{"id":1931,"date":"2016-11-02T22:23:20","date_gmt":"2016-11-02T22:23:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1931"},"modified":"2017-04-20T20:26:00","modified_gmt":"2017-04-20T20:26:00","slug":"radicals-as-inverse-polynomial-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/radicals-as-inverse-polynomial-functions\/","title":{"raw":"Radicals as Inverse Polynomial Functions","rendered":"Radicals as Inverse Polynomial Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Verify that a radical and a polynomial function are inverses of each other<\/li>\r\n \t<li>Find the inverse of a polynomial function<\/li>\r\n<\/ul>\r\n<\/div>\r\nTwo functions <em>f<\/em>\u00a0and <em>g<\/em>\u00a0are inverse functions if for every coordinate pair in <em>f<\/em>, (<em>a<\/em>, <em>b<\/em>), there exists a corresponding coordinate pair in the inverse function, <em>g<\/em>, (<em>b<\/em>, <em>a<\/em>). In other words, the coordinate pairs of the inverse functions have the input and output interchanged.\r\n\r\nFor a function to have an <strong>inverse function<\/strong> the function to create a new function that is <strong>one-to-one<\/strong> and would have an inverse function.\r\n\r\nFor example, suppose a water runoff collector is built in the shape of a parabolic trough as shown\u00a0below. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221704\/CNX_Precalc_Figure_03_08_0022.jpg\" alt=\"Diagram of a parabolic trough that is 18\" width=\"487\" height=\"279\" data-media-type=\"image\/jpg\" \/>\r\n\r\nBecause it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with <em>x<\/em>\u00a0measured horizontally and <em>y<\/em>\u00a0measured vertically, with the origin at the vertex of the parabola.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221706\/CNX_Precalc_Figure_03_08_0032.jpg\" alt=\"Graph of a parabola.\" width=\"487\" height=\"441\" data-media-type=\"image\/jpg\" \/>\r\n\r\nFrom this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form [latex]y\\left(x\\right)=a{x}^{2}[\/latex]. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor <em>a<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} 18=a{6}^{2}\\hfill \\\\ a=\\frac{18}{36}\\hfill \\\\ \\text{ }=\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nOur parabolic cross section has the equation\r\n<p style=\"text-align: center;\">[latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/p>\r\nWe are interested in the <strong>surface area<\/strong> of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth <em>y<\/em>\u00a0the width will be given by 2<em>x<\/em>, so we need to solve the equation above for <em>x<\/em>\u00a0and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.\r\n\r\nTo find an inverse, we can restrict our original function to a limited domain on which it <em data-effect=\"italics\">is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive <em>x<\/em>\u00a0values. On this domain, we can find an inverse by solving for the input variable:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=\\frac{1}{2}{x}^{2}\\hfill \\\\ 2y={x}^{2}\\hfill \\\\ \\text{ }x=\\pm \\sqrt{2y}\\hfill \\end{array}[\/latex]<\/p>\r\nThis is not a function as written. We are limiting ourselves to positive <em>x<\/em>\u00a0values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.\r\n<p style=\"text-align: center;\">[latex]y=\\frac{{x}^{2}}{2},\\text{ }x&gt;0[\/latex]<\/p>\r\nBecause <em>x<\/em>\u00a0is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2<em>x<\/em>. The trough is 3 feet (36 inches) long, so the surface area will then be:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{Area} &amp; =l\\cdot w\\hfill \\\\ \\text{ } &amp; =36\\cdot 2x\\hfill \\\\ \\text{ } &amp; =72x\\hfill \\\\ \\text{ } &amp; =72\\sqrt{2y}\\hfill \\end{array}[\/latex]<\/p>\r\nThis example illustrates two important points:\r\n<ol>\r\n \t<li>When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.<\/li>\r\n \t<li>The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.<\/li>\r\n<\/ol>\r\nFunctions involving roots are often called <strong>radical functions<\/strong>. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called <strong>invertible functions<\/strong>, and we use the notation [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n\r\nWarning: [latex]{f}^{-1}\\left(x\\right)[\/latex] is not the same as the reciprocal of the function [latex]f\\left(x\\right)[\/latex]. This use of \u20131 is reserved to denote inverse functions. To denote the reciprocal of a function [latex]f\\left(x\\right)[\/latex], we would need to write [latex]{\\left(f\\left(x\\right)\\right)}^{-1}=\\frac{1}{f\\left(x\\right)}[\/latex].\r\n\r\nAn important relationship between inverse functions is that they \"undo\" each other. If [latex]{f}^{-1}[\/latex] is the inverse of a function <em>f<\/em>,\u00a0then <em>f<\/em>\u00a0is the inverse of the function [latex]{f}^{-1}[\/latex]. In other words, whatever the function <em>f<\/em>\u00a0does to <em>x<\/em>, [latex]{f}^{-1}[\/latex] undoes it\u2014and vice-versa. More formally, we write\r\n<p style=\"text-align: center;\">[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }f[\/latex]<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }{f}^{-1}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Verifying Two Functions Are Inverses of One Another<\/h3>\r\nTwo functions, <em>f<\/em>\u00a0and <i>g<\/i>, are inverses of one another if for all <em>x<\/em>\u00a0in the domain of <em>f\u00a0<\/em>and <em>g<\/em>.\r\n\r\n[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.<\/h3>\r\n<ol>\r\n \t<li>Replace [latex]f\\left(x\\right)[\/latex] with <em>y<\/em>.<\/li>\r\n \t<li>Interchange <em>x<\/em>\u00a0and <em>y<\/em>.<\/li>\r\n \t<li>Solve for <em>y<\/em>, and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Verifying Inverse Functions<\/h3>\r\nShow that [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex] and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses, for [latex]x\\ne 0,-1[\/latex] .\r\n\r\n[reveal-answer q=\"631376\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"631376\"]\r\n\r\nWe must show that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{1}{x+1}\\right)\\hfill \\\\ \\text{ }=\\frac{1}{\\frac{1}{x+1}}-1\\hfill \\\\ \\text{ }=\\left(x+1\\right)-1\\hfill \\\\ \\text{ }=x\\hfill \\\\ f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(\\frac{1}{x}-1\\right)\\hfill \\\\ \\text{ }=\\frac{1}{\\left(\\frac{1}{x}-1\\right)+1}\\hfill \\\\ \\text{ }=\\frac{1}{\\frac{1}{x}}\\hfill \\\\ \\text{ }=x\\hfill \\end{array}[\/latex]<\/p>\r\nTherefore, [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nShow that [latex]f\\left(x\\right)=\\frac{x+5}{3}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=3x - 5[\/latex] are inverses.\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n[reveal-answer q=\"593015\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"593015\"][latex]{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{x+5}{3}\\right)=3\\left(\\frac{x+5}{3}\\right)-5=\\left(x - 5\\right)+5=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(3x - 5\\right)=\\frac{\\left(3x - 5\\right)+5}{3}=\\frac{3x}{3}=x[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Inverse of a Cubic Function<\/h3>\r\nFind the inverse of the function [latex]f\\left(x\\right)=5{x}^{3}+1[\/latex].\r\n\r\n[reveal-answer q=\"289537\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"289537\"]\r\n\r\nThis is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for <em>x<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }y=5{x}^{3}+1\\hfill \\\\ \\text{ }x=5{y}^{3}+1\\hfill \\\\ \\text{ }x - 1=5{y}^{3}\\hfill \\\\ \\text{ }\\frac{x - 1}{5}={y}^{3}\\hfill \\\\ {f}^{-1}\\left(x\\right)=\\sqrt[3]{\\frac{x - 1}{5}}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nLook at the graph of <em>f<\/em>\u00a0and [latex]{f}^{-1}[\/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[\/latex]. This is always the case when graphing a function and its inverse function.\r\n\r\nAlso, since the method involved interchanging <em>x<\/em>\u00a0and <em>y<\/em>, notice corresponding points. If [latex]\\left(a,b\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(b,a\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Since [latex]\\left(0,1\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(1,0\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Similarly, since [latex]\\left(1,6\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(6,1\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221708\/CNX_Precalc_Figure_03_08_0042.jpg\" alt=\"Graph of f(x)=5x^3+1 and its inverse, f^(-1)(x)=3sqrt((x-1)\/(5)).\" width=\"487\" height=\"554\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the inverse function of [latex]f\\left(x\\right)=\\sqrt[3]{x+4}[\/latex].\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n\r\n[reveal-answer q=\"987592\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"987592\"][latex]{f}^{-1}\\left(x\\right)={x}^{3}-4[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=15856&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\">\r\n<\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Verify that a radical and a polynomial function are inverses of each other<\/li>\n<li>Find the inverse of a polynomial function<\/li>\n<\/ul>\n<\/div>\n<p>Two functions <em>f<\/em>\u00a0and <em>g<\/em>\u00a0are inverse functions if for every coordinate pair in <em>f<\/em>, (<em>a<\/em>, <em>b<\/em>), there exists a corresponding coordinate pair in the inverse function, <em>g<\/em>, (<em>b<\/em>, <em>a<\/em>). In other words, the coordinate pairs of the inverse functions have the input and output interchanged.<\/p>\n<p>For a function to have an <strong>inverse function<\/strong> the function to create a new function that is <strong>one-to-one<\/strong> and would have an inverse function.<\/p>\n<p>For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown\u00a0below. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221704\/CNX_Precalc_Figure_03_08_0022.jpg\" alt=\"Diagram of a parabolic trough that is 18\" width=\"487\" height=\"279\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with <em>x<\/em>\u00a0measured horizontally and <em>y<\/em>\u00a0measured vertically, with the origin at the vertex of the parabola.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221706\/CNX_Precalc_Figure_03_08_0032.jpg\" alt=\"Graph of a parabola.\" width=\"487\" height=\"441\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form [latex]y\\left(x\\right)=a{x}^{2}[\/latex]. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor <em>a<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} 18=a{6}^{2}\\hfill \\\\ a=\\frac{18}{36}\\hfill \\\\ \\text{ }=\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>Our parabolic cross section has the equation<\/p>\n<p style=\"text-align: center;\">[latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/p>\n<p>We are interested in the <strong>surface area<\/strong> of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth <em>y<\/em>\u00a0the width will be given by 2<em>x<\/em>, so we need to solve the equation above for <em>x<\/em>\u00a0and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.<\/p>\n<p>To find an inverse, we can restrict our original function to a limited domain on which it <em data-effect=\"italics\">is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive <em>x<\/em>\u00a0values. On this domain, we can find an inverse by solving for the input variable:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=\\frac{1}{2}{x}^{2}\\hfill \\\\ 2y={x}^{2}\\hfill \\\\ \\text{ }x=\\pm \\sqrt{2y}\\hfill \\end{array}[\/latex]<\/p>\n<p>This is not a function as written. We are limiting ourselves to positive <em>x<\/em>\u00a0values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.<\/p>\n<p style=\"text-align: center;\">[latex]y=\\frac{{x}^{2}}{2},\\text{ }x>0[\/latex]<\/p>\n<p>Because <em>x<\/em>\u00a0is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2<em>x<\/em>. The trough is 3 feet (36 inches) long, so the surface area will then be:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{Area} & =l\\cdot w\\hfill \\\\ \\text{ } & =36\\cdot 2x\\hfill \\\\ \\text{ } & =72x\\hfill \\\\ \\text{ } & =72\\sqrt{2y}\\hfill \\end{array}[\/latex]<\/p>\n<p>This example illustrates two important points:<\/p>\n<ol>\n<li>When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.<\/li>\n<li>The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.<\/li>\n<\/ol>\n<p>Functions involving roots are often called <strong>radical functions<\/strong>. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called <strong>invertible functions<\/strong>, and we use the notation [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<p>Warning: [latex]{f}^{-1}\\left(x\\right)[\/latex] is not the same as the reciprocal of the function [latex]f\\left(x\\right)[\/latex]. This use of \u20131 is reserved to denote inverse functions. To denote the reciprocal of a function [latex]f\\left(x\\right)[\/latex], we would need to write [latex]{\\left(f\\left(x\\right)\\right)}^{-1}=\\frac{1}{f\\left(x\\right)}[\/latex].<\/p>\n<p>An important relationship between inverse functions is that they &#8220;undo&#8221; each other. If [latex]{f}^{-1}[\/latex] is the inverse of a function <em>f<\/em>,\u00a0then <em>f<\/em>\u00a0is the inverse of the function [latex]{f}^{-1}[\/latex]. In other words, whatever the function <em>f<\/em>\u00a0does to <em>x<\/em>, [latex]{f}^{-1}[\/latex] undoes it\u2014and vice-versa. More formally, we write<\/p>\n<p style=\"text-align: center;\">[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }f[\/latex]<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }{f}^{-1}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Verifying Two Functions Are Inverses of One Another<\/h3>\n<p>Two functions, <em>f<\/em>\u00a0and <i>g<\/i>, are inverses of one another if for all <em>x<\/em>\u00a0in the domain of <em>f\u00a0<\/em>and <em>g<\/em>.<\/p>\n<p>[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.<\/h3>\n<ol>\n<li>Replace [latex]f\\left(x\\right)[\/latex] with <em>y<\/em>.<\/li>\n<li>Interchange <em>x<\/em>\u00a0and <em>y<\/em>.<\/li>\n<li>Solve for <em>y<\/em>, and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Verifying Inverse Functions<\/h3>\n<p>Show that [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex] and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses, for [latex]x\\ne 0,-1[\/latex] .<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q631376\">Solution<\/span><\/p>\n<div id=\"q631376\" class=\"hidden-answer\" style=\"display: none\">\n<p>We must show that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{1}{x+1}\\right)\\hfill \\\\ \\text{ }=\\frac{1}{\\frac{1}{x+1}}-1\\hfill \\\\ \\text{ }=\\left(x+1\\right)-1\\hfill \\\\ \\text{ }=x\\hfill \\\\ f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(\\frac{1}{x}-1\\right)\\hfill \\\\ \\text{ }=\\frac{1}{\\left(\\frac{1}{x}-1\\right)+1}\\hfill \\\\ \\text{ }=\\frac{1}{\\frac{1}{x}}\\hfill \\\\ \\text{ }=x\\hfill \\end{array}[\/latex]<\/p>\n<p>Therefore, [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Show that [latex]f\\left(x\\right)=\\frac{x+5}{3}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=3x - 5[\/latex] are inverses.<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q593015\">Solution<\/span><\/p>\n<div id=\"q593015\" class=\"hidden-answer\" style=\"display: none\">[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{x+5}{3}\\right)=3\\left(\\frac{x+5}{3}\\right)-5=\\left(x - 5\\right)+5=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(3x - 5\\right)=\\frac{\\left(3x - 5\\right)+5}{3}=\\frac{3x}{3}=x[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Inverse of a Cubic Function<\/h3>\n<p>Find the inverse of the function [latex]f\\left(x\\right)=5{x}^{3}+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q289537\">Solution<\/span><\/p>\n<div id=\"q289537\" class=\"hidden-answer\" style=\"display: none\">\n<p>This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for <em>x<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }y=5{x}^{3}+1\\hfill \\\\ \\text{ }x=5{y}^{3}+1\\hfill \\\\ \\text{ }x - 1=5{y}^{3}\\hfill \\\\ \\text{ }\\frac{x - 1}{5}={y}^{3}\\hfill \\\\ {f}^{-1}\\left(x\\right)=\\sqrt[3]{\\frac{x - 1}{5}}\\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Look at the graph of <em>f<\/em>\u00a0and [latex]{f}^{-1}[\/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[\/latex]. This is always the case when graphing a function and its inverse function.<\/p>\n<p>Also, since the method involved interchanging <em>x<\/em>\u00a0and <em>y<\/em>, notice corresponding points. If [latex]\\left(a,b\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(b,a\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Since [latex]\\left(0,1\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(1,0\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Similarly, since [latex]\\left(1,6\\right)[\/latex] is on the graph of <em>f<\/em>, then [latex]\\left(6,1\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221708\/CNX_Precalc_Figure_03_08_0042.jpg\" alt=\"Graph of f(x)=5x^3+1 and its inverse, f^(-1)(x)=3sqrt((x-1)\/(5)).\" width=\"487\" height=\"554\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the inverse function of [latex]f\\left(x\\right)=\\sqrt[3]{x+4}[\/latex].<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q987592\">Solution<\/span><\/p>\n<div id=\"q987592\" class=\"hidden-answer\" style=\"display: none\">[latex]{f}^{-1}\\left(x\\right)={x}^{3}-4[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=15856&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><br \/>\n<\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1931\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Question ID 15856. <strong>Authored by<\/strong>: Sousa,James. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Question ID 15856\",\"author\":\"Sousa,James\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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