{"id":157,"date":"2023-06-21T13:22:39","date_gmt":"2023-06-21T13:22:39","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/define-and-graph-an-inverse\/"},"modified":"2023-07-04T04:05:14","modified_gmt":"2023-07-04T04:05:14","slug":"define-and-graph-an-inverse","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/define-and-graph-an-inverse\/","title":{"raw":"\u25aa   Define and Graph an Inverse","rendered":"\u25aa   Define and Graph an Inverse"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find and evaluate the inverse of a function in tabular form.<\/li>\r\n \t<li>Find and evaluate the inverse of a function given as an equation.<\/li>\r\n \t<li>Graph a function and its inverse.<\/li>\r\n<\/ul>\r\n<\/div>\r\nOnce we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.\r\n<h2>Inverting Tabular Functions<\/h2>\r\nSuppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.\r\n\r\nEach row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.\r\n<div class=\"textbox examples\">\r\n<h3>recall reading functions from tables<\/h3>\r\nLook to see which row or column of a table is labeled as the input and which is labeled as the output.\r\n\r\nThe tables below indicate one row represents \"[latex]t[\/latex]\" and the other, [latex]f(t)[\/latex].\r\n\r\nIt is common for input to be represented by time when output is a distance.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Interpreting the Inverse of a Tabular Function<\/h3>\r\nA function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes. Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].\r\n<table summary=\"Two rows and five columns. The first row is labeled\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\r\n<td>30<\/td>\r\n<td>50<\/td>\r\n<td>70<\/td>\r\n<td>90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\r\n<td>20<\/td>\r\n<td>40<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"61261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"61261\"]\r\n\r\nThe inverse function takes an output of [latex]f[\/latex] and returns an input for [latex]f[\/latex]. So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.\r\n\r\nAlternatively, recall that the definition of the inverse was that if [latex]f\\left(a\\right)=b[\/latex], then [latex]{f}^{-1}\\left(b\\right)=a[\/latex]. By this definition, if we are given [latex]{f}^{-1}\\left(70\\right)=a[\/latex], then we are looking for a value [latex]a[\/latex] so that [latex]f\\left(a\\right)=70[\/latex]. In this case, we are looking for a [latex]t[\/latex] so that [latex]f\\left(t\\right)=70[\/latex], which is when [latex]t=90[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the table below,\u00a0find and interpret (a) [latex]\\text{ }f\\left(60\\right)[\/latex], and (b) [latex]\\text{ }{f}^{-1}\\left(60\\right)[\/latex].\r\n<table summary=\"Two rows and five columns. The first row is labeled\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>[latex]t\\text{ (minutes)}[\/latex]<\/td>\r\n<td>30<\/td>\r\n<td>50<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<td>90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex]<\/td>\r\n<td>20<\/td>\r\n<td>40<\/td>\r\n<td>50<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"401025\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"401025\"]\r\n\r\na.\u00a0[latex]f\\left(60\\right)=50[\/latex]. In 60 minutes, 50 miles are traveled.\r\n\r\nb. [latex]{f}^{-1}\\left(60\\right)=70[\/latex]. To travel 60 miles, it will take 70 minutes.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/h2>\r\nThe domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the <em>vertical<\/em> extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the <em>horizontal<\/em> extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function\u2019s graph.\r\n<div class=\"textbox\">\r\n<h3>How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\r\n<ol>\r\n \t<li>Find the desired input of the inverse function on the [latex]y[\/latex]-axis of the given graph.<\/li>\r\n \t<li>Read the inverse function\u2019s output from the [latex]x[\/latex]-axis of the given graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/h3>\r\nA function [latex]g\\left(x\\right)[\/latex] is given\u00a0below. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205520\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/>\r\n\r\n[reveal-answer q=\"334632\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"334632\"]\r\n\r\nTo evaluate [latex]g\\left(3\\right)[\/latex], we find 3 on the <em>x<\/em>-axis and find the corresponding output value on the [latex]y[\/latex]-axis. The point [latex]\\left(3,1\\right)[\/latex] tells us that [latex]g\\left(3\\right)=1[\/latex].\r\n\r\nTo evaluate [latex]{g}^{-1}\\left(3\\right)[\/latex], recall that by definition [latex]{g}^{-1}\\left(3\\right)[\/latex] means the value of <em>x<\/em> for which [latex]g\\left(x\\right)=3[\/latex]. By looking for the output value 3 on the vertical axis, we find the point [latex]\\left(5,3\\right)[\/latex] on the graph, which means [latex]g\\left(5\\right)=3[\/latex], so by definition, [latex]{g}^{-1}\\left(3\\right)=5[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205521\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the graph in the previous example, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"350455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"350455\"]\r\n\r\na. 3; b. 5.6\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Inverses of Functions Represented by Formulas<\/h2>\r\nSometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula\u2014for example, [latex]y[\/latex] as a function of [latex]x-[\/latex] we can often find the inverse function by solving to obtain [latex]x[\/latex] as a function of [latex]y[\/latex].\r\n<div class=\"textbox\">\r\n<h3>How To: Given a function represented by a formula, find the inverse.<\/h3>\r\n<ol>\r\n \t<li>Verify that\u00a0[latex]f[\/latex] is a one-to-one function.<\/li>\r\n \t<li>Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex].<\/li>\r\n \t<li>Interchange [latex]x[\/latex]\u00a0and [latex]y[\/latex].<\/li>\r\n \t<li>Solve for [latex]y[\/latex], and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Inverting the Fahrenheit-to-Celsius Function<\/h3>\r\nFind a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.\r\n<p style=\"text-align: center;\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"625400\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"625400\"]\r\n<p style=\"text-align: center;\">[latex]{ C }=\\frac{5}{9}\\left(F - 32\\right)[\/latex]\r\n[latex]C\\cdot \\frac{9}{5}=F - 32[\/latex]\r\n[latex]F=\\frac{9}{5}C+32[\/latex]<\/p>\r\nBy solving in general, we have uncovered the inverse function. If\r\n<p style=\"text-align: center;\">[latex]C=h\\left(F\\right)=\\frac{5}{9}\\left(F - 32\\right)[\/latex],<\/p>\r\nthen\r\n<p style=\"text-align: center;\">[latex]F={h}^{-1}\\left(C\\right)=\\frac{9}{5}C+32[\/latex].<\/p>\r\nIn this case, we introduced a function [latex]h[\/latex] to represent the conversion because the input and output variables are descriptive, and writing [latex]{C}^{-1}[\/latex] could get confusing.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve for [latex]x[\/latex] in terms of [latex]y[\/latex] given [latex]y=\\frac{1}{3}\\left(x - 5\\right)[\/latex]\r\n\r\n[reveal-answer q=\"875458\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"875458\"]\r\n\r\n[latex]x=3y+5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving to Find an Inverse Function<\/h3>\r\nFind the inverse of the function [latex]f\\left(x\\right)=\\dfrac{2}{x - 3}+4[\/latex].\r\n<p style=\"text-align: left;\">[reveal-answer q=\"903464\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"903464\"]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=\\frac{2}{x - 3}+4 &amp;&amp; \\text{Change }f(x)\\text{ to }y. \\\\[1.5mm]&amp;x=\\frac{2}{y - 3}+4 &amp;&amp; \\text{Switch }x\\text{ and }y. \\\\[1.5mm] &amp;y - 4=\\frac{2}{x - 3} &amp;&amp; \\text{Subtract 4 from both sides}. \\\\[1.5mm] &amp;y - 3=\\frac{2}{x - 4} &amp;&amp; \\text{Multiply both sides by }y - 3\\text{ and divide by }x - 4. \\\\[1.5mm] &amp;y=\\frac{2}{x - 4}+3 &amp;&amp; \\text{Add 3 to both sides}.\\\\[-3mm]&amp;\\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">So [latex]{f}^{-1}\\left(x\\right)=\\dfrac{2}{x - 4}+3[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThe domain and range of [latex]f[\/latex] exclude the values 3 and 4, respectively. [latex]f[\/latex] and [latex]{f}^{-1}[\/latex] are equal at two points but are not the same function, as we can see by creating\u00a0the table below.\r\n<table id=\"Table_01_07_05\" summary=\"The values of f(x) are: f(1)=3, f(2)=2, and f(5)=5. So f^(-1)(y)=y.\">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>5<\/td>\r\n<td>[latex]{f}^{-1}\\left(y\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\r\n<td>3<\/td>\r\n<td>2<\/td>\r\n<td>5<\/td>\r\n<td>[latex]y[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving to Find an Inverse with Radicals<\/h3>\r\nFind the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt{x - 4}[\/latex].\r\n\r\n[reveal-answer q=\"444147\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"444147\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=2+\\sqrt{x - 4}\\\\[1.5mm]&amp;x=2+\\sqrt{y - 4}\\\\[1.5mm] &amp;{\\left(x - 2\\right)}^{2}=y - 4 \\\\[1.5mm] &amp;y={\\left(x- 2\\right)}^{2}+4 \\end{align}[\/latex]<\/p>\r\nSo [latex]{f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4[\/latex].\r\n\r\nThe domain of [latex]f[\/latex] is [latex]\\left[4,\\infty \\right)[\/latex]. Notice that the range of [latex]f[\/latex] is [latex]\\left[2,\\infty \\right)[\/latex], so this means that the domain of the inverse function [latex]{f}^{-1}[\/latex] is also [latex]\\left[2,\\infty \\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe formula we found for [latex]{f}^{-1}\\left(x\\right)[\/latex] looks like it would be valid for all real [latex]x[\/latex]. However, [latex]{f}^{-1}[\/latex] itself must have an inverse (namely, [latex]f[\/latex] ) so we have to restrict the domain of [latex]{f}^{-1}[\/latex] to [latex]\\left[2,\\infty \\right)[\/latex] in order to make [latex]{f}^{-1}[\/latex] a one-to-one function. This domain of [latex]{f}^{-1}[\/latex] is exactly the range of [latex]f[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWhat is the inverse of the function [latex]f\\left(x\\right)=2-\\sqrt{x}[\/latex]? State the domains of both the function and the inverse function.\r\nUse an online graphing calculator to graph the function, it's inverse, and [latex]f(x) = x[\/latex] to check whether you are correct.\r\n\r\n[reveal-answer q=\"327817\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"327817\"]\r\n\r\n[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2}[\/latex]; domain of\u00a0 [latex]f:\\left[0,\\infty \\right)[\/latex]; domain of [latex]{ f}^{-1}:\\left(-\\infty ,2\\right][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Graph a Function's Inverse<\/h2>\r\nNow that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205523\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/> Quadratic function with domain restricted to [0, \u221e).[\/caption]<strong>Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.\r\n\r\nWe already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]\r\n\r\nWe notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205525\/CNX_Precalc_Figure_01_07_0092.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"251\" \/> Square and square-root functions on the non-negative domain[\/caption]\r\n\r\nThis relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\r\nGiven the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205527\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/>\r\n[reveal-answer q=\"694066\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"694066\"]\r\n\r\nThis is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].\r\n\r\nIf we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in the graph below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205529\/CNX_Precalc_Figure_01_07_0112.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"363\" \/> The function and its inverse, showing reflection about the identity line[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n\r\n<strong>Q &amp; A<\/strong>\r\n\r\n<strong>Is there any function that is equal to its own inverse?<\/strong>\r\n\r\n<em>Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em>\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/p>\r\n<em>Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find and evaluate the inverse of a function in tabular form.<\/li>\n<li>Find and evaluate the inverse of a function given as an equation.<\/li>\n<li>Graph a function and its inverse.<\/li>\n<\/ul>\n<\/div>\n<p>Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.<\/p>\n<h2>Inverting Tabular Functions<\/h2>\n<p>Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.<\/p>\n<p>Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.<\/p>\n<div class=\"textbox examples\">\n<h3>recall reading functions from tables<\/h3>\n<p>Look to see which row or column of a table is labeled as the input and which is labeled as the output.<\/p>\n<p>The tables below indicate one row represents &#8220;[latex]t[\/latex]&#8221; and the other, [latex]f(t)[\/latex].<\/p>\n<p>It is common for input to be represented by time when output is a distance.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Interpreting the Inverse of a Tabular Function<\/h3>\n<p>A function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes. Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].<\/p>\n<table summary=\"Two rows and five columns. The first row is labeled\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\n<td>30<\/td>\n<td>50<\/td>\n<td>70<\/td>\n<td>90<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q61261\">Show Solution<\/span><\/p>\n<div id=\"q61261\" class=\"hidden-answer\" style=\"display: none\">\n<p>The inverse function takes an output of [latex]f[\/latex] and returns an input for [latex]f[\/latex]. So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.<\/p>\n<p>Alternatively, recall that the definition of the inverse was that if [latex]f\\left(a\\right)=b[\/latex], then [latex]{f}^{-1}\\left(b\\right)=a[\/latex]. By this definition, if we are given [latex]{f}^{-1}\\left(70\\right)=a[\/latex], then we are looking for a value [latex]a[\/latex] so that [latex]f\\left(a\\right)=70[\/latex]. In this case, we are looking for a [latex]t[\/latex] so that [latex]f\\left(t\\right)=70[\/latex], which is when [latex]t=90[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the table below,\u00a0find and interpret (a) [latex]\\text{ }f\\left(60\\right)[\/latex], and (b) [latex]\\text{ }{f}^{-1}\\left(60\\right)[\/latex].<\/p>\n<table summary=\"Two rows and five columns. The first row is labeled\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>[latex]t\\text{ (minutes)}[\/latex]<\/td>\n<td>30<\/td>\n<td>50<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<td>90<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex]<\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>50<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q401025\">Show Solution<\/span><\/p>\n<div id=\"q401025\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex]f\\left(60\\right)=50[\/latex]. In 60 minutes, 50 miles are traveled.<\/p>\n<p>b. [latex]{f}^{-1}\\left(60\\right)=70[\/latex]. To travel 60 miles, it will take 70 minutes.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/h2>\n<p>The domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the <em>vertical<\/em> extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the <em>horizontal<\/em> extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function\u2019s graph.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\n<ol>\n<li>Find the desired input of the inverse function on the [latex]y[\/latex]-axis of the given graph.<\/li>\n<li>Read the inverse function\u2019s output from the [latex]x[\/latex]-axis of the given graph.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/h3>\n<p>A function [latex]g\\left(x\\right)[\/latex] is given\u00a0below. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205520\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q334632\">Show Solution<\/span><\/p>\n<div id=\"q334632\" class=\"hidden-answer\" style=\"display: none\">\n<p>To evaluate [latex]g\\left(3\\right)[\/latex], we find 3 on the <em>x<\/em>-axis and find the corresponding output value on the [latex]y[\/latex]-axis. The point [latex]\\left(3,1\\right)[\/latex] tells us that [latex]g\\left(3\\right)=1[\/latex].<\/p>\n<p>To evaluate [latex]{g}^{-1}\\left(3\\right)[\/latex], recall that by definition [latex]{g}^{-1}\\left(3\\right)[\/latex] means the value of <em>x<\/em> for which [latex]g\\left(x\\right)=3[\/latex]. By looking for the output value 3 on the vertical axis, we find the point [latex]\\left(5,3\\right)[\/latex] on the graph, which means [latex]g\\left(5\\right)=3[\/latex], so by definition, [latex]{g}^{-1}\\left(3\\right)=5[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205521\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the graph in the previous example, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q350455\">Show Solution<\/span><\/p>\n<div id=\"q350455\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. 3; b. 5.6<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Inverses of Functions Represented by Formulas<\/h2>\n<p>Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula\u2014for example, [latex]y[\/latex] as a function of [latex]x-[\/latex] we can often find the inverse function by solving to obtain [latex]x[\/latex] as a function of [latex]y[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a function represented by a formula, find the inverse.<\/h3>\n<ol>\n<li>Verify that\u00a0[latex]f[\/latex] is a one-to-one function.<\/li>\n<li>Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex].<\/li>\n<li>Interchange [latex]x[\/latex]\u00a0and [latex]y[\/latex].<\/li>\n<li>Solve for [latex]y[\/latex], and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Inverting the Fahrenheit-to-Celsius Function<\/h3>\n<p>Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.<\/p>\n<p style=\"text-align: center;\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q625400\">Show Solution<\/span><\/p>\n<div id=\"q625400\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]{ C }=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<br \/>\n[latex]C\\cdot \\frac{9}{5}=F - 32[\/latex]<br \/>\n[latex]F=\\frac{9}{5}C+32[\/latex]<\/p>\n<p>By solving in general, we have uncovered the inverse function. If<\/p>\n<p style=\"text-align: center;\">[latex]C=h\\left(F\\right)=\\frac{5}{9}\\left(F - 32\\right)[\/latex],<\/p>\n<p>then<\/p>\n<p style=\"text-align: center;\">[latex]F={h}^{-1}\\left(C\\right)=\\frac{9}{5}C+32[\/latex].<\/p>\n<p>In this case, we introduced a function [latex]h[\/latex] to represent the conversion because the input and output variables are descriptive, and writing [latex]{C}^{-1}[\/latex] could get confusing.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve for [latex]x[\/latex] in terms of [latex]y[\/latex] given [latex]y=\\frac{1}{3}\\left(x - 5\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q875458\">Show Solution<\/span><\/p>\n<div id=\"q875458\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=3y+5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving to Find an Inverse Function<\/h3>\n<p>Find the inverse of the function [latex]f\\left(x\\right)=\\dfrac{2}{x - 3}+4[\/latex].<\/p>\n<p style=\"text-align: left;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q903464\">Show Solution<\/span><\/p>\n<div id=\"q903464\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=\\frac{2}{x - 3}+4 && \\text{Change }f(x)\\text{ to }y. \\\\[1.5mm]&x=\\frac{2}{y - 3}+4 && \\text{Switch }x\\text{ and }y. \\\\[1.5mm] &y - 4=\\frac{2}{x - 3} && \\text{Subtract 4 from both sides}. \\\\[1.5mm] &y - 3=\\frac{2}{x - 4} && \\text{Multiply both sides by }y - 3\\text{ and divide by }x - 4. \\\\[1.5mm] &y=\\frac{2}{x - 4}+3 && \\text{Add 3 to both sides}.\\\\[-3mm]&\\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">So [latex]{f}^{-1}\\left(x\\right)=\\dfrac{2}{x - 4}+3[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The domain and range of [latex]f[\/latex] exclude the values 3 and 4, respectively. [latex]f[\/latex] and [latex]{f}^{-1}[\/latex] are equal at two points but are not the same function, as we can see by creating\u00a0the table below.<\/p>\n<table id=\"Table_01_07_05\" summary=\"The values of f(x) are: f(1)=3, f(2)=2, and f(5)=5. So f^(-1)(y)=y.\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>5<\/td>\n<td>[latex]{f}^{-1}\\left(y\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\n<td>3<\/td>\n<td>2<\/td>\n<td>5<\/td>\n<td>[latex]y[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving to Find an Inverse with Radicals<\/h3>\n<p>Find the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt{x - 4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q444147\">Show Solution<\/span><\/p>\n<div id=\"q444147\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=2+\\sqrt{x - 4}\\\\[1.5mm]&x=2+\\sqrt{y - 4}\\\\[1.5mm] &{\\left(x - 2\\right)}^{2}=y - 4 \\\\[1.5mm] &y={\\left(x- 2\\right)}^{2}+4 \\end{align}[\/latex]<\/p>\n<p>So [latex]{f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4[\/latex].<\/p>\n<p>The domain of [latex]f[\/latex] is [latex]\\left[4,\\infty \\right)[\/latex]. Notice that the range of [latex]f[\/latex] is [latex]\\left[2,\\infty \\right)[\/latex], so this means that the domain of the inverse function [latex]{f}^{-1}[\/latex] is also [latex]\\left[2,\\infty \\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The formula we found for [latex]{f}^{-1}\\left(x\\right)[\/latex] looks like it would be valid for all real [latex]x[\/latex]. However, [latex]{f}^{-1}[\/latex] itself must have an inverse (namely, [latex]f[\/latex] ) so we have to restrict the domain of [latex]{f}^{-1}[\/latex] to [latex]\\left[2,\\infty \\right)[\/latex] in order to make [latex]{f}^{-1}[\/latex] a one-to-one function. This domain of [latex]{f}^{-1}[\/latex] is exactly the range of [latex]f[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>What is the inverse of the function [latex]f\\left(x\\right)=2-\\sqrt{x}[\/latex]? State the domains of both the function and the inverse function.<br \/>\nUse an online graphing calculator to graph the function, it&#8217;s inverse, and [latex]f(x) = x[\/latex] to check whether you are correct.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q327817\">Show Solution<\/span><\/p>\n<div id=\"q327817\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2}[\/latex]; domain of\u00a0 [latex]f:\\left[0,\\infty \\right)[\/latex]; domain of [latex]{ f}^{-1}:\\left(-\\infty ,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Graph a Function&#8217;s Inverse<\/h2>\n<p>Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as below.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205523\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\">Quadratic function with domain restricted to [0, \u221e).<\/p>\n<\/div>\n<p><strong>Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.<\/p>\n<p>We already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]<\/p>\n<p>We notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown below.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205525\/CNX_Precalc_Figure_01_07_0092.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"251\" \/><\/p>\n<p class=\"wp-caption-text\">Square and square-root functions on the non-negative domain<\/p>\n<\/div>\n<p>This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\n<p>Given the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205527\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q694066\">Show Solution<\/span><\/p>\n<div id=\"q694066\" class=\"hidden-answer\" style=\"display: none\">\n<p>This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<p>If we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in the graph below.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205529\/CNX_Precalc_Figure_01_07_0112.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\">The function and its inverse, showing reflection about the identity line<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<p><strong>Q &amp; A<\/strong><\/p>\n<p><strong>Is there any function that is equal to its own inverse?<\/strong><\/p>\n<p><em>Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em><\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/p>\n<p><em>Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em><\/p>\n<\/div>\n\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-157\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"}]","CANDELA_OUTCOMES_GUID":"7f503683-c266-49d1-b986-5ec4a3353e20","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-157","chapter","type-chapter","status-publish","hentry"],"part":251,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/157","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/157\/revisions"}],"predecessor-version":[{"id":790,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/157\/revisions\/790"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/251"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/157\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=157"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=157"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=157"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}