are the corresponding input and output values of the original function [latex] h? [\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">How To<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>Given two functions\u00a0[latex] f(x) [\/latex] and\u00a0[latex] g(x) [\/latex] test whether the functions are inverses of each other.<\/strong>\r\n<ol>\r\n \t<li>Determine whether\u00a0[latex] f(g(x))=x [\/latex] or [latex] g(f(x))=x. [\/latex]<\/li>\r\n \t<li>If either statement is true, then both are true, and\u00a0[latex] g=f^{-1} [\/latex] and\u00a0[latex] f=g^{-1}. [\/latex] If either statements is false, then both are false, and\u00a0[latex] g\\not=f^{-1} [\/latex] and [latex] f\\not=g^{-1}. [\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 2: Testing Inverse Relationships Algebraically<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIf [latex] f(x)=\\frac{1}{x+2} [\/latex] and [latex] g(x)=\\frac{1}{x}-2, [\/latex] is [latex] g=f^{-1}? [\/latex]\r\n\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rcl}g(f(x)) &amp;=&amp; \\frac{1}{\\left(\\frac{1}{x+2}\\right)} - 2 \\\\&amp;=&amp; x + 2 - 2 \\\\&amp;=&amp; x\\end{array}\u00a0[\/latex]<\/p>\r\nWe must also verify the other formula.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rcl}f(g(x)) &amp;=&amp; \\frac{1}{\\frac{1}{x} - 2 + 2} \\\\&amp;=&amp; \\frac{1}{\\frac{1}{x}} \\\\&amp;=&amp; x\\end{array} [\/latex]<\/p>\r\nso\r\n<p style=\"text-align: center;\">[latex] g=f^{-1} \\ \\ \\text{and} \\ \\ f=g^{-1} [\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\nNotice the inverse operations are in reverse order of the operations from the original function.\r\n\r\n<\/details><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIf\u00a0[latex] f(x)=x^3-4 [\/latex] and\u00a0[latex] g(x)=\\sqrt[3]{x+4}, [\/latex] is [latex] g=f^{-1}? [\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 3: Determining Inverse Relationships for Power Functions<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIf\u00a0[latex] f(x)=x^3 [\/latex] (the cube function) and\u00a0[latex] g(x)=\\frac{1}{3}x, [\/latex] is [latex] g=f^{-1}? [\/latex]\r\n\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>\r\n<p style=\"text-align: center;\">[latex] f(g(x))=\\frac{x^3}{27}\\not=x [\/latex]<\/p>\r\nNo, the functions are not inverses.\r\n<h3>Analysis<\/h3>\r\nThe correct inverse to the cube is, of course, the cube root\u00a0[latex] \\sqrt[3]{x}=x^{\\frac{1}{3}}, [\/latex] that is, the one-third is an exponent, not a multiplier.\r\n\r\n<\/details><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIf\u00a0[latex] f(x)=(x-1)^3 [\/latex] and\u00a0[latex] g(x)=\\sqrt[3]{x}+1, [\/latex] is [latex] g=f^{-1}? [\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1165137660004\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Finding Domain and Range of Inverse Functions<\/h2>\r\n<p id=\"fs-id1165137591020\">The outputs of the function\u00a0[latex] f [\/latex] are the inputs to\u00a0[latex] f^{-1}, [\/latex] so the range of\u00a0[latex] f [\/latex] is also the domain of\u00a0[latex] f^{-1}. [\/latex] Likewise, because the inputs to\u00a0[latex] f [\/latex] are the outputs of\u00a0[latex] f^{-1}, [\/latex] the domain of\u00a0[latex] f [\/latex] is the range of\u00a0[latex] f^{-1}. [\/latex]\u00a0We can visualize the situation as in Figure 3.<\/p>\r\n&nbsp;\r\n\r\n[caption id=\"attachment_628\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-628\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-3-300x88.jpeg\" alt=\"\" width=\"300\" height=\"88\" \/> Figure 3. 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\u00a0[latex] f(x)=\\sqrt{x} [\/latex] is\u00a0[latex] f^{-1}(x)=x^2, [\/latex] because a square \u201cundoes\u201d a square root; but the square is only the inverse of the square root on the domain\u00a0[latex] [0, \\infty), [\/latex] since that is the range of [latex] f(x)=\\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\u00a0[latex] f(x)=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 \u201cinverse\u201d 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\u00a0[latex] f(x)=x^2 [\/latex] with its domain limited to\u00a0[latex] [0, \\infty), [\/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\u00a0[latex] f(x)=(x-1)^2 [\/latex] on\u00a0[latex] [1, \\infty), [\/latex] then the inverse function is [latex] f^{-1}(x)=\\sqrt{x}+1. [\/latex]<\/p>\r\n\r\n<ul id=\"fs-id1165137851227\">\r\n \t<li>The domain of [latex] f= [\/latex]\u00a0range of [latex] f^{-1}=[1, \\infty). [\/latex]<\/li>\r\n \t<li>The domain of\u00a0[latex] f^{-1}= [\/latex] range of [latex] f=[0, \\infty). [\/latex]<\/li>\r\n<\/ul>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Q&amp;A<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>Q: Is it possible for a function to have more than one inverse?<\/strong>\r\n\r\n<em data-effect=\"italics\">A: No. If two supposedly different functions, say, [latex] g [\/latex] and\u00a0[latex] h, [\/latex] both meet the definition of being inverses of another function\u00a0[latex] f, [\/latex] then you can prove that\u00a0[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>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Domain and Range of Inverse Functions<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nThe range of a function\u00a0[latex] f(x) [\/latex] is the domain of the inverse function [latex] f^{-1}(x). [\/latex]\r\n\r\nThe domain of\u00a0[latex] f(x) [\/latex] is the range of [latex] f^{-1}(x). [\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">How To<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>Given a function, find the domain and range of its inverse.<\/strong>\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>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 4: Finding the Inverses of Toolkit Functions<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\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 in Table 2. We restrict the domain in such a fashion that the function assumes all [latex] [\/latex]values exactly once.\r\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\" border=\"0\"><caption>Table 2<\/caption>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 20%;\"><strong>Constant<\/strong><\/td>\r\n<td style=\"width: 20%;\"><strong>Identity<\/strong><\/td>\r\n<td style=\"width: 20%;\"><strong>Quadratic<\/strong><\/td>\r\n<td style=\"width: 20%;\"><strong>Cubic<\/strong><\/td>\r\n<td style=\"width: 20%;\"><strong>Reciprocal<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;\">[latex] f(x)=c [\/latex]<\/td>\r\n<td style=\"width: 20%;\">[latex] f(x)=x [\/latex]<\/td>\r\n<td style=\"width: 20%;\">[latex] f(x)-x^2 [\/latex]<\/td>\r\n<td style=\"width: 20%;\">[latex] f(x)-x^3 [\/latex]<\/td>\r\n<td style=\"width: 20%;\">[latex] f(x)=\\frac{1}{x} [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;\"><strong>Reciprocal Squared<\/strong><\/td>\r\n<td style=\"width: 20%;\"><strong>Cube Root<\/strong><\/td>\r\n<td style=\"width: 20%;\"><strong>Square Root<\/strong><\/td>\r\n<td style=\"width: 20%;\"><strong>Absolute Value<\/strong><\/td>\r\n<td style=\"width: 20%;\"><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;\">[latex] f(x)=\\frac{1}{x^2} [\/latex]<\/td>\r\n<td style=\"width: 20%;\">[latex] f(x)=\\sqrt[3]{x} [\/latex]<\/td>\r\n<td style=\"width: 20%;\">[latex] f(x)=\\sqrt{x} [\/latex]<\/td>\r\n<td style=\"width: 20%;\">[latex] f(x)=|x| [\/latex]<\/td>\r\n<td style=\"width: 20%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>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 inverse.\r\n\r\nThe absolute value function can be restricted to the domain\u00a0[latex] [0, \\infty), [\/latex] where it is equal to the identity function.\r\n\r\nThe reciprocal-squared function can be restricted to the domain [latex] (0, \\infty). [\/latex]\r\n<h3>Analysis<\/h3>\r\nWe can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure 4. They 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&nbsp;\r\n\r\n[caption id=\"attachment_629\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-629\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-4-300x146.jpeg\" alt=\"\" width=\"300\" height=\"146\" \/> Figure 4. (a) Absolute value (b) Reciprocal square[\/caption]\r\n\r\n<\/details><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nThe domain of function\u00a0[latex] f [\/latex] is\u00a0[latex] (1, \\infty) [\/latex] and the range of function\u00a0[latex] f [\/latex] is\u00a0[latex] (-\\infty, 2). [\/latex] Find the domain and range of the inverse function.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137619159\" data-depth=\"1\">\r\n<h2 data-type=\"title\">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\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Inverting Tabular Functions<\/h3>\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 class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 5: Interpreting the Inverse of a Tabular Function<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA function\u00a0[latex] f(t) [\/latex] is given in Table 3, showing the distance in miles that a car has traveled in\u00a0 minutes. Find and interpret [latex] f^{-1}(70). [\/latex]\r\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\" border=\"0\"><caption>Table 3<\/caption>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 20%;\">[latex] t [\/latex](minutes)<\/td>\r\n<td style=\"width: 20%;\">30<\/td>\r\n<td style=\"width: 20%;\">50<\/td>\r\n<td style=\"width: 20%;\">70<\/td>\r\n<td style=\"width: 20%;\">90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;\">[latex] f(t) [\/latex](miles)<\/td>\r\n<td style=\"width: 20%;\">20<\/td>\r\n<td style=\"width: 20%;\">40<\/td>\r\n<td style=\"width: 20%;\">60<\/td>\r\n<td style=\"width: 20%;\">80<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>The inverse function takes an output of\u00a0[latex] f [\/latex] and returns an input for\u00a0[latex] f. [\/latex] So in the expression\u00a0[latex] f^{-1}(70), [\/latex] 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function\u00a0[latex] f, [\/latex] 90 minutes, so\u00a0[latex] f^{-1}(70)=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\u00a0[latex] f(a)=b, [\/latex] then\u00a0[latex] f^{-1}(b)=a. [\/latex] By this definition, if we are given\u00a0[latex] f^{-1}(70)=a, [\/latex] then we are looking for a value\u00a0[latex]a [\/latex] so that\u00a0[latex] f(a)=70. [\/latex] In this case, we are looking for a [latex] t [\/latex] so that\u00a0[latex] f(t)=70, [\/latex] which is when [latex] t=90. [\/latex]\r\n\r\n<\/details><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #5<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nUsing Table 4, find and interpret (a)\u00a0[latex] f(60), [\/latex] and (b) [latex] f^{-1}(60). [\/latex]\r\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\" border=\"0\"><caption>Table 4<\/caption>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 16.6667%;\">[latex] t [\/latex](minutes)<\/td>\r\n<td style=\"width: 16.6667%;\">30<\/td>\r\n<td style=\"width: 16.6667%;\">50<\/td>\r\n<td style=\"width: 16.6667%;\">60<\/td>\r\n<td style=\"width: 16.6667%;\">70<\/td>\r\n<td style=\"width: 16.6667%;\">90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 16.6667%;\">[latex] f(t) [\/latex](miles)<\/td>\r\n<td style=\"width: 16.6667%;\">20<\/td>\r\n<td style=\"width: 16.6667%;\">40<\/td>\r\n<td style=\"width: 16.6667%;\">50<\/td>\r\n<td style=\"width: 16.6667%;\">60<\/td>\r\n<td style=\"width: 16.6667%;\">70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n<h2>Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/h2>\r\n<\/section><section id=\"fs-id1165137418615\" data-depth=\"2\">\r\n<p id=\"fs-id1165137400045\">We saw in Functions and Function Notation 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 data-effect=\"italics\">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 data-effect=\"italics\">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 class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">How To<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>Given the graph of a function, evaluate its inverse at specific points.<\/strong>\r\n<ol>\r\n \t<li>Find\u00a0the desired input on the [latex] y- [\/latex]axis of the given graph.<\/li>\r\n \t<li>Read\u00a0the inverse function\u2019s output from the [latex] x- [\/latex]axis of the given graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 6: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA function\u00a0[latex] g(x) [\/latex] is given in Figure 5. Find\u00a0[latex] g(3) [\/latex] and [latex] g^{-1}(3). [\/latex]\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_632\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-632\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-5-300x156.jpeg\" alt=\"\" width=\"300\" height=\"156\" \/> Figure 5[\/caption]\r\n\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>To evaluate\u00a0[latex] g(3), [\/latex] we find 3 on the [latex] y- [\/latex]axis and find the corresponding output value on the axis. The point\u00a0[latex] (3, 1) [\/latex] tells us that [latex] g(3)=1. [\/latex]\r\n\r\nTo evaluate\u00a0[latex] g^{-1}(3), [\/latex] recall that by definition\u00a0[latex] g^{-1}(3) [\/latex] means the value or x for which\u00a0[latex] g(x)=3. [\/latex] By looking for the output value 3 on the vertical axis, we find the point [latex] (5, 3) [\/latex] on the graph, which means\u00a0[latex] g(5)=3, [\/latex] so by definition\u00a0[latex] g^{-1}(3)=5. [\/latex] See Figure 6.\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_633\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-633\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-6-300x156.jpeg\" alt=\"\" width=\"300\" height=\"156\" \/> Figure 6[\/caption]\r\n\r\n<\/details><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #6<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nUsing the graph in Figure 5, (a) find\u00a0[latex] g^{-1}(1), [\/latex] and (b) estimate [latex] g^{-1}(4). [\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137605437\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Finding Inverses of Functions Represented by Formulas<\/h3>\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\u2014for example,\u00a0[latex] y [\/latex] as a function of [latex] x [\/latex] --\u00a0 we can often find the inverse function by solving to obtain\u00a0[latex] x [\/latex] as a function of [latex] y. [\/latex]<\/p>\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">How To<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>Given a function represented by a formula, find the inverse.<\/strong>\r\n<ol>\r\n \t<li>Make sure\u00a0[latex] f [\/latex] is a one-to-one function.<\/li>\r\n \t<li>Solve for [latex] x. [\/latex]<\/li>\r\n \t<li>Interchange\u00a0[latex] x [\/latex] and [latex] y. [\/latex]<\/li>\r\n \t<li>Replace\u00a0[latex] y [\/latex] with\u00a0[latex] f^{-1}(x). [\/latex] (Variables may be different in different cases, but the principle is the same.)<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7: Inverting the Fahrenheit-to-Celsius Function<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\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}(F-32) [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rcl}C &amp;=&amp; \\frac{5}{9}(F-32) \\\\C \\cdot \\frac{9}{5} &amp;=&amp; F-32 \\\\F &amp;=&amp; \\frac{9}{5}C + 32\\end{array}[\/latex]<\/p>\r\nBy solving in general, we have uncovered the inverse function. If\r\n<p style=\"text-align: center;\">[latex] C=h(F)=\\frac{5}{9}(F-32), [\/latex]<\/p>\r\nthen\r\n<p style=\"text-align: center;\">[latex] F=h^{-1}(C)=\\frac{9}{5}C+32 [\/latex]<\/p>\r\nIn this case, we introduced a function\u00a0[latex] h [\/latex] to represent the conversion because the input and output variables are descriptive, and writing\u00a0[latex] C^{-1} [\/latex] could get confusing.\r\n\r\n<\/details><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #7<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSolve for\u00a0[latex] x [\/latex] in terms of\u00a0[latex] y [\/latex] given [latex] y=\\frac{1}{3}(x-5). [\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 8: Solving to Find an Inverse Function<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the inverse of the function [latex] f(x)=\\frac{}{x-3}+4. [\/latex]\r\n\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rcll}y &amp;=&amp; \\frac{2}{x-3} + 4 &amp; \\quad \\text{Set up an equation.} \\\\y-4 &amp;=&amp; \\frac{2}{x-3} &amp; \\quad \\text{Subtract 4 from both sides.} \\\\x-3 &amp;=&amp; \\frac{2}{y-4} &amp; \\quad \\text{Multiply both sides by } x-3 \\text{ and divide by } y-4. \\\\x &amp;=&amp; \\frac{2}{y-4} + 3 &amp; \\quad \\text{Add 3 to both sides.}\\end{array} [\/latex]<\/p>\r\nSo\u00a0[latex] f^{-1}(y)=\\frac{2}{y-4}+3 [\/latex] or [latex] f^{-1}(x)=\\frac{2}{x-4}+3. [\/latex]\r\n<h3>Analysis<\/h3>\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 Table 5.\r\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\" border=\"0\"><caption>\u00a0<\/caption>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 20%;\">[latex] x [\/latex]<\/td>\r\n<td style=\"width: 20%;\">1<\/td>\r\n<td style=\"width: 20%;\">2<\/td>\r\n<td style=\"width: 20%;\">5<\/td>\r\n<td style=\"width: 20%;\">[latex] f^{1}(x) [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;\">[latex] f(x) [\/latex]<\/td>\r\n<td style=\"width: 20%;\">3<\/td>\r\n<td style=\"width: 20%;\">2<\/td>\r\n<td style=\"width: 20%;\">5<\/td>\r\n<td style=\"width: 20%;\">[latex] y [\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/details><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 9: Solving to Find an Inverse with Radicals<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the inverse of the function [latex] f(x)=2+\\sqrt{x-4}. [\/latex]\r\n\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{rcl}y &amp;=&amp; 2 + \\sqrt{x-4} \\\\(y-2)^2 &amp;=&amp; x-4 \\\\x &amp;=&amp; (y-2)^2 + 4\\end{array} [\/latex]<\/p>\r\nSo [latex] f^{-1}(x)=(x-2)^4. [\/latex]\r\n\r\nThe domain of\u00a0[latex] f [\/latex] is\u00a0[latex] [4, \\infty). [\/latex] Notice that the range of\u00a0[latex] f [\/latex] is\u00a0[latex] [2, \\infty), [\/latex] so this means that the domain of the inverse function\u00a0[latex] f^{-1} [\/latex] is also [latex] [2, \\infty). [\/latex]\r\n<h3>Analysis<\/h3>\r\nThe formula we found for\u00a0[latex] f^{-1}(x) [\/latex] looks like it would be valid for all real\u00a0[latex] x. [\/latex] However,\u00a0[latex] f^{-1} [\/latex] itself must have an inverse (namely,\u00a0[latex] f [\/latex]) so we have to restrict the domain of\u00a0[latex] f^{-1} [\/latex] to\u00a0[latex] [2, \\infty) [\/latex] in order to make\u00a0[latex] f^{-1} [\/latex] a one-to-one function. This domain of\u00a0[latex] f^{-1} [\/latex] is exactly the range of [latex] f. [\/latex]\r\n\r\n<\/details><\/div>\r\n<\/div>\r\n<section>\r\n<div class=\"body\">\r\n<div id=\"fs-id1165137603677\" class=\"unnumbered\" data-type=\"exercise\"><section>\r\n<div id=\"fs-id1165137667328\" data-type=\"commentary\">\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #8<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nWhat is the inverse of the function\u00a0[latex] f(x)=2-\\sqrt{x}? [\/latex] State the domains of both the function and the inverse function.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/section><\/section><section id=\"fs-id1165137473011\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Finding Inverse Functions and Their Graphs<\/h2>\r\n<p id=\"fs-id1165137463843\">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\u00a0[latex] f(x)=x^2 [\/latex] restricted to the domain\u00a0[latex] [0, \\infty), [\/latex] on which this function is one-to-one, and graph it as in Figure 7.<\/p>\r\n&nbsp;\r\n\r\n[caption id=\"attachment_634\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-634\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-7-300x156.jpeg\" alt=\"\" width=\"300\" height=\"156\" \/> Figure 7. Quadratic function with domain restricted to [latex] [0, \\infty). [\/latex][\/caption]\r\n<p id=\"fs-id1165137419977\">Restricting the domain to\u00a0[latex] [0, \\infty) [\/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,\u00a0[latex] f^{-1}(x)=\\sqrt{x}. [\/latex] What happens if we graph both\u00a0[latex] f [\/latex] and\u00a0[latex] f^{-1} [\/latex] on the same set of axes, using the\u00a0[latex] x- [\/latex]axis for the input to both\u00a0[latex] f [\/latex] and [latex] f^{-1}? [\/latex]<\/p>\r\n<p id=\"fs-id1165131968090\">We notice a distinct relationship: The graph of\u00a0[latex] f^{-1}(x) [\/latex] is the graph of\u00a0[latex] f(x) [\/latex] reflected about the diagonal line\u00a0[latex] y=x, [\/latex] which we will call the identity line, shown in Figure 8.<\/p>\r\n&nbsp;\r\n\r\n[caption id=\"attachment_635\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-635\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-8-300x155.jpeg\" alt=\"\" width=\"300\" height=\"155\" \/> Figure 8. 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 class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 10: Finding the Inverse of a Function Using Reflection about the Identity Line<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nGiven the graph of\u00a0[latex] f(x) [\/latex] in Figure 9, sketch a graph of [latex] f^{-1}(x). [\/latex]\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_636\" align=\"aligncenter\" width=\"288\"]<img class=\"size-medium wp-image-636\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-9-288x300.jpeg\" alt=\"\" width=\"288\" height=\"300\" \/> Figure 9[\/caption]\r\n\r\n&nbsp;\r\n\r\n<details><summary><strong>Solution (click to expand)<\/strong><\/summary>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\u00a0 of [latex] (0, \\infty) [\/latex] and range of\u00a0[latex] (-\\infty, \\infty), [\/latex] so the inverse will have a domain of\u00a0[latex] (-\\infty, \\infty) [\/latex] and range of [latex] (0, \\infty). [\/latex]\r\n\r\nIf we reflect this graph over the line\u00a0[latex] y=x, [\/latex] the point\u00a0[latex] (1, 0) [\/latex] reflects to\u00a0[latex] (0, 1) [\/latex] and the point\u00a0[latex] (4, 2) [\/latex] reflects to\u00a0[latex] (2, 4). [\/latex] Sketching the inverse on the same axes as the original graph gives Figure 10.\r\n\r\n[caption id=\"attachment_637\" align=\"aligncenter\" width=\"288\"]<img class=\"size-medium wp-image-637\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-10-288x300.jpeg\" alt=\"\" width=\"288\" height=\"300\" \/> Figure 10. The function and its inverse, showing reflection about the identity line[\/caption]\r\n\r\n<\/details><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Try It #9<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nDraw graphs of the functions\u00a0[latex] f [\/latex] and\u00a0[latex] f^{-1} [\/latex] from Example 8.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Q&amp;A<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>Q: Is there any function that is equal to its own inverse?<\/strong>\r\n\r\n<em>A:\u00a0Yes. If\u00a0[latex] f=f^{-1}, [\/latex] then\u00a0[latex] f(f(x))=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\u00a0[latex] f(x)=c-x [\/latex] where\u00a0[latex] c [\/latex] is a constant, is also equal to its own inverse.<\/em>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Media<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nAccess these online resources for additional instruction and practice with inverse functions.\r\n<ul>\r\n \t<li><a href=\"https:\/\/www.youtube.com\/watch?v=qgezKpQYH2w\">Inverse Functions<\/a><\/li>\r\n \t<li><a href=\"https:\/\/www.youtube.com\/watch?v=QFOJmevha_Y&amp;feature=youtu.be\/\">One-to-One Functions<\/a><\/li>\r\n \t<li><a href=\"https:\/\/www.youtube.com\/watch?v=FIF8SdZkJc8\">Inverse Function Values Using Graph<\/a><\/li>\r\n \t<li><a href=\"https:\/\/www.youtube.com\/watch?v=rsJ14O5-KDw\">Restricting the Domain and Finding the Inverse<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"os-eos os-section-exercises-container\" data-uuid-key=\".section-exercises\">\r\n<h2 data-type=\"document-title\" data-rex-keep=\"true\"><span class=\"os-text\">3.7 Section Exercises<\/span><\/h2>\r\n<section id=\"fs-id1165137871042\" class=\"section-exercises\" data-depth=\"1\"><section id=\"fs-id1165135187563\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Verbal<\/h3>\r\n<div id=\"fs-id1165137407341\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135193086\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137407341-solution\">1<\/a><span class=\"os-divider\">. <\/span>Describe why the horizontal line test is an effective way to determine whether a function is one-to-one?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137408636\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137408638\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">2<\/span><span class=\"os-divider\">. <\/span>Why do we restrict the domain of the function\u00a0[latex] f(x)=x^2 [\/latex] to find the function\u2019s inverse?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137389621\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135400199\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137389621-solution\">3<\/a><span class=\"os-divider\">. <\/span>Can a function be its own inverse? Explain.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135188794\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137564806\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">4<\/span><span class=\"os-divider\">. <\/span>Are one-to-one functions either always increasing or always decreasing? Why or why not?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137419050\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137932403\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137419050-solution\">5<\/a><span class=\"os-divider\">. <\/span>How do you find the inverse of a function algebraically?\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><section id=\"fs-id1165137836714\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Algebraic<\/h3>\r\n<div id=\"fs-id1165137422830\" class=\"material-set-2\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137806758\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">6<\/span><span class=\"os-divider\">. <\/span>Show that the function [latex] f(x)=a-x [\/latex] is its own inverse for all real numbers [latex] a. [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165137469451\">For the following exercises, find\u00a0[latex] f^{-1}(x) [\/latex] for each function.<\/p>\r\n\r\n<div id=\"fs-id1165134312158\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134312161\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165134312158-solution\">7<\/a><span class=\"os-divider\">.<\/span> [latex] f(x)=x+3 [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137634366\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137634368\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">8<\/span><span class=\"os-divider\">. <\/span> [latex] f(x)=x+5 [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137706135\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137422592\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137706135-solution\">9<\/a><span class=\"os-divider\">. <\/span> [latex] f(x)=2-x [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137561652\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137653456\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">10<\/span><span class=\"os-divider\">. <\/span> [latex] f(x)=3-x [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137600416\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135198605\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137600416-solution\">11<\/a><span class=\"os-divider\">. <\/span> [latex] f(x)=\\frac{x}{x+2} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137541591\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134043731\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">12<\/span><span class=\"os-divider\">. <\/span> [latex] f(x)=\\frac{2x+3}{5x+4} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165137410615\">For the following exercises, find a domain on which each function\u00a0[latex] f [\/latex] is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of\u00a0[latex] f [\/latex] restricted to that domain.<\/p>\r\n\r\n<div id=\"fs-id1165134148519\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165134148521\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165134148519-solution\">13<\/a><span class=\"os-divider\">. <\/span> [latex] f(x)=(x+7)^2 [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137862703\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137531119\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">14<\/span><span class=\"os-divider\">. <\/span> [latex] f(x)=(x-60^2 [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137938698\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137938700\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137938698-solution\">15<\/a><span class=\"os-divider\">. <\/span> [latex] f(x)=x^2-5 [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137734966\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137734969\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137734966-solution\">16<\/a><span class=\"os-divider\">. <\/span>Given\u00a0[latex] f(x)=\\frac{x}{2+x} [\/latex] and [latex] g(x)=\\frac{2x}{1-x}: [\/latex]\r\n\r\n(a) Find\u00a0[latex] f(g(x)) [\/latex] and [latex] g(f(x)). [\/latex]\r\n<div class=\"os-problem-container\">\r\n\r\n(b) What does the answer tell us about the relationship between\u00a0[latex] f(x) [\/latex] and [latex] g(x)? [\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165137444427\">For the following exercises, use function composition to verify that\u00a0[latex] f(x) [\/latex] and\u00a0[latex] g(x) [\/latex] are inverse functions.<\/p>\r\n\r\n<div id=\"fs-id1165137437578\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137619341\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137437578-solution\">17<\/a><span class=\"os-divider\">. <\/span>[latex] f(x)=\\sqrt[3]{x-1} \\ \\ \\text{and} \\ \\ g(x)=x^3+1 [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135538749\" class=\"material-set-2\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135538751\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">18<\/span><span class=\"os-divider\">. <\/span>[latex] f(x)=-3x+5 \\ \\ \\text{and} \\ \\ g(x)=\\frac{x-5}{-3} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><section id=\"fs-id1165135188614\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Graphical<\/h3>\r\n<p id=\"fs-id1165135176520\">For the following exercises, use a graphing utility to determine whether each function is one-to-one.<\/p>\r\n\r\n<div id=\"fs-id1165137645254\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135436660\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137645254-solution\">19<\/a><span class=\"os-divider\">. <\/span> [latex] f(x)=\\sqrt{x} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135173420\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135173423\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">20<\/span><span class=\"os-divider\">. <\/span> [latex] f(x)=\\sqrt[3]{3x+1} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135424683\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137408415\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135424683-solution\">21<\/a><span class=\"os-divider\">. <\/span> [latex] f(x)=-5x+1 [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137704606\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137704608\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">22<\/span><span class=\"os-divider\">. <\/span> [latex] f(x)=x^3-27 [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165137528556\">For the following exercises, determine whether the graph represents a one-to-one function.<\/p>\r\n\r\n<div id=\"fs-id1165137528559\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135251340\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137528559-solution\">23<\/a><span class=\"os-divider\">. <\/span>\r\n<div class=\"os-problem-container has-first-element\"><span id=\"fs-id1165135341390\" class=\"first-element\" data-type=\"media\" data-alt=\"Graph of a parabola.\" data-display=\"block\">\r\n<img class=\"alignnone size-full wp-image-638\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.23.jpeg\" alt=\"\" width=\"252\" height=\"253\" \/>\r\n<\/span><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135192971\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135192973\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">24<\/span><span class=\"os-divider\">. <\/span>\r\n<div class=\"os-problem-container has-first-element\">\r\n\r\n<span id=\"fs-id1165134379457\" class=\"first-element\" data-type=\"media\" data-alt=\"Graph of a step-function.\" data-display=\"block\">\r\n<img class=\"alignnone size-medium wp-image-639\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.24-300x300.jpeg\" alt=\"\" width=\"300\" height=\"300\" \/>\r\n<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165137849556\">For the following exercises, use the graph of\u00a0[latex] f [\/latex] shown in Figure 11.<\/p>\r\n&nbsp;\r\n\r\n[caption id=\"attachment_640\" align=\"alignnone\" width=\"288\"]<img class=\"size-medium wp-image-640\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-11-288x300.jpeg\" alt=\"\" width=\"288\" height=\"300\" \/> Figure 11[\/caption]\r\n\r\n<div id=\"fs-id1165137863913\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137854842\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137863913-solution\">25<\/a><span class=\"os-divider\">. <\/span>Find- [latex] f(0). [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137417814\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137417816\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">26<\/span><span class=\"os-divider\">. <\/span>Solve [latex] f(x)=0. [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165133093360\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165133093363\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165133093360-solution\">27<\/a><span class=\"os-divider\">. <\/span>Find [latex] f^{-1}(0). [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137451393\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137806036\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">28<\/span><span class=\"os-divider\">. <\/span>Solve [latex] f^{-1}(x)=0. [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165135639868\">For the following exercises, use the graph of the one-to-one function shown in Figure 12.<\/p>\r\n&nbsp;\r\n\r\n[caption id=\"attachment_641\" align=\"alignnone\" width=\"255\"]<img class=\"size-full wp-image-641\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-12.jpeg\" alt=\"\" width=\"255\" height=\"254\" \/> Figure 12[\/caption]\r\n\r\n<div id=\"fs-id1165137884386\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137812123\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137884386-solution\">29<\/a><span class=\"os-divider\">. <\/span>Sketch the graph of [latex] f^{-1}. [\/latex]\r\n<div class=\"os-problem-container\"><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137892244\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137603262\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">30<\/span><span class=\"os-divider\">. <\/span>Find\u00a0[latex] f(6) \\ \\ \\text{and} \\ \\ f^{-1}(2). [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137827763\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137827765\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137827763-solution\">31<\/a><span class=\"os-divider\">. <\/span>If the complete graph of\u00a0[latex] f [\/latex] is shown, find the domain of [latex] f. [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137644582\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137644584\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">32<\/span><span class=\"os-divider\">. <\/span>If the complete graph of\u00a0[latex] f [\/latex] is shown, find the range of [latex] f. [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><section id=\"fs-id1165135434734\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Numeric<\/h3>\r\n<p id=\"fs-id1165137781583\">For the following exercises, evaluate or solve, assuming that the function\u00a0[latex] f [\/latex] is one-to-one.<\/p>\r\n\r\n<div id=\"fs-id1165137771148\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135250622\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137771148-solution\">33<\/a><span class=\"os-divider\">. <\/span>If\u00a0[latex] f(6)=7, \\ \\ \\text{find} \\ \\ f^{-1}(7). [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137805757\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135190808\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">34<\/span><span class=\"os-divider\">. <\/span>If [latex] f(3)=2, \\ \\ \\text{find} \\ \\ f^{-1}(2). [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137640685\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137640687\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137640685-solution\">35<\/a><span class=\"os-divider\">. <\/span>If\u00a0[latex] f^{-1}(-4)=-8, \\ \\ \\text{find} \\ \\ f(-8). [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137404972\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137404974\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">36<\/span><span class=\"os-divider\">. <\/span>If\u00a0[latex] f^{-1}(-2)=-1, \\ \\ \\text{find} \\ \\ f(-1). [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1165135195398\">For the following exercises, use the values listed in Table 6 to evaluate or solve.<\/p>\r\n\r\n<div id=\"eip-151\" class=\"os-table\">\r\n<table class=\"grid\" data-id=\"eip-151\"><caption>Table 6<\/caption>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 109px;\">[latex] x [\/latex]<\/td>\r\n<td style=\"width: 60px;\">0<\/td>\r\n<td style=\"width: 60px;\">1<\/td>\r\n<td style=\"width: 57px;\">2<\/td>\r\n<td style=\"width: 57px;\">3<\/td>\r\n<td style=\"width: 57px;\">4<\/td>\r\n<td style=\"width: 58px;\">5<\/td>\r\n<td style=\"width: 58px;\">6<\/td>\r\n<td style=\"width: 57px;\">7<\/td>\r\n<td style=\"width: 58px;\">8<\/td>\r\n<td style=\"width: 58px;\">9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 109px;\" data-align=\"center\">[latex] f(x) [\/latex]<\/td>\r\n<td style=\"width: 60px;\">8<\/td>\r\n<td style=\"width: 60px;\">0<\/td>\r\n<td style=\"width: 57px;\">7<\/td>\r\n<td style=\"width: 57px;\">4<\/td>\r\n<td style=\"width: 57px;\">2<\/td>\r\n<td style=\"width: 58px;\">6<\/td>\r\n<td style=\"width: 58px;\">5<\/td>\r\n<td style=\"width: 57px;\">3<\/td>\r\n<td style=\"width: 58px;\">9<\/td>\r\n<td style=\"width: 58px;\">1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"os-caption-container\"><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137400581\" class=\"os-hasSolution\" data-type=\"exercise\"><header><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137400581-solution\">37<\/a><span class=\"os-divider\">.<\/span> Find [latex] f(1). [\/latex]<\/header><\/div>\r\n<div id=\"fs-id1165137807065\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137807068\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">38<\/span><span class=\"os-divider\">. <\/span>Solve [latex] f(x)=3. [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137736914\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137736916\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137736914-solution\">39<\/a><span class=\"os-divider\">. <\/span>Find [latex] f^{-1}(0). [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137686729\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137686731\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">40<\/span><span class=\"os-divider\">. <\/span>Solve [latex] f^{-1}(x)=7. [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135341429\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135209686\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135341429-solution\">41<\/a><span class=\"os-divider\">. <\/span>Use the tabular representation of\u00a0[latex] f [\/latex] in Table 7 to create a table for [latex] f^{-1}(x). [\/latex]\r\n<div class=\"os-problem-container\">\r\n<div id=\"Table_01_07_08\" class=\"os-table\">\r\n<table class=\"grid\" data-id=\"Table_01_07_08\"><caption>Table 7<\/caption>\r\n<tbody>\r\n<tr>\r\n<td data-align=\"center\">[latex] x [\/latex]<\/td>\r\n<td data-align=\"center\">3<\/td>\r\n<td data-align=\"center\">6<\/td>\r\n<td data-align=\"center\">9<\/td>\r\n<td data-align=\"center\">13<\/td>\r\n<td data-align=\"center\">14<\/td>\r\n<\/tr>\r\n<tr>\r\n<td data-align=\"center\">[latex] f(x) [\/latex]<\/td>\r\n<td data-align=\"center\">1<\/td>\r\n<td data-align=\"center\">4<\/td>\r\n<td data-align=\"center\">7<\/td>\r\n<td data-align=\"center\">12<\/td>\r\n<td data-align=\"center\">16<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h3 class=\"os-caption-container\">Technology<\/h3>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><section id=\"fs-id1165137641552\" data-depth=\"2\">\r\n<p id=\"fs-id1165137401884\">For the following exercises, find the inverse function. Then, graph the function and its inverse.<\/p>\r\n\r\n<div id=\"fs-id1165137401888\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137470994\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">42<\/span><span class=\"os-divider\">. <\/span> [latex] f(x)=\\frac{3}{x-2} [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135435667\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135435669\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135435667-solution\">43<\/a><span class=\"os-divider\">. <\/span> [latex] f(x)=x^3-1 [\/latex]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137419353\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135209709\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">44<\/span><span class=\"os-divider\">. <\/span>Find the inverse function of\u00a0[latex] f(x)=\\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\n<\/div>\r\n<\/section><\/div>\r\n<\/section><section id=\"fs-id1165137592239\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Real-World Applications<\/h3>\r\n<div id=\"fs-id1165135512365\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135512368\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135512365-solution\">45<\/a><span class=\"os-divider\">. <\/span>To convert from\u00a0[latex] x [\/latex] degrees Celsius to\u00a0[latex] y [\/latex] degrees Fahrenheit, we use the formula\u00a0[latex] f(x)=\\frac{9}{5}x+32. [\/latex] Find the inverse function, if it exists, and explain its meaning.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165137462831\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165137462833\" data-type=\"problem\">\r\n\r\n<span class=\"os-number\">46<\/span><span class=\"os-divider\">. <\/span>The circumference\u00a0[latex] C [\/latex] of a circle is a function of its radius given by\u00a0[latex] C(r)=2\\pi r. [\/latex] Express the radius of a circle as a function of its circumference. Call this function\u00a0[latex] r(c). [\/latex] Find\u00a0[latex] r(36\\pi) [\/latex] and interpret its meaning.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-id1165135532474\" class=\"os-hasSolution\" data-type=\"exercise\"><header><\/header><section>\r\n<div id=\"fs-id1165135532476\" data-type=\"problem\">\r\n\r\n<a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135532474-solution\">47<\/a><span class=\"os-divider\">. <\/span>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(t)=50t. [\/latex] Find the inverse function by ex4ressing the time of travel in terms of the distance traveled. Call this function [latex] t(d). [\/latex] Find [latex] t(180). [\/latex] and interpret its meaning.\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/section><\/div>","rendered":"<div id=\"main-content\" class=\"MainContent__ContentStyles-sc-6yy1if-0 NnXKu\" tabindex=\"-1\" data-dynamic-style=\"true\">\n<div id=\"page_f592aad0-19d8-42d6-94b9-086bdd84c2b5\" class=\"chapter-content-module\" data-type=\"page\" data-book-content=\"true\">\n<div class=\"textbox textbox--learning-objectives\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Learning Objectives<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>In this section, you will:<\/p>\n<ul>\n<li>Verify inverse functions.<\/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<\/div>\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 id=\"fs-id1165135701544\">If some physical machines can run in two directions, we might ask whether some of the function \u201cmachines\u201d we have been studying can also run backwards. Figure 1 provides a visual representation of this question. In this section, we will consider the reverse nature of functions.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_626\" aria-describedby=\"caption-attachment-626\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-626\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-1-300x125.jpeg\" alt=\"\" width=\"300\" height=\"125\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-1-300x125.jpeg 300w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-1-65x27.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-1-225x94.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-1-350x146.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-1.jpeg 731w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-626\" class=\"wp-caption-text\">Figure 1. Can a function \u201cmachine\u201d operate in reverse?<\/figcaption><\/figure>\n<section id=\"fs-id1165137725994\" data-depth=\"1\">\n<h2 data-type=\"title\">Verifying That Two Functions Are Inverse Functions<\/h2>\n<p id=\"fs-id1165135705795\">Betty is traveling to Milan for a fashion show and wants to know what the temperature will be. She is not familiar with the <span id=\"term-00007\" class=\"no-emphasis\" data-type=\"term\">Celsius<\/span> scale. To get an idea of how temperature measurements are related, Betty wants to convert 75 degrees <span id=\"term-00008\" class=\"no-emphasis\" data-type=\"term\">Fahrenheit<\/span> to degrees Celsius using the formula<\/p>\n<p style=\"text-align: center;\">[latex]C=\\frac{5}{9}(F-32)[\/latex]<\/p>\n<p id=\"fs-id1165135433486\">and substitutes 75 for\u00a0[latex]F[\/latex] to calculate<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{5}{9}(75-32)\\approx24^\\circ C[\/latex]<\/p>\n<p id=\"fs-id1165137409312\">Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, Betty gets the week\u2019s weather forecast from Figure 2 for Milan, and wants to convert all of the temperatures to degrees Fahrenheit.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_627\" aria-describedby=\"caption-attachment-627\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-627\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-2-300x93.jpeg\" alt=\"\" width=\"300\" height=\"93\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-2-300x93.jpeg 300w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-2-65x20.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-2-225x70.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-2-350x108.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-2.jpeg 731w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-627\" class=\"wp-caption-text\">Figure 2<\/figcaption><\/figure>\n<p id=\"fs-id1165137724415\">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\u00a0[latex]F[\/latex] after substituting a value for\u00a0[latex]C.[\/latex] For example, to convert 26 degrees Celsius, she could write<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}26 &=& \\frac{5}{9}(F-32) \\\\26 \\cdot \\frac{9}{5} &=& F-32 \\\\F &=& 26 \\cdot \\frac{9}{5} + 32 \\approx 79\\end{array}[\/latex]<\/p>\n<p id=\"fs-id1165137540705\">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 id=\"fs-id1165137827441\">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 id=\"fs-id1165135528385\">Given a function\u00a0[latex]f(x),[\/latex] we represent its inverse as\u00a0[latex]f^{-1}(x),[\/latex] read as\u00a0&#8220;[latex]f[\/latex] inverse of\u00a0[latex]x.[\/latex]&#8221; The raised\u00a0[latex]-1[\/latex] is part of the notation. It is not an exponent; it does not imply a power of\u00a0[latex]-1.[\/latex] \u00a0In other words, [latex]f^{-1}(x)[\/latex] does <em data-effect=\"italics\">not<\/em> mean\u00a0[latex]\\frac{1}{f(x)}[\/latex] because\u00a0[latex]\\frac{1}{f(x)}[\/latex] is the reciprocal of\u00a0[latex]f[\/latex] and not the inverse.<\/p>\n<p id=\"fs-id1165137724926\">The \u201cexponent-like\u201d notation comes from an analogy between function composition and multiplication: just as\u00a0[latex]- a^{-1}a=1[\/latex] (1 is the identity element for multiplication) for any nonzero number\u00a0[latex]a[\/latex] so\u00a0[latex]f^{-1}\\circ f[\/latex] equals the identity function, that is,<\/p>\n<p style=\"text-align: center;\">[latex](f^{-1}\\circ f)(x)=f^{-1}(f(x))=f^{-1}(y)=x[\/latex]<\/p>\n<p>This holds for all\u00a0[latex][\/latex] in the domain of\u00a0[latex][\/latex] Informally, this means that inverse functions \u201cundo\u201d each other. However, just as zero does not have a <span id=\"term-00009\" class=\"no-emphasis\" data-type=\"term\">reciprocal<\/span>, some functions do not have inverses.<\/p>\n<p id=\"fs-id1165137655153\">Given a function\u00a0[latex]f(x),[\/latex] we can verify whether some other function\u00a0[latex]g(x)[\/latex] is the inverse of\u00a0[latex]f(x)[\/latex] by checking if both\u00a0[latex]g(f(x))=x[\/latex] and\u00a0[latex]f(g(x))=x[\/latex] are true.<\/p>\n<p id=\"fs-id1165135397975\">For example,\u00a0[latex]y=4x[\/latex] and\u00a0[latex]y=\\frac{1}{4}x[\/latex] are inverse functions.<\/p>\n<p style=\"text-align: center;\">[latex]f^{-1}\\circ f)(x)=f^{-1}(4x)=\\frac{1}{4}(4x)=x[\/latex]<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex](f\\circ f^{-1})(x)=f(\\frac{1}{4}x)=4(\\frac{1}{4}x)=x[\/latex]<\/p>\n<p id=\"fs-id1165137438777\">A few coordinate pairs from the graph of the function\u00a0[latex]y=4x[\/latex] are\u00a0[latex](-2, -8), (0, 0),[\/latex] and\u00a0[latex](2, 8).[\/latex] A few coordinate pairs from the graph of the function\u00a0[latex]y=\\frac{1}{4}x[\/latex] are\u00a0[latex](-8, -2), (0, 0),[\/latex] and\u00a0[latex](8, 2).[\/latex] 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 textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Inverse Function<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>For any one-to-one function\u00a0[latex]f(x)=y,[\/latex] a function\u00a0 is an <strong>inverse function<\/strong> of\u00a0[latex]f[\/latex] if\u00a0[latex]f^{-1}(y)=x.[\/latex] This can also be written as\u00a0[latex]f^{-1}(f(x))=x[\/latex] for all\u00a0[latex]x[\/latex] in the domain of\u00a0[latex]f.[\/latex] It also follows that\u00a0[latex]f(f^{-1}(x))=x[\/latex] for all\u00a0[latex]x[\/latex] in the domain of\u00a0[latex]f^{-1}[\/latex] if\u00a0[latex]f^{-1}[\/latex] is the inverse of [latex]f.[\/latex]<\/p>\n<p>The notation\u00a0[latex]f^{-1}[\/latex] is read \u201c[latex]f[\/latex] inverse.&#8221; Like any other function, we can use any variable name as the input for\u00a0[latex]f^{-1},[\/latex] so we will often write\u00a0[latex]f^{-1}(x),[\/latex] which we read as &#8220;[latex]f[\/latex] inverse of [latex]x.[\/latex]&#8221; Keep in mind that<\/p>\n<p style=\"text-align: center;\">[latex]f^{-1}(x)\\not=\\frac{1}{f(x)}[\/latex]<\/p>\n<p>and not all functions have inverses.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 1: Identifying an Inverse Function for a Given Input-Output Pair<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>If for a particular one-to-one function\u00a0[latex]f(2)=4[\/latex] and\u00a0[latex]f(5)=12,[\/latex] what are the corresponding input and output values for the inverse function?<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>The inverse function reverses the input and output quantities, so if<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}f(2) &=& 4, \\, \\text{then } f^{-1}(4) = 2; \\\\f(5) &=& 12, \\, \\text{then } f^{-1}(12) = 5.\\end{array}[\/latex]<\/p>\n<p>Alternatively, if we want to name the inverse function\u00a0[latex]g,[\/latex] then\u00a0[latex]g(4)=2[\/latex] and [latex]g(12)=5.[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p>Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table 1.<\/p>\n<table class=\"grid\">\n<caption>Table 1<\/caption>\n<tbody>\n<tr>\n<td>[latex](x, f(x))[\/latex]<\/td>\n<td>[latex](x, g(x))[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex](2, 4)[\/latex]<\/td>\n<td>[latex](4, 2)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex](5, 12)[\/latex]<\/td>\n<td>[latex](12, 5)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/details>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Given that\u00a0[latex]h^{-1}(6)=2,[\/latex] what are the corresponding input and output values of the original function [latex]h?[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">How To<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>Given two functions\u00a0[latex]f(x)[\/latex] and\u00a0[latex]g(x)[\/latex] test whether the functions are inverses of each other.<\/strong><\/p>\n<ol>\n<li>Determine whether\u00a0[latex]f(g(x))=x[\/latex] or [latex]g(f(x))=x.[\/latex]<\/li>\n<li>If either statement is true, then both are true, and\u00a0[latex]g=f^{-1}[\/latex] and\u00a0[latex]f=g^{-1}.[\/latex] If either statements is false, then both are false, and\u00a0[latex]g\\not=f^{-1}[\/latex] and [latex]f\\not=g^{-1}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 2: Testing Inverse Relationships Algebraically<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>If [latex]f(x)=\\frac{1}{x+2}[\/latex] and [latex]g(x)=\\frac{1}{x}-2,[\/latex] is [latex]g=f^{-1}?[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}g(f(x)) &=& \\frac{1}{\\left(\\frac{1}{x+2}\\right)} - 2 \\\\&=& x + 2 - 2 \\\\&=& x\\end{array}\u00a0[\/latex]<\/p>\n<p>We must also verify the other formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}f(g(x)) &=& \\frac{1}{\\frac{1}{x} - 2 + 2} \\\\&=& \\frac{1}{\\frac{1}{x}} \\\\&=& x\\end{array}[\/latex]<\/p>\n<p>so<\/p>\n<p style=\"text-align: center;\">[latex]g=f^{-1} \\ \\ \\text{and} \\ \\ f=g^{-1}[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p>Notice the inverse operations are in reverse order of the operations from the original function.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>If\u00a0[latex]f(x)=x^3-4[\/latex] and\u00a0[latex]g(x)=\\sqrt[3]{x+4},[\/latex] is [latex]g=f^{-1}?[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 3: Determining Inverse Relationships for Power Functions<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>If\u00a0[latex]f(x)=x^3[\/latex] (the cube function) and\u00a0[latex]g(x)=\\frac{1}{3}x,[\/latex] is [latex]g=f^{-1}?[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p style=\"text-align: center;\">[latex]f(g(x))=\\frac{x^3}{27}\\not=x[\/latex]<\/p>\n<p>No, the functions are not inverses.<\/p>\n<h3>Analysis<\/h3>\n<p>The correct inverse to the cube is, of course, the cube root\u00a0[latex]\\sqrt[3]{x}=x^{\\frac{1}{3}},[\/latex] that is, the one-third is an exponent, not a multiplier.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>If\u00a0[latex]f(x)=(x-1)^3[\/latex] and\u00a0[latex]g(x)=\\sqrt[3]{x}+1,[\/latex] is [latex]g=f^{-1}?[\/latex]<\/p>\n<\/div>\n<\/div>\n<section id=\"fs-id1165137660004\" data-depth=\"1\">\n<h2 data-type=\"title\">Finding Domain and Range of Inverse Functions<\/h2>\n<p id=\"fs-id1165137591020\">The outputs of the function\u00a0[latex]f[\/latex] are the inputs to\u00a0[latex]f^{-1},[\/latex] so the range of\u00a0[latex]f[\/latex] is also the domain of\u00a0[latex]f^{-1}.[\/latex] Likewise, because the inputs to\u00a0[latex]f[\/latex] are the outputs of\u00a0[latex]f^{-1},[\/latex] the domain of\u00a0[latex]f[\/latex] is the range of\u00a0[latex]f^{-1}.[\/latex]\u00a0We can visualize the situation as in Figure 3.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_628\" aria-describedby=\"caption-attachment-628\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-628\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-3-300x88.jpeg\" alt=\"\" width=\"300\" height=\"88\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-3-300x88.jpeg 300w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-3-65x19.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-3-225x66.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-3-350x103.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-3.jpeg 487w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-628\" class=\"wp-caption-text\">Figure 3. Domain and range of a function and its inverse.<\/figcaption><\/figure>\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\u00a0[latex]f(x)=\\sqrt{x}[\/latex] is\u00a0[latex]f^{-1}(x)=x^2,[\/latex] because a square \u201cundoes\u201d a square root; but the square is only the inverse of the square root on the domain\u00a0[latex][0, \\infty),[\/latex] since that is the range of [latex]f(x)=\\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\u00a0[latex]f(x)=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 \u201cinverse\u201d 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\u00a0[latex]f(x)=x^2[\/latex] with its domain limited to\u00a0[latex][0, \\infty),[\/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\u00a0[latex]f(x)=(x-1)^2[\/latex] on\u00a0[latex][1, \\infty),[\/latex] then the inverse function is [latex]f^{-1}(x)=\\sqrt{x}+1.[\/latex]<\/p>\n<ul id=\"fs-id1165137851227\">\n<li>The domain of [latex]f=[\/latex]\u00a0range of [latex]f^{-1}=[1, \\infty).[\/latex]<\/li>\n<li>The domain of\u00a0[latex]f^{-1}=[\/latex] range of [latex]f=[0, \\infty).[\/latex]<\/li>\n<\/ul>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Q&amp;A<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>Q: Is it possible for a function to have more than one inverse?<\/strong><\/p>\n<p><em data-effect=\"italics\">A: No. If two supposedly different functions, say, [latex]g[\/latex] and\u00a0[latex]h,[\/latex] both meet the definition of being inverses of another function\u00a0[latex]f,[\/latex] then you can prove that\u00a0[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>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Domain and Range of Inverse Functions<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>The range of a function\u00a0[latex]f(x)[\/latex] is the domain of the inverse function [latex]f^{-1}(x).[\/latex]<\/p>\n<p>The domain of\u00a0[latex]f(x)[\/latex] is the range of [latex]f^{-1}(x).[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">How To<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>Given a function, find the domain and range of its inverse.<\/strong><\/p>\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>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 4: Finding the Inverses of Toolkit Functions<\/p>\n<\/header>\n<div class=\"textbox__content\">\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 in Table 2. We restrict the domain in such a fashion that the function assumes all [latex][\/latex]values exactly once.<\/p>\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\">\n<caption>Table 2<\/caption>\n<tbody>\n<tr>\n<td style=\"width: 20%;\"><strong>Constant<\/strong><\/td>\n<td style=\"width: 20%;\"><strong>Identity<\/strong><\/td>\n<td style=\"width: 20%;\"><strong>Quadratic<\/strong><\/td>\n<td style=\"width: 20%;\"><strong>Cubic<\/strong><\/td>\n<td style=\"width: 20%;\"><strong>Reciprocal<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;\">[latex]f(x)=c[\/latex]<\/td>\n<td style=\"width: 20%;\">[latex]f(x)=x[\/latex]<\/td>\n<td style=\"width: 20%;\">[latex]f(x)-x^2[\/latex]<\/td>\n<td style=\"width: 20%;\">[latex]f(x)-x^3[\/latex]<\/td>\n<td style=\"width: 20%;\">[latex]f(x)=\\frac{1}{x}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;\"><strong>Reciprocal Squared<\/strong><\/td>\n<td style=\"width: 20%;\"><strong>Cube Root<\/strong><\/td>\n<td style=\"width: 20%;\"><strong>Square Root<\/strong><\/td>\n<td style=\"width: 20%;\"><strong>Absolute Value<\/strong><\/td>\n<td style=\"width: 20%;\"><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;\">[latex]f(x)=\\frac{1}{x^2}[\/latex]<\/td>\n<td style=\"width: 20%;\">[latex]f(x)=\\sqrt[3]{x}[\/latex]<\/td>\n<td style=\"width: 20%;\">[latex]f(x)=\\sqrt{x}[\/latex]<\/td>\n<td style=\"width: 20%;\">[latex]f(x)=|x|[\/latex]<\/td>\n<td style=\"width: 20%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\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 inverse.<\/p>\n<p>The absolute value function can be restricted to the domain\u00a0[latex][0, \\infty),[\/latex] where it is equal to the identity function.<\/p>\n<p>The reciprocal-squared function can be restricted to the domain [latex](0, \\infty).[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p>We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure 4. They 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<p>&nbsp;<\/p>\n<figure id=\"attachment_629\" aria-describedby=\"caption-attachment-629\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-629\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-4-300x146.jpeg\" alt=\"\" width=\"300\" height=\"146\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-4-300x146.jpeg 300w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-4-65x32.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-4-225x109.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-4-350x170.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-4.jpeg 711w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-629\" class=\"wp-caption-text\">Figure 4. (a) Absolute value (b) Reciprocal square<\/figcaption><\/figure>\n<\/details>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>The domain of function\u00a0[latex]f[\/latex] is\u00a0[latex](1, \\infty)[\/latex] and the range of function\u00a0[latex]f[\/latex] is\u00a0[latex](-\\infty, 2).[\/latex] Find the domain and range of the inverse function.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137619159\" data-depth=\"1\">\n<h2 data-type=\"title\">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\" data-depth=\"2\">\n<h3 data-type=\"title\">Inverting Tabular Functions<\/h3>\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 class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 5: Interpreting the Inverse of a Tabular Function<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A function\u00a0[latex]f(t)[\/latex] is given in Table 3, showing the distance in miles that a car has traveled in\u00a0 minutes. Find and interpret [latex]f^{-1}(70).[\/latex]<\/p>\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\">\n<caption>Table 3<\/caption>\n<tbody>\n<tr>\n<td style=\"width: 20%;\">[latex]t[\/latex](minutes)<\/td>\n<td style=\"width: 20%;\">30<\/td>\n<td style=\"width: 20%;\">50<\/td>\n<td style=\"width: 20%;\">70<\/td>\n<td style=\"width: 20%;\">90<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;\">[latex]f(t)[\/latex](miles)<\/td>\n<td style=\"width: 20%;\">20<\/td>\n<td style=\"width: 20%;\">40<\/td>\n<td style=\"width: 20%;\">60<\/td>\n<td style=\"width: 20%;\">80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>The inverse function takes an output of\u00a0[latex]f[\/latex] and returns an input for\u00a0[latex]f.[\/latex] So in the expression\u00a0[latex]f^{-1}(70),[\/latex] 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function\u00a0[latex]f,[\/latex] 90 minutes, so\u00a0[latex]f^{-1}(70)=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\u00a0[latex]f(a)=b,[\/latex] then\u00a0[latex]f^{-1}(b)=a.[\/latex] By this definition, if we are given\u00a0[latex]f^{-1}(70)=a,[\/latex] then we are looking for a value\u00a0[latex]a[\/latex] so that\u00a0[latex]f(a)=70.[\/latex] In this case, we are looking for a [latex]t[\/latex] so that\u00a0[latex]f(t)=70,[\/latex] which is when [latex]t=90.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #5<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Using Table 4, find and interpret (a)\u00a0[latex]f(60),[\/latex] and (b) [latex]f^{-1}(60).[\/latex]<\/p>\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\">\n<caption>Table 4<\/caption>\n<tbody>\n<tr>\n<td style=\"width: 16.6667%;\">[latex]t[\/latex](minutes)<\/td>\n<td style=\"width: 16.6667%;\">30<\/td>\n<td style=\"width: 16.6667%;\">50<\/td>\n<td style=\"width: 16.6667%;\">60<\/td>\n<td style=\"width: 16.6667%;\">70<\/td>\n<td style=\"width: 16.6667%;\">90<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 16.6667%;\">[latex]f(t)[\/latex](miles)<\/td>\n<td style=\"width: 16.6667%;\">20<\/td>\n<td style=\"width: 16.6667%;\">40<\/td>\n<td style=\"width: 16.6667%;\">50<\/td>\n<td style=\"width: 16.6667%;\">60<\/td>\n<td style=\"width: 16.6667%;\">70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<h2>Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/h2>\n<\/section>\n<section id=\"fs-id1165137418615\" data-depth=\"2\">\n<p id=\"fs-id1165137400045\">We saw in Functions and Function Notation 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 data-effect=\"italics\">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 data-effect=\"italics\">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 textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">How To<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>Given the graph of a function, evaluate its inverse at specific points.<\/strong><\/p>\n<ol>\n<li>Find\u00a0the desired input on the [latex]y-[\/latex]axis of the given graph.<\/li>\n<li>Read\u00a0the inverse function\u2019s output from the [latex]x-[\/latex]axis of the given graph.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 6: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A function\u00a0[latex]g(x)[\/latex] is given in Figure 5. Find\u00a0[latex]g(3)[\/latex] and [latex]g^{-1}(3).[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_632\" aria-describedby=\"caption-attachment-632\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-632\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-5-300x156.jpeg\" alt=\"\" width=\"300\" height=\"156\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-5-300x156.jpeg 300w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-5-65x34.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-5-225x117.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-5-350x183.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-5.jpeg 487w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-632\" class=\"wp-caption-text\">Figure 5<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p>To evaluate\u00a0[latex]g(3),[\/latex] we find 3 on the [latex]y-[\/latex]axis and find the corresponding output value on the axis. The point\u00a0[latex](3, 1)[\/latex] tells us that [latex]g(3)=1.[\/latex]<\/p>\n<p>To evaluate\u00a0[latex]g^{-1}(3),[\/latex] recall that by definition\u00a0[latex]g^{-1}(3)[\/latex] means the value or x for which\u00a0[latex]g(x)=3.[\/latex] By looking for the output value 3 on the vertical axis, we find the point [latex](5, 3)[\/latex] on the graph, which means\u00a0[latex]g(5)=3,[\/latex] so by definition\u00a0[latex]g^{-1}(3)=5.[\/latex] See Figure 6.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_633\" aria-describedby=\"caption-attachment-633\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-633\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-6-300x156.jpeg\" alt=\"\" width=\"300\" height=\"156\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-6-300x156.jpeg 300w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-6-65x34.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-6-225x117.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-6-350x183.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-6.jpeg 487w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-633\" class=\"wp-caption-text\">Figure 6<\/figcaption><\/figure>\n<\/details>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #6<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Using the graph in Figure 5, (a) find\u00a0[latex]g^{-1}(1),[\/latex] and (b) estimate [latex]g^{-1}(4).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137605437\" data-depth=\"2\">\n<h3 data-type=\"title\">Finding Inverses of Functions Represented by Formulas<\/h3>\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\u2014for example,\u00a0[latex]y[\/latex] as a function of [latex]x[\/latex] &#8212;\u00a0 we can often find the inverse function by solving to obtain\u00a0[latex]x[\/latex] as a function of [latex]y.[\/latex]<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">How To<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>Given a function represented by a formula, find the inverse.<\/strong><\/p>\n<ol>\n<li>Make sure\u00a0[latex]f[\/latex] is a one-to-one function.<\/li>\n<li>Solve for [latex]x.[\/latex]<\/li>\n<li>Interchange\u00a0[latex]x[\/latex] and [latex]y.[\/latex]<\/li>\n<li>Replace\u00a0[latex]y[\/latex] with\u00a0[latex]f^{-1}(x).[\/latex] (Variables may be different in different cases, but the principle is the same.)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7: Inverting the Fahrenheit-to-Celsius Function<\/p>\n<\/header>\n<div class=\"textbox__content\">\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}(F-32)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}C &=& \\frac{5}{9}(F-32) \\\\C \\cdot \\frac{9}{5} &=& F-32 \\\\F &=& \\frac{9}{5}C + 32\\end{array}[\/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(F)=\\frac{5}{9}(F-32),[\/latex]<\/p>\n<p>then<\/p>\n<p style=\"text-align: center;\">[latex]F=h^{-1}(C)=\\frac{9}{5}C+32[\/latex]<\/p>\n<p>In this case, we introduced a function\u00a0[latex]h[\/latex] to represent the conversion because the input and output variables are descriptive, and writing\u00a0[latex]C^{-1}[\/latex] could get confusing.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #7<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Solve for\u00a0[latex]x[\/latex] in terms of\u00a0[latex]y[\/latex] given [latex]y=\\frac{1}{3}(x-5).[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 8: Solving to Find an Inverse Function<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the inverse of the function [latex]f(x)=\\frac{}{x-3}+4.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcll}y &=& \\frac{2}{x-3} + 4 & \\quad \\text{Set up an equation.} \\\\y-4 &=& \\frac{2}{x-3} & \\quad \\text{Subtract 4 from both sides.} \\\\x-3 &=& \\frac{2}{y-4} & \\quad \\text{Multiply both sides by } x-3 \\text{ and divide by } y-4. \\\\x &=& \\frac{2}{y-4} + 3 & \\quad \\text{Add 3 to both sides.}\\end{array}[\/latex]<\/p>\n<p>So\u00a0[latex]f^{-1}(y)=\\frac{2}{y-4}+3[\/latex] or [latex]f^{-1}(x)=\\frac{2}{x-4}+3.[\/latex]<\/p>\n<h3>Analysis<\/h3>\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 Table 5.<\/p>\n<table class=\"grid\" style=\"border-collapse: collapse; width: 100%;\">\n<caption>\u00a0<\/caption>\n<tbody>\n<tr>\n<td style=\"width: 20%;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 20%;\">1<\/td>\n<td style=\"width: 20%;\">2<\/td>\n<td style=\"width: 20%;\">5<\/td>\n<td style=\"width: 20%;\">[latex]f^{1}(x)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;\">[latex]f(x)[\/latex]<\/td>\n<td style=\"width: 20%;\">3<\/td>\n<td style=\"width: 20%;\">2<\/td>\n<td style=\"width: 20%;\">5<\/td>\n<td style=\"width: 20%;\">[latex]y[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/details>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 9: Solving to Find an Inverse with Radicals<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the inverse of the function [latex]f(x)=2+\\sqrt{x-4}.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}y &=& 2 + \\sqrt{x-4} \\\\(y-2)^2 &=& x-4 \\\\x &=& (y-2)^2 + 4\\end{array}[\/latex]<\/p>\n<p>So [latex]f^{-1}(x)=(x-2)^4.[\/latex]<\/p>\n<p>The domain of\u00a0[latex]f[\/latex] is\u00a0[latex][4, \\infty).[\/latex] Notice that the range of\u00a0[latex]f[\/latex] is\u00a0[latex][2, \\infty),[\/latex] so this means that the domain of the inverse function\u00a0[latex]f^{-1}[\/latex] is also [latex][2, \\infty).[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p>The formula we found for\u00a0[latex]f^{-1}(x)[\/latex] looks like it would be valid for all real\u00a0[latex]x.[\/latex] However,\u00a0[latex]f^{-1}[\/latex] itself must have an inverse (namely,\u00a0[latex]f[\/latex]) so we have to restrict the domain of\u00a0[latex]f^{-1}[\/latex] to\u00a0[latex][2, \\infty)[\/latex] in order to make\u00a0[latex]f^{-1}[\/latex] a one-to-one function. This domain of\u00a0[latex]f^{-1}[\/latex] is exactly the range of [latex]f.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<section>\n<div class=\"body\">\n<div id=\"fs-id1165137603677\" class=\"unnumbered\" data-type=\"exercise\">\n<section>\n<div id=\"fs-id1165137667328\" data-type=\"commentary\">\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #8<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>What is the inverse of the function\u00a0[latex]f(x)=2-\\sqrt{x}?[\/latex] State the domains of both the function and the inverse function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/section>\n<\/section>\n<section id=\"fs-id1165137473011\" data-depth=\"1\">\n<h2 data-type=\"title\">Finding Inverse Functions and Their Graphs<\/h2>\n<p id=\"fs-id1165137463843\">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\u00a0[latex]f(x)=x^2[\/latex] restricted to the domain\u00a0[latex][0, \\infty),[\/latex] on which this function is one-to-one, and graph it as in Figure 7.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_634\" aria-describedby=\"caption-attachment-634\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-634\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-7-300x156.jpeg\" alt=\"\" width=\"300\" height=\"156\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-7-300x156.jpeg 300w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-7-65x34.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-7-225x117.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-7-350x183.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-7.jpeg 487w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-634\" class=\"wp-caption-text\">Figure 7. Quadratic function with domain restricted to [latex] [0, \\infty). [\/latex]<\/figcaption><\/figure>\n<p id=\"fs-id1165137419977\">Restricting the domain to\u00a0[latex][0, \\infty)[\/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,\u00a0[latex]f^{-1}(x)=\\sqrt{x}.[\/latex] What happens if we graph both\u00a0[latex]f[\/latex] and\u00a0[latex]f^{-1}[\/latex] on the same set of axes, using the\u00a0[latex]x-[\/latex]axis for the input to both\u00a0[latex]f[\/latex] and [latex]f^{-1}?[\/latex]<\/p>\n<p id=\"fs-id1165131968090\">We notice a distinct relationship: The graph of\u00a0[latex]f^{-1}(x)[\/latex] is the graph of\u00a0[latex]f(x)[\/latex] reflected about the diagonal line\u00a0[latex]y=x,[\/latex] which we will call the identity line, shown in Figure 8.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_635\" aria-describedby=\"caption-attachment-635\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-635\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-8-300x155.jpeg\" alt=\"\" width=\"300\" height=\"155\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-8-300x155.jpeg 300w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-8-65x34.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-8-225x116.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-8-350x180.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-8.jpeg 487w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-635\" class=\"wp-caption-text\">Figure 8. Square and square-root functions on the non-negative domain<\/figcaption><\/figure>\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 class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 10: Finding the Inverse of a Function Using Reflection about the Identity Line<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Given the graph of\u00a0[latex]f(x)[\/latex] in Figure 9, sketch a graph of [latex]f^{-1}(x).[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_636\" aria-describedby=\"caption-attachment-636\" style=\"width: 288px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-636\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-9-288x300.jpeg\" alt=\"\" width=\"288\" height=\"300\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-9-288x300.jpeg 288w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-9-65x68.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-9-225x234.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-9-350x365.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-9.jpeg 357w\" sizes=\"auto, (max-width: 288px) 100vw, 288px\" \/><figcaption id=\"caption-attachment-636\" class=\"wp-caption-text\">Figure 9<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<details>\n<summary><strong>Solution (click to expand)<\/strong><\/summary>\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\u00a0 of [latex](0, \\infty)[\/latex] and range of\u00a0[latex](-\\infty, \\infty),[\/latex] so the inverse will have a domain of\u00a0[latex](-\\infty, \\infty)[\/latex] and range of [latex](0, \\infty).[\/latex]<\/p>\n<p>If we reflect this graph over the line\u00a0[latex]y=x,[\/latex] the point\u00a0[latex](1, 0)[\/latex] reflects to\u00a0[latex](0, 1)[\/latex] and the point\u00a0[latex](4, 2)[\/latex] reflects to\u00a0[latex](2, 4).[\/latex] Sketching the inverse on the same axes as the original graph gives Figure 10.<\/p>\n<figure id=\"attachment_637\" aria-describedby=\"caption-attachment-637\" style=\"width: 288px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-637\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-10-288x300.jpeg\" alt=\"\" width=\"288\" height=\"300\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-10-288x300.jpeg 288w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-10-65x68.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-10-225x234.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-10-350x365.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-10.jpeg 357w\" sizes=\"auto, (max-width: 288px) 100vw, 288px\" \/><figcaption id=\"caption-attachment-637\" class=\"wp-caption-text\">Figure 10. The function and its inverse, showing reflection about the identity line<\/figcaption><\/figure>\n<\/details>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Try It #9<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Draw graphs of the functions\u00a0[latex]f[\/latex] and\u00a0[latex]f^{-1}[\/latex] from Example 8.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Q&amp;A<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>Q: Is there any function that is equal to its own inverse?<\/strong><\/p>\n<p><em>A:\u00a0Yes. If\u00a0[latex]f=f^{-1},[\/latex] then\u00a0[latex]f(f(x))=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\u00a0[latex]f(x)=c-x[\/latex] where\u00a0[latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Media<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Access these online resources for additional instruction and practice with inverse functions.<\/p>\n<ul>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=qgezKpQYH2w\">Inverse Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=QFOJmevha_Y&amp;feature=youtu.be\/\">One-to-One Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=FIF8SdZkJc8\">Inverse Function Values Using Graph<\/a><\/li>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=rsJ14O5-KDw\">Restricting the Domain and Finding the Inverse<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"os-eos os-section-exercises-container\" data-uuid-key=\".section-exercises\">\n<h2 data-type=\"document-title\" data-rex-keep=\"true\"><span class=\"os-text\">3.7 Section Exercises<\/span><\/h2>\n<section id=\"fs-id1165137871042\" class=\"section-exercises\" data-depth=\"1\">\n<section id=\"fs-id1165135187563\" data-depth=\"2\">\n<h3 data-type=\"title\">Verbal<\/h3>\n<div id=\"fs-id1165137407341\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135193086\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137407341-solution\">1<\/a><span class=\"os-divider\">. <\/span>Describe why the horizontal line test is an effective way to determine whether a function is one-to-one?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137408636\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137408638\" data-type=\"problem\">\n<p><span class=\"os-number\">2<\/span><span class=\"os-divider\">. <\/span>Why do we restrict the domain of the function\u00a0[latex]f(x)=x^2[\/latex] to find the function\u2019s inverse?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137389621\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135400199\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137389621-solution\">3<\/a><span class=\"os-divider\">. <\/span>Can a function be its own inverse? Explain.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135188794\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137564806\" data-type=\"problem\">\n<p><span class=\"os-number\">4<\/span><span class=\"os-divider\">. <\/span>Are one-to-one functions either always increasing or always decreasing? Why or why not?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137419050\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137932403\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137419050-solution\">5<\/a><span class=\"os-divider\">. <\/span>How do you find the inverse of a function algebraically?<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137836714\" data-depth=\"2\">\n<h3 data-type=\"title\">Algebraic<\/h3>\n<div id=\"fs-id1165137422830\" class=\"material-set-2\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137806758\" data-type=\"problem\">\n<p><span class=\"os-number\">6<\/span><span class=\"os-divider\">. <\/span>Show that the function [latex]f(x)=a-x[\/latex] is its own inverse for all real numbers [latex]a.[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165137469451\">For the following exercises, find\u00a0[latex]f^{-1}(x)[\/latex] for each function.<\/p>\n<div id=\"fs-id1165134312158\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134312161\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165134312158-solution\">7<\/a><span class=\"os-divider\">.<\/span> [latex]f(x)=x+3[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137634366\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137634368\" data-type=\"problem\">\n<p><span class=\"os-number\">8<\/span><span class=\"os-divider\">. <\/span> [latex]f(x)=x+5[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137706135\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137422592\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137706135-solution\">9<\/a><span class=\"os-divider\">. <\/span> [latex]f(x)=2-x[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137561652\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137653456\" data-type=\"problem\">\n<p><span class=\"os-number\">10<\/span><span class=\"os-divider\">. <\/span> [latex]f(x)=3-x[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137600416\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135198605\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137600416-solution\">11<\/a><span class=\"os-divider\">. <\/span> [latex]f(x)=\\frac{x}{x+2}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137541591\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134043731\" data-type=\"problem\">\n<p><span class=\"os-number\">12<\/span><span class=\"os-divider\">. <\/span> [latex]f(x)=\\frac{2x+3}{5x+4}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165137410615\">For the following exercises, find a domain on which each function\u00a0[latex]f[\/latex] is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of\u00a0[latex]f[\/latex] restricted to that domain.<\/p>\n<div id=\"fs-id1165134148519\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165134148521\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165134148519-solution\">13<\/a><span class=\"os-divider\">. <\/span> [latex]f(x)=(x+7)^2[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137862703\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137531119\" data-type=\"problem\">\n<p><span class=\"os-number\">14<\/span><span class=\"os-divider\">. <\/span> [latex]f(x)=(x-60^2[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137938698\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137938700\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137938698-solution\">15<\/a><span class=\"os-divider\">. <\/span> [latex]f(x)=x^2-5[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137734966\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137734969\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137734966-solution\">16<\/a><span class=\"os-divider\">. <\/span>Given\u00a0[latex]f(x)=\\frac{x}{2+x}[\/latex] and [latex]g(x)=\\frac{2x}{1-x}:[\/latex]<\/p>\n<p>(a) Find\u00a0[latex]f(g(x))[\/latex] and [latex]g(f(x)).[\/latex]<\/p>\n<div class=\"os-problem-container\">\n<p>(b) What does the answer tell us about the relationship between\u00a0[latex]f(x)[\/latex] and [latex]g(x)?[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165137444427\">For the following exercises, use function composition to verify that\u00a0[latex]f(x)[\/latex] and\u00a0[latex]g(x)[\/latex] are inverse functions.<\/p>\n<div id=\"fs-id1165137437578\" class=\"material-set-2 os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137619341\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137437578-solution\">17<\/a><span class=\"os-divider\">. <\/span>[latex]f(x)=\\sqrt[3]{x-1} \\ \\ \\text{and} \\ \\ g(x)=x^3+1[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135538749\" class=\"material-set-2\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135538751\" data-type=\"problem\">\n<p><span class=\"os-number\">18<\/span><span class=\"os-divider\">. <\/span>[latex]f(x)=-3x+5 \\ \\ \\text{and} \\ \\ g(x)=\\frac{x-5}{-3}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<section id=\"fs-id1165135188614\" data-depth=\"2\">\n<h3 data-type=\"title\">Graphical<\/h3>\n<p id=\"fs-id1165135176520\">For the following exercises, use a graphing utility to determine whether each function is one-to-one.<\/p>\n<div id=\"fs-id1165137645254\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135436660\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137645254-solution\">19<\/a><span class=\"os-divider\">. <\/span> [latex]f(x)=\\sqrt{x}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135173420\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135173423\" data-type=\"problem\">\n<p><span class=\"os-number\">20<\/span><span class=\"os-divider\">. <\/span> [latex]f(x)=\\sqrt[3]{3x+1}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135424683\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137408415\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135424683-solution\">21<\/a><span class=\"os-divider\">. <\/span> [latex]f(x)=-5x+1[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137704606\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137704608\" data-type=\"problem\">\n<p><span class=\"os-number\">22<\/span><span class=\"os-divider\">. <\/span> [latex]f(x)=x^3-27[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165137528556\">For the following exercises, determine whether the graph represents a one-to-one function.<\/p>\n<div id=\"fs-id1165137528559\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135251340\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137528559-solution\">23<\/a><span class=\"os-divider\">. <\/span><\/p>\n<div class=\"os-problem-container has-first-element\"><span id=\"fs-id1165135341390\" class=\"first-element\" data-type=\"media\" data-alt=\"Graph of a parabola.\" data-display=\"block\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-638\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.23.jpeg\" alt=\"\" width=\"252\" height=\"253\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.23.jpeg 252w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.23-150x150.jpeg 150w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.23-65x65.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.23-225x226.jpeg 225w\" sizes=\"auto, (max-width: 252px) 100vw, 252px\" \/><br \/>\n<\/span><\/div>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135192971\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135192973\" data-type=\"problem\">\n<p><span class=\"os-number\">24<\/span><span class=\"os-divider\">. <\/span><\/p>\n<div class=\"os-problem-container has-first-element\">\n<p><span id=\"fs-id1165134379457\" class=\"first-element\" data-type=\"media\" data-alt=\"Graph of a step-function.\" data-display=\"block\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-639\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.24-300x300.jpeg\" alt=\"\" width=\"300\" height=\"300\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.24-300x300.jpeg 300w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.24-150x150.jpeg 150w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.24-65x65.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.24-225x226.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.24-350x351.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7.24.jpeg 375w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\n<\/span><\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165137849556\">For the following exercises, use the graph of\u00a0[latex]f[\/latex] shown in Figure 11.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_640\" aria-describedby=\"caption-attachment-640\" style=\"width: 288px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-640\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-11-288x300.jpeg\" alt=\"\" width=\"288\" height=\"300\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-11-288x300.jpeg 288w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-11-65x68.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-11-225x234.jpeg 225w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-11-350x365.jpeg 350w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-11.jpeg 357w\" sizes=\"auto, (max-width: 288px) 100vw, 288px\" \/><figcaption id=\"caption-attachment-640\" class=\"wp-caption-text\">Figure 11<\/figcaption><\/figure>\n<div id=\"fs-id1165137863913\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137854842\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137863913-solution\">25<\/a><span class=\"os-divider\">. <\/span>Find- [latex]f(0).[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137417814\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137417816\" data-type=\"problem\">\n<p><span class=\"os-number\">26<\/span><span class=\"os-divider\">. <\/span>Solve [latex]f(x)=0.[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165133093360\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165133093363\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165133093360-solution\">27<\/a><span class=\"os-divider\">. <\/span>Find [latex]f^{-1}(0).[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137451393\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137806036\" data-type=\"problem\">\n<p><span class=\"os-number\">28<\/span><span class=\"os-divider\">. <\/span>Solve [latex]f^{-1}(x)=0.[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165135639868\">For the following exercises, use the graph of the one-to-one function shown in Figure 12.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_641\" aria-describedby=\"caption-attachment-641\" style=\"width: 255px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-641\" src=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-12.jpeg\" alt=\"\" width=\"255\" height=\"254\" srcset=\"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-12.jpeg 255w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-12-150x150.jpeg 150w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-12-65x65.jpeg 65w, https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-content\/uploads\/sites\/234\/2025\/04\/3.7-fig-12-225x224.jpeg 225w\" sizes=\"auto, (max-width: 255px) 100vw, 255px\" \/><figcaption id=\"caption-attachment-641\" class=\"wp-caption-text\">Figure 12<\/figcaption><\/figure>\n<div id=\"fs-id1165137884386\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137812123\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137884386-solution\">29<\/a><span class=\"os-divider\">. <\/span>Sketch the graph of [latex]f^{-1}.[\/latex]<\/p>\n<div class=\"os-problem-container\"><\/div>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137892244\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137603262\" data-type=\"problem\">\n<p><span class=\"os-number\">30<\/span><span class=\"os-divider\">. <\/span>Find\u00a0[latex]f(6) \\ \\ \\text{and} \\ \\ f^{-1}(2).[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137827763\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137827765\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137827763-solution\">31<\/a><span class=\"os-divider\">. <\/span>If the complete graph of\u00a0[latex]f[\/latex] is shown, find the domain of [latex]f.[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137644582\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137644584\" data-type=\"problem\">\n<p><span class=\"os-number\">32<\/span><span class=\"os-divider\">. <\/span>If the complete graph of\u00a0[latex]f[\/latex] is shown, find the range of [latex]f.[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<section id=\"fs-id1165135434734\" data-depth=\"2\">\n<h3 data-type=\"title\">Numeric<\/h3>\n<p id=\"fs-id1165137781583\">For the following exercises, evaluate or solve, assuming that the function\u00a0[latex]f[\/latex] is one-to-one.<\/p>\n<div id=\"fs-id1165137771148\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135250622\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137771148-solution\">33<\/a><span class=\"os-divider\">. <\/span>If\u00a0[latex]f(6)=7, \\ \\ \\text{find} \\ \\ f^{-1}(7).[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137805757\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135190808\" data-type=\"problem\">\n<p><span class=\"os-number\">34<\/span><span class=\"os-divider\">. <\/span>If [latex]f(3)=2, \\ \\ \\text{find} \\ \\ f^{-1}(2).[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137640685\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137640687\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137640685-solution\">35<\/a><span class=\"os-divider\">. <\/span>If\u00a0[latex]f^{-1}(-4)=-8, \\ \\ \\text{find} \\ \\ f(-8).[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137404972\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137404974\" data-type=\"problem\">\n<p><span class=\"os-number\">36<\/span><span class=\"os-divider\">. <\/span>If\u00a0[latex]f^{-1}(-2)=-1, \\ \\ \\text{find} \\ \\ f(-1).[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1165135195398\">For the following exercises, use the values listed in Table 6 to evaluate or solve.<\/p>\n<div id=\"eip-151\" class=\"os-table\">\n<table class=\"grid\" data-id=\"eip-151\">\n<caption>Table 6<\/caption>\n<tbody>\n<tr>\n<td style=\"width: 109px;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 60px;\">0<\/td>\n<td style=\"width: 60px;\">1<\/td>\n<td style=\"width: 57px;\">2<\/td>\n<td style=\"width: 57px;\">3<\/td>\n<td style=\"width: 57px;\">4<\/td>\n<td style=\"width: 58px;\">5<\/td>\n<td style=\"width: 58px;\">6<\/td>\n<td style=\"width: 57px;\">7<\/td>\n<td style=\"width: 58px;\">8<\/td>\n<td style=\"width: 58px;\">9<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 109px;\" data-align=\"center\">[latex]f(x)[\/latex]<\/td>\n<td style=\"width: 60px;\">8<\/td>\n<td style=\"width: 60px;\">0<\/td>\n<td style=\"width: 57px;\">7<\/td>\n<td style=\"width: 57px;\">4<\/td>\n<td style=\"width: 57px;\">2<\/td>\n<td style=\"width: 58px;\">6<\/td>\n<td style=\"width: 58px;\">5<\/td>\n<td style=\"width: 57px;\">3<\/td>\n<td style=\"width: 58px;\">9<\/td>\n<td style=\"width: 58px;\">1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"os-caption-container\"><\/div>\n<\/div>\n<div id=\"fs-id1165137400581\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137400581-solution\">37<\/a><span class=\"os-divider\">.<\/span> Find [latex]f(1).[\/latex]<\/header>\n<\/div>\n<div id=\"fs-id1165137807065\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137807068\" data-type=\"problem\">\n<p><span class=\"os-number\">38<\/span><span class=\"os-divider\">. <\/span>Solve [latex]f(x)=3.[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137736914\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137736916\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165137736914-solution\">39<\/a><span class=\"os-divider\">. <\/span>Find [latex]f^{-1}(0).[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137686729\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137686731\" data-type=\"problem\">\n<p><span class=\"os-number\">40<\/span><span class=\"os-divider\">. <\/span>Solve [latex]f^{-1}(x)=7.[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135341429\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135209686\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135341429-solution\">41<\/a><span class=\"os-divider\">. <\/span>Use the tabular representation of\u00a0[latex]f[\/latex] in Table 7 to create a table for [latex]f^{-1}(x).[\/latex]<\/p>\n<div class=\"os-problem-container\">\n<div id=\"Table_01_07_08\" class=\"os-table\">\n<table class=\"grid\" data-id=\"Table_01_07_08\">\n<caption>Table 7<\/caption>\n<tbody>\n<tr>\n<td data-align=\"center\">[latex]x[\/latex]<\/td>\n<td data-align=\"center\">3<\/td>\n<td data-align=\"center\">6<\/td>\n<td data-align=\"center\">9<\/td>\n<td data-align=\"center\">13<\/td>\n<td data-align=\"center\">14<\/td>\n<\/tr>\n<tr>\n<td data-align=\"center\">[latex]f(x)[\/latex]<\/td>\n<td data-align=\"center\">1<\/td>\n<td data-align=\"center\">4<\/td>\n<td data-align=\"center\">7<\/td>\n<td data-align=\"center\">12<\/td>\n<td data-align=\"center\">16<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3 class=\"os-caption-container\">Technology<\/h3>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137641552\" data-depth=\"2\">\n<p id=\"fs-id1165137401884\">For the following exercises, find the inverse function. Then, graph the function and its inverse.<\/p>\n<div id=\"fs-id1165137401888\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137470994\" data-type=\"problem\">\n<p><span class=\"os-number\">42<\/span><span class=\"os-divider\">. <\/span> [latex]f(x)=\\frac{3}{x-2}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135435667\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135435669\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135435667-solution\">43<\/a><span class=\"os-divider\">. <\/span> [latex]f(x)=x^3-1[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137419353\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135209709\" data-type=\"problem\">\n<p><span class=\"os-number\">44<\/span><span class=\"os-divider\">. <\/span>Find the inverse function of\u00a0[latex]f(x)=\\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<\/div>\n<\/section>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137592239\" data-depth=\"2\">\n<h3 data-type=\"title\">Real-World Applications<\/h3>\n<div id=\"fs-id1165135512365\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135512368\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135512365-solution\">45<\/a><span class=\"os-divider\">. <\/span>To convert from\u00a0[latex]x[\/latex] degrees Celsius to\u00a0[latex]y[\/latex] degrees Fahrenheit, we use the formula\u00a0[latex]f(x)=\\frac{9}{5}x+32.[\/latex] Find the inverse function, if it exists, and explain its meaning.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165137462831\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165137462833\" data-type=\"problem\">\n<p><span class=\"os-number\">46<\/span><span class=\"os-divider\">. <\/span>The circumference\u00a0[latex]C[\/latex] of a circle is a function of its radius given by\u00a0[latex]C(r)=2\\pi r.[\/latex] Express the radius of a circle as a function of its circumference. Call this function\u00a0[latex]r(c).[\/latex] Find\u00a0[latex]r(36\\pi)[\/latex] and interpret its meaning.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-id1165135532474\" class=\"os-hasSolution\" data-type=\"exercise\">\n<header><\/header>\n<section>\n<div id=\"fs-id1165135532476\" data-type=\"problem\">\n<p><a class=\"os-number\" href=\"chapter-3\" data-page-slug=\"chapter-3\" data-page-uuid=\"9b8c8027-99b3-55ba-86c1-ba45e9ff3eff\" data-page-fragment=\"fs-id1165135532474-solution\">47<\/a><span class=\"os-divider\">. <\/span>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(t)=50t.[\/latex] Find the inverse function by ex4ressing the time of travel in terms of the distance traveled. Call this function [latex]t(d).[\/latex] Find [latex]t(180).[\/latex] and interpret its meaning.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/section>\n<\/div>\n","protected":false},"author":158,"menu_order":7,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-138","chapter","type-chapter","status-publish","hentry"],"part":105,"_links":{"self":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters\/138","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/users\/158"}],"version-history":[{"count":27,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters\/138\/revisions"}],"predecessor-version":[{"id":625,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters\/138\/revisions\/625"}],"part":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/parts\/105"}],"metadata":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapters\/138\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/media?parent=138"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=138"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/contributor?post=138"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/ccacollegealgebra\/wp-json\/wp\/v2\/license?post=138"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}