{"id":1149,"date":"2024-04-11T17:28:43","date_gmt":"2024-04-11T17:28:43","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=1149"},"modified":"2024-08-31T05:49:33","modified_gmt":"2024-08-31T05:49:33","slug":"inverse-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/inverse-functions-learn-it-3\/","title":{"raw":"Inverse Functions: Learn It 3","rendered":"Inverse Functions: Learn It 3"},"content":{"raw":"<h2>Graphing Inverse Functions<\/h2>\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.<\/p>\r\n<p>Consider the graph of [latex]f[\/latex] shown in Figure 9(a) and a point [latex](a,b)[\/latex] on the graph.\u00a0<\/p>\r\n\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\r\n\r\n<p>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<section class=\"textbox questionHelp\">\r\n<p><b>How to: Graph the Inverse of a Function<br \/>\r\n<br \/>\r\n<\/b><\/p>\r\n<ol>\r\n\t<li><strong>Plot the Function<\/strong>: Graph the original function [latex]f(x)[\/latex]\u00a0and plot a few key points.<\/li>\r\n\t<li><strong>Reflect Over Line<\/strong>: Reflect these points over the line [latex]y=x[\/latex] to find the corresponding points on [latex]f^{-1}[\/latex].<\/li>\r\n\t<li><strong>Draw the Inverse<\/strong>: Connect these reflected points to graph the inverse function.<\/li>\r\n\t<li><strong>Check<\/strong>: Ensure that each point [latex](a,b)[\/latex] on the original function corresponds to the point [latex](b,a)[\/latex] on the inverse function.<\/li>\r\n\t<li><strong>Line of Symmetry<\/strong>: The line [latex]y=x[\/latex]\u00a0should act as a line of symmetry between the function and its inverse.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<div id=\"fs-id1170572213184\" class=\"textbook exercises\">\r\n<section class=\"textbox example\">\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\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]\r\n<span id=\"fs-id1170572222895\"><\/span><br \/>\r\n[reveal-answer q=\"fs-id1170572222910\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170572222910\"]\r\n\r\n\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\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]\r\n<span id=\"fs-id1170572480252\"><\/span>[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]170589[\/ohm_question]<\/p>\r\n<\/section>\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>.<\/p>\r\n<p>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.<\/p>\r\n<p>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].<\/p>\r\n<p>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\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\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>restricted domain<\/h3>\r\n<p>Some functions don\u2019t have inverses over their full domains because they\u2019re not one-to-one. By restricting the domain, we ensure the function is one-to-one. Once the domain is restricted, we can define an inverse.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170572551608\">Consider the function [latex]f(x)=(x+1)^2[\/latex].<\/p>\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<p>[reveal-answer q=\"fs-id1170572477860\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170572477860\"]<\/p>\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.<br \/>\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]\r\n<\/li>\r\n\t<li>On the interval [latex][-1,\\infty), \\, f[\/latex] is one-to-one.<br \/>\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\n<br \/>\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]. <br \/>\r\n<br \/>\r\nTo 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\r\n\r\nWatch the following video to see the worked solution to this example.<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\" data-mce-fragment=\"1\"><\/iframe><\/center>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\n\r\n<p>You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.4InverseFunctions564to791_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]197109[\/ohm_question]<\/p>\r\n<\/section>\r\n<\/div>","rendered":"<h2>Graphing Inverse Functions<\/h2>\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.<\/p>\n<p>Consider the graph of [latex]f[\/latex] shown in Figure 9(a) and a point [latex](a,b)[\/latex] on the graph.\u00a0<\/p>\n<figure style=\"width: 710px\" 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\" \/><figcaption 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].<\/figcaption><\/figure>\n<p>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<section class=\"textbox questionHelp\">\n<p><b>How to: Graph the Inverse of a Function<\/p>\n<p><\/b><\/p>\n<ol>\n<li><strong>Plot the Function<\/strong>: Graph the original function [latex]f(x)[\/latex]\u00a0and plot a few key points.<\/li>\n<li><strong>Reflect Over Line<\/strong>: Reflect these points over the line [latex]y=x[\/latex] to find the corresponding points on [latex]f^{-1}[\/latex].<\/li>\n<li><strong>Draw the Inverse<\/strong>: Connect these reflected points to graph the inverse function.<\/li>\n<li><strong>Check<\/strong>: Ensure that each point [latex](a,b)[\/latex] on the original function corresponds to the point [latex](b,a)[\/latex] on the inverse function.<\/li>\n<li><strong>Line of Symmetry<\/strong>: The line [latex]y=x[\/latex]\u00a0should act as a line of symmetry between the function and its inverse.<\/li>\n<\/ol>\n<\/section>\n<div id=\"fs-id1170572213184\" class=\"textbook exercises\">\n<section class=\"textbox example\">\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<figure style=\"width: 325px\" 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\" \/><figcaption class=\"wp-caption-text\">Figure 10. Graph of [latex]f(x)[\/latex].<\/figcaption><\/figure>\n<p><span id=\"fs-id1170572222895\"><\/span><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572222910\">Show Solution<\/button><\/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<figure style=\"width: 325px\" 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\" \/><figcaption class=\"wp-caption-text\">Figure 11. Graph of [latex]f(x)[\/latex] and its inverse.<\/figcaption><\/figure>\n<p><span id=\"fs-id1170572480252\"><\/span><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm170589\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=170589&theme=lumen&iframe_resize_id=ohm170589&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\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>.<\/p>\n<p>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.<\/p>\n<p>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].<\/p>\n<p>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<figure style=\"width: 656px\" 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\" \/><figcaption 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].<\/figcaption><\/figure>\n<section class=\"textbox keyTakeaway\">\n<h3>restricted domain<\/h3>\n<p>Some functions don\u2019t have inverses over their full domains because they\u2019re not one-to-one. By restricting the domain, we ensure the function is one-to-one. Once the domain is restricted, we can define an inverse.<\/p>\n<\/section>\n<section class=\"textbox example\">\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<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572477860\">Show Solution<\/button><\/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.<br \/>\n<figure style=\"width: 325px\" 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\" \/><figcaption class=\"wp-caption-text\">Figure 14. Graph of the function\u00a0[latex]f(x)=(x+1)^2[\/latex].<\/figcaption><\/figure>\n<\/li>\n<li>On the interval [latex][-1,\\infty), \\, f[\/latex] is one-to-one.<br \/>\n<figure style=\"width: 325px\" 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\" \/><figcaption class=\"wp-caption-text\">Figure 15. Looking at a restricted domain graph of a function.<\/figcaption><\/figure>\n<p>\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]. <\/p>\n<p>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<p>Watch the following video to see the worked solution to this example.<\/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\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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 transcript for this video using <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.4InverseFunctions564to791_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm197109\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=197109&theme=lumen&iframe_resize_id=ohm197109&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<\/div>\n","protected":false},"author":15,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":759,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1149"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":16,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1149\/revisions"}],"predecessor-version":[{"id":4650,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1149\/revisions\/4650"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/759"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1149\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=1149"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=1149"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=1149"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=1149"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}