{"id":149,"date":"2021-02-03T22:01:02","date_gmt":"2021-02-03T22:01:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=149"},"modified":"2022-03-11T21:44:19","modified_gmt":"2022-03-11T21:44:19","slug":"inverse-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/inverse-functions\/","title":{"raw":"Inverse Functions","rendered":"Inverse Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine the conditions for when a function has an inverse<\/li>\r\n \t<li>Use the horizontal line test to recognize when a function is one-to-one<\/li>\r\n \t<li>Find the inverse of a given function<\/li>\r\n \t<li>Draw the graph of an inverse function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Existence of an Inverse Function<\/h2>\r\nWe begin with an example. Given a function [latex]f[\/latex] and an output [latex]y=f(x)[\/latex], we are often interested in finding what value or values [latex]x[\/latex] were mapped to [latex]y[\/latex] by [latex]f[\/latex]. For example, consider the function [latex]f(x)=x^3+4[\/latex]. Since any output [latex]y=x^3+4[\/latex], we can solve this equation for [latex]x[\/latex] to find that the input is [latex]x=\\sqrt[3]{y-4}[\/latex]. This equation defines [latex]x[\/latex] as a function of [latex]y[\/latex]. Denoting this function as [latex]f^{-1}[\/latex], and writing [latex]x=f^{-1}(y)=\\sqrt[3]{y-4}[\/latex], we see that for any [latex]x[\/latex] in the domain of [latex]f, \\, f^{-1}(f(x))=f^{-1}(x^3+4)=x[\/latex]. Thus, this new function, [latex]f^{-1}[\/latex], \u201cundid\u201d what the original function [latex]f[\/latex] did. A function with this property is called the inverse function of the original function.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572110489\">Given a function [latex]f[\/latex] with domain [latex]D[\/latex] and range [latex]R[\/latex], its <strong>inverse function<\/strong> (if it exists) is the function [latex]f^{-1}[\/latex] with domain [latex]R[\/latex] and range [latex]D[\/latex] such that [latex]f^{-1}(y)=x[\/latex] if [latex]f(x)=y[\/latex]. In other words, for a function [latex]f[\/latex] and its inverse [latex]f^{-1}[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1170572141883\" class=\"equation\" style=\"text-align: center;\">[latex]f^{-1}(f(x))=x[\/latex] for all [latex]x[\/latex] in [latex]D[\/latex], and [latex]f(f^{-1}(y))=y[\/latex] for all [latex]y[\/latex] in [latex]R[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\nNote that [latex]f^{-1}[\/latex] is read as \u201cf inverse.\u201d Here, the -1 is not used as an exponent and [latex]f^{-1}(x) \\ne 1\/f(x)[\/latex]. Figure 1 shows the relationship between the domain and range of [latex]f[\/latex] and the domain and range of [latex]f^{-1}[\/latex].\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\/2332\/2018\/01\/11202523\/CNX_Calc_Figure_01_04_001.jpg\" alt=\"An image of two bubbles. The first bubble is orange and has two labels: the top label is \u201cDomain of f\u201d and the bottom label is \u201cRange of f inverse\u201d. Within this bubble is the variable \u201cx\u201d. An orange arrow with the label \u201cf\u201d points from this bubble to the second bubble. The second bubble is blue and has two labels: the top label is \u201crange of f\u201d and the bottom label is \u201cdomain of f inverse\u201d. Within this bubble is the variable \u201cy\u201d. A blue arrow with the label \u201cf inverse\u201d points from this bubble to the first bubble.\" width=\"487\" height=\"160\" \/> Figure 1. Given a function [latex]f[\/latex] and its inverse [latex]f^{-1}, \\, f^{-1}(y)=x[\/latex] if and only if [latex]f(x)=y[\/latex]. The range of [latex]f[\/latex] becomes the domain of [latex]f^{-1}[\/latex] and the domain of [latex]f[\/latex] becomes the range of [latex]f^{-1}[\/latex].[\/caption]\r\n<p id=\"fs-id1170572141586\">Recall that a function has exactly one output for each input. Remember the vertical line test?<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Recall: THe vertical line test<\/h3>\r\nWe can identify whether the graph of a relation represents a function\u00a0because for each [latex]x[\/latex]-coordinate there will be exactly one [latex]y[\/latex]-coordinate.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/04\/18232417\/image001.jpg\" alt=\"A graph of a semicircle. Four vertical lines cross the semicircle at one point each.\" width=\"304\" height=\"307\" \/> Figure 2. Semicircle graph undergoing the vertical line test.[\/caption]\r\n\r\nWhen a vertical line is placed across the plot of this relation, it does not intersect the graph more than once for any values of [latex]x[\/latex]. This is a graph of a function.\r\n\r\nIf, on the other hand, a graph shows two or more intersections with a vertical line, then an input ([latex]x[\/latex]-coordinate) can have more than one output ([latex]y[\/latex]-coordinate), and [latex]y[\/latex]\u00a0is not a function of [latex]x[\/latex]. Examining the graph of a relation to determine if a vertical line would intersect with more than one point is a quick way to determine if the relation shown by the graph is a function.\r\n\r\n<\/div>\r\nTherefore, to define an inverse function, we need to map each input to exactly one output. For example, let\u2019s try to find the inverse function for [latex]f(x)=x^2[\/latex]. Solving the equation [latex]y=x^2[\/latex] for [latex]x[\/latex], we arrive at the equation [latex]x= \\pm \\sqrt{y}[\/latex]. This equation does not describe [latex]x[\/latex] as a function of [latex]y[\/latex] because there are two solutions to this equation for every [latex]y&gt;0[\/latex]. The problem with trying to find an inverse function for [latex]f(x)=x^2[\/latex] is that two inputs are sent to the same output for each output [latex]y&gt;0[\/latex]. The function [latex]f(x)=x^3+4[\/latex] discussed earlier did not have this problem. For that function, each input was sent to a different output. A function that sends each input to a <em>different<\/em> output is called a one-to-one function.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]4062[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nWe say a [latex]f[\/latex] is a <strong>one-to-one function<\/strong> if [latex]f(x_1) \\ne f(x_2)[\/latex] when [latex]x_1 \\ne x_2[\/latex].\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572103424\">One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the [latex]xy[\/latex]-plane, according to the <strong>horizontal line test<\/strong>, it cannot intersect the graph more than once. We note that the horizontal line test is different from the vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line test determines whether a function is one-to-one (Figure 3).<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Horizontal Line Test<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572137262\">A function [latex]f[\/latex] is one-to-one if and only if every horizontal line intersects the graph of [latex]f[\/latex] no more than once.<\/p>\r\n\r\n<\/div>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202527\/CNX_Calc_Figure_01_04_002.jpg\" alt=\"An image of two graphs. Both graphs have an x axis that runs from -3 to 3 and a y axis that runs from -3 to 4. The first graph is of the function \u201cf(x) = x squared\u201d, which is a parabola. The function decreases until it hits the origin, where it begins to increase. The x intercept and y intercept are both at the origin. There are two orange horizontal lines also plotted on the graph, both of which run through the function at two points each. The second graph is of the function \u201cf(x) = x cubed\u201d, which is an increasing curved function. The x intercept and y intercept are both at the origin. There are three orange lines also plotted on the graph, each of which only intersects the function at one point.\" width=\"487\" height=\"313\" \/> Figure 3. (a) The function [latex]f(x)=x^2[\/latex] is not one-to-one because it fails the horizontal line test. (b) The function [latex]f(x)=x^3[\/latex] is one-to-one because it passes the horizontal line test.[\/caption]\r\n<div id=\"fs-id1170572228079\" class=\"textbook exercises\">\r\n<h3>Example: Determining Whether a Function Is One-to-One<\/h3>\r\n<p id=\"fs-id1170572150756\">For each of the following functions, use the horizontal line test to determine whether it is one-to-one.<\/p>\r\na.\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"420\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202531\/CNX_Calc_Figure_01_04_003.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 11 and the y axis runs from -3 to 11. The graph is of a step function which contains 10 horizontal steps. Each steps starts with a closed circle and ends with an open circle. The first step starts at the origin and ends at the point (1, 0). The second step starts at the point (1, 1) and ends at the point (1, 2). Each of the following 8 steps starts 1 unit higher in the y direction than where the previous step ended. The tenth and final step starts at the point (9, 9) and ends at the point (10, 9)\" width=\"420\" height=\"428\" \/> Figure 4. Is this function one-to-one?[\/caption]\r\n\r\nb.\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"420\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202534\/CNX_Calc_Figure_01_04_004.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 6 and the y axis runs from -3 to 6. The graph is of the function \u201cf(x) = (1\/x)\u201d, a curved decreasing function. The graph of the function starts right below the x axis in the 4th quadrant and begins to decreases until it comes close to the y axis. The graph keeps decreasing as it gets closer and closer to the y axis, but never touches it due to the vertical asymptote. In the first quadrant, the graph of the function starts close to the y axis and keeps decreasing until it gets close to the x axis. As the function continues to decreases it gets closer and closer to the x axis without touching it, where there is a horizontal asymptote.\" width=\"420\" height=\"438\" \/> Figure 5.\u00a0Is this function one-to-one?[\/caption]\r\n\r\n[reveal-answer q=\"fs-id1170572241370\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572241370\"]\r\n<ol id=\"fs-id1170572241370\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since the horizontal line [latex]y=n[\/latex] for any integer [latex]n\\ge 0[\/latex] intersects the graph more than once, this function is not one-to-one.\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"410\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202536\/CNX_Calc_Figure_01_04_005.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 11 and the y axis runs from -3 to 11. The graph is of a step function which contains 10 horizontal steps. Each steps starts with a closed circle and ends with an open circle. The first step starts at the origin and ends at the point (1, 0). The second step starts at the point (1, 1) and ends at the point (1, 2). Each of the following 8 steps starts 1 unit higher in the y direction than where the previous step ended. The tenth and final step starts at the point (9, 9) and ends at the point (10, 9). There are also two horizontal orange lines plotted on the graph, each of which run through an entire step of the function.\" width=\"410\" height=\"418\" \/> Figure 6. This is not a one-to-one function.[\/caption]<\/li>\r\n \t<li>Since every horizontal line intersects the graph once (at most), this function is one-to-one.\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"410\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202538\/CNX_Calc_Figure_01_04_006.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 6 and the y axis runs from -3 to 6. The graph is of the function \u201cf(x) = (1\/x)\u201d, a curved decreasing function. The graph of the function starts right below the x axis in the 4th quadrant and begins to decreases until it comes close to the y axis. The graph keeps decreasing as it gets closer and closer to the y axis, but never touches it due to the vertical asymptote. In the first quadrant, the graph of the function starts close to the y axis and keeps decreasing until it gets close to the x axis. As the function continues to decreases it gets closer and closer to the x axis without touching it, where there is a horizontal asymptote. There are also three horizontal orange lines plotted on the graph, each of which only runs through the function at one point.\" width=\"410\" height=\"427\" \/> Figure 7. This is a one-to-one function.[\/caption]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572151745\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572294776\">Is the function [latex]f[\/latex] graphed in the following image one-to-one?<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202541\/CNX_Calc_Figure_01_04_007.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of the function \u201cf(x) = (x cubed) - x\u201d which is a curved function. The function increases, decreases, then increases again. The x intercepts are at the points (-1, 0), (0,0), and (1, 0). The y intercept is at the origin.\" width=\"325\" height=\"366\" \/> Figure 8. Is this function one-to-one?[\/caption]\r\n\r\n[reveal-answer q=\"880346\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"880346\"]\r\n<p id=\"fs-id1165042094151\">Use the horizontal line test.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572216730\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572216730\"]\r\n<p id=\"fs-id1170572216730\">No.<\/p>\r\n<span style=\"font-size: 1rem; font-weight: normal; orphans: 1; text-align: initial;\">[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]169058[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Finding a Function\u2019s Inverse<\/h2>\r\n<p id=\"fs-id1170572150503\">We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of [latex]f[\/latex] to elements in the range of [latex]f[\/latex]. The inverse function maps each element from the range of [latex]f[\/latex] back to its corresponding element from the domain of [latex]f[\/latex]. Therefore, to find the inverse function of a one-to-one function [latex]f[\/latex], given any [latex]y[\/latex] in the range of [latex]f[\/latex], we need to determine which [latex]x[\/latex] in the domain of [latex]f[\/latex] satisfies [latex]f(x)=y[\/latex]. Since [latex]f[\/latex] is one-to-one, there is exactly one such value [latex]x[\/latex]. We can find that value [latex]x[\/latex] by solving the equation [latex]f(x)=y[\/latex] for [latex]x[\/latex]. Doing so, we are able to write [latex]x[\/latex] as a function of [latex]y[\/latex] where the domain of this function is the range of [latex]f[\/latex] and the range of this new function is the domain of [latex]f[\/latex]. Consequently, this function is the inverse of [latex]f[\/latex], and we write [latex]x=f^{-1}(y)[\/latex]. Since we typically use the variable [latex]x[\/latex] to denote the independent variable and [latex]y[\/latex] to denote the dependent variable, we often interchange the roles of [latex]x[\/latex] and [latex]y[\/latex], and write [latex]y=f^{-1}(x)[\/latex]. Representing the inverse function in this way is also helpful later when we graph a function [latex]f[\/latex] and its inverse [latex]f^{-1}[\/latex] on the same axes.<\/p>\r\n\r\n<div id=\"fs-id1170572552427\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy, Finding an Inverse Function<\/h3>\r\n<ol id=\"fs-id1170572450945\">\r\n \t<li>Solve the equation [latex]y=f(x)[\/latex] for [latex]x[\/latex].<\/li>\r\n \t<li>Interchange the variables [latex]x[\/latex] and [latex]y[\/latex] and write [latex]y=f^{-1}(x)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nTo complete the first step to finding an inverse function, we recall isolating a variable in a given equation.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: isolate a variable in a formula<\/h3>\r\nIsolate the variable for width [latex]w[\/latex]<i>\u00a0<\/i>from the formula for the perimeter of a rectangle<em>: \u00a0<\/em>\r\n<p style=\"text-align: center;\">[latex]{P}=2\\left({l}\\right)+2\\left({w}\\right)[\/latex].<\/p>\r\nFirst, isolate the term with\u00a0[latex]w[\/latex]\u00a0by subtracting [latex]2l[\/latex] from both sides of the equation.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,P\\,=\\,\\,\\,\\,2l+2w\\\\\\underline{\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}\\\\P-2l=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2w\\end{array}[\/latex]<i> <\/i><\/p>\r\nNext, clear the coefficient of\u00a0[latex]w[\/latex]<i>\u00a0<\/i>by dividing both sides of the equation by [latex]2[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\underline{P-2l}=\\underline{2w}\\\\\\,\\,\\,\\,\\,\\,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\\\ \\,\\,\\,\\dfrac{P-2l}{2}\\,\\,=\\,\\,w\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\dfrac{P-2l}{2}\\end{array}[\/latex]<\/p>\r\nYou can rewrite the equation so the isolated variable is on the left side.\r\n<p style=\"text-align: center;\">[latex]w=\\Large\\frac{P-2l}{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572208522\" class=\"textbook exercises\">\r\n<h3>Example: Finding an Inverse Function<\/h3>\r\n<p id=\"fs-id1170572546520\">Find the inverse for the function [latex]f(x)=3x-4[\/latex]. State the domain and range of the inverse function. Verify that [latex]f^{-1}(f(x))=x[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572481500\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572481500\"]\r\n<p id=\"fs-id1170572481500\">Follow the steps outlined in the strategy.<\/p>\r\n<p id=\"fs-id1170572548820\">Step 1. If [latex]y=3x-4[\/latex], then [latex]3x=y+4[\/latex] and [latex]x=\\frac{1}{3}y+\\frac{4}{3}[\/latex].<\/p>\r\n<p id=\"fs-id1170572479776\">Step 2. Rewrite as [latex]y=\\frac{1}{3}x+\\frac{4}{3}[\/latex] and let [latex]y=f^{-1}(x)[\/latex].<\/p>\r\n<p id=\"fs-id1170572453043\">Therefore, [latex]f^{-1}(x)=\\frac{1}{3}x+\\frac{4}{3}[\/latex].<\/p>\r\n<p id=\"fs-id1170572240611\">Since the domain of [latex]f[\/latex] is [latex](\u2212\\infty ,\\infty)[\/latex], the range of [latex]f^{-1}[\/latex] is [latex](\u2212\\infty ,\\infty)[\/latex]. Since the range of [latex]f[\/latex] is [latex](\u2212\\infty ,\\infty)[\/latex], the domain of [latex]f^{-1}[\/latex] is [latex](\u2212\\infty ,\\infty)[\/latex].<\/p>\r\n<p id=\"fs-id1170572222969\">You can verify that [latex]f^{-1}(f(x))=x[\/latex] by writing<\/p>\r\n\r\n<div id=\"fs-id1170572453591\" class=\"equation unnumbered\">[latex]f^{-1}(f(x))=f^{-1}(3x-4)=\\frac{1}{3}(3x-4)+\\frac{4}{3}=x-\\frac{4}{3}+\\frac{4}{3}=x[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572546429\">Note that for [latex]f^{-1}(x)[\/latex] to be the inverse of [latex]f(x)[\/latex], both [latex]f^{-1}(f(x))=x[\/latex] and [latex]f(f^{-1}(x))=x[\/latex] for all [latex]x[\/latex] in the domain of the inside function.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Finding an Inverse Function[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/JFQ8maupdT8?controls=0&amp;start=215&amp;end=405&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.4InverseFunctions215to405_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.4 Inverse Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572478768\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572478776\">Find the inverse of the function [latex]f(x)=\\dfrac{3x}{(x-2)}[\/latex]. State the domain and range of the inverse function.<\/p>\r\n[reveal-answer q=\"466233\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"466233\"]\r\n<p id=\"fs-id1165042050880\">Use the Problem-Solving Strategy above for finding inverse functions.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572479045\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572479045\"]\r\n<p id=\"fs-id1170572479045\">[latex]f^{-1}(x)=\\dfrac{2x}{x-3}[\/latex]. The domain of [latex]f^{-1}[\/latex] is [latex]\\{x|x \\ne 3\\}[\/latex]. The range of [latex]f^{-1}[\/latex] is [latex]\\{y|y \\ne 2\\}[\/latex].<\/p>\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]217478[\/ohm_question]\r\n\r\n<\/div>\r\n<h3>Graphing Inverse Functions<\/h3>\r\n<p id=\"fs-id1170572363375\">Let\u2019s consider the relationship between the graph of a function [latex]f[\/latex] and the graph of its inverse. Consider the graph of [latex]f[\/latex] shown in Figure 9(a) and a point [latex](a,b)[\/latex] on the graph. Since [latex]b=f(a)[\/latex], then [latex]f^{-1}(b)=a[\/latex]. Therefore, when we graph [latex]f^{-1}[\/latex], the point [latex](b,a)[\/latex] is on the graph. As a result, the graph of [latex]f^{-1}[\/latex] is a reflection of the graph of [latex]f[\/latex] about the line [latex]y=x[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"710\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202543\/CNX_Calc_Figure_01_04_017.jpg\" alt=\"An image of two graphs. The first graph is of \u201cy = f(x)\u201d, which is a curved increasing function, that increases at a faster rate as x increases. The point (a, b) is on the graph of the function in the first quadrant. The second graph also graphs \u201cy = f(x)\u201d with the point (a, b), but also graphs the function \u201cy = f inverse (x)\u201d, an increasing curved function, that increases at a slower rate as x increases. This function includes the point (b, a). In addition to the two functions, there is a diagonal dotted line potted with the equation \u201cy =x\u201d, which shows that \u201cf(x)\u201d and \u201cf inverse (x)\u201d are mirror images about the line \u201cy =x\u201d.\" width=\"710\" height=\"387\" \/> Figure 9. (a) The graph of this function [latex]f[\/latex] shows point [latex](a,b)[\/latex] on the graph of [latex]f[\/latex]. (b) Since [latex](a,b)[\/latex] is on the graph of [latex]f[\/latex], the point [latex](b,a)[\/latex] is on the graph of [latex]f^{-1}[\/latex]. The graph of [latex]f^{-1}[\/latex] is a reflection of the graph of [latex]f[\/latex] about the line [latex]y=x[\/latex].[\/caption]\r\n<div id=\"fs-id1170572213184\" class=\"textbook exercises\">\r\n<h3>Example: Sketching Graphs of Inverse Functions<\/h3>\r\n<p id=\"fs-id1170572213194\">For the graph of [latex]f[\/latex] in the following image, sketch a graph of [latex]f^{-1}[\/latex] by sketching the line [latex]y=x[\/latex] and using symmetry. Identify the domain and range of [latex]f^{-1}[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202545\/CNX_Calc_Figure_01_04_009.jpg\" alt=\"An image of a graph. The x axis runs from -2 to 2 and the y axis runs from 0 to 2. The graph is of the function \u201cf(x) = square root of (x +2)\u201d, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4).\" width=\"325\" height=\"321\" \/> Figure 10. Graph of [latex]f(x)[\/latex].[\/caption]<span id=\"fs-id1170572222895\"><\/span>\r\n[reveal-answer q=\"fs-id1170572222910\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572222910\"]\r\n<p id=\"fs-id1170572222910\">Reflect the graph about the line [latex]y=x[\/latex]. The domain of [latex]f^{-1}[\/latex] is [latex][0,\\infty)[\/latex]. The range of [latex]f^{-1}[\/latex] is [latex][-2,\\infty)[\/latex]. By using the preceding strategy for finding inverse functions, we can verify that the inverse function is [latex]f^{-1}(x)=x^2-2[\/latex], as shown in the graph.<\/p>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202548\/CNX_Calc_Figure_01_04_010.jpg\" alt=\"An image of a graph. The x axis runs from -2 to 2 and the y axis runs from -2 to 2. The graph is of two functions. The first function is \u201cf(x) = square root of (x +2)\u201d, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4). The second function is \u201cf inverse (x) = (x squared) -2\u201d, an increasing curved function that starts at the point (0, -2). The x intercept is at the approximate point (1.4, 0) and the y intercept is at the point (0, -2). In addition to the two functions, there is a diagonal dotted line potted with the equation \u201cy =x\u201d, which shows that \u201cf(x)\u201d and \u201cf inverse (x)\u201d are mirror images about the line \u201cy =x\u201d.\" width=\"325\" height=\"324\" \/> Figure 11. Graph of [latex]f(x)[\/latex] and its inverse.[\/caption]<span id=\"fs-id1170572480252\"><\/span>[\/hidden-answer]<\/div>\r\n<div id=\"fs-id1170572480264\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572480271\">Sketch the graph of [latex]f(x)=2x+3[\/latex] and the graph of its inverse using the symmetry property of inverse functions.<\/p>\r\n[reveal-answer q=\"708256\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"708256\"]\r\n<p id=\"fs-id1165041823836\">The graphs are symmetric about the line [latex]y=x[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572481068\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572481068\"]\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202551\/CNX_Calc_Figure_01_04_011.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of two functions. The first function is \u201cf(x) = 2x +3\u201d, an increasing straight line function. The function has an x intercept at (-1.5, 0) and a y intercept at (0, 3). The second function is \u201cf inverse (x) = (x - 3)\/2\u201d, an increasing straight line function, which increases at a slower rate than the first function. The function has an x intercept at (3, 0) and a y intercept at (0, -1.5). In addition to the two functions, there is a diagonal dotted line potted with the equation \u201cy =x\u201d, which shows that \u201cf(x)\u201d and \u201cf inverse (x)\u201d are mirror images about the line \u201cy =x\u201d.\" width=\"325\" height=\"358\" \/> Figure 12. Graph of [latex]f(x)=2x+3[\/latex] and its inverse.[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]170589[\/ohm_question]\r\n\r\n<\/div>\r\n<h3>Restricting Domains<\/h3>\r\n<p id=\"fs-id1170572481095\">As we have seen, [latex]f(x)=x^2[\/latex] does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of [latex]f[\/latex] such that the function is one-to-one. This subset is called a <strong>restricted domain<\/strong>. By restricting the domain of [latex]f[\/latex], we can define a new function [latex]g[\/latex] such that the domain of [latex]g[\/latex] is the restricted domain of [latex]f[\/latex] and [latex]g(x)=f(x)[\/latex] for all [latex]x[\/latex] in the domain of [latex]g[\/latex]. Then we can define an inverse function for [latex]g[\/latex] on that domain. For example, since [latex]f(x)=x^2[\/latex] is one-to-one on the interval [latex][0,\\infty)[\/latex], we can define a new function [latex]g[\/latex] such that the domain of [latex]g[\/latex] is [latex][0,\\infty)[\/latex] and [latex]g(x)=x^2[\/latex] for all [latex]x[\/latex] in its domain. Since [latex]g[\/latex] is a one-to-one function, it has an inverse function, given by the formula [latex]g^{-1}(x)=\\sqrt{x}[\/latex]. On the other hand, the function [latex]f(x)=x^2[\/latex] is also one-to-one on the domain [latex](\u2212\\infty,0][\/latex]. Therefore, we could also define a new function [latex]h[\/latex] such that the domain of [latex]h[\/latex] is [latex](\u2212\\infty,0][\/latex] and [latex]h(x)=x^2[\/latex] for all [latex]x[\/latex] in the domain of [latex]h[\/latex]. Then [latex]h[\/latex] is a one-to-one function and must also have an inverse. Its inverse is given by the formula [latex]h^{-1}(x)=\u2212\\sqrt{x}[\/latex] (Figure 13).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"656\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202554\/CNX_Calc_Figure_01_04_012.jpg\" alt=\"An image of two graphs. Both graphs have an x axis that runs from -2 to 5 and a y axis that runs from -2 to 5. The first graph is of two functions. The first function is \u201cg(x) = x squared\u201d, an increasing curved function that starts at the point (0, 0). This function increases at a faster rate for larger values of x. The second function is \u201cg inverse (x) = square root of x\u201d, an increasing curved function that starts at the point (0, 0). This function increases at a slower rate for larger values of x. The first function is \u201ch(x) = x squared\u201d, a decreasing curved function that ends at the point (0, 0). This function decreases at a slower rate for larger values of x. The second function is \u201ch inverse (x) = -(square root of x)\u201d, an increasing curved function that starts at the point (0, 0). This function decreases at a slower rate for larger values of x. In addition to the two functions, there is a diagonal dotted line potted with the equation \u201cy =x\u201d, which shows that \u201cf(x)\u201d and \u201cf inverse (x)\u201d are mirror images about the line \u201cy =x\u201d.\" width=\"656\" height=\"353\" \/> Figure 13. (a) For [latex]g(x)=x^2[\/latex] restricted to [latex][0,\\infty), \\, g^{-1}(x)=\\sqrt{x}[\/latex]. (b) For [latex]h(x)=x^2[\/latex] restricted to [latex](\u2212\\infty,0], \\, h^{-1}(x)=\u2212\\sqrt{x}[\/latex].[\/caption]\r\n<div id=\"fs-id1170572551599\" class=\"textbook exercises\">\r\n<h3>Example: Restricting the Domain<\/h3>\r\n<p id=\"fs-id1170572551608\">Consider the function [latex]f(x)=(x+1)^2[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170572551645\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Sketch the graph of [latex]f[\/latex] and use the horizontal line test to show that [latex]f[\/latex] is not one-to-one.<\/li>\r\n \t<li>Show that [latex]f[\/latex] is one-to-one on the restricted domain [latex][-1,\\infty)[\/latex]. Determine the domain and range for the inverse of [latex]f[\/latex] on this restricted domain and find a formula for [latex]f^{-1}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572477860\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572477860\"]\r\n<ol id=\"fs-id1170572477860\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>The graph of [latex]f[\/latex] is the graph of [latex]y=x^2[\/latex] shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, [latex]f[\/latex] is not one-to-one.\r\n[caption id=\"\" align=\"alignnone\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202557\/CNX_Calc_Figure_01_04_013.jpg\" alt=\"An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function \u201cf(x) = (x+ 1) squared\u201d, which is a parabola. The function decreases until the point (-1, 0), where it begins it increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1). There is also a horizontal dotted line plotted on the graph, which crosses through the function at two points.\" width=\"325\" height=\"286\" \/> Figure 14. Graph of the function\u00a0[latex]f(x)=(x+1)^2[\/latex].[\/caption]<\/li>\r\n \t<li>On the interval [latex][-1,\\infty), \\, f[\/latex] is one-to-one.\r\n[caption id=\"\" align=\"alignnone\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202559\/CNX_Calc_Figure_01_04_014.jpg\" alt=\"An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function \u201cf(x) = (x+ 1) squared\u201d, on the interval [1, infinity). The function starts from the point (-1, 0) and increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1).\" width=\"325\" height=\"511\" \/> Figure 15. Looking at a restricted domain graph of a function.[\/caption]\r\nThe domain and range of [latex]f^{-1}[\/latex] are given by the range and domain of [latex]f[\/latex], respectively. Therefore, the domain of [latex]f^{-1}[\/latex] is [latex][0,\\infty)[\/latex] and the range of [latex]f^{-1}[\/latex] is [latex][-1,\\infty)[\/latex]. To find a formula for [latex]f^{-1}[\/latex], solve the equation [latex]y=(x+1)^2[\/latex] for [latex]x[\/latex]. If [latex]y=(x+1)^2[\/latex], then [latex]x=-1 \\pm \\sqrt{y}[\/latex]. Since we are restricting the domain to the interval where [latex]x \\ge -1[\/latex], we need [latex]\\pm \\sqrt{y} \\ge 0[\/latex]. Therefore, [latex]x=-1+\\sqrt{y}[\/latex]. Interchanging [latex]x[\/latex] and [latex]y[\/latex], we write [latex]y=-1+\\sqrt{x}[\/latex] and conclude that [latex]f^{-1}(x)=-1+\\sqrt{x}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Restricting the Domain[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/JFQ8maupdT8?controls=0&amp;start=564&amp;end=791&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.4InverseFunctions564to791_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.4 Inverse Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572551765\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572551773\">Consider [latex]f(x)=\\dfrac{1}{x^2}[\/latex] restricted to the domain [latex](\u2212\\infty ,0)[\/latex]. Verify that [latex]f[\/latex] is one-to-one on this domain. Determine the domain and range of the inverse of [latex]f[\/latex] and find a formula for [latex]f^{-1}[\/latex].<\/p>\r\n[reveal-answer q=\"118844\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"118844\"]\r\n<p id=\"fs-id1165042134545\">The domain and range of [latex]f^{-1}[\/latex] is given by the range and domain of [latex]f[\/latex], respectively. To find [latex]f^{-1}[\/latex], solve [latex]y=\\dfrac{1}{x^2}[\/latex] for [latex]x[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572141203\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572141203\"]\r\n<p id=\"fs-id1170572141203\">The domain of [latex]f^{-1}[\/latex] is [latex](0,\\infty)[\/latex]. The range of [latex]f^{-1}[\/latex] is [latex](\u2212\\infty ,0)[\/latex]. The inverse function is given by the formula [latex]f^{-1}(x)=\\dfrac{-1\r\n}{\\sqrt{x}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine the conditions for when a function has an inverse<\/li>\n<li>Use the horizontal line test to recognize when a function is one-to-one<\/li>\n<li>Find the inverse of a given function<\/li>\n<li>Draw the graph of an inverse function<\/li>\n<\/ul>\n<\/div>\n<h2>Existence of an Inverse Function<\/h2>\n<p>We begin with an example. Given a function [latex]f[\/latex] and an output [latex]y=f(x)[\/latex], we are often interested in finding what value or values [latex]x[\/latex] were mapped to [latex]y[\/latex] by [latex]f[\/latex]. For example, consider the function [latex]f(x)=x^3+4[\/latex]. Since any output [latex]y=x^3+4[\/latex], we can solve this equation for [latex]x[\/latex] to find that the input is [latex]x=\\sqrt[3]{y-4}[\/latex]. This equation defines [latex]x[\/latex] as a function of [latex]y[\/latex]. Denoting this function as [latex]f^{-1}[\/latex], and writing [latex]x=f^{-1}(y)=\\sqrt[3]{y-4}[\/latex], we see that for any [latex]x[\/latex] in the domain of [latex]f, \\, f^{-1}(f(x))=f^{-1}(x^3+4)=x[\/latex]. Thus, this new function, [latex]f^{-1}[\/latex], \u201cundid\u201d what the original function [latex]f[\/latex] did. A function with this property is called the inverse function of the original function.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170572110489\">Given a function [latex]f[\/latex] with domain [latex]D[\/latex] and range [latex]R[\/latex], its <strong>inverse function<\/strong> (if it exists) is the function [latex]f^{-1}[\/latex] with domain [latex]R[\/latex] and range [latex]D[\/latex] such that [latex]f^{-1}(y)=x[\/latex] if [latex]f(x)=y[\/latex]. In other words, for a function [latex]f[\/latex] and its inverse [latex]f^{-1}[\/latex],<\/p>\n<div id=\"fs-id1170572141883\" class=\"equation\" style=\"text-align: center;\">[latex]f^{-1}(f(x))=x[\/latex] for all [latex]x[\/latex] in [latex]D[\/latex], and [latex]f(f^{-1}(y))=y[\/latex] for all [latex]y[\/latex] in [latex]R[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<p>Note that [latex]f^{-1}[\/latex] is read as \u201cf inverse.\u201d Here, the -1 is not used as an exponent and [latex]f^{-1}(x) \\ne 1\/f(x)[\/latex]. Figure 1 shows the relationship between the domain and range of [latex]f[\/latex] and the domain and range 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\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202523\/CNX_Calc_Figure_01_04_001.jpg\" alt=\"An image of two bubbles. The first bubble is orange and has two labels: the top label is \u201cDomain of f\u201d and the bottom label is \u201cRange of f inverse\u201d. Within this bubble is the variable \u201cx\u201d. An orange arrow with the label \u201cf\u201d points from this bubble to the second bubble. The second bubble is blue and has two labels: the top label is \u201crange of f\u201d and the bottom label is \u201cdomain of f inverse\u201d. Within this bubble is the variable \u201cy\u201d. A blue arrow with the label \u201cf inverse\u201d points from this bubble to the first bubble.\" width=\"487\" height=\"160\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Given a function [latex]f[\/latex] and its inverse [latex]f^{-1}, \\, f^{-1}(y)=x[\/latex] if and only if [latex]f(x)=y[\/latex]. The range of [latex]f[\/latex] becomes the domain of [latex]f^{-1}[\/latex] and the domain of [latex]f[\/latex] becomes the range of [latex]f^{-1}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170572141586\">Recall that a function has exactly one output for each input. Remember the vertical line test?<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: THe vertical line test<\/h3>\n<p>We can identify whether the graph of a relation represents a function\u00a0because for each [latex]x[\/latex]-coordinate there will be exactly one [latex]y[\/latex]-coordinate.<\/p>\n<div style=\"width: 314px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/04\/18232417\/image001.jpg\" alt=\"A graph of a semicircle. Four vertical lines cross the semicircle at one point each.\" width=\"304\" height=\"307\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Semicircle graph undergoing the vertical line test.<\/p>\n<\/div>\n<p>When a vertical line is placed across the plot of this relation, it does not intersect the graph more than once for any values of [latex]x[\/latex]. This is a graph of a function.<\/p>\n<p>If, on the other hand, a graph shows two or more intersections with a vertical line, then an input ([latex]x[\/latex]-coordinate) can have more than one output ([latex]y[\/latex]-coordinate), and [latex]y[\/latex]\u00a0is not a function of [latex]x[\/latex]. Examining the graph of a relation to determine if a vertical line would intersect with more than one point is a quick way to determine if the relation shown by the graph is a function.<\/p>\n<\/div>\n<p>Therefore, to define an inverse function, we need to map each input to exactly one output. For example, let\u2019s try to find the inverse function for [latex]f(x)=x^2[\/latex]. Solving the equation [latex]y=x^2[\/latex] for [latex]x[\/latex], we arrive at the equation [latex]x= \\pm \\sqrt{y}[\/latex]. This equation does not describe [latex]x[\/latex] as a function of [latex]y[\/latex] because there are two solutions to this equation for every [latex]y>0[\/latex]. The problem with trying to find an inverse function for [latex]f(x)=x^2[\/latex] is that two inputs are sent to the same output for each output [latex]y>0[\/latex]. The function [latex]f(x)=x^3+4[\/latex] discussed earlier did not have this problem. For that function, each input was sent to a different output. A function that sends each input to a <em>different<\/em> output is called a one-to-one function.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm4062\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4062&theme=oea&iframe_resize_id=ohm4062&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p>We say a [latex]f[\/latex] is a <strong>one-to-one function<\/strong> if [latex]f(x_1) \\ne f(x_2)[\/latex] when [latex]x_1 \\ne x_2[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170572103424\">One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the [latex]xy[\/latex]-plane, according to the <strong>horizontal line test<\/strong>, it cannot intersect the graph more than once. We note that the horizontal line test is different from the vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line test determines whether a function is one-to-one (Figure 3).<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Horizontal Line Test<\/h3>\n<hr \/>\n<p id=\"fs-id1170572137262\">A function [latex]f[\/latex] is one-to-one if and only if every horizontal line intersects the graph of [latex]f[\/latex] no more than once.<\/p>\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\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202527\/CNX_Calc_Figure_01_04_002.jpg\" alt=\"An image of two graphs. Both graphs have an x axis that runs from -3 to 3 and a y axis that runs from -3 to 4. The first graph is of the function \u201cf(x) = x squared\u201d, which is a parabola. The function decreases until it hits the origin, where it begins to increase. The x intercept and y intercept are both at the origin. There are two orange horizontal lines also plotted on the graph, both of which run through the function at two points each. The second graph is of the function \u201cf(x) = x cubed\u201d, which is an increasing curved function. The x intercept and y intercept are both at the origin. There are three orange lines also plotted on the graph, each of which only intersects the function at one point.\" width=\"487\" height=\"313\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. (a) The function [latex]f(x)=x^2[\/latex] is not one-to-one because it fails the horizontal line test. (b) The function [latex]f(x)=x^3[\/latex] is one-to-one because it passes the horizontal line test.<\/p>\n<\/div>\n<div id=\"fs-id1170572228079\" class=\"textbook exercises\">\n<h3>Example: Determining Whether a Function Is One-to-One<\/h3>\n<p id=\"fs-id1170572150756\">For each of the following functions, use the horizontal line test to determine whether it is one-to-one.<\/p>\n<p>a.<\/p>\n<div style=\"width: 430px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202531\/CNX_Calc_Figure_01_04_003.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 11 and the y axis runs from -3 to 11. The graph is of a step function which contains 10 horizontal steps. Each steps starts with a closed circle and ends with an open circle. The first step starts at the origin and ends at the point (1, 0). The second step starts at the point (1, 1) and ends at the point (1, 2). Each of the following 8 steps starts 1 unit higher in the y direction than where the previous step ended. The tenth and final step starts at the point (9, 9) and ends at the point (10, 9)\" width=\"420\" height=\"428\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Is this function one-to-one?<\/p>\n<\/div>\n<p>b.<\/p>\n<div style=\"width: 430px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202534\/CNX_Calc_Figure_01_04_004.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 6 and the y axis runs from -3 to 6. The graph is of the function \u201cf(x) = (1\/x)\u201d, a curved decreasing function. The graph of the function starts right below the x axis in the 4th quadrant and begins to decreases until it comes close to the y axis. The graph keeps decreasing as it gets closer and closer to the y axis, but never touches it due to the vertical asymptote. In the first quadrant, the graph of the function starts close to the y axis and keeps decreasing until it gets close to the x axis. As the function continues to decreases it gets closer and closer to the x axis without touching it, where there is a horizontal asymptote.\" width=\"420\" height=\"438\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5.\u00a0Is this function one-to-one?<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572241370\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572241370\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572241370\" style=\"list-style-type: lower-alpha;\">\n<li>Since the horizontal line [latex]y=n[\/latex] for any integer [latex]n\\ge 0[\/latex] intersects the graph more than once, this function is not one-to-one.\n<div style=\"width: 420px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202536\/CNX_Calc_Figure_01_04_005.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 11 and the y axis runs from -3 to 11. The graph is of a step function which contains 10 horizontal steps. Each steps starts with a closed circle and ends with an open circle. The first step starts at the origin and ends at the point (1, 0). The second step starts at the point (1, 1) and ends at the point (1, 2). Each of the following 8 steps starts 1 unit higher in the y direction than where the previous step ended. The tenth and final step starts at the point (9, 9) and ends at the point (10, 9). There are also two horizontal orange lines plotted on the graph, each of which run through an entire step of the function.\" width=\"410\" height=\"418\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. This is not a one-to-one function.<\/p>\n<\/div>\n<\/li>\n<li>Since every horizontal line intersects the graph once (at most), this function is one-to-one.\n<div style=\"width: 420px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202538\/CNX_Calc_Figure_01_04_006.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 6 and the y axis runs from -3 to 6. The graph is of the function \u201cf(x) = (1\/x)\u201d, a curved decreasing function. The graph of the function starts right below the x axis in the 4th quadrant and begins to decreases until it comes close to the y axis. The graph keeps decreasing as it gets closer and closer to the y axis, but never touches it due to the vertical asymptote. In the first quadrant, the graph of the function starts close to the y axis and keeps decreasing until it gets close to the x axis. As the function continues to decreases it gets closer and closer to the x axis without touching it, where there is a horizontal asymptote. There are also three horizontal orange lines plotted on the graph, each of which only runs through the function at one point.\" width=\"410\" height=\"427\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. This is a one-to-one function.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572151745\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572294776\">Is the function [latex]f[\/latex] graphed in the following image one-to-one?<\/p>\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202541\/CNX_Calc_Figure_01_04_007.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of the function \u201cf(x) = (x cubed) - x\u201d which is a curved function. The function increases, decreases, then increases again. The x intercepts are at the points (-1, 0), (0,0), and (1, 0). The y intercept is at the origin.\" width=\"325\" height=\"366\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. Is this function one-to-one?<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q880346\">Hint<\/span><\/p>\n<div id=\"q880346\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042094151\">Use the horizontal line test.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572216730\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572216730\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572216730\">No.<\/p>\n<p><span style=\"font-size: 1rem; font-weight: normal; orphans: 1; text-align: initial;\"><\/div>\n<\/div>\n<p><\/span><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169058\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169058&theme=oea&iframe_resize_id=ohm169058&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Finding a Function\u2019s Inverse<\/h2>\n<p id=\"fs-id1170572150503\">We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of [latex]f[\/latex] to elements in the range of [latex]f[\/latex]. The inverse function maps each element from the range of [latex]f[\/latex] back to its corresponding element from the domain of [latex]f[\/latex]. Therefore, to find the inverse function of a one-to-one function [latex]f[\/latex], given any [latex]y[\/latex] in the range of [latex]f[\/latex], we need to determine which [latex]x[\/latex] in the domain of [latex]f[\/latex] satisfies [latex]f(x)=y[\/latex]. Since [latex]f[\/latex] is one-to-one, there is exactly one such value [latex]x[\/latex]. We can find that value [latex]x[\/latex] by solving the equation [latex]f(x)=y[\/latex] for [latex]x[\/latex]. Doing so, we are able to write [latex]x[\/latex] as a function of [latex]y[\/latex] where the domain of this function is the range of [latex]f[\/latex] and the range of this new function is the domain of [latex]f[\/latex]. Consequently, this function is the inverse of [latex]f[\/latex], and we write [latex]x=f^{-1}(y)[\/latex]. Since we typically use the variable [latex]x[\/latex] to denote the independent variable and [latex]y[\/latex] to denote the dependent variable, we often interchange the roles of [latex]x[\/latex] and [latex]y[\/latex], and write [latex]y=f^{-1}(x)[\/latex]. Representing the inverse function in this way is also helpful later when we graph a function [latex]f[\/latex] and its inverse [latex]f^{-1}[\/latex] on the same axes.<\/p>\n<div id=\"fs-id1170572552427\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy, Finding an Inverse Function<\/h3>\n<ol id=\"fs-id1170572450945\">\n<li>Solve the equation [latex]y=f(x)[\/latex] for [latex]x[\/latex].<\/li>\n<li>Interchange the variables [latex]x[\/latex] and [latex]y[\/latex] and write [latex]y=f^{-1}(x)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>To complete the first step to finding an inverse function, we recall isolating a variable in a given equation.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: isolate a variable in a formula<\/h3>\n<p>Isolate the variable for width [latex]w[\/latex]<i>\u00a0<\/i>from the formula for the perimeter of a rectangle<em>: \u00a0<\/em><\/p>\n<p style=\"text-align: center;\">[latex]{P}=2\\left({l}\\right)+2\\left({w}\\right)[\/latex].<\/p>\n<p>First, isolate the term with\u00a0[latex]w[\/latex]\u00a0by subtracting [latex]2l[\/latex] from both sides of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,P\\,=\\,\\,\\,\\,2l+2w\\\\\\underline{\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}\\\\P-2l=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2w\\end{array}[\/latex]<i> <\/i><\/p>\n<p>Next, clear the coefficient of\u00a0[latex]w[\/latex]<i>\u00a0<\/i>by dividing both sides of the equation by [latex]2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\underline{P-2l}=\\underline{2w}\\\\\\,\\,\\,\\,\\,\\,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\\\ \\,\\,\\,\\dfrac{P-2l}{2}\\,\\,=\\,\\,w\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\dfrac{P-2l}{2}\\end{array}[\/latex]<\/p>\n<p>You can rewrite the equation so the isolated variable is on the left side.<\/p>\n<p style=\"text-align: center;\">[latex]w=\\Large\\frac{P-2l}{2}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170572208522\" class=\"textbook exercises\">\n<h3>Example: Finding an Inverse Function<\/h3>\n<p id=\"fs-id1170572546520\">Find the inverse for the function [latex]f(x)=3x-4[\/latex]. State the domain and range of the inverse function. Verify that [latex]f^{-1}(f(x))=x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572481500\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572481500\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572481500\">Follow the steps outlined in the strategy.<\/p>\n<p id=\"fs-id1170572548820\">Step 1. If [latex]y=3x-4[\/latex], then [latex]3x=y+4[\/latex] and [latex]x=\\frac{1}{3}y+\\frac{4}{3}[\/latex].<\/p>\n<p id=\"fs-id1170572479776\">Step 2. Rewrite as [latex]y=\\frac{1}{3}x+\\frac{4}{3}[\/latex] and let [latex]y=f^{-1}(x)[\/latex].<\/p>\n<p id=\"fs-id1170572453043\">Therefore, [latex]f^{-1}(x)=\\frac{1}{3}x+\\frac{4}{3}[\/latex].<\/p>\n<p id=\"fs-id1170572240611\">Since the domain of [latex]f[\/latex] is [latex](\u2212\\infty ,\\infty)[\/latex], the range of [latex]f^{-1}[\/latex] is [latex](\u2212\\infty ,\\infty)[\/latex]. Since the range of [latex]f[\/latex] is [latex](\u2212\\infty ,\\infty)[\/latex], the domain of [latex]f^{-1}[\/latex] is [latex](\u2212\\infty ,\\infty)[\/latex].<\/p>\n<p id=\"fs-id1170572222969\">You can verify that [latex]f^{-1}(f(x))=x[\/latex] by writing<\/p>\n<div id=\"fs-id1170572453591\" class=\"equation unnumbered\">[latex]f^{-1}(f(x))=f^{-1}(3x-4)=\\frac{1}{3}(3x-4)+\\frac{4}{3}=x-\\frac{4}{3}+\\frac{4}{3}=x[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572546429\">Note that for [latex]f^{-1}(x)[\/latex] to be the inverse of [latex]f(x)[\/latex], both [latex]f^{-1}(f(x))=x[\/latex] and [latex]f(f^{-1}(x))=x[\/latex] for all [latex]x[\/latex] in the domain of the inside function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding an Inverse Function<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/JFQ8maupdT8?controls=0&amp;start=215&amp;end=405&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.4InverseFunctions215to405_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.4 Inverse Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572478768\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572478776\">Find the inverse of the function [latex]f(x)=\\dfrac{3x}{(x-2)}[\/latex]. State 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=\"q466233\">Hint<\/span><\/p>\n<div id=\"q466233\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042050880\">Use the Problem-Solving Strategy above for finding inverse functions.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572479045\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572479045\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572479045\">[latex]f^{-1}(x)=\\dfrac{2x}{x-3}[\/latex]. The domain of [latex]f^{-1}[\/latex] is [latex]\\{x|x \\ne 3\\}[\/latex]. The range of [latex]f^{-1}[\/latex] is [latex]\\{y|y \\ne 2\\}[\/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=\"ohm217478\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=217478&theme=oea&iframe_resize_id=ohm217478&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3>Graphing Inverse Functions<\/h3>\n<p id=\"fs-id1170572363375\">Let\u2019s consider the relationship between the graph of a function [latex]f[\/latex] and the graph of its inverse. Consider the graph of [latex]f[\/latex] shown in Figure 9(a) and a point [latex](a,b)[\/latex] on the graph. Since [latex]b=f(a)[\/latex], then [latex]f^{-1}(b)=a[\/latex]. Therefore, when we graph [latex]f^{-1}[\/latex], the point [latex](b,a)[\/latex] is on the graph. As a result, the graph of [latex]f^{-1}[\/latex] is a reflection of the graph of [latex]f[\/latex] about the line [latex]y=x[\/latex].<\/p>\n<div style=\"width: 720px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202543\/CNX_Calc_Figure_01_04_017.jpg\" alt=\"An image of two graphs. The first graph is of \u201cy = f(x)\u201d, which is a curved increasing function, that increases at a faster rate as x increases. The point (a, b) is on the graph of the function in the first quadrant. The second graph also graphs \u201cy = f(x)\u201d with the point (a, b), but also graphs the function \u201cy = f inverse (x)\u201d, an increasing curved function, that increases at a slower rate as x increases. This function includes the point (b, a). In addition to the two functions, there is a diagonal dotted line potted with the equation \u201cy =x\u201d, which shows that \u201cf(x)\u201d and \u201cf inverse (x)\u201d are mirror images about the line \u201cy =x\u201d.\" width=\"710\" height=\"387\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9. (a) The graph of this function [latex]f[\/latex] shows point [latex](a,b)[\/latex] on the graph of [latex]f[\/latex]. (b) Since [latex](a,b)[\/latex] is on the graph of [latex]f[\/latex], the point [latex](b,a)[\/latex] is on the graph of [latex]f^{-1}[\/latex]. The graph of [latex]f^{-1}[\/latex] is a reflection of the graph of [latex]f[\/latex] about the line [latex]y=x[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1170572213184\" class=\"textbook exercises\">\n<h3>Example: Sketching Graphs of Inverse Functions<\/h3>\n<p id=\"fs-id1170572213194\">For the graph of [latex]f[\/latex] in the following image, sketch a graph of [latex]f^{-1}[\/latex] by sketching the line [latex]y=x[\/latex] and using symmetry. Identify the domain and range of [latex]f^{-1}[\/latex].<\/p>\n<div style=\"width: 335px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202545\/CNX_Calc_Figure_01_04_009.jpg\" alt=\"An image of a graph. The x axis runs from -2 to 2 and the y axis runs from 0 to 2. The graph is of the function \u201cf(x) = square root of (x +2)\u201d, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4).\" width=\"325\" height=\"321\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 10. Graph of [latex]f(x)[\/latex].<\/p>\n<\/div>\n<p><span id=\"fs-id1170572222895\"><\/span><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572222910\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572222910\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572222910\">Reflect the graph about the line [latex]y=x[\/latex]. The domain of [latex]f^{-1}[\/latex] is [latex][0,\\infty)[\/latex]. The range of [latex]f^{-1}[\/latex] is [latex][-2,\\infty)[\/latex]. By using the preceding strategy for finding inverse functions, we can verify that the inverse function is [latex]f^{-1}(x)=x^2-2[\/latex], as shown in the graph.<\/p>\n<div style=\"width: 335px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202548\/CNX_Calc_Figure_01_04_010.jpg\" alt=\"An image of a graph. The x axis runs from -2 to 2 and the y axis runs from -2 to 2. The graph is of two functions. The first function is \u201cf(x) = square root of (x +2)\u201d, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4). The second function is \u201cf inverse (x) = (x squared) -2\u201d, an increasing curved function that starts at the point (0, -2). The x intercept is at the approximate point (1.4, 0) and the y intercept is at the point (0, -2). In addition to the two functions, there is a diagonal dotted line potted with the equation \u201cy =x\u201d, which shows that \u201cf(x)\u201d and \u201cf inverse (x)\u201d are mirror images about the line \u201cy =x\u201d.\" width=\"325\" height=\"324\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 11. Graph of [latex]f(x)[\/latex] and its inverse.<\/p>\n<\/div>\n<p><span id=\"fs-id1170572480252\"><\/span><\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572480264\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572480271\">Sketch the graph of [latex]f(x)=2x+3[\/latex] and the graph of its inverse using the symmetry property of inverse functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q708256\">Hint<\/span><\/p>\n<div id=\"q708256\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165041823836\">The graphs are symmetric about the line [latex]y=x[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572481068\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572481068\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"width: 335px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202551\/CNX_Calc_Figure_01_04_011.jpg\" alt=\"An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of two functions. The first function is \u201cf(x) = 2x +3\u201d, an increasing straight line function. The function has an x intercept at (-1.5, 0) and a y intercept at (0, 3). The second function is \u201cf inverse (x) = (x - 3)\/2\u201d, an increasing straight line function, which increases at a slower rate than the first function. The function has an x intercept at (3, 0) and a y intercept at (0, -1.5). In addition to the two functions, there is a diagonal dotted line potted with the equation \u201cy =x\u201d, which shows that \u201cf(x)\u201d and \u201cf inverse (x)\u201d are mirror images about the line \u201cy =x\u201d.\" width=\"325\" height=\"358\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 12. Graph of [latex]f(x)=2x+3[\/latex] and its inverse.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm170589\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=170589&theme=oea&iframe_resize_id=ohm170589&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3>Restricting Domains<\/h3>\n<p id=\"fs-id1170572481095\">As we have seen, [latex]f(x)=x^2[\/latex] does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of [latex]f[\/latex] such that the function is one-to-one. This subset is called a <strong>restricted domain<\/strong>. By restricting the domain of [latex]f[\/latex], we can define a new function [latex]g[\/latex] such that the domain of [latex]g[\/latex] is the restricted domain of [latex]f[\/latex] and [latex]g(x)=f(x)[\/latex] for all [latex]x[\/latex] in the domain of [latex]g[\/latex]. Then we can define an inverse function for [latex]g[\/latex] on that domain. For example, since [latex]f(x)=x^2[\/latex] is one-to-one on the interval [latex][0,\\infty)[\/latex], we can define a new function [latex]g[\/latex] such that the domain of [latex]g[\/latex] is [latex][0,\\infty)[\/latex] and [latex]g(x)=x^2[\/latex] for all [latex]x[\/latex] in its domain. Since [latex]g[\/latex] is a one-to-one function, it has an inverse function, given by the formula [latex]g^{-1}(x)=\\sqrt{x}[\/latex]. On the other hand, the function [latex]f(x)=x^2[\/latex] is also one-to-one on the domain [latex](\u2212\\infty,0][\/latex]. Therefore, we could also define a new function [latex]h[\/latex] such that the domain of [latex]h[\/latex] is [latex](\u2212\\infty,0][\/latex] and [latex]h(x)=x^2[\/latex] for all [latex]x[\/latex] in the domain of [latex]h[\/latex]. Then [latex]h[\/latex] is a one-to-one function and must also have an inverse. Its inverse is given by the formula [latex]h^{-1}(x)=\u2212\\sqrt{x}[\/latex] (Figure 13).<\/p>\n<div style=\"width: 666px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202554\/CNX_Calc_Figure_01_04_012.jpg\" alt=\"An image of two graphs. Both graphs have an x axis that runs from -2 to 5 and a y axis that runs from -2 to 5. The first graph is of two functions. The first function is \u201cg(x) = x squared\u201d, an increasing curved function that starts at the point (0, 0). This function increases at a faster rate for larger values of x. The second function is \u201cg inverse (x) = square root of x\u201d, an increasing curved function that starts at the point (0, 0). This function increases at a slower rate for larger values of x. The first function is \u201ch(x) = x squared\u201d, a decreasing curved function that ends at the point (0, 0). This function decreases at a slower rate for larger values of x. The second function is \u201ch inverse (x) = -(square root of x)\u201d, an increasing curved function that starts at the point (0, 0). This function decreases at a slower rate for larger values of x. In addition to the two functions, there is a diagonal dotted line potted with the equation \u201cy =x\u201d, which shows that \u201cf(x)\u201d and \u201cf inverse (x)\u201d are mirror images about the line \u201cy =x\u201d.\" width=\"656\" height=\"353\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 13. (a) For [latex]g(x)=x^2[\/latex] restricted to [latex][0,\\infty), \\, g^{-1}(x)=\\sqrt{x}[\/latex]. (b) For [latex]h(x)=x^2[\/latex] restricted to [latex](\u2212\\infty,0], \\, h^{-1}(x)=\u2212\\sqrt{x}[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1170572551599\" class=\"textbook exercises\">\n<h3>Example: Restricting the Domain<\/h3>\n<p id=\"fs-id1170572551608\">Consider the function [latex]f(x)=(x+1)^2[\/latex].<\/p>\n<ol id=\"fs-id1170572551645\" style=\"list-style-type: lower-alpha;\">\n<li>Sketch the graph of [latex]f[\/latex] and use the horizontal line test to show that [latex]f[\/latex] is not one-to-one.<\/li>\n<li>Show that [latex]f[\/latex] is one-to-one on the restricted domain [latex][-1,\\infty)[\/latex]. Determine the domain and range for the inverse of [latex]f[\/latex] on this restricted domain and find a formula for [latex]f^{-1}[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572477860\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572477860\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572477860\" style=\"list-style-type: lower-alpha;\">\n<li>The graph of [latex]f[\/latex] is the graph of [latex]y=x^2[\/latex] shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, [latex]f[\/latex] is not one-to-one.\n<div style=\"width: 335px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202557\/CNX_Calc_Figure_01_04_013.jpg\" alt=\"An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function \u201cf(x) = (x+ 1) squared\u201d, which is a parabola. The function decreases until the point (-1, 0), where it begins it increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1). There is also a horizontal dotted line plotted on the graph, which crosses through the function at two points.\" width=\"325\" height=\"286\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 14. Graph of the function\u00a0[latex]f(x)=(x+1)^2[\/latex].<\/p>\n<\/div>\n<\/li>\n<li>On the interval [latex][-1,\\infty), \\, f[\/latex] is one-to-one.\n<div style=\"width: 335px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202559\/CNX_Calc_Figure_01_04_014.jpg\" alt=\"An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function \u201cf(x) = (x+ 1) squared\u201d, on the interval &#091;1, infinity). The function starts from the point (-1, 0) and increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1).\" width=\"325\" height=\"511\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 15. Looking at a restricted domain graph of a function.<\/p>\n<\/div>\n<p>The domain and range of [latex]f^{-1}[\/latex] are given by the range and domain of [latex]f[\/latex], respectively. Therefore, the domain of [latex]f^{-1}[\/latex] is [latex][0,\\infty)[\/latex] and the range of [latex]f^{-1}[\/latex] is [latex][-1,\\infty)[\/latex]. To find a formula for [latex]f^{-1}[\/latex], solve the equation [latex]y=(x+1)^2[\/latex] for [latex]x[\/latex]. If [latex]y=(x+1)^2[\/latex], then [latex]x=-1 \\pm \\sqrt{y}[\/latex]. Since we are restricting the domain to the interval where [latex]x \\ge -1[\/latex], we need [latex]\\pm \\sqrt{y} \\ge 0[\/latex]. Therefore, [latex]x=-1+\\sqrt{y}[\/latex]. Interchanging [latex]x[\/latex] and [latex]y[\/latex], we write [latex]y=-1+\\sqrt{x}[\/latex] and conclude that [latex]f^{-1}(x)=-1+\\sqrt{x}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Restricting the Domain<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/JFQ8maupdT8?controls=0&amp;start=564&amp;end=791&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.4InverseFunctions564to791_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.4 Inverse Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572551765\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572551773\">Consider [latex]f(x)=\\dfrac{1}{x^2}[\/latex] restricted to the domain [latex](\u2212\\infty ,0)[\/latex]. Verify that [latex]f[\/latex] is one-to-one on this domain. Determine the domain and range of the inverse of [latex]f[\/latex] and find a formula for [latex]f^{-1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q118844\">Hint<\/span><\/p>\n<div id=\"q118844\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042134545\">The domain and range of [latex]f^{-1}[\/latex] is given by the range and domain of [latex]f[\/latex], respectively. To find [latex]f^{-1}[\/latex], solve [latex]y=\\dfrac{1}{x^2}[\/latex] for [latex]x[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572141203\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572141203\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572141203\">The domain of [latex]f^{-1}[\/latex] is [latex](0,\\infty)[\/latex]. The range of [latex]f^{-1}[\/latex] is [latex](\u2212\\infty ,0)[\/latex]. The inverse function is given by the formula [latex]f^{-1}(x)=\\dfrac{-1  }{\\sqrt{x}}[\/latex].<\/p>\n<\/div>\n<\/div>\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-149\">\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>1.4 Inverse Functions. <strong>Authored by<\/strong>: Ryan Melton. <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>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/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":17533,"menu_order":18,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"1.4 Inverse Functions\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-149","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/149","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":37,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/149\/revisions"}],"predecessor-version":[{"id":4752,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/149\/revisions\/4752"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/149\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=149"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=149"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=149"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=149"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}