{"id":13771,"date":"2018-08-24T18:08:16","date_gmt":"2018-08-24T18:08:16","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalcone\/?post_type=chapter&#038;p=13771"},"modified":"2020-11-22T19:57:33","modified_gmt":"2020-11-22T19:57:33","slug":"inverse-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/chapter\/inverse-functions\/","title":{"raw":"Walkthrough of Unit 5: Inverse Functions","rendered":"Walkthrough of Unit 5: Inverse Functions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\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<p id=\"fs-id1165135358875\">A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.<\/p>\r\nIf some physical machines can run in two directions, we might ask whether some of the function \"machines\" we have been studying can also run backwards. Figure 1\u00a0provides a visual representation of this question. In this section, we will consider the reverse nature of functions.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010622\/CNX_Precalc_Figure_01_07_0012.jpg\" alt=\"Diagram of a function and what would be its inverse.\" width=\"731\" height=\"305\" \/> <b>Figure 1.<\/b> Can a function \"machine\" operate in reverse?[\/caption]\r\n<h2>Verifying That Two Functions Are Inverse Functions<\/h2>\r\n<p id=\"fs-id1165135705795\">Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the <strong>Celsius<\/strong> scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees <strong>Fahrenheit<\/strong> to degrees Celsius. She finds the formula<\/p>\r\n\r\n<div id=\"fs-id1165137807176\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165135433486\">and substitutes 75 for [latex]F[\/latex] to calculate<\/p>\r\n\r\n<div id=\"fs-id1165137911210\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\frac{5}{9}\\left(75 - 32\\right)\\approx {24}^{ \\circ} {C}[\/latex].<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010623\/CNX_Precalc_Figure_01_07_0022.jpg\" alt=\"A forecast of Monday\u2019s through Thursday\u2019s weather.\" width=\"731\" height=\"226\" \/> <b>Figure 2<\/b>[\/caption]\r\n<p id=\"fs-id1165137409312\">Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week\u2019s weather forecast\u00a0for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.<span id=\"fs-id1165137414400\">\r\n<\/span><\/p>\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 [latex]F[\/latex] after substituting a value for [latex]C[\/latex]. For example, to convert 26 degrees Celsius, she could write<\/p>\r\n\r\n<div id=\"fs-id1165135548255\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{gathered}26=\\frac{5}{9}\\left(F - 32\\right) \\\\ 26\\cdot \\frac{9}{5}=F - 32 \\\\ F=26\\cdot \\frac{9}{5}+32\\approx 79 \\end{gathered}[\/latex]<\/div>\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 [latex]f\\left(x\\right)[\/latex], we represent its inverse as [latex]{f}^{-1}\\left(x\\right)[\/latex], read as [latex]\"f[\/latex] inverse of [latex]x.\\text{\"}[\/latex] The raised [latex]-1[\/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[\/latex] . In other words, [latex]{f}^{-1}\\left(x\\right)[\/latex] does <em>not<\/em> mean [latex]\\frac{1}{f\\left(x\\right)}[\/latex] because [latex]\\frac{1}{f\\left(x\\right)}[\/latex] is the reciprocal of [latex]f[\/latex] and not the inverse.<\/p>\r\n<p id=\"fs-id1165137724926\">The \"exponent-like\" notation comes from an analogy between function composition and multiplication: just as [latex]{a}^{-1}a=1[\/latex] (1 is the identity element for multiplication) for any nonzero number [latex]a[\/latex], so [latex]{f}^{-1}\\circ f[\/latex] equals the identity function, that is,<\/p>\r\n\r\n<div id=\"fs-id1165134302408\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(y\\right)=x[\/latex]<\/div>\r\n<p id=\"fs-id1165135667832\">This holds for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. Informally, this means that inverse functions \"undo\" each other. However, just as zero does not have a <strong>reciprocal<\/strong>, some functions do not have inverses.<\/p>\r\n<p id=\"fs-id1165137655153\">Given a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether either [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] or [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)<\/p>\r\n<p id=\"fs-id1165135397975\">For example, [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.<\/p>\r\n\r\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/div>\r\n<p id=\"fs-id1165137767233\">and<\/p>\r\n\r\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left({f}^{}\\circ {f}^{-1}\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/div>\r\n<p id=\"fs-id1165137438777\">A few coordinate pairs from the graph of the function [latex]y=4x[\/latex] are (\u22122, \u22128), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function [latex]y=\\frac{1}{4}x[\/latex] are (\u22128, \u22122), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.<\/p>\r\n\r\n<div id=\"fs-id1165137933105\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Inverse Function<\/h3>\r\n<p id=\"fs-id1165137473076\">For any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex]. This can also be written as [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. It also follows that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex] if [latex]{f}^{-1}[\/latex] is the inverse of [latex]f[\/latex].<\/p>\r\n<p id=\"fs-id1165137444821\">The notation [latex]{f}^{-1}[\/latex] is read [latex]\\text{\"}f[\/latex] inverse.\" Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x.\"[\/latex]\r\nKeep in mind that<\/p>\r\n<p style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)\\ne \\frac{1}{f\\left(x\\right)}[\/latex]<\/p>\r\n<p id=\"fs-id1165135194095\">and not all functions have inverses.<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_01_07_01\" class=\"example\">\r\n<div id=\"fs-id1165137656641\" class=\"exercise\">\r\n<div id=\"fs-id1165137922642\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Identifying an Inverse Function for a Given Input-Output Pair<\/h3>\r\n<p id=\"fs-id1165137659325\">If for a particular one-to-one function [latex]f\\left(2\\right)=4[\/latex] and [latex]f\\left(5\\right)=12[\/latex], what are the corresponding input and output values for the inverse function?<\/p>\r\n[reveal-answer q=\"177916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"177916\"]\r\n<p id=\"fs-id1165137737081\">The inverse function reverses the input and output quantities, so if [latex]f\\left(2\\right)=4 [\/latex], then [latex]{f}^{-1}\\left(4\\right)=2[\/latex] and if [latex]f\\left(5\\right)=12[\/latex], then [latex] {f}^{-1}\\left(12\\right)=5[\/latex].<\/p>\r\n<p id=\"fs-id1165137659464\">Alternatively, if we want to name the inverse function [latex]g[\/latex], then [latex]g\\left(4\\right)=2[\/latex] and [latex]g\\left(12\\right)=5[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135508518\">Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.<\/p>\r\n\r\n<table id=\"Table_01_07_01\" style=\"width: 295px\" summary=\"For (x,f(x)) we have the values (2, 4) and (5, 12); for (x, g(x)), we have the values (4, 2) and (12, 5).\">\r\n<thead>\r\n<tr>\r\n<th style=\"width: 155px\">[latex]\\left(x,f\\left(x\\right)\\right)[\/latex]<\/th>\r\n<th style=\"width: 140px\">[latex]\\left(x,g\\left(x\\right)\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 155px\">[latex]\\left(2,4\\right)[\/latex]<\/td>\r\n<td style=\"width: 140px\">[latex]\\left(4,2\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 155px\">[latex]\\left(5,12\\right)[\/latex]<\/td>\r\n<td style=\"width: 140px\">[latex]\\left(12,5\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137659089\">Given that [latex]{h}^{-1}\\left(6\\right)=2[\/latex], what are the corresponding input and output values of the original function [latex]h?[\/latex]<\/p>\r\n[reveal-answer q=\"749671\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"749671\"]\r\n\r\n[latex]h\\left(2\\right)=6[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/TSztRfzmk0M\r\n<div id=\"fs-id1165134357354\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135434077\">How To: Given two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex], test whether the functions are inverses of each other.<\/h3>\r\n<ol id=\"fs-id1165137452358\">\r\n \t<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] or [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\r\n \t<li>If both statements are true, then [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_07_02\" class=\"example\">\r\n<div id=\"fs-id1165137557051\" class=\"exercise\">\r\n<div id=\"fs-id1165137679032\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Testing Inverse Relationships Algebraically<\/h3>\r\n<p id=\"fs-id1165135519417\">If [latex]f\\left(x\\right)=\\frac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\frac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\r\n[reveal-answer q=\"619704\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"619704\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align} g\\left(f\\left(x\\right)\\right)&amp;=\\frac{1}{\\left(\\frac{1}{x+2}\\right)}{-2 } \\\\ &amp;= x+2 - 2 \\\\ &amp;= x \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137611481\">and<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)&amp;=\\frac{1}{\\frac{1}{x}-2+2}\\\\ &amp;=\\frac{1}{\\frac{1}{x}} \\\\ &amp;=x \\end{align}[\/latex]<\/p>\r\nSo\r\n<p style=\"text-align: center\">[latex]g={f}^{-1}\\text{ and }f={g}^{-1}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135389000\">Notice the inverse operations are in reverse order of the operations from the original function.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135160550\">If [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\r\n[reveal-answer q=\"196142\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"196142\"]\r\n\r\nYes\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_01_07_03\" class=\"example\">\r\n<div id=\"fs-id1165135259560\" class=\"exercise\">\r\n<div id=\"fs-id1165134042918\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Determining Inverse Relationships for Power Functions<\/h3>\r\n<p id=\"fs-id1165137441834\">If [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\r\n[reveal-answer q=\"506\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"506\"]\r\n<p style=\"text-align: center\">[latex]f\\left(g\\left(x\\right)\\right)=\\frac{{x}^{3}}{27}\\ne x[\/latex]<\/p>\r\n<p id=\"fs-id1165137694053\">No, the functions are not inverses.<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165134192978\">The correct inverse to the cube is the cube root [latex]\\sqrt[\\leftroot{-1}\\uproot{2}3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137573532\">If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}\\text{and}g\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\r\n[reveal-answer q=\"746859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"746859\"]\r\n\r\nYes\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]166520[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Finding Domain and Range of Inverse Functions<\/h2>\r\nThe outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010624\/CNX_Precalc_Figure_01_07_0032.jpg\" alt=\"Domain and range of a function and its inverse.\" width=\"487\" height=\"143\" \/> <b>Figure 3.<\/b> Domain and range of a function and its inverse[\/caption]\r\n<p id=\"fs-id1165135557891\">When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of [latex]f\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}[\/latex] is [latex]{f}^{-1}\\left(x\\right)={x}^{2}[\/latex], because a square \"undoes\" a square root; but the square is only the inverse of the square root on the domain [latex]\\left[0,\\infty \\right)[\/latex], since that is the range of [latex]f\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}[\/latex].<\/p>\r\n<p id=\"fs-id1165137730185\">We can look at this problem from the other side, starting with the square (toolkit quadratic) function [latex]f\\left(x\\right)={x}^{2}[\/latex]. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and \u20133. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the \"inverse\" is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.<\/p>\r\n<p id=\"fs-id1165137823552\">In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).<\/p>\r\n<p id=\"fs-id1165132037000\">If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}[\/latex] on [latex]\\left[1,\\infty \\right)[\/latex], then the inverse function is [latex]{f}^{-1}\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}+1[\/latex].<\/p>\r\n\r\n<ul id=\"fs-id1165137851227\">\r\n \t<li>The domain of [latex]f[\/latex] = range of [latex]{f}^{-1}[\/latex] = [latex]\\left[1,\\infty \\right)[\/latex].<\/li>\r\n \t<li>The domain of [latex]{f}^{-1}[\/latex] = range of [latex]f[\/latex] = [latex]\\left[0,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1165137733804\" class=\"note precalculus qa textbox\">\r\n<p id=\"fs-id1165137723526\"><strong>Q &amp; A<\/strong><\/p>\r\n<strong>Is it possible for a function to have more than one inverse?<\/strong>\r\n<p id=\"fs-id1165137456608\"><em>No. If two supposedly different functions, say, [latex]g[\/latex] and [latex]h[\/latex], both meet the definition of being inverses of another function [latex]f[\/latex], then you can prove that [latex]g=h[\/latex]. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137704938\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Domain and Range of Inverse Functions<\/h3>\r\n<p id=\"fs-id1165135319550\">The range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165137673886\">The domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135308785\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137605040\"><strong>How To: Given a function, find the domain and range of its inverse.\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id1165137530434\">\r\n \t<li>If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.<\/li>\r\n \t<li>If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_07_05\" class=\"example\">\r\n<div id=\"fs-id1165137667922\" class=\"exercise\">\r\n<div id=\"fs-id1165135511321\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Finding the Inverses of Toolkit Functions<\/h3>\r\n<p id=\"fs-id1165137448020\">Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed below. We restrict the domain in such a fashion that the function assumes all <em>y<\/em>-values exactly once.<\/p>\r\n\r\n<table id=\"Table_01_07_02\" summary=\"A list of the toolkit function. The constant function is f(x) = c where c is the constant; the identity function is f(x) = x; the absolute function is f(x)=|x|; the quadratic function is f(x) = x^2; the cubic function is f(x)=x^3; the reciprocal function is f(x)=1\/x; the reciprocal squared function is f(x)=1\/x^2; the square root function is f(x)=sqrt(x); the cube root function is f(x) = x^(1\/3).\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>Constant<\/td>\r\n<td>Identity<\/td>\r\n<td>Quadratic<\/td>\r\n<td>Cubic<\/td>\r\n<td>Reciprocal<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(x\\right)=c[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=x[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)={x}^{2}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)={x}^{3}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Reciprocal squared<\/td>\r\n<td>Cube root<\/td>\r\n<td>Square root<\/td>\r\n<td>Absolute value<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}3]{x}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=|x|[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"552567\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"552567\"]\r\n<p id=\"fs-id1165132988445\">The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.<\/p>\r\n<p id=\"fs-id1165134080947\">The absolute value function can be restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], where it is equal to the identity function.<\/p>\r\n<p id=\"fs-id1165137642849\">The reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\r\nWe can see that these functions (if unrestricted) are not one-to-one by looking at their graphs.\u00a0They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010624\/CNX_Precalc_Figure_01_07_004ab2.jpg\" alt=\"Graph of an absolute function.\" width=\"975\" height=\"404\" \/> <b>Figure 4.<\/b> (a) Absolute value (b) Reciprocal squared[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137507853\">The domain of function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of function [latex]f[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex]. Find the domain and range of the inverse function.<\/p>\r\n[reveal-answer q=\"333382\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"333382\"]\r\n\r\nThe domain of function [latex]{f}^{-1}[\/latex] is [latex]\\left(-\\infty \\text{,}-2\\right)[\/latex] and the range of function [latex]{f}^{-1}[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding and Evaluating Inverse Functions<\/h2>\r\n<p id=\"fs-id1165137761017\">Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.<\/p>\r\n\r\n<section id=\"fs-id1165135466392\">\r\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Inverting Tabular Functions<\/span><\/h2>\r\n<p id=\"fs-id1165135190714\">Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.<\/p>\r\n<p id=\"fs-id1165137422578\">Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.<\/p>\r\n\r\n<div id=\"Example_01_07_06\" class=\"example\">\r\n<div id=\"fs-id1165135544995\" class=\"exercise\">\r\n<div id=\"fs-id1165137698262\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Interpreting the Inverse of a Tabular Function<\/h3>\r\n<p id=\"fs-id1165135435474\">A function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes. Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].<\/p>\r\n\r\n<table id=\"Table_01_07_03\" summary=\"Two rows and five columns. The first row is labeled \"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\r\n<td>30<\/td>\r\n<td>50<\/td>\r\n<td>70<\/td>\r\n<td>90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\r\n<td>20<\/td>\r\n<td>40<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"736190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736190\"]\r\n<p id=\"fs-id1165137640334\">The inverse function takes an output of [latex]f[\/latex] and returns an input for [latex]f[\/latex]. So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.<\/p>\r\n<p id=\"fs-id1165135181841\">Alternatively, recall that the definition of the inverse was that if [latex]f\\left(a\\right)=b[\/latex], then [latex]{f}^{-1}\\left(b\\right)=a[\/latex]. By this definition, if we are given [latex]{f}^{-1}\\left(70\\right)=a[\/latex], then we are looking for a value [latex]a[\/latex] so that [latex]f\\left(a\\right)=70[\/latex]. In this case, we are looking for a [latex]t[\/latex] so that [latex]f\\left(t\\right)=70[\/latex], which is when [latex]t=90[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134108483\">Using the table below,\u00a0find and interpret (a) [latex]\\text{ }f\\left(60\\right)[\/latex], and (b) [latex]\\text{ }{f}^{-1}\\left(60\\right)[\/latex].<\/p>\r\n\r\n<table id=\"Table_01_07_04\" summary=\"Two rows and five columns. The first row is labeled \"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>[latex]t\\text{ (minutes)}[\/latex]<\/td>\r\n<td>30<\/td>\r\n<td>50<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<td>90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex]<\/td>\r\n<td>20<\/td>\r\n<td>40<\/td>\r\n<td>50<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"245501\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"245501\"]\r\n\r\na.\u00a0[latex]f\\left(60\\right)=50[\/latex]. In 60 minutes, 50 miles are traveled.\r\n\r\nb. [latex]{f}^{-1}\\left(60\\right)=70[\/latex]. To travel 60 miles, it will take 70 minutes.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137418615\">\r\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/span><\/h2>\r\n<p id=\"fs-id1165137400045\">We saw in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/precalcone\/chapter\/introduction\/\" target=\"_blank\" rel=\"noopener\">Functions and Function Notation<\/a> that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the <em>vertical<\/em> extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the <em>horizontal<\/em> extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function\u2019s graph.<\/p>\r\n\r\n<div id=\"fs-id1165133045388\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135333128\">How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\r\n<ol id=\"fs-id1165137464840\">\r\n \t<li>Find the desired input on the <em>y<\/em>-axis of the given graph.<\/li>\r\n \t<li>Read the inverse function\u2019s output from the <em>x<\/em>-axis of the given graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_07_07\" class=\"example\">\r\n<div id=\"fs-id1165135434803\" class=\"exercise\">\r\n<div id=\"fs-id1165135434805\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/h3>\r\nA function [latex]g\\left(x\\right)[\/latex] is given in Figure 5. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"56740\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"56740\"]\r\n<p id=\"fs-id1165137468842\">To evaluate [latex]g\\left(3\\right)[\/latex], we find 3 on the <em>x<\/em>-axis and find the corresponding output value on the <em>y<\/em>-axis. The point [latex]\\left(3,1\\right)[\/latex] tells us that [latex]g\\left(3\\right)=1[\/latex].<\/p>\r\nTo evaluate [latex]{g}^{-1}\\left(3\\right)[\/latex], recall that by definition [latex]{g}^{-1}\\left(3\\right)[\/latex] means the value of <em>x<\/em> for which [latex]g\\left(x\\right)=3[\/latex]. By looking for the output value 3 on the vertical axis, we find the point [latex]\\left(5,3\\right)[\/latex] on the graph, which means [latex]g\\left(5\\right)=3[\/latex], so by definition, [latex]{g}^{-1}\\left(3\\right)=5[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n[\/hidden-answer]<b><\/b>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137812560\">Using the graph in Example 6, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"307093\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"307093\"]\r\n\r\na. 3; b. 5.6\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137605437\">\r\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Finding Inverses of Functions Represented by Formulas<\/span><\/h2>\r\n<p id=\"fs-id1165137433184\">Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula\u2014 for example, [latex]y[\/latex] as a function of [latex]x\\text{-\\hspace{0.17em}}[\/latex] we can often find the inverse function by solving to obtain [latex]x[\/latex] as a function of [latex]y[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165137652548\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135195849\">How To: Given a function represented by a formula, find the inverse.<\/h3>\r\n<ol id=\"fs-id1165135443898\">\r\n \t<li>Make sure [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 [latex]x[\/latex] and [latex]y[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_07_08\" class=\"example\">\r\n<div id=\"fs-id1165135186316\" class=\"exercise\">\r\n<div id=\"fs-id1165135186318\" class=\"problem textbox shaded\">\r\n<h3>Example 7: Inverting the Fahrenheit-to-Celsius Function<\/h3>\r\n<p id=\"fs-id1165137596585\">Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.<\/p>\r\n\r\n<div id=\"fs-id1165133306998\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/div>\r\n<div>[reveal-answer q=\"386193\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"386193\"]\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} C =\\frac{5}{9}\\left(F - 32\\right) \\\\ C\\cdot \\frac{9}{5}=F - 32 \\\\ F=\\frac{9}{5}C+32 \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165137819987\">By solving in general, we have uncovered the inverse function. If<\/p>\r\n<p style=\"text-align: center\">[latex]C=h\\left(F\\right)=\\frac{5}{9}\\left(F - 32\\right)[\/latex],<\/p>\r\n<p id=\"fs-id1165135435603\">then<\/p>\r\n<p style=\"text-align: center\">[latex]F={h}^{-1}\\left(C\\right)=\\frac{9}{5}C+32[\/latex].<\/p>\r\n<p id=\"fs-id1165137573279\">In this case, we introduced a function [latex]h[\/latex] to represent the conversion because the input and output variables are descriptive, and writing [latex]{C}^{-1}[\/latex] could get confusing.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135563331\">Solve for [latex]x[\/latex] in terms of [latex]y[\/latex] given [latex]y=\\frac{1}{3}\\left(x - 5\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"320735\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"320735\"]\r\n\r\n[latex]x=3y+5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_01_07_09\" class=\"example\">\r\n<div id=\"fs-id1165134065146\" class=\"exercise\">\r\n<div id=\"fs-id1165137409366\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Solving to Find an Inverse Function<\/h3>\r\n<p id=\"fs-id1165137891504\">Find the inverse of the function [latex]f\\left(x\\right)=\\frac{2}{x - 3}+4[\/latex].<\/p>\r\n[reveal-answer q=\"91472\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"91472\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;y=\\frac{2}{x - 3}+4 &amp;&amp; \\text{Set up an equation}. \\\\ &amp;y - 4=\\frac{2}{x - 3} &amp;&amp; \\text{Subtract 4 from both sides}. \\\\ &amp;x - 3=\\frac{2}{y - 4} &amp;&amp; \\text{Multiply both sides by }x - 3\\text{ and divide by }y - 4. \\\\ &amp;x=\\frac{2}{y - 4}+3 &amp;&amp; \\text{Add 3 to both sides}. \\end{align}[\/latex]<\/p>\r\nSo [latex]{f}^{-1}\\left(y\\right)=\\frac{2}{y - 4}+3[\/latex] or [latex]{f}^{-1}\\left(x\\right)=\\frac{2}{x - 4}+3[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135394231\">The domain and range of [latex]f[\/latex] exclude the values 3 and 4, respectively. [latex]f[\/latex] and [latex]{f}^{-1}[\/latex] are equal at two points but are not the same function, as we can see by creating\u00a0the table below.<\/p>\r\n\r\n<table id=\"Table_01_07_05\" summary=\"The values of f(x) are: f(1)=3, f(2)=2, and f(5)=5. So f^(-1)(y)=y.\">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>5<\/td>\r\n<td>[latex]{f}^{-1}\\left(y\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\r\n<td>3<\/td>\r\n<td>2<\/td>\r\n<td>5<\/td>\r\n<td>[latex]y[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_01_07_10\" class=\"example\">\r\n<div id=\"fs-id1165137603677\" class=\"exercise\">\r\n<div id=\"fs-id1165137547656\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Solving to Find an Inverse with Radicals<\/h3>\r\n<p id=\"fs-id1165137841687\">Find the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt[\\leftroot{-1}\\uproot{2}]{x - 4}[\/latex].<\/p>\r\n[reveal-answer q=\"109918\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"109918\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;y=2+\\sqrt{x - 4} \\\\ &amp;{\\left(y - 2\\right)}^{2}=x - 4 \\\\ &amp;x={\\left(y - 2\\right)}^{2}+4 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135160183\">So [latex]{f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4[\/latex].<\/p>\r\nThe domain of [latex]f[\/latex] is [latex]\\left[4,\\infty \\right)[\/latex]. Notice that the range of [latex]f[\/latex] is [latex]\\left[2,\\infty \\right)[\/latex], so this means that the domain of the inverse function [latex]{f}^{-1}[\/latex] is also [latex]\\left[2,\\infty \\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135546050\">The formula we found for [latex]{f}^{-1}\\left(x\\right)[\/latex] looks like it would be valid for all real [latex]x[\/latex]. However, [latex]{f}^{-1}[\/latex] itself must have an inverse (namely, [latex]f[\/latex] ) so we have to restrict the domain of [latex]{f}^{-1}[\/latex] to [latex]\\left[2,\\infty \\right)[\/latex] in order to make [latex]{f}^{-1}[\/latex] a one-to-one function. This domain of [latex]{f}^{-1}[\/latex] is exactly the range of [latex]f[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137756074\">What is the inverse of the function [latex]f\\left(x\\right)=2-\\sqrt[\\leftroot{-1}\\uproot{2}]{x}?[\/latex] State the domains of both the function and the inverse function.<\/p>\r\n[reveal-answer q=\"479951\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"479951\"]\r\n\r\n[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2}[\/latex]; domain of [latex]f:\\left[0,\\infty \\right)[\/latex]; domain of [latex]{f}^{-1}:\\left(-\\infty ,2\\right][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]166546[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nNow that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as in Figure 7.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/> <b>Figure 7.<\/b> Quadratic function with domain restricted to [0, \u221e).[\/caption]\r\n<p id=\"fs-id1165137419977\"><strong>Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.<\/p>\r\n<p id=\"fs-id1165137656093\">We already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]<\/p>\r\nWe notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown in Figure 8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0092.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"251\" \/> <b>Figure 8.<\/b> Square and square-root functions on the non-negative domain[\/caption]\r\n<p id=\"fs-id1165137393212\">This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.<\/p>\r\n\r\n<div id=\"Example_01_07_11\" class=\"example\">\r\n<div id=\"fs-id1165134430460\" class=\"exercise\">\r\n<div id=\"fs-id1165134430463\" class=\"problem textbox shaded\">\r\n<h3>Example 10: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\r\nGiven the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"990585\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"990585\"]\r\n<p id=\"fs-id1165137407660\">This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\r\nIf we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in Figure 10.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0112.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"363\" \/> <b>Figure 10.<\/b> The function and its inverse, showing reflection about the identity line[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]166540[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137911739\" class=\"solution\">\r\n<div id=\"fs-id1165137627081\" class=\"note precalculus qa textbox\">\r\n<p id=\"fs-id1165134388228\"><strong>Q &amp; A <\/strong><\/p>\r\n<strong>Is there any function that is equal to its own inverse?<\/strong>\r\n<p id=\"fs-id1165137602656\"><em>Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em><\/p>\r\n<p style=\"text-align: center\">[latex]\\frac{1}{1\/x}=x[\/latex]<\/p>\r\n<p id=\"fs-id1165137897050\"><em>Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em><\/p>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the inverse of a polynomial function.<\/li>\r\n \t<li>Restrict the domain to find the inverse of a polynomial function.<\/li>\r\n \t<li>Find or evaluate the inverse of a function.<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010800\/CNX_Precalc_Figure_03_08_0012.jpg\" alt=\"Gravel in the shape of a cone.\" width=\"487\" height=\"410\" \/> <b>Figure 1<\/b>[\/caption]\r\n<p id=\"fs-id1165137793975\">A mound of gravel is in the shape of a cone with the height equal to twice the radius.<span id=\"fs-id1165137939558\">\r\n<\/span><\/p>\r\n<p id=\"fs-id1165137411369\">The volume is found using a formula from elementary geometry.<\/p>\r\n\r\n<div id=\"eip-854\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}V&amp;=\\frac{1}{3}\\pi {r}^{2}h \\\\ &amp;=\\frac{1}{3}\\pi {r}^{2}\\left(2r\\right) \\\\ &amp;=\\frac{2}{3}\\pi {r}^{3} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137727278\">We have written the volume <em>V<\/em>\u00a0in terms of the radius <em>r<\/em>. However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula<\/p>\r\n\r\n<div id=\"eip-931\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]r=\\sqrt[3]{\\dfrac{3V}{2\\pi }\\\\}[\/latex]<\/div>\r\n<p id=\"fs-id1165134129769\">This function is the inverse of the formula for <em>V<\/em>\u00a0in terms of <em>r<\/em>.<\/p>\r\n\r\n<h2>Find the inverse of a polynomial function<\/h2>\r\n<p id=\"fs-id1165137439029\">Two functions <em>f<\/em>\u00a0and <em>g<\/em>\u00a0are inverse functions if for every coordinate pair in <em>f<\/em>, (<em>a<\/em>, <em>b<\/em>), there exists a corresponding coordinate pair in the inverse function, <em>g<\/em>, (<em>b<\/em>, <em>a<\/em>). In other words, the coordinate pairs of the inverse functions have the input and output interchanged.<\/p>\r\n<p id=\"fs-id1165137475924\">For a function to have an <strong>inverse function<\/strong> the function to create a new function that is <strong>one-to-one<\/strong> and would have an inverse function.<\/p>\r\n<p id=\"fs-id1165137448308\">For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown\u00a0below. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010801\/CNX_Precalc_Figure_03_08_0022.jpg\" alt=\"Diagram of a parabolic trough that is 18\" width=\"487\" height=\"279\" \/> <b>Figure 2<\/b>[\/caption]\r\n<p id=\"fs-id1165137793665\">Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with <em>x<\/em>\u00a0measured horizontally and <em>y<\/em>\u00a0measured vertically, with the origin at the vertex of the parabola.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010801\/CNX_Precalc_Figure_03_08_0032.jpg\" alt=\"Graph of a parabola.\" width=\"487\" height=\"441\" \/> <b>Figure 3<\/b>[\/caption]\r\n<p id=\"fs-id1165137771677\">From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form [latex]y\\left(x\\right)=a{x}^{2}[\/latex]. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor <em>a<\/em>.<\/p>\r\n\r\n<div id=\"eip-893\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align} 18&amp;=a{6}^{2} \\\\ a&amp;=\\frac{18}{36} \\\\ &amp;=\\frac{1}{2} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137633973\">Our parabolic cross section has the equation<\/p>\r\n\r\n<div id=\"eip-420\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/div>\r\n<p id=\"fs-id1165137770004\">We are interested in the <strong>surface area<\/strong> of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth <em>y<\/em>\u00a0the width will be given by 2<em>x<\/em>, so we need to solve the equation above for <em>x<\/em>\u00a0and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.<\/p>\r\n<p id=\"fs-id1165137638570\">To find an inverse, we can restrict our original function to a limited domain on which it <em>is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive <em>x<\/em>\u00a0values. On this domain, we can find an inverse by solving for the input variable:<\/p>\r\n\r\n<div id=\"eip-598\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}y&amp;=\\frac{1}{2}{x}^{2} \\\\ 2y&amp;={x}^{2} \\\\ x&amp;=\\pm \\sqrt{2y} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137453965\">This is not a function as written. We are limiting ourselves to positive <em>x<\/em>\u00a0values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.<\/p>\r\n\r\n<div id=\"eip-793\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]y=\\frac{{x}^{2}}{2},\\text{ }x&gt;0[\/latex]<\/div>\r\n<p id=\"fs-id1165137643958\">Because <em>x<\/em>\u00a0is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2<em>x<\/em>. The trough is 3 feet (36 inches) long, so the surface area will then be:<\/p>\r\n\r\n<div id=\"eip-491\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\text{Area} &amp; =l\\cdot w \\\\ &amp; =36\\cdot 2x \\\\ &amp; =72x \\\\ &amp; =72\\sqrt{2y} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137407432\">This example illustrates two important points:<\/p>\r\n\r\n<ol id=\"fs-id1165135545666\">\r\n \t<li>When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.<\/li>\r\n \t<li>The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165137618975\">Functions involving roots are often called <strong>radical functions<\/strong>. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called <strong>invertible functions<\/strong>, and we use the notation [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165135185952\">Warning: [latex]{f}^{-1}\\left(x\\right)[\/latex] is not the same as the reciprocal of the function [latex]f\\left(x\\right)[\/latex]. This use of \u20131 is reserved to denote inverse functions. To denote the reciprocal of a function [latex]f\\left(x\\right)[\/latex], we would need to write [latex]{\\left(f\\left(x\\right)\\right)}^{-1}=\\frac{1}{f\\left(x\\right)}[\/latex].<\/p>\r\n<p id=\"fs-id1165137561919\">An important relationship between inverse functions is that they \"undo\" each other. If [latex]{f}^{-1}[\/latex] is the inverse of a function <em>f<\/em>,\u00a0then <em>f<\/em>\u00a0is the inverse of the function [latex]{f}^{-1}[\/latex]. In other words, whatever the function <em>f<\/em>\u00a0does to <em>x<\/em>, [latex]{f}^{-1}[\/latex] undoes it\u2014and vice-versa. More formally, we write<\/p>\r\n\r\n<div id=\"eip-519\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }f[\/latex]<\/div>\r\n<p id=\"fs-id1165135503755\">and<\/p>\r\n\r\n<div id=\"eip-590\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }{f}^{-1}[\/latex]<\/div>\r\n<div id=\"fs-id1165137735698\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Verifying Two Functions Are Inverses of One Another<\/h3>\r\n<p id=\"fs-id1165137852132\">Two functions, <em>f<\/em>\u00a0and <i>g<\/i>, are inverses of one another if for all <em>x<\/em>\u00a0in the domain of <em>f\u00a0<\/em>and <em>g<\/em>.<\/p>\r\n\r\n<div id=\"eip-973\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137646263\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137501372\">How To: Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.<\/h3>\r\n<ol id=\"fs-id1165133276230\">\r\n \t<li>Replace [latex]f\\left(x\\right)[\/latex] with <em>y<\/em>.<\/li>\r\n \t<li>Interchange <em>x<\/em>\u00a0and <em>y<\/em>.<\/li>\r\n \t<li>Solve for <em>y<\/em>, and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_08_01\" class=\"example\">\r\n<div id=\"fs-id1165135150650\" class=\"exercise\">\r\n<div id=\"fs-id1165135620877\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Verifying Inverse Functions<\/h3>\r\n<p id=\"fs-id1165134148383\">Show that [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex] and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses, for [latex]x\\ne 0,-1[\/latex] .<\/p>\r\n[reveal-answer q=\"28284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"28284\"]\r\n<p id=\"fs-id1165137834138\">We must show that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex].<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}{f}^{-1}\\left(f\\left(x\\right)\\right)&amp;={f}^{-1}\\left(\\frac{1}{x+1}\\right) \\\\ &amp;=\\frac{1}{\\frac{1}{x+1}}-1 \\\\ &amp;=\\left(x+1\\right)-1 \\\\ &amp;=x\\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}f\\left({f}^{-1}\\left(x\\right)\\right)&amp;=f\\left(\\frac{1}{x}-1\\right) \\\\ &amp;=\\frac{1}{\\left(\\frac{1}{x}-1\\right)+1} \\\\ &amp;=\\frac{1}{\\frac{1}{x}} \\\\ &amp;=x \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135168183\">Therefore, [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137426116\">Show that [latex]f\\left(x\\right)=\\frac{x+5}{3}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=3x - 5[\/latex] are inverses.<\/p>\r\n[reveal-answer q=\"692921\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"692921\"]\r\n\r\n[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{x+5}{3}\\right)=3\\left(\\frac{x+5}{3}\\right)-5=\\left(x - 5\\right)+5=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(3x - 5\\right)=\\frac{\\left(3x - 5\\right)+5}{3}=\\frac{3x}{3}=x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_03_08_02\" class=\"example\">\r\n<div id=\"fs-id1165137600799\" class=\"exercise\">\r\n<div id=\"fs-id1165135160775\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Finding the Inverse of a Cubic Function<\/h3>\r\n<p id=\"fs-id1165137569920\">Find the inverse of the function [latex]f\\left(x\\right)=5{x}^{3}+1[\/latex].<\/p>\r\n[reveal-answer q=\"790658\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"790658\"]\r\n<p id=\"fs-id1165135412872\">This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for <em>x<\/em>.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}y=5{x}^{3}+1 \\\\ x=5{y}^{3}+1 \\\\ x - 1=5{y}^{3} \\\\ \\frac{x - 1}{5}={y}^{3} \\\\ {f}^{-1}\\left(x\\right)=\\sqrt[3]{\\frac{x - 1}{5}} \\end{gathered}[\/latex]<\/p>\r\n\r\n<div id=\"Example_03_08_02\" class=\"example\">\r\n<div id=\"fs-id1165137600799\" class=\"exercise\">\r\n<div id=\"fs-id1165137635322\" class=\"commentary\">\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137641602\">Look at the graph of <em>f<\/em>\u00a0and [latex]{f}^{-1}[\/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[\/latex]. This is always the case when graphing a function and its inverse function.<\/p>\r\n<p id=\"fs-id1165137793468\">Also, since the method involved interchanging <em>x<\/em>\u00a0and <em>y<\/em>, notice corresponding points. If [latex]\\left(a,b\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(b,a\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Since [latex]\\left(0,1\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(1,0\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Similarly, since [latex]\\left(1,6\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(6,1\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010801\/CNX_Precalc_Figure_03_08_0042.jpg\" alt=\"Graph of f(x)=5x^3+1 and its inverse, f^(-1)(x)=3sqrt((x-1)\/(5)).\" width=\"487\" height=\"554\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137635322\" class=\"commentary\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165133047522\">Find the inverse function of [latex]f\\left(x\\right)=\\sqrt[3]{x+4}[\/latex].<\/p>\r\n[reveal-answer q=\"242169\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"242169\"]\r\n\r\n[latex]{f}^{-1}\\left(x\\right)={x}^{3}-4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]1620[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Restrict the domain to find the inverse of a polynomial function<\/h2>\r\n<p id=\"fs-id1165137471808\">So far, we have been able to find the inverse functions of <strong>cubic functions<\/strong> without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an <strong>inverse function<\/strong>. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.<\/p>\r\n\r\n<div id=\"fs-id1165137434585\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Restricting the Domain<\/h3>\r\n<p id=\"fs-id1165137409777\">If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137431545\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137656706\">How To: Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.<\/h3>\r\n<ol id=\"fs-id1165137532171\">\r\n \t<li>Restrict the domain by determining a domain on which the original function is one-to-one.<\/li>\r\n \t<li>Replace <em>f<\/em>(<em>x<\/em>)\u00a0with <em>y<\/em>.<\/li>\r\n \t<li>Interchange <em>x<\/em>\u00a0and <em>y<\/em>.<\/li>\r\n \t<li>Solve for <em>y<\/em>, and rename the function or pair of function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\r\n \t<li>Revise the formula for [latex]{f}^{-1}\\left(x\\right)[\/latex] by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_08_03\" class=\"example\">\r\n<div id=\"fs-id1165137737103\" class=\"exercise\">\r\n<div id=\"fs-id1165137668130\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Restricting the Domain to Find the Inverse of a Polynomial Function<\/h3>\r\n<p id=\"fs-id1165137482766\">Find the inverse function of <em>f<\/em>:<\/p>\r\n\r\n<ol id=\"fs-id1165137638318\">\r\n \t<li>[latex]f\\left(x\\right)={\\left(x - 4\\right)}^{2}, x\\ge 4[\/latex]<\/li>\r\n \t<li>[latex]f\\left(x\\right)={\\left(x - 4\\right)}^{2}, x\\le 4[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"546705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"546705\"]\r\n<p id=\"fs-id1165137506731\">The original function [latex]f\\left(x\\right)={\\left(x - 4\\right)}^{2}[\/latex] is not one-to-one, but the function is restricted to a domain of [latex]x\\ge 4[\/latex] or [latex]x\\le 4[\/latex] on which it is one-to-one.<\/p>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010801\/CNX_Precalc_Figure_03_08_0052.jpg\" alt=\"&quot;Two\" \/>\r\n<p id=\"fs-id1165137706306\">To find the inverse, start by replacing [latex]f\\left(x\\right)[\/latex] with the simple variable <em>y<\/em>.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;y={\\left(x - 4\\right)}^{2} &amp;&amp; \\text{Interchange } x \\text{ and }y. \\\\ &amp;x={\\left(y - 4\\right)}^{2} &amp;&amp; \\text{Take the square root}. \\\\ &amp;\\pm \\sqrt{x}=y - 4 &amp;&amp; \\text{Add } 4 \\text{ to both sides}. \\\\ &amp;4\\pm \\sqrt{x}=y \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137444285\">This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of <em>x<\/em>\u00a0and <em>y<\/em>\u00a0for the original <em>f<\/em>(<em>x<\/em>), we looked at the domain: the values <em>x<\/em>\u00a0could assume. When we reversed the roles of <em>x<\/em>\u00a0and <em>y<\/em>,\u00a0this gave us the values <em>y<\/em>\u00a0could assume. For this function, [latex]x\\ge 4[\/latex], so for the inverse, we should have [latex]y\\ge 4[\/latex], which is what our inverse function gives.<\/p>\r\n\r\n<ol id=\"fs-id1165137735027\">\r\n \t<li>The domain of the original function was restricted to [latex]x\\ge 4[\/latex], so the outputs of the inverse need to be the same, [latex]f\\left(x\\right)\\ge 4[\/latex], and we must use the + case:\r\n<div id=\"eip-id1165134294825\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)=4+\\sqrt{x}[\/latex]<\/div><\/li>\r\n \t<li>The domain of the original function was restricted to [latex]x\\le 4[\/latex], so the outputs of the inverse need to be the same, [latex]f\\left(x\\right)\\le 4[\/latex], and we must use the \u2013 case:\r\n<div id=\"eip-id1165137482501\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)=4-\\sqrt{x}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137534054\">On the graphs below, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line [latex]y=x[\/latex]. The coordinate pair [latex]\\left(4, 0\\right)[\/latex] is on the graph of <em>f<\/em>\u00a0and the coordinate pair [latex]\\left(0, 4\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. For any coordinate pair, if (<em>a<\/em>, <em>b<\/em>) is on the graph of <em>f<\/em>, then (<em>b<\/em>, <em>a<\/em>) is on the graph of [latex]{f}^{-1}[\/latex]. Finally, observe that the graph of <em>f<\/em>\u00a0intersects the graph of [latex]{f}^{-1}[\/latex] on the line <em>y\u00a0<\/em>= <em>x<\/em>. Points of intersection for the graphs of <em>f<\/em>\u00a0and [latex]{f}^{-1}[\/latex] will always lie on the line <em>y\u00a0<\/em>= <em>x<\/em>.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010802\/CNX_Precalc_Figure_03_08_0062.jpg\" alt=\"Two graphs of a parabolic function with half of its inverse.\" width=\"975\" height=\"442\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137736620\" class=\"commentary\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_03_08_04\" class=\"example\">\r\n<div id=\"fs-id1165137786597\" class=\"exercise\">\r\n<div id=\"fs-id1165135481235\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified<\/h3>\r\n<p id=\"fs-id1165137410909\">Restrict the domain and then find the inverse of<\/p>\r\n<p style=\"text-align: center\">[latex]f\\left(x\\right)={\\left(x - 2\\right)}^{2}-3[\/latex].<\/p>\r\n[reveal-answer q=\"368262\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"368262\"]\r\n<p id=\"fs-id1165137414415\">We can see this is a parabola with vertex at [latex]\\left(2, -3\\right)[\/latex] that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to [latex]x\\ge 2[\/latex].<\/p>\r\n<p id=\"fs-id1165137842529\">To find the inverse, we will use the vertex form of the quadratic. We start by replacing <em>f<\/em>(<em>x<\/em>) with a simple variable, <em>y<\/em>, then solve for <em>x<\/em>.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;y={\\left(x - 2\\right)}^{2}-3 &amp;&amp; \\text{Interchange } x \\text{ and } y. \\\\ &amp;x={\\left(y - 2\\right)}^{2}-3 &amp;&amp; \\text{Add 3 to both sides}. \\\\ &amp;x+3={\\left(y - 2\\right)}^{2} &amp;&amp; \\text{Take the square root}. \\\\ &amp;\\pm \\sqrt{x+3}=y - 2 &amp;&amp; \\text{Add 2 to both sides}. \\\\ &amp;2\\pm \\sqrt{x+3}=y &amp;&amp; \\text{Rename the function}. \\\\ &amp;{f}^{-1}\\left(x\\right)=2\\pm \\sqrt{x+3} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137419504\">Now we need to determine which case to use. Because we restricted our original function to a domain of [latex]x\\ge 2[\/latex], the outputs of the inverse should be the same, telling us to utilize the + case<\/p>\r\n<p style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)=2+\\sqrt{x+3}[\/latex]<\/p>\r\n<p id=\"fs-id1165137827988\">If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.<\/p>\r\n\r\n<div id=\"Example_03_08_04\" class=\"example\">\r\n<div id=\"fs-id1165137786597\" class=\"exercise\">\r\n<div id=\"fs-id1165134362839\" class=\"commentary\">\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135259538\">Notice that we arbitrarily decided to restrict the domain on [latex]x\\ge 2[\/latex]. We could just have easily opted to restrict the domain on [latex]x\\le 2[\/latex], in which case [latex]{f}^{-1}\\left(x\\right)=2-\\sqrt{x+3}[\/latex]. Observe the original function graphed on the same set of axes as its inverse function in the graph below. Notice that both graphs show symmetry about the line <em>y<\/em> =\u00a0<em>x<\/em>. The coordinate pair [latex]\\left(2,\\text{ }-3\\right)[\/latex] is on the graph of <em>f<\/em>\u00a0and the coordinate pair [latex]\\left(-3,\\text{ }2\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Observe from the graph of both functions on the same set of axes that<\/p>\r\n\r\n<div id=\"eip-id1165134122215\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\text{domain of }f=\\text{range of } {f}^{-1}=\\left[2,\\infty \\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165137642128\">and<\/p>\r\n\r\n<div id=\"eip-id1165134279478\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\text{domain of }{f}^{-1}=\\text{range of } f=\\left[-3,\\infty \\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165137723812\">Finally, observe that the graph of <em>f<\/em>\u00a0intersects the graph of [latex]{f}^{-1}[\/latex] along the line <em>y\u00a0<\/em>= <em>x<\/em>.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010802\/CNX_Precalc_Figure_03_08_0072.jpg\" alt=\"Graph of a parabolic function with half of its inverse.\" width=\"487\" height=\"487\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]<span style=\"background-color: #ffffff;font-size: 1em\">\u00a0<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135596379\">Find the inverse of the function [latex]f\\left(x\\right)={x}^{2}+1[\/latex], on the domain [latex]x\\ge 0[\/latex].<\/p>\r\n[reveal-answer q=\"455466\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455466\"]\r\n\r\n[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x - 1}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<section id=\"fs-id1165137894462\">\r\n<h2>Solving Applications of Radical Functions<\/h2>\r\n<p id=\"fs-id1165137696560\">Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the <strong>inverse of a radical function<\/strong>, we will need to restrict the domain of the answer because the range of the original function is limited.<\/p>\r\n\r\n<div id=\"fs-id1165137415876\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137455923\">How To: Given a radical function, find the inverse.<\/h3>\r\n<ol id=\"fs-id1165137542989\">\r\n \t<li>Determine the range of the original function.<\/li>\r\n \t<li>Replace <em>f<\/em>(<em>x<\/em>)\u00a0with <em>y<\/em>, then solve for <em>x<\/em>.<\/li>\r\n \t<li>If necessary, restrict the domain of the inverse function to the range of the original function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_08_05\" class=\"example\">\r\n<div id=\"fs-id1165135173813\" class=\"exercise\">\r\n<div id=\"fs-id1165137399685\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Finding the Inverse of a Radical Function<\/h3>\r\n<p id=\"fs-id1165135570491\">Restrict the domain and then find the inverse of the function [latex]f\\left(x\\right)=\\sqrt{x - 4}[\/latex].<\/p>\r\n[reveal-answer q=\"727309\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"727309\"]\r\n<p id=\"fs-id1165135545767\">Note that the original function has range [latex]f\\left(x\\right)\\ge 0[\/latex]. Replace [latex]f\\left(x\\right)[\/latex] with <em>y<\/em>, then solve for <em>x<\/em>.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align} &amp;y=\\sqrt{x - 4} &amp;&amp; \\text{Replace}f\\left(x\\right)\\text{with}y.\\\\ &amp;x=\\sqrt{y - 4} &amp;&amp; \\text{Interchange}x\\text{and}y. \\\\ &amp;x =\\sqrt{y - 4} &amp;&amp; \\text{Square each side}. \\\\ &amp;{x}^{2} =y - 4 &amp;&amp; \\text{Add 4}. \\\\ &amp;{x}^{2}+4 =y &amp;&amp; \\text{Rename the function}{f}^{-1}\\left(x\\right). \\\\ &amp;{f}^{-1}\\left(x\\right) ={x}^{2}+4 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135209570\">Recall that the domain of this function must be limited to the range of the original function.<\/p>\r\n<p style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)={x}^{2}+4,x\\ge 0[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135173239\">Notice in the graph below\u00a0that the inverse is a reflection of the original function over the line <em>y\u00a0<\/em>= <em>x<\/em>. Because the original function has only positive outputs, the inverse function has only positive inputs.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010802\/CNX_Precalc_Figure_03_08_0082.jpg\" alt=\"Graph of f(x)=sqrt(x-4) and its inverse, f^(-1)(x)=x^2+4.\" width=\"487\" height=\"444\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137784775\">Restrict the domain and then find the inverse of the function [latex]f\\left(x\\right)=\\sqrt{2x+3}[\/latex].<\/p>\r\n[reveal-answer q=\"107257\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"107257\"]\r\n\r\n[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-3}{2},x\\ge 0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137761571\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174218[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving Applications of Radical Functions<\/h2>\r\n<p id=\"fs-id1165135435831\">Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section.<\/p>\r\n\r\n<div id=\"Example_03_08_06\" class=\"example\">\r\n<div id=\"fs-id1165137634475\" class=\"exercise\">\r\n<div id=\"fs-id1165137531120\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Solving an Application with a Cubic Function<\/h3>\r\n<p id=\"fs-id1165137771982\">A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by<\/p>\r\n\r\n<div id=\"eip-id1165132187568\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]V=\\frac{2}{3}\\pi {r}^{3}[\/latex]<\/div>\r\n<p id=\"fs-id1165135181305\">Find the inverse of the function [latex]V=\\frac{2}{3}\\pi {r}^{3}[\/latex] that determines the volume <em>V<\/em>\u00a0of a cone and is a function of the radius <em>r<\/em>. Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use [latex]\\pi =3.14[\/latex].<\/p>\r\n[reveal-answer q=\"525201\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"525201\"]\r\n<p id=\"fs-id1165137405144\">Start with the given function for <em>V<\/em>. Notice that the meaningful domain for the function is [latex]r\\ge 0[\/latex] since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is [latex]V\\ge 0[\/latex]. Solve for <em>r<\/em>\u00a0in terms of <em>V<\/em>, using the method outlined previously.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align} V&amp;=\\frac{2}{3}\\pi {r}^{3} \\\\ {r}^{3}&amp;=\\frac{3V}{2\\pi } &amp;&amp; \\text{Solve for }{r}^{3}. \\\\ r&amp;=\\sqrt[3]{\\frac{3V}{2\\pi }} &amp;&amp; \\text{Solve for }r. \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137730324\">This is the result stated in the section opener. Now evaluate this for <em>V<\/em> = 100 and [latex]\\pi =3.14[\/latex].<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}r &amp; =\\sqrt[3]{\\frac{3V}{2\\pi }} \\\\ &amp; =\\sqrt[3]{\\frac{3\\cdot 100}{2\\cdot 3.14}} \\\\ &amp; \\approx \\sqrt[3]{47.7707} \\\\ &amp; \\approx 3.63 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137706279\">Therefore, the radius is about 3.63 ft.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137580023\">\r\n<h2>Determining the Domain of a Radical Function Composed with Other Functions<\/h2>\r\n<p id=\"fs-id1165134042947\">When radical functions are composed with other functions, determining domain can become more complicated.<\/p>\r\n\r\n<div id=\"Example_03_08_07\" class=\"example\">\r\n<div id=\"fs-id1165135378774\" class=\"exercise\">\r\n<div id=\"fs-id1165135378776\" class=\"problem textbox shaded\">\r\n<h3>Example 7: Finding the Domain of a Radical Function Composed with a Rational Function<\/h3>\r\n<p id=\"fs-id1165137658778\">Find the domain of the function [latex]f\\left(x\\right)=\\sqrt{\\frac{\\left(x+2\\right)\\left(x - 3\\right)}{\\left(x - 1\\right)}}[\/latex].<\/p>\r\n[reveal-answer q=\"429624\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"429624\"]\r\n<p id=\"fs-id1165137665549\">Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where [latex]\\frac{\\left(x+2\\right)\\left(x - 3\\right)}{\\left(x - 1\\right)}\\ge 0[\/latex]. The output of a rational function can change signs (change from positive to negative or vice versa) at <em>x<\/em>-intercepts and at vertical asymptotes. For this equation, the graph could change signs at <em>x<\/em>\u00a0= \u20132, 1, and 3.<\/p>\r\n<p id=\"fs-id1165135686721\">To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010802\/CNX_Precalc_Figure_03_08_0092.jpg\" alt=\"Graph of a radical function that shows where the outputs are nonnegative.\" width=\"731\" height=\"439\" \/> <b>Figure 9<\/b>[\/caption]\r\n<p id=\"fs-id1165137694167\">This function has two <em>x<\/em>-intercepts, both of which exhibit linear behavior near the <em>x<\/em>-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a <em>y<\/em>-intercept at (0, 6).<\/p>\r\n<p id=\"fs-id1165135333589\">From the <em>y<\/em>-intercept and <em>x<\/em>-intercept at <em>x\u00a0<\/em>= \u20132, we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.<\/p>\r\n<p id=\"fs-id1165137664081\">From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function <em>f<\/em>(<em>x<\/em>) will be defined. <em>f<\/em>(<em>x<\/em>) has domain [latex]-2\\le x&lt;1\\text{or}x\\ge 3[\/latex], or in interval notation, [latex]\\left[-2,1\\right)\\cup \\left[3,\\infty \\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165135173899\">\r\n<h2>Finding Inverses of Rational Functions<\/h2>\r\n<p id=\"fs-id1165135475898\">As with finding inverses of quadratic functions, it is sometimes desirable to find the <strong>inverse of a rational function<\/strong>, particularly of rational functions that are the ratio of linear functions, such as in concentration applications.<\/p>\r\n\r\n<div id=\"Example_03_08_08\" class=\"example\">\r\n<div id=\"fs-id1165137642525\" class=\"exercise\">\r\n<div id=\"fs-id1165137642528\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Finding the Inverse of a Rational Function<\/h3>\r\n<p id=\"fs-id1165135332364\">The function [latex]C=\\frac{20+0.4n}{100+n}[\/latex] represents the concentration <em>C<\/em>\u00a0of an acid solution after <em>n<\/em>\u00a0mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for <em>n<\/em>\u00a0in terms of <em>C<\/em>. Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.<\/p>\r\n[reveal-answer q=\"623836\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"623836\"]\r\n<p id=\"fs-id1165134223203\">We first want the inverse of the function. We will solve for <em>n<\/em>\u00a0in terms of <em>C<\/em>.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}C=\\frac{20+0.4n}{100+n}\\\\ C\\left(100+n\\right)=20+0.4n\\\\ 100C+Cn=20+0.4n\\\\ 100C - 20=0.4n-Cn\\\\ 100C - 20=\\left(0.4-C\\right)n\\\\ n=\\frac{100C - 20}{0.4-C}\\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165137637474\">Now evaluate this function for C=0.35 (35%).<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}n&amp;=\\frac{100\\left(0.35\\right)-20}{0.4 - 0.35}\\\\ &amp;=\\frac{15}{0.05}\\\\ &amp;=300\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137706154\">We can conclude that 300 mL of the 40% solution should be added.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134042923\">Find the inverse of the function [latex]f\\left(x\\right)=\\frac{x+3}{x - 2}[\/latex].<\/p>\r\n[reveal-answer q=\"457099\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"457099\"]\r\n\r\n[latex]{f}^{-1}\\left(x\\right)=\\frac{2x+3}{x - 1}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]34476[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>If [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex], then<\/li>\r\n \t<li>[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex].<\/li>\r\n \t<li>Each of the toolkit functions has an inverse.<\/li>\r\n \t<li>For a function to have an inverse, it must be one-to-one (pass the horizontal line test).<\/li>\r\n \t<li>A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.<\/li>\r\n \t<li>For a tabular function, exchange the input and output rows to obtain the inverse.<\/li>\r\n \t<li>The inverse of a function can be determined at specific points on its graph.<\/li>\r\n \t<li>To find the inverse of a formula, solve the equation [latex]y=f\\left(x\\right)[\/latex] for [latex]x[\/latex] as a function of\u00a0[latex]y[\/latex]. Then exchange the labels [latex]x[\/latex] and [latex]y[\/latex].<\/li>\r\n \t<li>The graph of an inverse function is the reflection of the graph of the original function across the line [latex]y=x[\/latex].<\/li>\r\n \t<li>The inverse of a quadratic function is a square root function.<\/li>\r\n \t<li>If [latex]{f}^{-1}[\/latex]\u00a0is the inverse of a function <em>f<\/em>,\u00a0then <em>f<\/em>\u00a0is the inverse of the function [latex]{f}^{-1}[\/latex].<\/li>\r\n \t<li>While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible.<\/li>\r\n \t<li>To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one.<\/li>\r\n \t<li>When finding the inverse of a radical function, we need a restriction on the domain of the answer.<\/li>\r\n \t<li>Inverse and radical and functions can be used to solve application problems.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137441703\" class=\"definition\">\r\n \t<dt>inverse function<\/dt>\r\n \t<dd id=\"fs-id1165137441708\">for any one-to-one function [latex]f\\left(x\\right)[\/latex], the inverse is a function [latex]{f}^{-1}\\left(x\\right)[\/latex] such that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]; this also implies that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex]<\/dd>\r\n<\/dl>\r\n<\/div>\r\n<\/section>\r\n<dl id=\"fs-id1165135169260\" class=\"definition\">\r\n \t<dt><strong>invertible function<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135169263\">any function that has an inverse function<\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\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<p id=\"fs-id1165135358875\">A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.<\/p>\n<p>If some physical machines can run in two directions, we might ask whether some of the function &#8220;machines&#8221; we have been studying can also run backwards. Figure 1\u00a0provides a visual representation of this question. In this section, we will consider the reverse nature of functions.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010622\/CNX_Precalc_Figure_01_07_0012.jpg\" alt=\"Diagram of a function and what would be its inverse.\" width=\"731\" height=\"305\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> Can a function &#8220;machine&#8221; operate in reverse?<\/p>\n<\/div>\n<h2>Verifying That Two Functions Are Inverse Functions<\/h2>\n<p id=\"fs-id1165135705795\">Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the <strong>Celsius<\/strong> scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees <strong>Fahrenheit<\/strong> to degrees Celsius. She finds the formula<\/p>\n<div id=\"fs-id1165137807176\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/div>\n<p id=\"fs-id1165135433486\">and substitutes 75 for [latex]F[\/latex] to calculate<\/p>\n<div id=\"fs-id1165137911210\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\frac{5}{9}\\left(75 - 32\\right)\\approx {24}^{ \\circ} {C}[\/latex].<\/div>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010623\/CNX_Precalc_Figure_01_07_0022.jpg\" alt=\"A forecast of Monday\u2019s through Thursday\u2019s weather.\" width=\"731\" height=\"226\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137409312\">Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week\u2019s weather forecast\u00a0for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.<span id=\"fs-id1165137414400\"><br \/>\n<\/span><\/p>\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 [latex]F[\/latex] after substituting a value for [latex]C[\/latex]. For example, to convert 26 degrees Celsius, she could write<\/p>\n<div id=\"fs-id1165135548255\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{gathered}26=\\frac{5}{9}\\left(F - 32\\right) \\\\ 26\\cdot \\frac{9}{5}=F - 32 \\\\ F=26\\cdot \\frac{9}{5}+32\\approx 79 \\end{gathered}[\/latex]<\/div>\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 [latex]f\\left(x\\right)[\/latex], we represent its inverse as [latex]{f}^{-1}\\left(x\\right)[\/latex], read as [latex]\"f[\/latex] inverse of [latex]x.\\text{\"}[\/latex] The raised [latex]-1[\/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[\/latex] . In other words, [latex]{f}^{-1}\\left(x\\right)[\/latex] does <em>not<\/em> mean [latex]\\frac{1}{f\\left(x\\right)}[\/latex] because [latex]\\frac{1}{f\\left(x\\right)}[\/latex] is the reciprocal of [latex]f[\/latex] and not the inverse.<\/p>\n<p id=\"fs-id1165137724926\">The &#8220;exponent-like&#8221; notation comes from an analogy between function composition and multiplication: just as [latex]{a}^{-1}a=1[\/latex] (1 is the identity element for multiplication) for any nonzero number [latex]a[\/latex], so [latex]{f}^{-1}\\circ f[\/latex] equals the identity function, that is,<\/p>\n<div id=\"fs-id1165134302408\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(y\\right)=x[\/latex]<\/div>\n<p id=\"fs-id1165135667832\">This holds for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. Informally, this means that inverse functions &#8220;undo&#8221; each other. However, just as zero does not have a <strong>reciprocal<\/strong>, some functions do not have inverses.<\/p>\n<p id=\"fs-id1165137655153\">Given a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether either [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] or [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)<\/p>\n<p id=\"fs-id1165135397975\">For example, [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.<\/p>\n<div id=\"fs-id1165137756798\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/div>\n<p id=\"fs-id1165137767233\">and<\/p>\n<div id=\"fs-id1165137755853\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left({f}^{}\\circ {f}^{-1}\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/div>\n<p id=\"fs-id1165137438777\">A few coordinate pairs from the graph of the function [latex]y=4x[\/latex] are (\u22122, \u22128), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function [latex]y=\\frac{1}{4}x[\/latex] are (\u22128, \u22122), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.<\/p>\n<div id=\"fs-id1165137933105\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Inverse Function<\/h3>\n<p id=\"fs-id1165137473076\">For any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex]. This can also be written as [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. It also follows that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex] if [latex]{f}^{-1}[\/latex] is the inverse of [latex]f[\/latex].<\/p>\n<p id=\"fs-id1165137444821\">The notation [latex]{f}^{-1}[\/latex] is read [latex]\\text{\"}f[\/latex] inverse.&#8221; Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x.\"[\/latex]<br \/>\nKeep in mind that<\/p>\n<p style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)\\ne \\frac{1}{f\\left(x\\right)}[\/latex]<\/p>\n<p id=\"fs-id1165135194095\">and not all functions have inverses.<\/p>\n<\/div>\n<div id=\"Example_01_07_01\" class=\"example\">\n<div id=\"fs-id1165137656641\" class=\"exercise\">\n<div id=\"fs-id1165137922642\" class=\"problem textbox shaded\">\n<h3>Example 1: Identifying an Inverse Function for a Given Input-Output Pair<\/h3>\n<p id=\"fs-id1165137659325\">If for a particular one-to-one function [latex]f\\left(2\\right)=4[\/latex] and [latex]f\\left(5\\right)=12[\/latex], what are the corresponding input and output values for the inverse function?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q177916\">Show Solution<\/span><\/p>\n<div id=\"q177916\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137737081\">The inverse function reverses the input and output quantities, so if [latex]f\\left(2\\right)=4[\/latex], then [latex]{f}^{-1}\\left(4\\right)=2[\/latex] and if [latex]f\\left(5\\right)=12[\/latex], then [latex]{f}^{-1}\\left(12\\right)=5[\/latex].<\/p>\n<p id=\"fs-id1165137659464\">Alternatively, if we want to name the inverse function [latex]g[\/latex], then [latex]g\\left(4\\right)=2[\/latex] and [latex]g\\left(12\\right)=5[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135508518\">Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.<\/p>\n<table id=\"Table_01_07_01\" style=\"width: 295px\" summary=\"For (x,f(x)) we have the values (2, 4) and (5, 12); for (x, g(x)), we have the values (4, 2) and (12, 5).\">\n<thead>\n<tr>\n<th style=\"width: 155px\">[latex]\\left(x,f\\left(x\\right)\\right)[\/latex]<\/th>\n<th style=\"width: 140px\">[latex]\\left(x,g\\left(x\\right)\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"width: 155px\">[latex]\\left(2,4\\right)[\/latex]<\/td>\n<td style=\"width: 140px\">[latex]\\left(4,2\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 155px\">[latex]\\left(5,12\\right)[\/latex]<\/td>\n<td style=\"width: 140px\">[latex]\\left(12,5\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137659089\">Given that [latex]{h}^{-1}\\left(6\\right)=2[\/latex], what are the corresponding input and output values of the original function [latex]h?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q749671\">Show Solution<\/span><\/p>\n<div id=\"q749671\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]h\\left(2\\right)=6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Find an Inverse Function From a Table\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TSztRfzmk0M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"fs-id1165134357354\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135434077\">How To: Given two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex], test whether the functions are inverses of each other.<\/h3>\n<ol id=\"fs-id1165137452358\">\n<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] or [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\n<li>If both statements are true, then [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_07_02\" class=\"example\">\n<div id=\"fs-id1165137557051\" class=\"exercise\">\n<div id=\"fs-id1165137679032\" class=\"problem textbox shaded\">\n<h3>Example 2: Testing Inverse Relationships Algebraically<\/h3>\n<p id=\"fs-id1165135519417\">If [latex]f\\left(x\\right)=\\frac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\frac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q619704\">Show Solution<\/span><\/p>\n<div id=\"q619704\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align} g\\left(f\\left(x\\right)\\right)&=\\frac{1}{\\left(\\frac{1}{x+2}\\right)}{-2 } \\\\ &= x+2 - 2 \\\\ &= x \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137611481\">and<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)&=\\frac{1}{\\frac{1}{x}-2+2}\\\\ &=\\frac{1}{\\frac{1}{x}} \\\\ &=x \\end{align}[\/latex]<\/p>\n<p>So<\/p>\n<p style=\"text-align: center\">[latex]g={f}^{-1}\\text{ and }f={g}^{-1}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135389000\">Notice the inverse operations are in reverse order of the operations from the original function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135160550\">If [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q196142\">Show Solution<\/span><\/p>\n<div id=\"q196142\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_01_07_03\" class=\"example\">\n<div id=\"fs-id1165135259560\" class=\"exercise\">\n<div id=\"fs-id1165134042918\" class=\"problem textbox shaded\">\n<h3>Example 3: Determining Inverse Relationships for Power Functions<\/h3>\n<p id=\"fs-id1165137441834\">If [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q506\">Show Solution<\/span><\/p>\n<div id=\"q506\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]f\\left(g\\left(x\\right)\\right)=\\frac{{x}^{3}}{27}\\ne x[\/latex]<\/p>\n<p id=\"fs-id1165137694053\">No, the functions are not inverses.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165134192978\">The correct inverse to the cube is the cube root [latex]\\sqrt[\\leftroot{-1}\\uproot{2}3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137573532\">If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}\\text{and}g\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q746859\">Show Solution<\/span><\/p>\n<div id=\"q746859\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm166520\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=166520&theme=oea&iframe_resize_id=ohm166520\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Finding Domain and Range of Inverse Functions<\/h2>\n<p>The outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010624\/CNX_Precalc_Figure_01_07_0032.jpg\" alt=\"Domain and range of a function and its inverse.\" width=\"487\" height=\"143\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3.<\/b> Domain and range of a function and its inverse<\/p>\n<\/div>\n<p id=\"fs-id1165135557891\">When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of [latex]f\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}[\/latex] is [latex]{f}^{-1}\\left(x\\right)={x}^{2}[\/latex], because a square &#8220;undoes&#8221; a square root; but the square is only the inverse of the square root on the domain [latex]\\left[0,\\infty \\right)[\/latex], since that is the range of [latex]f\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}[\/latex].<\/p>\n<p id=\"fs-id1165137730185\">We can look at this problem from the other side, starting with the square (toolkit quadratic) function [latex]f\\left(x\\right)={x}^{2}[\/latex]. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and \u20133. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the &#8220;inverse&#8221; is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.<\/p>\n<p id=\"fs-id1165137823552\">In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).<\/p>\n<p id=\"fs-id1165132037000\">If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}[\/latex] on [latex]\\left[1,\\infty \\right)[\/latex], then the inverse function is [latex]{f}^{-1}\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}+1[\/latex].<\/p>\n<ul id=\"fs-id1165137851227\">\n<li>The domain of [latex]f[\/latex] = range of [latex]{f}^{-1}[\/latex] = [latex]\\left[1,\\infty \\right)[\/latex].<\/li>\n<li>The domain of [latex]{f}^{-1}[\/latex] = range of [latex]f[\/latex] = [latex]\\left[0,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<div id=\"fs-id1165137733804\" class=\"note precalculus qa textbox\">\n<p id=\"fs-id1165137723526\"><strong>Q &amp; A<\/strong><\/p>\n<p><strong>Is it possible for a function to have more than one inverse?<\/strong><\/p>\n<p id=\"fs-id1165137456608\"><em>No. If two supposedly different functions, say, [latex]g[\/latex] and [latex]h[\/latex], both meet the definition of being inverses of another function [latex]f[\/latex], then you can prove that [latex]g=h[\/latex]. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.<\/em><\/p>\n<\/div>\n<div id=\"fs-id1165137704938\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Domain and Range of Inverse Functions<\/h3>\n<p id=\"fs-id1165135319550\">The range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<p id=\"fs-id1165137673886\">The domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165135308785\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137605040\"><strong>How To: Given a function, find the domain and range of its inverse.<br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id1165137530434\">\n<li>If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.<\/li>\n<li>If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_07_05\" class=\"example\">\n<div id=\"fs-id1165137667922\" class=\"exercise\">\n<div id=\"fs-id1165135511321\" class=\"problem textbox shaded\">\n<h3>Example 4: Finding the Inverses of Toolkit Functions<\/h3>\n<p id=\"fs-id1165137448020\">Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed below. We restrict the domain in such a fashion that the function assumes all <em>y<\/em>-values exactly once.<\/p>\n<table id=\"Table_01_07_02\" summary=\"A list of the toolkit function. The constant function is f(x) = c where c is the constant; the identity function is f(x) = x; the absolute function is f(x)=|x|; the quadratic function is f(x) = x^2; the cubic function is f(x)=x^3; the reciprocal function is f(x)=1\/x; the reciprocal squared function is f(x)=1\/x^2; the square root function is f(x)=sqrt(x); the cube root function is f(x) = x^(1\/3).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>Constant<\/td>\n<td>Identity<\/td>\n<td>Quadratic<\/td>\n<td>Cubic<\/td>\n<td>Reciprocal<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(x\\right)=c[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=x[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)={x}^{2}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)={x}^{3}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Reciprocal squared<\/td>\n<td>Cube root<\/td>\n<td>Square root<\/td>\n<td>Absolute value<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}3]{x}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=|x|[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q552567\">Show Solution<\/span><\/p>\n<div id=\"q552567\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165132988445\">The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.<\/p>\n<p id=\"fs-id1165134080947\">The absolute value function can be restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], where it is equal to the identity function.<\/p>\n<p id=\"fs-id1165137642849\">The reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<p>We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs.\u00a0They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010624\/CNX_Precalc_Figure_01_07_004ab2.jpg\" alt=\"Graph of an absolute function.\" width=\"975\" height=\"404\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4.<\/b> (a) Absolute value (b) Reciprocal squared<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137507853\">The domain of function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of function [latex]f[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex]. Find the domain and range of the inverse function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q333382\">Show Solution<\/span><\/p>\n<div id=\"q333382\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain of function [latex]{f}^{-1}[\/latex] is [latex]\\left(-\\infty \\text{,}-2\\right)[\/latex] and the range of function [latex]{f}^{-1}[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding and Evaluating Inverse Functions<\/h2>\n<p id=\"fs-id1165137761017\">Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.<\/p>\n<section id=\"fs-id1165135466392\">\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Inverting Tabular Functions<\/span><\/h2>\n<p id=\"fs-id1165135190714\">Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.<\/p>\n<p id=\"fs-id1165137422578\">Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.<\/p>\n<div id=\"Example_01_07_06\" class=\"example\">\n<div id=\"fs-id1165135544995\" class=\"exercise\">\n<div id=\"fs-id1165137698262\" class=\"problem textbox shaded\">\n<h3>Example 5: Interpreting the Inverse of a Tabular Function<\/h3>\n<p id=\"fs-id1165135435474\">A function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes. Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].<\/p>\n<table id=\"Table_01_07_03\" summary=\"Two rows and five columns. The first row is labeled\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\n<td>30<\/td>\n<td>50<\/td>\n<td>70<\/td>\n<td>90<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q736190\">Show Solution<\/span><\/p>\n<div id=\"q736190\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137640334\">The inverse function takes an output of [latex]f[\/latex] and returns an input for [latex]f[\/latex]. So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.<\/p>\n<p id=\"fs-id1165135181841\">Alternatively, recall that the definition of the inverse was that if [latex]f\\left(a\\right)=b[\/latex], then [latex]{f}^{-1}\\left(b\\right)=a[\/latex]. By this definition, if we are given [latex]{f}^{-1}\\left(70\\right)=a[\/latex], then we are looking for a value [latex]a[\/latex] so that [latex]f\\left(a\\right)=70[\/latex]. In this case, we are looking for a [latex]t[\/latex] so that [latex]f\\left(t\\right)=70[\/latex], which is when [latex]t=90[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134108483\">Using the table below,\u00a0find and interpret (a) [latex]\\text{ }f\\left(60\\right)[\/latex], and (b) [latex]\\text{ }{f}^{-1}\\left(60\\right)[\/latex].<\/p>\n<table id=\"Table_01_07_04\" summary=\"Two rows and five columns. The first row is labeled\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>[latex]t\\text{ (minutes)}[\/latex]<\/td>\n<td>30<\/td>\n<td>50<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<td>90<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex]<\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>50<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q245501\">Show Solution<\/span><\/p>\n<div id=\"q245501\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex]f\\left(60\\right)=50[\/latex]. In 60 minutes, 50 miles are traveled.<\/p>\n<p>b. [latex]{f}^{-1}\\left(60\\right)=70[\/latex]. To travel 60 miles, it will take 70 minutes.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137418615\">\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/span><\/h2>\n<p id=\"fs-id1165137400045\">We saw in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/precalcone\/chapter\/introduction\/\" target=\"_blank\" rel=\"noopener\">Functions and Function Notation<\/a> that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the <em>vertical<\/em> extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the <em>horizontal<\/em> extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function\u2019s graph.<\/p>\n<div id=\"fs-id1165133045388\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135333128\">How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\n<ol id=\"fs-id1165137464840\">\n<li>Find the desired input on the <em>y<\/em>-axis of the given graph.<\/li>\n<li>Read the inverse function\u2019s output from the <em>x<\/em>-axis of the given graph.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_07_07\" class=\"example\">\n<div id=\"fs-id1165135434803\" class=\"exercise\">\n<div id=\"fs-id1165135434805\" class=\"problem textbox shaded\">\n<h3>Example 6: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/h3>\n<p>A function [latex]g\\left(x\\right)[\/latex] is given in Figure 5. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q56740\">Show Solution<\/span><\/p>\n<div id=\"q56740\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137468842\">To evaluate [latex]g\\left(3\\right)[\/latex], we find 3 on the <em>x<\/em>-axis and find the corresponding output value on the <em>y<\/em>-axis. The point [latex]\\left(3,1\\right)[\/latex] tells us that [latex]g\\left(3\\right)=1[\/latex].<\/p>\n<p>To evaluate [latex]{g}^{-1}\\left(3\\right)[\/latex], recall that by definition [latex]{g}^{-1}\\left(3\\right)[\/latex] means the value of <em>x<\/em> for which [latex]g\\left(x\\right)=3[\/latex]. By looking for the output value 3 on the vertical axis, we find the point [latex]\\left(5,3\\right)[\/latex] on the graph, which means [latex]g\\left(5\\right)=3[\/latex], so by definition, [latex]{g}^{-1}\\left(3\\right)=5[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><b><\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137812560\">Using the graph in Example 6, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q307093\">Show Solution<\/span><\/p>\n<div id=\"q307093\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. 3; b. 5.6<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137605437\">\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Finding Inverses of Functions Represented by Formulas<\/span><\/h2>\n<p id=\"fs-id1165137433184\">Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula\u2014 for example, [latex]y[\/latex] as a function of [latex]x\\text{-\\hspace{0.17em}}[\/latex] we can often find the inverse function by solving to obtain [latex]x[\/latex] as a function of [latex]y[\/latex].<\/p>\n<div id=\"fs-id1165137652548\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135195849\">How To: Given a function represented by a formula, find the inverse.<\/h3>\n<ol id=\"fs-id1165135443898\">\n<li>Make sure [latex]f[\/latex] is a one-to-one function.<\/li>\n<li>Solve for [latex]x[\/latex].<\/li>\n<li>Interchange [latex]x[\/latex] and [latex]y[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_07_08\" class=\"example\">\n<div id=\"fs-id1165135186316\" class=\"exercise\">\n<div id=\"fs-id1165135186318\" class=\"problem textbox shaded\">\n<h3>Example 7: Inverting the Fahrenheit-to-Celsius Function<\/h3>\n<p id=\"fs-id1165137596585\">Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.<\/p>\n<div id=\"fs-id1165133306998\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/div>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q386193\">Show Solution<\/span><\/p>\n<div id=\"q386193\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{gathered} C =\\frac{5}{9}\\left(F - 32\\right) \\\\ C\\cdot \\frac{9}{5}=F - 32 \\\\ F=\\frac{9}{5}C+32 \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165137819987\">By solving in general, we have uncovered the inverse function. If<\/p>\n<p style=\"text-align: center\">[latex]C=h\\left(F\\right)=\\frac{5}{9}\\left(F - 32\\right)[\/latex],<\/p>\n<p id=\"fs-id1165135435603\">then<\/p>\n<p style=\"text-align: center\">[latex]F={h}^{-1}\\left(C\\right)=\\frac{9}{5}C+32[\/latex].<\/p>\n<p id=\"fs-id1165137573279\">In this case, we introduced a function [latex]h[\/latex] to represent the conversion because the input and output variables are descriptive, and writing [latex]{C}^{-1}[\/latex] could get confusing.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135563331\">Solve for [latex]x[\/latex] in terms of [latex]y[\/latex] given [latex]y=\\frac{1}{3}\\left(x - 5\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q320735\">Show Solution<\/span><\/p>\n<div id=\"q320735\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=3y+5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_01_07_09\" class=\"example\">\n<div id=\"fs-id1165134065146\" class=\"exercise\">\n<div id=\"fs-id1165137409366\" class=\"problem textbox shaded\">\n<h3>Example 8: Solving to Find an Inverse Function<\/h3>\n<p id=\"fs-id1165137891504\">Find the inverse of the function [latex]f\\left(x\\right)=\\frac{2}{x - 3}+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q91472\">Show Solution<\/span><\/p>\n<div id=\"q91472\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align}&y=\\frac{2}{x - 3}+4 && \\text{Set up an equation}. \\\\ &y - 4=\\frac{2}{x - 3} && \\text{Subtract 4 from both sides}. \\\\ &x - 3=\\frac{2}{y - 4} && \\text{Multiply both sides by }x - 3\\text{ and divide by }y - 4. \\\\ &x=\\frac{2}{y - 4}+3 && \\text{Add 3 to both sides}. \\end{align}[\/latex]<\/p>\n<p>So [latex]{f}^{-1}\\left(y\\right)=\\frac{2}{y - 4}+3[\/latex] or [latex]{f}^{-1}\\left(x\\right)=\\frac{2}{x - 4}+3[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135394231\">The domain and range of [latex]f[\/latex] exclude the values 3 and 4, respectively. [latex]f[\/latex] and [latex]{f}^{-1}[\/latex] are equal at two points but are not the same function, as we can see by creating\u00a0the table below.<\/p>\n<table id=\"Table_01_07_05\" summary=\"The values of f(x) are: f(1)=3, f(2)=2, and f(5)=5. So f^(-1)(y)=y.\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>5<\/td>\n<td>[latex]{f}^{-1}\\left(y\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\n<td>3<\/td>\n<td>2<\/td>\n<td>5<\/td>\n<td>[latex]y[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_01_07_10\" class=\"example\">\n<div id=\"fs-id1165137603677\" class=\"exercise\">\n<div id=\"fs-id1165137547656\" class=\"problem textbox shaded\">\n<h3>Example 9: Solving to Find an Inverse with Radicals<\/h3>\n<p id=\"fs-id1165137841687\">Find the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt[\\leftroot{-1}\\uproot{2}]{x - 4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q109918\">Show Solution<\/span><\/p>\n<div id=\"q109918\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align}&y=2+\\sqrt{x - 4} \\\\ &{\\left(y - 2\\right)}^{2}=x - 4 \\\\ &x={\\left(y - 2\\right)}^{2}+4 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135160183\">So [latex]{f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4[\/latex].<\/p>\n<p>The domain of [latex]f[\/latex] is [latex]\\left[4,\\infty \\right)[\/latex]. Notice that the range of [latex]f[\/latex] is [latex]\\left[2,\\infty \\right)[\/latex], so this means that the domain of the inverse function [latex]{f}^{-1}[\/latex] is also [latex]\\left[2,\\infty \\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135546050\">The formula we found for [latex]{f}^{-1}\\left(x\\right)[\/latex] looks like it would be valid for all real [latex]x[\/latex]. However, [latex]{f}^{-1}[\/latex] itself must have an inverse (namely, [latex]f[\/latex] ) so we have to restrict the domain of [latex]{f}^{-1}[\/latex] to [latex]\\left[2,\\infty \\right)[\/latex] in order to make [latex]{f}^{-1}[\/latex] a one-to-one function. This domain of [latex]{f}^{-1}[\/latex] is exactly the range of [latex]f[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137756074\">What is the inverse of the function [latex]f\\left(x\\right)=2-\\sqrt[\\leftroot{-1}\\uproot{2}]{x}?[\/latex] State the domains of both the function and the inverse function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q479951\">Show Solution<\/span><\/p>\n<div id=\"q479951\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2}[\/latex]; domain of [latex]f:\\left[0,\\infty \\right)[\/latex]; domain of [latex]{f}^{-1}:\\left(-\\infty ,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm166546\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=166546&theme=oea&iframe_resize_id=ohm166546\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as in Figure 7.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010625\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7.<\/b> Quadratic function with domain restricted to [0, \u221e).<\/p>\n<\/div>\n<p id=\"fs-id1165137419977\"><strong>Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.<\/p>\n<p id=\"fs-id1165137656093\">We already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt[\\leftroot{-1}\\uproot{2}]{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]<\/p>\n<p>We notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown in Figure 8.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0092.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"251\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8.<\/b> Square and square-root functions on the non-negative domain<\/p>\n<\/div>\n<p id=\"fs-id1165137393212\">This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.<\/p>\n<div id=\"Example_01_07_11\" class=\"example\">\n<div id=\"fs-id1165134430460\" class=\"exercise\">\n<div id=\"fs-id1165134430463\" class=\"problem textbox shaded\">\n<h3>Example 10: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\n<p>Given the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q990585\">Show Solution<\/span><\/p>\n<div id=\"q990585\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137407660\">This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<p>If we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in Figure 10.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010626\/CNX_Precalc_Figure_01_07_0112.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10.<\/b> The function and its inverse, showing reflection about the identity line<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm166540\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=166540&theme=oea&iframe_resize_id=ohm166540\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1165137911739\" class=\"solution\">\n<div id=\"fs-id1165137627081\" class=\"note precalculus qa textbox\">\n<p id=\"fs-id1165134388228\"><strong>Q &amp; A <\/strong><\/p>\n<p><strong>Is there any function that is equal to its own inverse?<\/strong><\/p>\n<p id=\"fs-id1165137602656\"><em>Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em><\/p>\n<p style=\"text-align: center\">[latex]\\frac{1}{1\/x}=x[\/latex]<\/p>\n<p id=\"fs-id1165137897050\"><em>Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em><\/p>\n<\/div>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the inverse of a polynomial function.<\/li>\n<li>Restrict the domain to find the inverse of a polynomial function.<\/li>\n<li>Find or evaluate the inverse of a function.<\/li>\n<\/ul>\n<\/div>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010800\/CNX_Precalc_Figure_03_08_0012.jpg\" alt=\"Gravel in the shape of a cone.\" width=\"487\" height=\"410\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137793975\">A mound of gravel is in the shape of a cone with the height equal to twice the radius.<span id=\"fs-id1165137939558\"><br \/>\n<\/span><\/p>\n<p id=\"fs-id1165137411369\">The volume is found using a formula from elementary geometry.<\/p>\n<div id=\"eip-854\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}V&=\\frac{1}{3}\\pi {r}^{2}h \\\\ &=\\frac{1}{3}\\pi {r}^{2}\\left(2r\\right) \\\\ &=\\frac{2}{3}\\pi {r}^{3} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137727278\">We have written the volume <em>V<\/em>\u00a0in terms of the radius <em>r<\/em>. However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula<\/p>\n<div id=\"eip-931\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]r=\\sqrt[3]{\\dfrac{3V}{2\\pi }\\\\}[\/latex]<\/div>\n<p id=\"fs-id1165134129769\">This function is the inverse of the formula for <em>V<\/em>\u00a0in terms of <em>r<\/em>.<\/p>\n<h2>Find the inverse of a polynomial function<\/h2>\n<p id=\"fs-id1165137439029\">Two functions <em>f<\/em>\u00a0and <em>g<\/em>\u00a0are inverse functions if for every coordinate pair in <em>f<\/em>, (<em>a<\/em>, <em>b<\/em>), there exists a corresponding coordinate pair in the inverse function, <em>g<\/em>, (<em>b<\/em>, <em>a<\/em>). In other words, the coordinate pairs of the inverse functions have the input and output interchanged.<\/p>\n<p id=\"fs-id1165137475924\">For a function to have an <strong>inverse function<\/strong> the function to create a new function that is <strong>one-to-one<\/strong> and would have an inverse function.<\/p>\n<p id=\"fs-id1165137448308\">For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown\u00a0below. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010801\/CNX_Precalc_Figure_03_08_0022.jpg\" alt=\"Diagram of a parabolic trough that is 18\" width=\"487\" height=\"279\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137793665\">Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with <em>x<\/em>\u00a0measured horizontally and <em>y<\/em>\u00a0measured vertically, with the origin at the vertex of the parabola.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010801\/CNX_Precalc_Figure_03_08_0032.jpg\" alt=\"Graph of a parabola.\" width=\"487\" height=\"441\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137771677\">From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form [latex]y\\left(x\\right)=a{x}^{2}[\/latex]. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor <em>a<\/em>.<\/p>\n<div id=\"eip-893\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align} 18&=a{6}^{2} \\\\ a&=\\frac{18}{36} \\\\ &=\\frac{1}{2} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137633973\">Our parabolic cross section has the equation<\/p>\n<div id=\"eip-420\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/div>\n<p id=\"fs-id1165137770004\">We are interested in the <strong>surface area<\/strong> of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth <em>y<\/em>\u00a0the width will be given by 2<em>x<\/em>, so we need to solve the equation above for <em>x<\/em>\u00a0and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.<\/p>\n<p id=\"fs-id1165137638570\">To find an inverse, we can restrict our original function to a limited domain on which it <em>is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive <em>x<\/em>\u00a0values. On this domain, we can find an inverse by solving for the input variable:<\/p>\n<div id=\"eip-598\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}y&=\\frac{1}{2}{x}^{2} \\\\ 2y&={x}^{2} \\\\ x&=\\pm \\sqrt{2y} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137453965\">This is not a function as written. We are limiting ourselves to positive <em>x<\/em>\u00a0values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.<\/p>\n<div id=\"eip-793\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]y=\\frac{{x}^{2}}{2},\\text{ }x>0[\/latex]<\/div>\n<p id=\"fs-id1165137643958\">Because <em>x<\/em>\u00a0is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2<em>x<\/em>. The trough is 3 feet (36 inches) long, so the surface area will then be:<\/p>\n<div id=\"eip-491\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\text{Area} & =l\\cdot w \\\\ & =36\\cdot 2x \\\\ & =72x \\\\ & =72\\sqrt{2y} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137407432\">This example illustrates two important points:<\/p>\n<ol id=\"fs-id1165135545666\">\n<li>When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.<\/li>\n<li>The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.<\/li>\n<\/ol>\n<p id=\"fs-id1165137618975\">Functions involving roots are often called <strong>radical functions<\/strong>. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called <strong>invertible functions<\/strong>, and we use the notation [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<p id=\"fs-id1165135185952\">Warning: [latex]{f}^{-1}\\left(x\\right)[\/latex] is not the same as the reciprocal of the function [latex]f\\left(x\\right)[\/latex]. This use of \u20131 is reserved to denote inverse functions. To denote the reciprocal of a function [latex]f\\left(x\\right)[\/latex], we would need to write [latex]{\\left(f\\left(x\\right)\\right)}^{-1}=\\frac{1}{f\\left(x\\right)}[\/latex].<\/p>\n<p id=\"fs-id1165137561919\">An important relationship between inverse functions is that they &#8220;undo&#8221; each other. If [latex]{f}^{-1}[\/latex] is the inverse of a function <em>f<\/em>,\u00a0then <em>f<\/em>\u00a0is the inverse of the function [latex]{f}^{-1}[\/latex]. In other words, whatever the function <em>f<\/em>\u00a0does to <em>x<\/em>, [latex]{f}^{-1}[\/latex] undoes it\u2014and vice-versa. More formally, we write<\/p>\n<div id=\"eip-519\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }f[\/latex]<\/div>\n<p id=\"fs-id1165135503755\">and<\/p>\n<div id=\"eip-590\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }{f}^{-1}[\/latex]<\/div>\n<div id=\"fs-id1165137735698\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Verifying Two Functions Are Inverses of One Another<\/h3>\n<p id=\"fs-id1165137852132\">Two functions, <em>f<\/em>\u00a0and <i>g<\/i>, are inverses of one another if for all <em>x<\/em>\u00a0in the domain of <em>f\u00a0<\/em>and <em>g<\/em>.<\/p>\n<div id=\"eip-973\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137646263\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137501372\">How To: Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.<\/h3>\n<ol id=\"fs-id1165133276230\">\n<li>Replace [latex]f\\left(x\\right)[\/latex] with <em>y<\/em>.<\/li>\n<li>Interchange <em>x<\/em>\u00a0and <em>y<\/em>.<\/li>\n<li>Solve for <em>y<\/em>, and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_08_01\" class=\"example\">\n<div id=\"fs-id1165135150650\" class=\"exercise\">\n<div id=\"fs-id1165135620877\" class=\"problem textbox shaded\">\n<h3>Example 1: Verifying Inverse Functions<\/h3>\n<p id=\"fs-id1165134148383\">Show that [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex] and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses, for [latex]x\\ne 0,-1[\/latex] .<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q28284\">Show Solution<\/span><\/p>\n<div id=\"q28284\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137834138\">We must show that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{f}^{-1}\\left(f\\left(x\\right)\\right)&={f}^{-1}\\left(\\frac{1}{x+1}\\right) \\\\ &=\\frac{1}{\\frac{1}{x+1}}-1 \\\\ &=\\left(x+1\\right)-1 \\\\ &=x\\end{align}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}f\\left({f}^{-1}\\left(x\\right)\\right)&=f\\left(\\frac{1}{x}-1\\right) \\\\ &=\\frac{1}{\\left(\\frac{1}{x}-1\\right)+1} \\\\ &=\\frac{1}{\\frac{1}{x}} \\\\ &=x \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135168183\">Therefore, [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137426116\">Show that [latex]f\\left(x\\right)=\\frac{x+5}{3}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=3x - 5[\/latex] are inverses.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q692921\">Show Solution<\/span><\/p>\n<div id=\"q692921\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{x+5}{3}\\right)=3\\left(\\frac{x+5}{3}\\right)-5=\\left(x - 5\\right)+5=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(3x - 5\\right)=\\frac{\\left(3x - 5\\right)+5}{3}=\\frac{3x}{3}=x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_08_02\" class=\"example\">\n<div id=\"fs-id1165137600799\" class=\"exercise\">\n<div id=\"fs-id1165135160775\" class=\"problem textbox shaded\">\n<h3>Example 2: Finding the Inverse of a Cubic Function<\/h3>\n<p id=\"fs-id1165137569920\">Find the inverse of the function [latex]f\\left(x\\right)=5{x}^{3}+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q790658\">Show Solution<\/span><\/p>\n<div id=\"q790658\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135412872\">This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for <em>x<\/em>.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}y=5{x}^{3}+1 \\\\ x=5{y}^{3}+1 \\\\ x - 1=5{y}^{3} \\\\ \\frac{x - 1}{5}={y}^{3} \\\\ {f}^{-1}\\left(x\\right)=\\sqrt[3]{\\frac{x - 1}{5}} \\end{gathered}[\/latex]<\/p>\n<div id=\"Example_03_08_02\" class=\"example\">\n<div id=\"fs-id1165137600799\" class=\"exercise\">\n<div id=\"fs-id1165137635322\" class=\"commentary\">\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137641602\">Look at the graph of <em>f<\/em>\u00a0and [latex]{f}^{-1}[\/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[\/latex]. This is always the case when graphing a function and its inverse function.<\/p>\n<p id=\"fs-id1165137793468\">Also, since the method involved interchanging <em>x<\/em>\u00a0and <em>y<\/em>, notice corresponding points. If [latex]\\left(a,b\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(b,a\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Since [latex]\\left(0,1\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(1,0\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Similarly, since [latex]\\left(1,6\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(6,1\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010801\/CNX_Precalc_Figure_03_08_0042.jpg\" alt=\"Graph of f(x)=5x^3+1 and its inverse, f^(-1)(x)=3sqrt((x-1)\/(5)).\" width=\"487\" height=\"554\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137635322\" class=\"commentary\"><\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165133047522\">Find the inverse function of [latex]f\\left(x\\right)=\\sqrt[3]{x+4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q242169\">Show Solution<\/span><\/p>\n<div id=\"q242169\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{f}^{-1}\\left(x\\right)={x}^{3}-4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm1620\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1620&theme=oea&iframe_resize_id=ohm1620\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Restrict the domain to find the inverse of a polynomial function<\/h2>\n<p id=\"fs-id1165137471808\">So far, we have been able to find the inverse functions of <strong>cubic functions<\/strong> without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an <strong>inverse function<\/strong>. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.<\/p>\n<div id=\"fs-id1165137434585\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Restricting the Domain<\/h3>\n<p id=\"fs-id1165137409777\">If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.<\/p>\n<\/div>\n<div id=\"fs-id1165137431545\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137656706\">How To: Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.<\/h3>\n<ol id=\"fs-id1165137532171\">\n<li>Restrict the domain by determining a domain on which the original function is one-to-one.<\/li>\n<li>Replace <em>f<\/em>(<em>x<\/em>)\u00a0with <em>y<\/em>.<\/li>\n<li>Interchange <em>x<\/em>\u00a0and <em>y<\/em>.<\/li>\n<li>Solve for <em>y<\/em>, and rename the function or pair of function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\n<li>Revise the formula for [latex]{f}^{-1}\\left(x\\right)[\/latex] by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_08_03\" class=\"example\">\n<div id=\"fs-id1165137737103\" class=\"exercise\">\n<div id=\"fs-id1165137668130\" class=\"problem textbox shaded\">\n<h3>Example 3: Restricting the Domain to Find the Inverse of a Polynomial Function<\/h3>\n<p id=\"fs-id1165137482766\">Find the inverse function of <em>f<\/em>:<\/p>\n<ol id=\"fs-id1165137638318\">\n<li>[latex]f\\left(x\\right)={\\left(x - 4\\right)}^{2}, x\\ge 4[\/latex]<\/li>\n<li>[latex]f\\left(x\\right)={\\left(x - 4\\right)}^{2}, x\\le 4[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q546705\">Show Solution<\/span><\/p>\n<div id=\"q546705\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137506731\">The original function [latex]f\\left(x\\right)={\\left(x - 4\\right)}^{2}[\/latex] is not one-to-one, but the function is restricted to a domain of [latex]x\\ge 4[\/latex] or [latex]x\\le 4[\/latex] on which it is one-to-one.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010801\/CNX_Precalc_Figure_03_08_0052.jpg\" alt=\"&quot;Two\" \/><\/p>\n<p id=\"fs-id1165137706306\">To find the inverse, start by replacing [latex]f\\left(x\\right)[\/latex] with the simple variable <em>y<\/em>.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&y={\\left(x - 4\\right)}^{2} && \\text{Interchange } x \\text{ and }y. \\\\ &x={\\left(y - 4\\right)}^{2} && \\text{Take the square root}. \\\\ &\\pm \\sqrt{x}=y - 4 && \\text{Add } 4 \\text{ to both sides}. \\\\ &4\\pm \\sqrt{x}=y \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137444285\">This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of <em>x<\/em>\u00a0and <em>y<\/em>\u00a0for the original <em>f<\/em>(<em>x<\/em>), we looked at the domain: the values <em>x<\/em>\u00a0could assume. When we reversed the roles of <em>x<\/em>\u00a0and <em>y<\/em>,\u00a0this gave us the values <em>y<\/em>\u00a0could assume. For this function, [latex]x\\ge 4[\/latex], so for the inverse, we should have [latex]y\\ge 4[\/latex], which is what our inverse function gives.<\/p>\n<ol id=\"fs-id1165137735027\">\n<li>The domain of the original function was restricted to [latex]x\\ge 4[\/latex], so the outputs of the inverse need to be the same, [latex]f\\left(x\\right)\\ge 4[\/latex], and we must use the + case:\n<div id=\"eip-id1165134294825\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)=4+\\sqrt{x}[\/latex]<\/div>\n<\/li>\n<li>The domain of the original function was restricted to [latex]x\\le 4[\/latex], so the outputs of the inverse need to be the same, [latex]f\\left(x\\right)\\le 4[\/latex], and we must use the \u2013 case:\n<div id=\"eip-id1165137482501\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)=4-\\sqrt{x}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137534054\">On the graphs below, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line [latex]y=x[\/latex]. The coordinate pair [latex]\\left(4, 0\\right)[\/latex] is on the graph of <em>f<\/em>\u00a0and the coordinate pair [latex]\\left(0, 4\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. For any coordinate pair, if (<em>a<\/em>, <em>b<\/em>) is on the graph of <em>f<\/em>, then (<em>b<\/em>, <em>a<\/em>) is on the graph of [latex]{f}^{-1}[\/latex]. Finally, observe that the graph of <em>f<\/em>\u00a0intersects the graph of [latex]{f}^{-1}[\/latex] on the line <em>y\u00a0<\/em>= <em>x<\/em>. Points of intersection for the graphs of <em>f<\/em>\u00a0and [latex]{f}^{-1}[\/latex] will always lie on the line <em>y\u00a0<\/em>= <em>x<\/em>.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010802\/CNX_Precalc_Figure_03_08_0062.jpg\" alt=\"Two graphs of a parabolic function with half of its inverse.\" width=\"975\" height=\"442\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137736620\" class=\"commentary\"><\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_08_04\" class=\"example\">\n<div id=\"fs-id1165137786597\" class=\"exercise\">\n<div id=\"fs-id1165135481235\" class=\"problem textbox shaded\">\n<h3>Example 4: Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified<\/h3>\n<p id=\"fs-id1165137410909\">Restrict the domain and then find the inverse of<\/p>\n<p style=\"text-align: center\">[latex]f\\left(x\\right)={\\left(x - 2\\right)}^{2}-3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q368262\">Show Solution<\/span><\/p>\n<div id=\"q368262\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137414415\">We can see this is a parabola with vertex at [latex]\\left(2, -3\\right)[\/latex] that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to [latex]x\\ge 2[\/latex].<\/p>\n<p id=\"fs-id1165137842529\">To find the inverse, we will use the vertex form of the quadratic. We start by replacing <em>f<\/em>(<em>x<\/em>) with a simple variable, <em>y<\/em>, then solve for <em>x<\/em>.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&y={\\left(x - 2\\right)}^{2}-3 && \\text{Interchange } x \\text{ and } y. \\\\ &x={\\left(y - 2\\right)}^{2}-3 && \\text{Add 3 to both sides}. \\\\ &x+3={\\left(y - 2\\right)}^{2} && \\text{Take the square root}. \\\\ &\\pm \\sqrt{x+3}=y - 2 && \\text{Add 2 to both sides}. \\\\ &2\\pm \\sqrt{x+3}=y && \\text{Rename the function}. \\\\ &{f}^{-1}\\left(x\\right)=2\\pm \\sqrt{x+3} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137419504\">Now we need to determine which case to use. Because we restricted our original function to a domain of [latex]x\\ge 2[\/latex], the outputs of the inverse should be the same, telling us to utilize the + case<\/p>\n<p style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)=2+\\sqrt{x+3}[\/latex]<\/p>\n<p id=\"fs-id1165137827988\">If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.<\/p>\n<div id=\"Example_03_08_04\" class=\"example\">\n<div id=\"fs-id1165137786597\" class=\"exercise\">\n<div id=\"fs-id1165134362839\" class=\"commentary\">\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135259538\">Notice that we arbitrarily decided to restrict the domain on [latex]x\\ge 2[\/latex]. We could just have easily opted to restrict the domain on [latex]x\\le 2[\/latex], in which case [latex]{f}^{-1}\\left(x\\right)=2-\\sqrt{x+3}[\/latex]. Observe the original function graphed on the same set of axes as its inverse function in the graph below. Notice that both graphs show symmetry about the line <em>y<\/em> =\u00a0<em>x<\/em>. The coordinate pair [latex]\\left(2,\\text{ }-3\\right)[\/latex] is on the graph of <em>f<\/em>\u00a0and the coordinate pair [latex]\\left(-3,\\text{ }2\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Observe from the graph of both functions on the same set of axes that<\/p>\n<div id=\"eip-id1165134122215\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\text{domain of }f=\\text{range of } {f}^{-1}=\\left[2,\\infty \\right)[\/latex]<\/div>\n<p id=\"fs-id1165137642128\">and<\/p>\n<div id=\"eip-id1165134279478\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\text{domain of }{f}^{-1}=\\text{range of } f=\\left[-3,\\infty \\right)[\/latex]<\/div>\n<p id=\"fs-id1165137723812\">Finally, observe that the graph of <em>f<\/em>\u00a0intersects the graph of [latex]{f}^{-1}[\/latex] along the line <em>y\u00a0<\/em>= <em>x<\/em>.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010802\/CNX_Precalc_Figure_03_08_0072.jpg\" alt=\"Graph of a parabolic function with half of its inverse.\" width=\"487\" height=\"487\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"background-color: #ffffff;font-size: 1em\">\u00a0<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135596379\">Find the inverse of the function [latex]f\\left(x\\right)={x}^{2}+1[\/latex], on the domain [latex]x\\ge 0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455466\">Show Solution<\/span><\/p>\n<div id=\"q455466\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x - 1}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1165137894462\">\n<h2>Solving Applications of Radical Functions<\/h2>\n<p id=\"fs-id1165137696560\">Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the <strong>inverse of a radical function<\/strong>, we will need to restrict the domain of the answer because the range of the original function is limited.<\/p>\n<div id=\"fs-id1165137415876\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137455923\">How To: Given a radical function, find the inverse.<\/h3>\n<ol id=\"fs-id1165137542989\">\n<li>Determine the range of the original function.<\/li>\n<li>Replace <em>f<\/em>(<em>x<\/em>)\u00a0with <em>y<\/em>, then solve for <em>x<\/em>.<\/li>\n<li>If necessary, restrict the domain of the inverse function to the range of the original function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_08_05\" class=\"example\">\n<div id=\"fs-id1165135173813\" class=\"exercise\">\n<div id=\"fs-id1165137399685\" class=\"problem textbox shaded\">\n<h3>Example 5: Finding the Inverse of a Radical Function<\/h3>\n<p id=\"fs-id1165135570491\">Restrict the domain and then find the inverse of the function [latex]f\\left(x\\right)=\\sqrt{x - 4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q727309\">Show Solution<\/span><\/p>\n<div id=\"q727309\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135545767\">Note that the original function has range [latex]f\\left(x\\right)\\ge 0[\/latex]. Replace [latex]f\\left(x\\right)[\/latex] with <em>y<\/em>, then solve for <em>x<\/em>.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} &y=\\sqrt{x - 4} && \\text{Replace}f\\left(x\\right)\\text{with}y.\\\\ &x=\\sqrt{y - 4} && \\text{Interchange}x\\text{and}y. \\\\ &x =\\sqrt{y - 4} && \\text{Square each side}. \\\\ &{x}^{2} =y - 4 && \\text{Add 4}. \\\\ &{x}^{2}+4 =y && \\text{Rename the function}{f}^{-1}\\left(x\\right). \\\\ &{f}^{-1}\\left(x\\right) ={x}^{2}+4 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135209570\">Recall that the domain of this function must be limited to the range of the original function.<\/p>\n<p style=\"text-align: center\">[latex]{f}^{-1}\\left(x\\right)={x}^{2}+4,x\\ge 0[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135173239\">Notice in the graph below\u00a0that the inverse is a reflection of the original function over the line <em>y\u00a0<\/em>= <em>x<\/em>. Because the original function has only positive outputs, the inverse function has only positive inputs.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010802\/CNX_Precalc_Figure_03_08_0082.jpg\" alt=\"Graph of f(x)=sqrt(x-4) and its inverse, f^(-1)(x)=x^2+4.\" width=\"487\" height=\"444\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137784775\">Restrict the domain and then find the inverse of the function [latex]f\\left(x\\right)=\\sqrt{2x+3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q107257\">Show Solution<\/span><\/p>\n<div id=\"q107257\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-3}{2},x\\ge 0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137761571\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174218\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174218&theme=oea&iframe_resize_id=ohm174218\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Applications of Radical Functions<\/h2>\n<p id=\"fs-id1165135435831\">Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section.<\/p>\n<div id=\"Example_03_08_06\" class=\"example\">\n<div id=\"fs-id1165137634475\" class=\"exercise\">\n<div id=\"fs-id1165137531120\" class=\"problem textbox shaded\">\n<h3>Example 6: Solving an Application with a Cubic Function<\/h3>\n<p id=\"fs-id1165137771982\">A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by<\/p>\n<div id=\"eip-id1165132187568\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]V=\\frac{2}{3}\\pi {r}^{3}[\/latex]<\/div>\n<p id=\"fs-id1165135181305\">Find the inverse of the function [latex]V=\\frac{2}{3}\\pi {r}^{3}[\/latex] that determines the volume <em>V<\/em>\u00a0of a cone and is a function of the radius <em>r<\/em>. Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use [latex]\\pi =3.14[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q525201\">Show Solution<\/span><\/p>\n<div id=\"q525201\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137405144\">Start with the given function for <em>V<\/em>. Notice that the meaningful domain for the function is [latex]r\\ge 0[\/latex] since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is [latex]V\\ge 0[\/latex]. Solve for <em>r<\/em>\u00a0in terms of <em>V<\/em>, using the method outlined previously.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} V&=\\frac{2}{3}\\pi {r}^{3} \\\\ {r}^{3}&=\\frac{3V}{2\\pi } && \\text{Solve for }{r}^{3}. \\\\ r&=\\sqrt[3]{\\frac{3V}{2\\pi }} && \\text{Solve for }r. \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137730324\">This is the result stated in the section opener. Now evaluate this for <em>V<\/em> = 100 and [latex]\\pi =3.14[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}r & =\\sqrt[3]{\\frac{3V}{2\\pi }} \\\\ & =\\sqrt[3]{\\frac{3\\cdot 100}{2\\cdot 3.14}} \\\\ & \\approx \\sqrt[3]{47.7707} \\\\ & \\approx 3.63 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137706279\">Therefore, the radius is about 3.63 ft.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137580023\">\n<h2>Determining the Domain of a Radical Function Composed with Other Functions<\/h2>\n<p id=\"fs-id1165134042947\">When radical functions are composed with other functions, determining domain can become more complicated.<\/p>\n<div id=\"Example_03_08_07\" class=\"example\">\n<div id=\"fs-id1165135378774\" class=\"exercise\">\n<div id=\"fs-id1165135378776\" class=\"problem textbox shaded\">\n<h3>Example 7: Finding the Domain of a Radical Function Composed with a Rational Function<\/h3>\n<p id=\"fs-id1165137658778\">Find the domain of the function [latex]f\\left(x\\right)=\\sqrt{\\frac{\\left(x+2\\right)\\left(x - 3\\right)}{\\left(x - 1\\right)}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q429624\">Show Solution<\/span><\/p>\n<div id=\"q429624\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137665549\">Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where [latex]\\frac{\\left(x+2\\right)\\left(x - 3\\right)}{\\left(x - 1\\right)}\\ge 0[\/latex]. The output of a rational function can change signs (change from positive to negative or vice versa) at <em>x<\/em>-intercepts and at vertical asymptotes. For this equation, the graph could change signs at <em>x<\/em>\u00a0= \u20132, 1, and 3.<\/p>\n<p id=\"fs-id1165135686721\">To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010802\/CNX_Precalc_Figure_03_08_0092.jpg\" alt=\"Graph of a radical function that shows where the outputs are nonnegative.\" width=\"731\" height=\"439\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137694167\">This function has two <em>x<\/em>-intercepts, both of which exhibit linear behavior near the <em>x<\/em>-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a <em>y<\/em>-intercept at (0, 6).<\/p>\n<p id=\"fs-id1165135333589\">From the <em>y<\/em>-intercept and <em>x<\/em>-intercept at <em>x\u00a0<\/em>= \u20132, we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.<\/p>\n<p id=\"fs-id1165137664081\">From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function <em>f<\/em>(<em>x<\/em>) will be defined. <em>f<\/em>(<em>x<\/em>) has domain [latex]-2\\le x<1\\text{or}x\\ge 3[\/latex], or in interval notation, [latex]\\left[-2,1\\right)\\cup \\left[3,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165135173899\">\n<h2>Finding Inverses of Rational Functions<\/h2>\n<p id=\"fs-id1165135475898\">As with finding inverses of quadratic functions, it is sometimes desirable to find the <strong>inverse of a rational function<\/strong>, particularly of rational functions that are the ratio of linear functions, such as in concentration applications.<\/p>\n<div id=\"Example_03_08_08\" class=\"example\">\n<div id=\"fs-id1165137642525\" class=\"exercise\">\n<div id=\"fs-id1165137642528\" class=\"problem textbox shaded\">\n<h3>Example 8: Finding the Inverse of a Rational Function<\/h3>\n<p id=\"fs-id1165135332364\">The function [latex]C=\\frac{20+0.4n}{100+n}[\/latex] represents the concentration <em>C<\/em>\u00a0of an acid solution after <em>n<\/em>\u00a0mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for <em>n<\/em>\u00a0in terms of <em>C<\/em>. Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q623836\">Show Solution<\/span><\/p>\n<div id=\"q623836\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134223203\">We first want the inverse of the function. We will solve for <em>n<\/em>\u00a0in terms of <em>C<\/em>.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}C=\\frac{20+0.4n}{100+n}\\\\ C\\left(100+n\\right)=20+0.4n\\\\ 100C+Cn=20+0.4n\\\\ 100C - 20=0.4n-Cn\\\\ 100C - 20=\\left(0.4-C\\right)n\\\\ n=\\frac{100C - 20}{0.4-C}\\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165137637474\">Now evaluate this function for C=0.35 (35%).<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}n&=\\frac{100\\left(0.35\\right)-20}{0.4 - 0.35}\\\\ &=\\frac{15}{0.05}\\\\ &=300\\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137706154\">We can conclude that 300 mL of the 40% solution should be added.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134042923\">Find the inverse of the function [latex]f\\left(x\\right)=\\frac{x+3}{x - 2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q457099\">Show Solution<\/span><\/p>\n<div id=\"q457099\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{f}^{-1}\\left(x\\right)=\\frac{2x+3}{x - 1}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm34476\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=34476&theme=oea&iframe_resize_id=ohm34476\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>If [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex], then<\/li>\n<li>[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex].<\/li>\n<li>Each of the toolkit functions has an inverse.<\/li>\n<li>For a function to have an inverse, it must be one-to-one (pass the horizontal line test).<\/li>\n<li>A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.<\/li>\n<li>For a tabular function, exchange the input and output rows to obtain the inverse.<\/li>\n<li>The inverse of a function can be determined at specific points on its graph.<\/li>\n<li>To find the inverse of a formula, solve the equation [latex]y=f\\left(x\\right)[\/latex] for [latex]x[\/latex] as a function of\u00a0[latex]y[\/latex]. Then exchange the labels [latex]x[\/latex] and [latex]y[\/latex].<\/li>\n<li>The graph of an inverse function is the reflection of the graph of the original function across the line [latex]y=x[\/latex].<\/li>\n<li>The inverse of a quadratic function is a square root function.<\/li>\n<li>If [latex]{f}^{-1}[\/latex]\u00a0is the inverse of a function <em>f<\/em>,\u00a0then <em>f<\/em>\u00a0is the inverse of the function [latex]{f}^{-1}[\/latex].<\/li>\n<li>While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible.<\/li>\n<li>To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one.<\/li>\n<li>When finding the inverse of a radical function, we need a restriction on the domain of the answer.<\/li>\n<li>Inverse and radical and functions can be used to solve application problems.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137441703\" class=\"definition\">\n<dt>inverse function<\/dt>\n<dd id=\"fs-id1165137441708\">for any one-to-one function [latex]f\\left(x\\right)[\/latex], the inverse is a function [latex]{f}^{-1}\\left(x\\right)[\/latex] such that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]; this also implies that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex]<\/dd>\n<\/dl>\n<\/div>\n<\/section>\n<dl id=\"fs-id1165135169260\" class=\"definition\">\n<dt><strong>invertible function<\/strong><\/dt>\n<dd id=\"fs-id1165135169263\">any function that has an inverse function<\/dd>\n<\/dl>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-13771\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":359553,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["racv2"],"pb_section_license":""},"chapter-type":[],"contributor":[64],"license":[],"class_list":["post-13771","chapter","type-chapter","status-publish","hentry","contributor-racv2"],"part":15999,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13771","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/users\/359553"}],"version-history":[{"count":17,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13771\/revisions"}],"predecessor-version":[{"id":15998,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13771\/revisions\/15998"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/parts\/15999"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13771\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/media?parent=13771"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=13771"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/contributor?post=13771"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/license?post=13771"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}