{"id":91,"date":"2019-08-20T17:02:19","date_gmt":"2019-08-20T21:02:19","guid":{"rendered":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/chapter\/inverses-and-radical-functions\/"},"modified":"2022-06-01T10:39:27","modified_gmt":"2022-06-01T14:39:27","slug":"inverses-and-radical-functions","status":"publish","type":"chapter","link":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/chapter\/inverses-and-radical-functions\/","title":{"raw":"Inverses and Radical Functions","rendered":"Inverses and Radical Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\nIn this section, you will:\n<ul>\n \t<li>Find the inverse of an invertible polynomial function.<\/li>\n \t<li>Restrict the domain to find the inverse of a polynomial function.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137793975\">A mound of gravel is in the shape of a cone with the height equal to twice the radius.<span id=\"fs-id1165137939558\"><\/span><\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/cnx.org\/resources\/b91ae06dad3cdb0412db8cff73515f372ea917b5\/CNX_Precalc_Figure_03_08_001.jpg\" alt=\"Gravel in the shape of a cone.\" width=\"487\" height=\"410\"> <strong>Figure 1.<\/strong>[\/caption]\n<p id=\"fs-id1165137411369\">The volume is found using a formula from elementary geometry.<\/p>\n\n<div id=\"eip-854\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}V&amp; \\hfill =&amp; \\frac{1}{3}\\pi {r}^{2}h\\hfill \\\\ &amp; =&amp; \\frac{1}{3}\\pi {r}^{2}\\left(2r\\right)\\hfill \\\\ &amp; =&amp; \\frac{2}{3}\\pi {r}^{3}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137727278\">We have written the volume[latex]\\,V\\,[\/latex]in terms of the radius[latex]\\,r.\\,[\/latex]However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula<\/p>\n\n<div id=\"eip-931\" class=\"unnumbered aligncenter\">[latex]r=\\sqrt[3]{\\frac{3V}{2\\pi }}[\/latex]<\/div>\n<p id=\"fs-id1165134129769\">This function is the inverse of the formula for[latex]\\,V\\,[\/latex]in terms of[latex]\\,r.[\/latex]<\/p>\n<p id=\"fs-id1165137656509\">In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process.<\/p>\n\n<div id=\"fs-id1165135500723\" class=\"bc-section section\">\n<h3>Finding the Inverse of a Polynomial Function<\/h3>\n<p id=\"fs-id1165137439029\">Two functions[latex]\\,f\\,[\/latex]and[latex]\\,g\\,[\/latex]are inverse functions if for every coordinate pair in[latex]\\,f,\\left(a,b\\right),\\,[\/latex]there exists a corresponding coordinate pair in the inverse function,[latex]\\,g,\\left(b,\\,a\\right).\\,[\/latex]In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Only one-to-one functions have inverses. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test.<\/p>\n<p id=\"fs-id1165137448308\">For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_08_002\">(Figure)<\/a>. 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\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/cnx.org\/resources\/1d361de6ef84f0703aaa4fe3b93dff47afbdb8a9\/CNX_Precalc_Figure_03_08_002.jpg\" alt=\"Diagram of a parabolic trough that is 18\u201d in height, 3\u2019 in length, and 12\u201d in width.\" width=\"487\" height=\"279\"> <strong>Figure 2.<\/strong>[\/caption]\n<p id=\"fs-id1165137793665\">Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with[latex]\\,x\\,[\/latex]measured horizontally and[latex]\\,y\\,[\/latex]measured vertically, with the origin at the vertex of the parabola. See <a class=\"autogenerated-content\" href=\"#Figure_03_08_003\">(Figure)<\/a>.<\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/cnx.org\/resources\/a1157fe5e3ce318e6eb3be33ccbd74ea983d961f\/CNX_Precalc_Figure_03_08_003.jpg\" alt=\"Graph of a parabola.\" width=\"487\" height=\"441\"> <strong>Figure 3.<\/strong>[\/caption]\n<p id=\"fs-id1165137771677\">From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form[latex]\\,y\\left(x\\right)=a{x}^{2}.\\,[\/latex]Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor[latex]\\,a.[\/latex]<\/p>\n\n<div id=\"eip-893\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill 18&amp; =&amp; a{6}^{2}\\hfill \\\\ \\hfill a&amp; =&amp; \\frac{18}{36}\\hfill \\\\ &amp; =&amp; \\frac{1}{2}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137633973\">Our parabolic cross section has the equation<\/p>\n\n<div id=\"eip-420\" class=\"unnumbered aligncenter\">[latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/div>\n<p id=\"fs-id1165137770004\">We are interested in the <span class=\"no-emphasis\">surface area<\/span> 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[latex]\\,y,\\,[\/latex]the width will be given by[latex]\\,2x,\\,[\/latex]so we need to solve the equation above for[latex]\\,x\\,[\/latex]and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.<\/p>\n<p id=\"fs-id1165137638570\">To find an inverse, we can restrict our original function to a limited domain on which it <em>is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive[latex]\\,x\\,[\/latex]values. On this domain, we can find an inverse by solving for the input variable:<\/p>\n\n<div id=\"eip-598\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill y&amp; =&amp; \\frac{1}{2}{x}^{2}\\hfill \\\\ \\hfill 2y&amp; =&amp; {x}^{2}\\hfill \\\\ \\hfill x&amp; =&amp; \u00b1\\sqrt{2y}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137453965\">This is not a function as written. We are limiting ourselves to positive[latex]\\,x\\,[\/latex]values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.<\/p>\n\n<div class=\"unnumbered\">[latex]y=\\frac{{x}^{2}}{2},\\text{ }x&gt;0[\/latex]<\/div>\n<p id=\"fs-id1165137643958\">Because[latex]\\,x\\,[\/latex]is the distance from the center of the parabola to either side, the entire width of the water at the top will be[latex]\\,2x.\\,[\/latex]The trough is 3 feet (36 inches) long, so the surface area will then be:<\/p>\n\n<div class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill \\text{Area}&amp; =&amp; l\\cdot w\\hfill \\\\ &amp; =&amp; 36\\cdot 2x\\hfill \\\\ &amp; =&amp; 72x\\hfill \\\\ &amp; =&amp; 72\\sqrt{2y}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137407432\">This example illustrates two important points:<\/p>\n\n<ol id=\"fs-id1165135545666\" type=\"1\">\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>\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>\n<\/ol>\n<p id=\"fs-id1165137618975\">Functions involving roots are often called <span class=\"no-emphasis\">radical functions<\/span>. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation[latex]\\,{f}^{-1}\\left(x\\right).[\/latex]<\/p>\n<p id=\"fs-id1165135185952\">Warning:[latex]\\,{f}^{-1}\\left(x\\right)\\,[\/latex]is not the same as the reciprocal of the function[latex]\\,f\\left(x\\right).\\,[\/latex]This use of \u201c\u20131\u201d is reserved to denote inverse functions. To denote the reciprocal of a function[latex]\\,f\\left(x\\right),\\,[\/latex]we would need to write[latex]\\,{\\left(f\\left(x\\right)\\right)}^{-1}=\\frac{1}{f\\left(x\\right)}.[\/latex]<\/p>\n<p id=\"fs-id1165137561919\">An important relationship between inverse functions is that they \u201cundo\u201d each other. If[latex]\\,{f}^{-1}\\,[\/latex]is the inverse of a function[latex]\\,f,\\,[\/latex]\nthen[latex]\\,f\\,[\/latex]is the inverse of the function[latex]\\,{f}^{-1}.\\,[\/latex]In other words, whatever the function[latex]\\,f\\,[\/latex]does to[latex]\\,x,[\/latex]\n[latex]\\,{f}^{-1}\\,[\/latex]undoes it\u2014and vice-versa.<\/p>\n\n<div id=\"eip-519\" class=\"unnumbered aligncenter\">[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x,\\,\\text{for all }x\\text{ in the domain of }f[\/latex]<\/div>\n<p id=\"fs-id1165135503755\">and<\/p>\n\n<div id=\"eip-590\" class=\"unnumbered aligncenter\">[latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x,\\,\\text{for all }x\\text{ in the domain of }{f}^{-1}[\/latex]<\/div>\n<p id=\"eip-457\">Note that the inverse switches the domain and range of the original function.<\/p>\n\n<div id=\"fs-id1165137735698\" class=\"textbox key-takeaways\">\n<h3>Verifying Two Functions Are Inverses of One Another<\/h3>\n<p id=\"fs-id1165137852132\">Two functions,[latex]\\,f\\,[\/latex]\nand[latex]\\,g,\\,[\/latex]are inverses of one another if for all[latex]\\,x\\,[\/latex]in the domain of[latex]\\,f\\,[\/latex]\nand[latex]\\,g.[\/latex]<\/p>\n\n<div id=\"eip-973\" class=\"unnumbered aligncenter\">[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137646263\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137501372\"><strong>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.<\/strong><\/p>\n\n<ol type=\"1\">\n \t<li>Replace[latex]\\,f\\left(x\\right)\\,[\/latex]with[latex]\\,y.[\/latex]<\/li>\n \t<li>Interchange[latex]\\,x\\,[\/latex]and[latex]\\,y.[\/latex]<\/li>\n \t<li>Solve for[latex]\\,y,\\,[\/latex]and rename the function[latex]\\,{f}^{-1}\\left(x\\right).[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_08_01\" class=\"textbox examples\">\n<div id=\"fs-id1165135150650\">\n<div id=\"fs-id1165135620877\">\n<h3>Verifying Inverse Functions<\/h3>\n<p id=\"fs-id1165134148383\">Show that[latex]\\,f\\left(x\\right)=\\frac{1}{x+1}\\,[\/latex]and[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1\\,[\/latex]are inverses, for[latex]\\,x\\ne 0,-1[\/latex].<\/p>\n\n<\/div>\n<div id=\"fs-id1165137724950\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137724950\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137724950\"]\n<p id=\"fs-id1165137834138\">We must show that[latex]\\,{f}^{-1}\\left(f\\left(x\\right)\\right)=x\\,[\/latex]and[latex]\\,f\\left({f}^{-1}\\left(x\\right)\\right)=x.[\/latex]<\/p>\n\n<div id=\"eip-id1165131967953\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill {f}^{-1}\\left(f\\left(x\\right)\\right)&amp; =&amp; {f}^{-1}\\left(\\frac{1}{x+1}\\right)\\hfill \\\\ &amp; =&amp; \\frac{1}{\\frac{1}{x+1}}-1\\hfill \\\\ &amp; =&amp; \\left(x+1\\right)-1\\\\ &amp; =&amp; x\\hfill \\\\ \\hfill f\\left({f}^{-1}\\left(x\\right)\\right)&amp; =&amp; f\\left(\\frac{1}{x}-1\\right)\\hfill \\\\ &amp; =&amp; \\frac{1}{\\left(\\frac{1}{x}-1\\right)+1}\\hfill \\\\ &amp; =&amp; \\frac{1}{\\frac{1}{x}}\\hfill \\\\ &amp; =&amp; x\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135168183\">Therefore,[latex]\\,f\\left(x\\right)=\\frac{1}{x+1}\\,[\/latex]\nand[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1\\,[\/latex]are inverses.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137594492\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_01\">\n<div id=\"fs-id1165137552094\">\n<p id=\"fs-id1165137426116\">Show that[latex]\\,f\\left(x\\right)=\\frac{x+5}{3}\\,[\/latex]\nand[latex]\\,{f}^{-1}\\left(x\\right)=3x-5\\,[\/latex]are inverses.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135502962\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135502962\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135502962\"]\n<p id=\"fs-id1165137696615\">[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{x+5}{3}\\right)=3\\left(\\frac{x+5}{3}\\right)-5=\\left(x-5\\right)+5=x\\,[\/latex]and[latex]\\,f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(3x-5\\right)=\\frac{\\left(3x-5\\right)+5}{3}=\\frac{3x}{3}=x[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_08_02\" class=\"textbox examples\">\n<div id=\"fs-id1165137600799\">\n<div id=\"fs-id1165135160775\">\n<h3>Finding the Inverse of a Cubic Function<\/h3>\n<p id=\"fs-id1165137569920\">Find the inverse of the function[latex]\\,f\\left(x\\right)=5{x}^{3}+1.[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137676384\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137676384\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137676384\"]\n<p id=\"fs-id1165135412872\">This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for[latex]\\,x.[\/latex]<\/p>\n\n<div id=\"eip-id1165133077998\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill y&amp; =&amp; 5{x}^{3}+1\\hfill \\\\ \\hfill x&amp; =&amp; 5{y}^{3}+1\\hfill \\\\ \\hfill x-1&amp; =&amp; 5{y}^{3}\\hfill \\\\ \\hfill \\frac{x-1}{5}&amp; =&amp; {y}^{3}\\hfill \\\\ \\hfill {f}^{-1}\\left(x\\right)&amp; =&amp; \\sqrt[3]{\\frac{x-1}{5}}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<div id=\"fs-id1165137635322\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165137641602\">Look at the graph of[latex]\\,f\\,[\/latex]and[latex]\\,{f}^{\u20131}.\\,[\/latex]Notice that one graph is the reflection of the other about the line[latex]\\,y=x.\\,[\/latex]This is always the case when graphing a function and its inverse function.<\/p>\n<p id=\"fs-id1165137793468\">Also, since the method involved interchanging[latex]\\,x\\,[\/latex]and[latex]\\,y,\\,[\/latex]notice corresponding points. If[latex]\\,\\left(a,b\\right)\\,[\/latex]is on the graph of[latex]\\,f,[\/latex]then[latex]\\,\\left(b,a\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{\u20131}.\\,[\/latex]Since[latex]\\,\\left(0,1\\right)\\,[\/latex]is on the graph of[latex]\\,f,\\,[\/latex]then[latex]\\,\\left(1,0\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{\u20131}.\\,[\/latex]Similarly, since[latex]\\,\\left(1,6\\right)\\,[\/latex]is on the graph of[latex]\\,f,[\/latex]then[latex]\\,\\left(6,1\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{\u20131}.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Figure_03_08_004\">(Figure)<\/a>.<\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/cnx.org\/resources\/90a0ca823116c8647b783db6d540dad88b0b04f0\/CNX_Precalc_Figure_03_08_004.jpg\" alt=\"Graph of f(x)=5x^3+1 and its inverse, f^(-1)(x)=3sqrt((x-1)\/(5)).\" width=\"487\" height=\"554\"> <strong>Figure 4.<\/strong>[\/caption]\n\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_02\">\n<div id=\"fs-id1165137704564\">\n<p id=\"fs-id1165133047522\">Find the inverse function of[latex]\\,f\\left(x\\right)=\\sqrt[3]{x+4}.[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137737307\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137737307\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137737307\"]\n<p id=\"fs-id1165137529408\">[latex]{f}^{-1}\\left(x\\right)={x}^{3}-4[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137847291\" class=\"bc-section section\">\n<h3>Restricting the Domain to Find the Inverse of a Polynomial Function<\/h3>\n<p id=\"fs-id1165137471808\">So far, we have been able to find the inverse functions of <span class=\"no-emphasis\">cubic functions<\/span> without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an <span class=\"no-emphasis\">inverse function<\/span>. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.<\/p>\n\n<div id=\"fs-id1165137434585\" class=\"textbox key-takeaways\">\n<h3>Restricting the Domain<\/h3>\n<p id=\"fs-id1165137409777\">If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.<\/p>\n\n<\/div>\n<div id=\"fs-id1165137431545\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137656706\"><strong>Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.<\/strong><\/p>\n\n<ol id=\"fs-id1165137532171\" type=\"1\">\n \t<li>Restrict the domain by determining a domain on which the original function is one-to-one.<\/li>\n \t<li>Replace[latex]\\,f\\left(x\\right)\\,[\/latex]with[latex]\\,y.[\/latex]<\/li>\n \t<li>Interchange[latex]\\,x\\,[\/latex]and[latex]\\,y.[\/latex]<\/li>\n \t<li>Solve for[latex]\\,y,[\/latex]and rename the function or pair of function[latex]\\,{f}^{-1}\\left(x\\right).[\/latex]<\/li>\n \t<li>Revise the formula for[latex]\\,{f}^{-1}\\left(x\\right)\\,[\/latex]by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_08_03\" class=\"textbox examples\">\n<div>\n<div id=\"fs-id1165137668130\">\n<h3>Restricting the Domain to Find the Inverse of a Polynomial Function<\/h3>\n<p id=\"fs-id1165137482766\">Find the inverse function of[latex]\\,f\\text{:}[\/latex]<\/p>\n\n<ol type=\"a\">\n \t<li>[latex]f\\left(x\\right)={\\left(x-4\\right)}^{2}, x\\ge 4[\/latex]<\/li>\n \t<li>[latex]f\\left(x\\right)={\\left(x-4\\right)}^{2}, x\\le 4[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165137606152\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137606152\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137606152\"]\n<p id=\"fs-id1165137506731\">The original function[latex]\\,f\\left(x\\right)={\\left(x-4\\right)}^{2}\\,[\/latex]is not one-to-one, but the function is restricted to a domain of[latex]\\,x\\ge 4\\,[\/latex]or[latex]\\,x\\le 4\\,[\/latex]on which it is one-to-one. See <a class=\"autogenerated-content\" href=\"#Figure_03_08_005\">(Figure)<\/a>.<\/p>\n\n<div id=\"Figure_03_08_005\" class=\"medium\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/cnx.org\/resources\/53d5170d583ead5b559b1d76266a29bcaa40c18b\/CNX_Precalc_Figure_03_08_005.jpg\" alt=\"Two graphs of f(x)=(x-4)^2 where the first is when x>=4 and the second is when x<=4.\" width=\"731\" height=\"365\"> <strong>Figure 5.<\/strong>[\/caption]\n\n<span id=\"fs-id1165137532798\"><\/span><\/div>\n<p id=\"fs-id1165137706306\">To find the inverse, start by replacing[latex]\\,f\\left(x\\right)\\,[\/latex]with the simple variable[latex]\\,y.[\/latex]<\/p>\n\n<div id=\"eip-id1165133259004\" class=\"unnumbered\">[latex]\\begin{array}{cccc}\\hfill y&amp; =&amp; {\\left(x-4\\right)}^{2}\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Interchange}x\\text{and }y.\\hfill \\\\ \\hfill x&amp; =&amp; {\\left(y-4\\right)}^{2}\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Take the square root}.\\hfill \\\\ \\hfill \u00b1\\sqrt{x}&amp; =&amp; y-4\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Add} 4 \\text{to both sides}.\\hfill \\\\ \\hfill 4\u00b1\\sqrt{x}&amp; =&amp; y\\hfill &amp; \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137444285\">This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of[latex]\\,x\\,[\/latex]and[latex]\\,y\\,[\/latex]for the original[latex]\\,f\\left(x\\right),\\,[\/latex]we looked at the domain: the values[latex]\\,x\\,[\/latex]\ncould assume. When we reversed the roles of[latex]\\,x\\,[\/latex]and[latex]\\,y,\\,[\/latex]\nthis gave us the values[latex]\\,y\\,[\/latex]could assume. For this function,[latex]\\,x\\ge 4,\\,[\/latex]so for the inverse, we should have[latex]\\,y\\ge 4,\\,[\/latex]which is what our inverse function gives.<\/p>\n\n<ol id=\"fs-id1165137735027\" type=\"a\">\n \t<li>The domain of the original function was restricted to[latex]\\,x\\ge 4,\\,[\/latex]so the outputs of the inverse need to be the same,[latex]\\,f\\left(x\\right)\\ge 4,\\,[\/latex]and we must use the + case:\n<div id=\"eip-id1165134294825\" class=\"unnumbered\">[latex]{f}^{-1}\\left(x\\right)=4+\\sqrt{x}[\/latex]<\/div><\/li>\n \t<li>The domain of the original function was restricted to[latex]\\,x\\le 4,\\,[\/latex]so the outputs of the inverse need to be the same,[latex]\\,f\\left(x\\right)\\le 4,\\,[\/latex]and we must use the \u2013 case:\n<div id=\"eip-id1165137482501\" class=\"unnumbered\">[latex]{f}^{-1}\\left(x\\right)=4-\\sqrt{x}[\/latex][\/hidden-answer]<\/div><\/li>\n<\/ol>\n<\/div>\n<div>\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165137534054\">On the graphs in <a class=\"autogenerated-content\" href=\"#Figure_03_08_006\">(Figure)<\/a>, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line[latex]\\,y=x.\\,[\/latex]The coordinate pair[latex]\\,\\left(4, 0\\right)\\,[\/latex]is on the graph of[latex]\\,f\\,[\/latex]and the coordinate pair[latex]\\,\\left(0, 4\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{-1}.\\,[\/latex]For any coordinate pair, if[latex]\\,\\left(a,\\text{ }b\\right)\\,[\/latex]is on the graph of[latex]\\,f,\\,[\/latex]then[latex]\\,\\left(b,\\text{ }a\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{-1}.\\,[\/latex]Finally, observe that the graph of[latex]\\,f\\,[\/latex]intersects the graph of[latex]\\,{f}^{-1}[\/latex]on the line[latex]\\,y=x.\\,[\/latex]Points of intersection for the graphs of[latex]\\,f\\,[\/latex]and[latex]\\,{f}^{-1}\\,[\/latex]will always lie on the line[latex]\\,y=x.[\/latex]<span id=\"fs-id1165137471020\"><\/span><\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/cnx.org\/resources\/6ee202f7074c245ac7cf28a92a363acbf8e6d22a\/CNX_Precalc_Figure_03_08_006.jpg\" alt=\"Two graphs of a parabolic function with half of its inverse.\" width=\"975\" height=\"442\"> <strong>Figure 6.<\/strong>[\/caption]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_08_04\" class=\"textbox examples\">\n<div id=\"fs-id1165137786597\">\n<div id=\"fs-id1165135481235\">\n<h3>Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified<\/h3>\n<p id=\"fs-id1165137410909\">Restrict the domain and then find the inverse of<\/p>\n\n<div id=\"eip-id1165133355860\" class=\"unnumbered\">[latex]f\\left(x\\right)={\\left(x-2\\right)}^{2}-3.[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165135394334\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135394334\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135394334\"]\n<p id=\"fs-id1165137414415\">We can see this is a parabola with vertex at[latex]\\,\\left(2,\u20133\\right)\\,[\/latex]that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to[latex]\\,x\\ge 2.[\/latex]<\/p>\n<p id=\"fs-id1165137842529\">To find the inverse, we will use the vertex form of the quadratic. We start by replacing[latex]\\,f\\left(x\\right)\\,[\/latex]with a simple variable,[latex]\\,y,\\,[\/latex]then solve for[latex]\\,x.[\/latex]<\/p>\n\n<div id=\"eip-id1165133305354\" class=\"unnumbered\">[latex]\\begin{array}{cccc}\\hfill y&amp; =&amp; {\\left(x-2\\right)}^{2}-3\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Interchange }x\\text{ and }y.\\hfill \\\\ \\hfill x&amp; =&amp; {\\left(y-2\\right)}^{2}-3\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Add 3 to both sides}.\\hfill \\\\ \\hfill x+3&amp; =&amp; {\\left(y-2\\right)}^{2}\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Take the square root}.\\hfill \\\\ \\hfill \u00b1\\sqrt{x+3}&amp; =&amp; y-2\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Add 2 to both sides}.\\hfill \\\\ \\hfill 2\u00b1\\sqrt{x+3}&amp; =&amp; y\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Rename the function}.\\hfill \\\\ \\hfill {f}^{-1}\\left(x\\right)&amp; =&amp; 2\u00b1\\sqrt{x+3}\\hfill &amp; \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137419504\">Now we need to determine which case to use. Because we restricted our original function to a domain of[latex]\\,x\\ge 2,\\,[\/latex]the outputs of the inverse should be the same, telling us to utilize the + case<\/p>\n\n<div id=\"eip-id1165132193808\" class=\"unnumbered\">[latex]{f}^{-1}\\left(x\\right)=2+\\sqrt{x+3}[\/latex]<\/div>\nIf the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.<span id=\"fs-id1165135194279\"><\/span>[\/hidden-answer]\n\n<\/div>\n<div id=\"fs-id1165134362839\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165135259538\">Notice that we arbitrarily decided to restrict the domain on[latex]\\,x\\ge 2.\\,[\/latex]We could just have easily opted to restrict the domain on[latex]\\,x\\le 2,\\,[\/latex]in which case[latex]\\,{f}^{-1}\\left(x\\right)=2-\\sqrt{x+3}.\\,[\/latex]Observe the original function graphed on the same set of axes as its inverse function in <a class=\"autogenerated-content\" href=\"#Figure_03_08_007\">(Figure)<\/a>. Notice that both graphs show symmetry about the line[latex]\\,y=x.\\,[\/latex]The coordinate pair[latex]\\,\\left(2,\\text{ }-3\\right)\\,[\/latex]is on the graph of[latex]\\,f\\,[\/latex]and the coordinate pair[latex]\\,\\left(-3,\\text{ }2\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{-1}.\\,[\/latex]Observe from the graph of both functions on the same set of axes that<\/p>\n\n<div id=\"eip-id1165134122215\" class=\"unnumbered\">[latex]\\text{domain of }f=\\text{range of} {f}^{\u20131}=\\left[2,\\infty \\right)[\/latex]<\/div>\n<p id=\"fs-id1165137642128\">and<\/p>\n\n<div id=\"eip-id1165134279478\" class=\"unnumbered\">[latex]\\text{domain of }{f}^{\u20131}=\\text{range of} f=\\left[\u20133,\\infty \\right).[\/latex]<\/div>\nFinally, observe that the graph of[latex]\\,f\\,[\/latex]intersects the graph of[latex]\\,{f}^{-1}\\,[\/latex]along the line[latex]\\,y=x.[\/latex]\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/cnx.org\/resources\/93673f4f9c9538c5e8b351f6535b7a9f4886a8f6\/CNX_Precalc_Figure_03_08_007.jpg\" alt=\"Graph of a parabolic function with half of its inverse.\" width=\"487\" height=\"487\"> <strong>Figure 7.<\/strong>[\/caption]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137419950\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_03\">\n<div id=\"fs-id1165137600895\">\n<p id=\"fs-id1165135596379\">Find the inverse of the function[latex]\\,f\\left(x\\right)={x}^{2}+1,\\,[\/latex]on the domain[latex]\\,x\\ge 0.[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137737548\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137737548\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137737548\"]\n<p id=\"fs-id1165137737550\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x-1}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137894462\" class=\"bc-section section\">\n<h4>Solving Applications of Radical Functions<\/h4>\n<p id=\"fs-id1165137696560\">Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the <span class=\"no-emphasis\">inverse of a radical function<\/span>, we will need to restrict the domain of the answer because the range of the original function is limited.<\/p>\n\n<div id=\"fs-id1165137415876\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137455923\"><strong>Given a radical function, find the inverse.<\/strong><\/p>\n\n<ol id=\"fs-id1165137542989\" type=\"1\">\n \t<li>Determine the range of the original function.<\/li>\n \t<li>Replace[latex]\\,f\\left(x\\right)\\,[\/latex] with[latex]\\,y,\\,[\/latex]then solve for[latex]\\,x.[\/latex]<\/li>\n \t<li>If necessary, restrict the domain of the inverse function to the range of the original function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_08_05\" class=\"textbox examples\">\n<div id=\"fs-id1165135173813\">\n<div id=\"fs-id1165137399685\">\n<h3>Finding the Inverse of a Radical Function<\/h3>\n<p id=\"fs-id1165135570491\">Restrict the domain of the function[latex]\\,f\\left(x\\right)=\\sqrt{x-4}\\,[\/latex]and then find the inverse.<\/p>\n\n<\/div>\n<div id=\"fs-id1165137766917\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137766917\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137766917\"]\n<p id=\"fs-id1165135545767\">Note that the original function has range[latex]\\,f\\left(x\\right)\\ge 0.\\,[\/latex]Replace[latex]\\,f\\left(x\\right)\\,[\/latex]with[latex]\\,y,\\,[\/latex]then solve for[latex]\\,x.[\/latex]<\/p>\n\n<div id=\"eip-id1165135508330\" class=\"unnumbered\">[latex]\\begin{array}{cccc}\\hfill y&amp; =&amp; \\sqrt{x-4}\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Replace }f\\left(x\\right)\\text{ with }y.\\hfill \\\\ \\hfill x&amp; =&amp; \\sqrt{y-4}\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Interchange }x\\text{ and }y.\\hfill \\\\ \\hfill x&amp; =&amp; \\sqrt{y-4}\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Square each side}.\\hfill \\\\ \\hfill {x}^{2}&amp; =&amp; y-4\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Add 4}.\\hfill \\\\ \\hfill {x}^{2}+4&amp; =&amp; y\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Rename the function }{f}^{-1}\\left(x\\right).\\hfill \\\\ \\hfill {f}^{-1}\\left(x\\right)&amp; =&amp; {x}^{2}+4\\hfill &amp; \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135209570\">Recall that the domain of this function must be limited to the range of the original function.<\/p>\n\n<div id=\"eip-id1165135452086\" class=\"unnumbered\">[latex]{f}^{-1}\\left(x\\right)={x}^{2}+4,x\\ge 0[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<div id=\"fs-id1165137750058\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165135173239\">Notice in <a class=\"autogenerated-content\" href=\"#Figure_03_08_008\">(Figure)<\/a> that the inverse is a reflection of the original function over the line[latex]\\,y=x.\\,[\/latex]Because the original function has only positive outputs, the inverse function has only positive inputs.<\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/cnx.org\/resources\/278b5e9547f105e6c6d64031e7d5c208c55c0996\/CNX_Precalc_Figure_03_08_008.jpg\" alt=\"Graph of f(x)=sqrt(x-4) and its inverse, f^(-1)(x)=x^2+4.\" width=\"487\" height=\"444\"> <strong>Figure 8.<\/strong>[\/caption]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137662048\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_04\">\n<div id=\"fs-id1165135524557\">\n<p id=\"fs-id1165137784775\">Restrict the domain and then find the inverse of the function[latex]\\,f\\left(x\\right)=\\sqrt{2x+3}.[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137784778\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137784778\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137784778\"]\n<p id=\"fs-id1165137419726\">[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-3}{2},x\\ge 0[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137761571\" class=\"bc-section section\">\n<h4>Solving Applications of Radical Functions<\/h4>\n<p id=\"fs-id1165135435831\">Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section.<\/p>\n\n<div id=\"Example_03_08_06\" class=\"textbox examples\">\n<div>\n<div id=\"fs-id1165137531120\">\n<h3>Solving an Application with a Cubic Function<\/h3>\n<p id=\"fs-id1165137771982\">A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by<\/p>\n\n<div id=\"eip-id1165132187568\" class=\"unnumbered\">[latex]V=\\frac{2}{3}\\pi {r}^{3}[\/latex]<\/div>\n<p id=\"fs-id1165135181305\">Find the inverse of the function[latex]\\,V=\\frac{2}{3}\\pi {r}^{3}\\,[\/latex]that determines the volume[latex]\\,V\\,[\/latex]of a cone and is a function of the radius[latex]\\,r.\\,[\/latex]Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use[latex]\\,\\pi =3.14.[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137405142\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137405142\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137405142\"]\n<p id=\"fs-id1165137405144\">Start with the given function for[latex]\\,V.\\,[\/latex]Notice that the meaningful domain for the function is[latex]\\,r&gt;0\\,[\/latex]since negative radii would not make sense in this context nor would a radius of 0. Also note the range of the function (hence, the domain of the inverse function) is[latex]\\,V&gt;0.\\,[\/latex]Solve for[latex]\\,r\\,[\/latex]in terms of[latex]\\,V,\\,[\/latex]using the method outlined previously. Note that in real-world applications, we do not swap the variables when finding inverses. Instead, we change which variable is considered to be the independent variable.<\/p>\n\n<div id=\"eip-id1165133447890\" class=\"unnumbered\">[latex]\\begin{array}{cccc}\\hfill V&amp; =&amp; \\frac{2}{3}\\pi {r}^{3}\\hfill &amp; \\\\ \\hfill {r}^{3}&amp; =&amp; \\frac{3V}{2\\pi }\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Solve for }{r}^{3}.\\hfill \\\\ \\hfill r&amp; =&amp; \\sqrt[3]{\\frac{3V}{2\\pi }}\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Solve for }r.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137730324\">This is the result stated in the section opener. Now evaluate this for[latex]\\,V=100\\,[\/latex]and[latex]\\,\\pi =3.14.[\/latex]<\/p>\n\n<div id=\"eip-id1165135203234\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill r&amp; =&amp; \\sqrt[3]{\\frac{3V}{2\\pi }}\\hfill \\\\ &amp; =&amp; \\sqrt[3]{\\frac{3\\cdot 100}{2\\cdot 3.14}}\\hfill \\\\ &amp; \\approx &amp; \\sqrt[3]{47.7707}\\hfill \\\\ &amp; \\approx \\hfill &amp; 3.63\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137706279\">Therefore, the radius is about 3.63 ft.[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137580023\" class=\"bc-section section\">\n<h4>Determining the Domain of a Radical Function Composed with Other Functions<\/h4>\n<p id=\"fs-id1165134042947\">When radical functions are composed with other functions, determining domain can become more complicated.<\/p>\n\n<div id=\"Example_03_08_07\" class=\"textbox examples\">\n<div id=\"fs-id1165135378774\">\n<div id=\"fs-id1165135378776\">\n<h3>Finding the Domain of a Radical Function Composed with a Rational Function<\/h3>\n<p id=\"fs-id1165137658778\">Find the domain of the function[latex]\\,f\\left(x\\right)=\\sqrt{\\frac{\\left(x+2\\right)\\left(x-3\\right)}{\\left(x-1\\right)}}.[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137550072\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137550072\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137550072\"]\n<p id=\"fs-id1165137665549\">Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where[latex]\\,\\frac{\\left(x+2\\right)\\left(x-3\\right)}{\\left(x-1\\right)}\\ge 0.\\,[\/latex]The output of a rational function can change signs (change from positive to negative or vice versa) at <em>x<\/em>-intercepts and at vertical asymptotes. For this equation, the graph could change signs at[latex]\\,x=\u20132, 1, \\text{and} 3.[\/latex]<\/p>\n<p id=\"fs-id1165135686721\">To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_08_009\">(Figure)<\/a>.<\/p>\n\n<div id=\"Figure_03_08_009\" class=\"medium\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140154\/CNX_Precalc_Figure_03_08_009.jpg\" alt=\"Graph of a radical function that shows where the outputs are nonnegative.\" width=\"731\" height=\"439\"> <strong>Figure 9.<\/strong>[\/caption]\n\n<\/div>\n<p id=\"fs-id1165137694167\">This function has two <em>x<\/em>-intercepts, both of which exhibit linear behavior near the <em>x<\/em>-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a <em>y<\/em>-intercept at[latex]\\,\\left(0,\\sqrt{6}\\right).[\/latex]<\/p>\n<p id=\"fs-id1165135333589\">From the <em>y<\/em>-intercept and <em>x<\/em>-intercept at[latex]\\,x=-2,\\,[\/latex]we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.<\/p>\n<p id=\"fs-id1165137664081\">From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function[latex]\\,f\\left(x\\right)\\,[\/latex]will be defined.[latex]\\,f\\left(x\\right)\\,[\/latex]has domain[latex]\\,-2\\le x&lt;1\\,\\text{or}\\,x\\ge 3,\\,[\/latex]or in interval notation,[latex]\\,\\left[-2,1\\right)\\cup \\left[3,\\infty \\right).[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\">\n<h4>Finding Inverses of Rational Functions<\/h4>\n<p id=\"fs-id1165135475898\">As with finding inverses of quadratic functions, it is sometimes desirable to find the <span class=\"no-emphasis\">inverse of a rational function<\/span>, particularly of rational functions that are the ratio of linear functions, such as in concentration applications.<\/p>\n\n<div id=\"Example_03_08_08\" class=\"textbox examples\">\n<div id=\"fs-id1165137642525\">\n<div id=\"fs-id1165137642528\">\n<h3>Finding the Inverse of a Rational Function<\/h3>\n<p id=\"fs-id1165135332364\">The function[latex]\\,C=\\frac{20+0.4n}{100+n}\\,[\/latex]represents the concentration[latex]\\,C\\,[\/latex]of an acid solution after[latex]\\,n\\,[\/latex]mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for[latex]\\,n\\,[\/latex]in terms of[latex]\\,C.\\,[\/latex]Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.<\/p>\n\n<\/div>\n<div id=\"fs-id1165134223201\" class=\"solution textbox shaded\">\n\n[reveal-answer q=\"450987\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"450987\"]\n\nWe first want the inverse of the function in order to determine how many mL we need for a given concentration. We will solve for[latex]\\,n\\,[\/latex]in terms of[latex]\\,C.[\/latex]\n<div id=\"eip-id1165132957110\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill C&amp; =&amp; \\frac{20+0.4n}{100+n}\\hfill \\\\ \\hfill C\\left(100+n\\right)&amp; =&amp; 20+0.4n\\hfill \\\\ \\hfill 100C+Cn&amp; =&amp; 20+0.4n\\hfill \\\\ \\hfill 100C-20&amp; =&amp; 0.4n-Cn\\hfill \\\\ \\hfill 100C-20&amp; =&amp; \\left(0.4n-C\\right)n\\hfill \\\\ \\hfill n&amp; =&amp; \\frac{100C-20}{0.4-C}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137637474\">Now evaluate this function at 35%, which is[latex]\\,C=0.35.[\/latex]<\/p>\n\n<div id=\"eip-id1165131986931\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill n&amp; =&amp; \\frac{100\\left(0.35\\right)-20}{0.4-0.35}\\hfill \\\\ &amp; =&amp; \\frac{15}{0.05}\\hfill \\\\ &amp; =&amp; 300\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137706154\">We can conclude that 300 mL of the 40% solution should be added.<\/p>\n<p id=\"fs-id1165137706154\">[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137697102\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_05\">\n<div id=\"fs-id1165134042922\">\n<p id=\"fs-id1165134042923\">Find the inverse of the function[latex]\\,f\\left(x\\right)=\\frac{x+3}{x-2}.[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137731924\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137731924\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137731924\"]\n<p id=\"fs-id1165137731925\">[latex]{f}^{-1}\\left(x\\right)=\\frac{2x+3}{x-1}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137551968\" class=\"precalculus media\">\n<p id=\"fs-id1165137761312\">Access these online resources for additional instruction and practice with inverses and radical functions.<\/p>\n\n<ul id=\"fs-id1165134306731\">\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/graphsquareroot\">Graphing the Basic Square Root Function<\/a><\/li>\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/inversesquare\">Find the Inverse of a Square Root Function<\/a><\/li>\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/inverserational\">Find the Inverse of a Rational Function<\/a><\/li>\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/rationalinverse\">Find the Inverse of a Rational Function and an Inverse Function Value<\/a><\/li>\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/inversefunction\">Inverse Functions<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135192373\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165135528386\">\n \t<li>The inverse of a quadratic function is a square root function.<\/li>\n \t<li>If[latex]\\,{f}^{-1}\\,[\/latex] is the inverse of a function[latex]\\,f,\\,[\/latex] then[latex]\\,f\\,[\/latex] is the inverse of the function[latex]\\,{f}^{-1}.\\,[\/latex] See <a class=\"autogenerated-content\" href=\"#Example_03_08_01\">(Figure)<\/a>.<\/li>\n \t<li>While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible. See <a class=\"autogenerated-content\" href=\"#Example_03_08_02\">(Figure)<\/a>.<\/li>\n \t<li>To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one. See <a class=\"autogenerated-content\" href=\"#Example_03_08_03\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_03_08_04\">(Figure)<\/a>.<\/li>\n \t<li>When finding the inverse of a radical function, we need a restriction on the domain of the answer. See <a class=\"autogenerated-content\" href=\"#Example_03_08_05\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_03_08_07\">(Figure)<\/a>.<\/li>\n \t<li>Inverse and radical and functions can be used to solve application problems. See <a class=\"autogenerated-content\" href=\"#Example_03_08_06\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_03_08_08\">(Figure)<\/a>.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Section Exercises<\/h3>\n<div id=\"fs-id1165137414778\" class=\"bc-section section\">\n<h4>Verbal<\/h4>\n<div id=\"fs-id1165137678194\">\n<div id=\"fs-id1165137452992\">\n<p id=\"fs-id1165137452993\">Explain why we cannot find inverse functions for all polynomial functions.<\/p>\n\n<\/div>\n<div id=\"fs-id1165137452997\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137452997\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137452997\"]\n<p id=\"fs-id1165137723132\">It can be too difficult or impossible to solve for[latex]\\,x\\,[\/latex]in terms of[latex]\\,y.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137648362\">\n<div id=\"fs-id1165137854841\">\n\nWhy must we restrict the domain of a quadratic function when finding its inverse?\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137854845\">\n<div id=\"fs-id1165135356600\">\n<p id=\"fs-id1165135356601\">When finding the inverse of a radical function, what restriction will we need to make?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135356605\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135356605\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135356605\"]\n<p id=\"fs-id1165137590162\">We will need a restriction on the domain of the answer.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137590165\">\n<div id=\"fs-id1165137590166\">\n<p id=\"fs-id1165137661780\">The inverse of a quadratic function will always take what form?<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137661784\" class=\"bc-section section\">\n<h4>Algebraic<\/h4>\n<p id=\"fs-id1165137936804\">For the following exercises, find the inverse of the function on the given domain.<\/p>\n\n<div id=\"fs-id1165135193880\">\n<div id=\"fs-id1165135193881\">\n<p id=\"fs-id1165135193882\">[latex]f\\left(x\\right)={\\left(x-4\\right)}^{2}, \\left[4,\\infty \\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137472528\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137472528\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137472528\"]\n<p id=\"fs-id1165137472529\">[latex]\\,\\,{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137706126\">\n<div id=\"fs-id1165137706127\">\n<p id=\"fs-id1165137706128\">[latex]f\\left(x\\right)={\\left(x+2\\right)}^{2}, \\left[-2,\\infty \\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137405850\">\n<div id=\"fs-id1165137405851\">\n<p id=\"fs-id1165137405852\">[latex]f\\left(x\\right)={\\left(x+1\\right)}^{2}-3, \\left[-1,\\infty \\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135572106\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135572106\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135572106\"]\n<p id=\"fs-id1165135572107\">[latex]\\,\\,{f}^{-1}\\left(x\\right)=\\sqrt{x+3}-1[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135209972\">\n<div id=\"fs-id1165135209973\">\n<p id=\"fs-id1165137425962\">[latex]f\\left(x\\right)=3{x}^{2}+5,\\,\\,\\left(\\infty ,0\\right][\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135436476\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135436476\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135436476\"]\n<p id=\"fs-id1165135436478\">[latex]{f}^{-1}\\left(x\\right)=-\\sqrt{\\frac{x-5}{3}}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134042512\">\n<div id=\"fs-id1165137696696\">\n<p id=\"fs-id1165137696698\">[latex]f\\left(x\\right)=12-{x}^{2}, \\left[0,\\infty \\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137579624\">\n<div>[latex]f\\left(x\\right)=9-{x}^{2}, \\left[0,\\infty \\right)[\/latex]<\/div>\n<div id=\"fs-id1165135194244\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135194244\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135194244\"]\n<p id=\"fs-id1165137862827\">[latex]f\\left(x\\right)=\\sqrt{9-x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134555020\">\n<div id=\"fs-id1165134555021\">\n<p id=\"fs-id1165134555022\">[latex]f\\left(x\\right)=2{x}^{2}+4, \\left[0,\\infty \\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<p id=\"fs-id1165137603367\">For the following exercises, find the inverse of the functions.<\/p>\n\n<div id=\"fs-id1165137725343\">\n<div id=\"fs-id1165137725344\">\n<p id=\"fs-id1165137725345\">[latex]f\\left(x\\right)={x}^{3}+5[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137885943\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137885943\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137885943\"]\n<p id=\"fs-id1165137885944\">[latex]\\,\\,{f}^{-1}\\left(x\\right)=\\sqrt[3]{x-5}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137728086\">\n<div id=\"fs-id1165137728087\">\n<p id=\"fs-id1165137728088\">[latex]f\\left(x\\right)=3{x}^{3}+1[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135530666\">\n<div id=\"fs-id1165133035997\">\n<p id=\"fs-id1165133035998\">[latex]f\\left(x\\right)=4-{x}^{3}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137767092\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137767092\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137767092\"]\n<p id=\"fs-id1165137767093\">[latex]\\,{f}^{-1}\\left(x\\right)=\\sqrt[3]{4-x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135500747\">\n<div id=\"fs-id1165135500748\">\n<p id=\"fs-id1165135500749\">[latex]f\\left(x\\right)=4-2{x}^{3}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<p id=\"fs-id1165135530341\">For the following exercises, find the inverse of the functions.<\/p>\n\n<div id=\"fs-id1165135530344\">\n<div id=\"fs-id1165135530345\">\n<p id=\"fs-id1165137656262\">[latex]f\\left(x\\right)=\\sqrt{2x+1}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135250674\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135250674\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135250674\"]\n<p id=\"fs-id1165135250675\">[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-1}{2},\\,\\,\\left[0,\\infty \\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165137463046\">\n<p id=\"fs-id1165137463047\">[latex]f\\left(x\\right)=\\sqrt{3-4x}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137502900\">\n<div id=\"fs-id1165137502901\">\n<p id=\"fs-id1165137502902\">[latex]f\\left(x\\right)=9+\\sqrt{4x-4}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137871173\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137871173\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137871173\"]\n<p id=\"fs-id1165137871174\">[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{{\\left(x-9\\right)}^{2}+4}{4},\\,\\,\\left[9,\\infty \\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135319426\">\n<div id=\"fs-id1165137662665\">\n<p id=\"fs-id1165137662666\">[latex]f\\left(x\\right)=\\sqrt{6x-8}+5[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137471620\">\n<div id=\"fs-id1165137471621\">\n<p id=\"fs-id1165137471622\">[latex]f\\left(x\\right)=9+2\\sqrt[3]{x}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137735210\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137735210\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137735210\"]\n<p id=\"fs-id1165137735211\">[latex]{f}^{-1}\\left(x\\right)={\\left(\\frac{x-9}{2}\\right)}^{3}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135699114\">\n<div id=\"fs-id1165135699115\">\n<p id=\"fs-id1165137461567\">[latex]f\\left(x\\right)=3-\\sqrt[3]{x}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137544702\">\n<div id=\"fs-id1165137544703\">\n<p id=\"fs-id1165137660257\">[latex]f\\left(x\\right)=\\frac{2}{x+8}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137737152\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137737152\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137737152\"]\n<p id=\"fs-id1165137737153\">[latex]{f}^{-1}\\left(x\\right)={\\frac{2-8x}{x}}^{}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137436932\">\n<div id=\"fs-id1165137436933\">\n<p id=\"fs-id1165137436934\">[latex]f\\left(x\\right)=\\frac{3}{x-4}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137442675\">\n<div id=\"fs-id1165137442676\">\n<p id=\"fs-id1165137442677\">[latex]f\\left(x\\right)=\\frac{x+3}{x+7}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137466246\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137466246\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137466246\"]\n<p id=\"fs-id1165137466248\">[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{7x-3}{1-x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135701622\">\n<div id=\"fs-id1165135701623\">\n<p id=\"fs-id1165135701624\">[latex]f\\left(x\\right)=\\frac{x-2}{x+7}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135333251\">\n<div id=\"fs-id1165137443976\">\n<p id=\"fs-id1165137443977\">[latex]f\\left(x\\right)=\\frac{3x+4}{5-4x}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137668289\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137668289\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137668289\"]\n<p id=\"fs-id1165137668290\">[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{5x-4}{4x+3}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137642936\">\n<div id=\"fs-id1165137642938\">\n<p id=\"fs-id1165137642939\">[latex]f\\left(x\\right)=\\frac{5x+1}{2-5x}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137760837\">\n<div id=\"fs-id1165137408974\">\n<p id=\"fs-id1165137408975\">[latex]f\\left(x\\right)={x}^{2}+2x, \\left[-1,\\infty \\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135176553\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135176553\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135176553\"]\n<p id=\"fs-id1165135176554\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+1}-1[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137662088\">\n<div id=\"fs-id1165137662089\">\n<p id=\"fs-id1165137436907\">[latex]f\\left(x\\right)={x}^{2}+4x+1, \\left[-2,\\infty \\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165133184252\">\n<div id=\"fs-id1165133184253\">\n<p id=\"fs-id1165137675697\">[latex]f\\left(x\\right)={x}^{2}-6x+3, \\left[3,\\infty \\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137760888\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137760888\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137760888\"]\n<p id=\"fs-id1165137760889\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+6}+3[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135193360\" class=\"bc-section section\">\n<h4>Graphical<\/h4>\n<p id=\"fs-id1165135193366\">For the following exercises, find the inverse of the function and graph both the function and its inverse.<\/p>\n\n<div id=\"fs-id1165137715400\">\n<div id=\"fs-id1165137715401\">\n<p id=\"fs-id1165137715402\">[latex]f\\left(x\\right)={x}^{2}+2,\\,x\\ge 0[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165137640500\">\n<p id=\"fs-id1165137640501\">[latex]f\\left(x\\right)=4-{x}^{2},\\,x\\ge 0[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137502874\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137502874\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137502874\"]\n<p id=\"fs-id1165137437197\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{4-x}[\/latex]<\/p>\n<span id=\"fs-id1165137462220\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140157\/CNX_Precalc_Figure_03_08_202.jpg\" alt=\"Graph of f(x)=4- x^2 and its inverse, f^(-1)(x)= sqrt(4-x).\"><\/span>[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135547140\">\n<div id=\"fs-id1165135547141\">\n<p id=\"fs-id1165135547142\">[latex]f\\left(x\\right)={\\left(x+3\\right)}^{2},\\,x\\ge -3[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137452181\">\n<div>\n<p id=\"fs-id1165137452184\">[latex]f\\left(x\\right)={\\left(x-4\\right)}^{2},\\,x\\ge 4[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137459688\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137459688\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137459688\"]\n<p id=\"fs-id1165137459689\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]<\/p>\n<span id=\"fs-id1165137811172\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140207\/CNX_Precalc_Figure_03_08_204.jpg\" alt=\"Graph of f(x)= (x-4)^2 and its inverse, f^(-1)(x)= sqrt(x)+4.\"><\/span>[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135187190\">\n<div id=\"fs-id1165135187191\">\n<p id=\"fs-id1165135187192\">[latex]f\\left(x\\right)={x}^{3}+3[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135651499\">\n<div id=\"fs-id1165135651500\">\n<p id=\"fs-id1165135651501\">[latex]f\\left(x\\right)=1-{x}^{3}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137462309\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137462309\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137462309\"]\n<p id=\"fs-id1165137462310\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{1-x}[\/latex]<\/p>\n<span id=\"fs-id1165135694369\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140209\/CNX_Precalc_Figure_03_08_206.jpg\" alt=\"Graph of f(x)= 1-x^3 and its inverse, f^(-1)(x)= (1-x)^(1\/3).\"><\/span>[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135368483\">\n<div id=\"fs-id1165135368484\">[latex]f\\left(x\\right)={x}^{2}+4x,\\,x\\ge -2[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137454354\">\n<div id=\"fs-id1165137454356\">\n<p id=\"fs-id1165137454357\">[latex]f\\left(x\\right)={x}^{2}-6x+1,\\,x\\ge 3[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137742528\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137742528\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137742528\"]\n[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+8}+3[\/latex]<span id=\"fs-id1165135678741\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140232\/CNX_Precalc_Figure_03_08_208.jpg\" alt=\"Graph of f(x)= x^2-6x+1 and its inverse, f^(-1)(x)= sqrt(x+8)+3.\"><\/span>[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135206104\">\n<div id=\"fs-id1165135206105\">\n<p id=\"fs-id1165135206106\">[latex]f\\left(x\\right)=\\frac{2}{x}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137651509\">\n<div id=\"fs-id1165137651510\">\n<p id=\"fs-id1165137651511\">[latex]f\\left(x\\right)=\\frac{1}{{x}^{2}},\\,x\\ge 0[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165134557344\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134557344\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134557344\"]\n<p id=\"fs-id1165137634193\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{\\frac{1}{x}}[\/latex]<\/p>\n<img src=\"https:\/\/cnx.org\/resources\/c5037767394b50c474c6620703ea24eb45d7a1c8\/CNX_Precalc_Figure_03_08_210.jpg\" alt=\"Graph of f(x)= 1\/x^2 and its inverse, f^(-1)(x)= sqrt(1\/x).\">[\/hidden-answer]\n\n<\/div>\n<\/div>\n<p id=\"fs-id1165137451122\">For the following exercises, use a graph to help determine the domain of the functions.<\/p>\n\n<div id=\"fs-id1165137480291\">\n<div id=\"fs-id1165137480292\">\n<p id=\"fs-id1165137480293\">[latex]f\\left(x\\right)=\\sqrt{\\frac{\\left(x+1\\right)\\left(x-1\\right)}{x}}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137422286\">\n<div id=\"fs-id1165137422287\">\n<p id=\"fs-id1165137422288\">[latex]f\\left(x\\right)=\\sqrt{\\frac{\\left(x+2\\right)\\left(x-3\\right)}{x-1}}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137407593\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137407593\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137407593\"]\n<p id=\"fs-id1165137407594\">[latex]\\left[-2,1\\right)\\cup \\left[3,\\infty \\right)[\/latex]<\/p>\n<img src=\"https:\/\/cnx.org\/resources\/31d07c963aa8e9a0703c8cd298ead16cbe2e1845\/CNX_Precalc_Figure_03_08_212.jpg\" alt=\"Graph of f(x)= sqrt((x+2)(x-3)\/(x-1)).\">[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135173418\">\n<div id=\"fs-id1165135173419\">\n<p id=\"fs-id1165137772203\">[latex]f\\left(x\\right)=\\sqrt{\\frac{x\\left(x+3\\right)}{x-4}}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137550583\">\n<div id=\"fs-id1165137550584\">\n<p id=\"fs-id1165137550585\">[latex]f\\left(x\\right)=\\sqrt{\\frac{{x}^{2}-x-20}{x-2}}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137758690\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137758690\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137758690\"]\n<p id=\"fs-id1165137539722\">[latex]\\left[-4,2\\right)\\cup \\left[5,\\infty \\right)[\/latex]<\/p>\n<img src=\"https:\/\/cnx.org\/resources\/259c69be03f0d36862a3fceb79891f7db338332b\/CNX_Precalc_Figure_03_08_214.jpg\" alt=\"Graph of f(x)= sqrt((x^2-x-20)\/(x-2)).\">[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137898053\">\n<div id=\"fs-id1165137898054\">\n<p id=\"fs-id1165137898056\">[latex]f\\left(x\\right)=\\sqrt{\\frac{9-{x}^{2}}{x+4}}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137676077\" class=\"bc-section section\">\n<h4>Technology<\/h4>\n<p id=\"fs-id1165137551626\">For the following exercises, use a calculator to graph the function. Then, using the graph, give three points on the graph of the inverse with <em>y<\/em>-coordinates given.<\/p>\n\n<div id=\"fs-id1165137855040\">\n<div id=\"fs-id1165137575507\">\n<p id=\"fs-id1165137575508\">[latex]f\\left(x\\right)={x}^{3}-x-2,y=1,2,3[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137724899\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137724899\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137724899\"]\n<p id=\"fs-id1165137724900\">[latex]\\left(\u20132, 0\\right); \\left(4, 2\\right); \\left(22, 3\\right)[\/latex]<\/p>\n<img src=\"https:\/\/cnx.org\/resources\/4d856c01c1cc568f1f7ff472813acee8aa8776bf\/CNX_Precalc_Figure_03_08_216.jpg\" alt=\"Graph of f(x)= x^3-x-2.\">[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137571412\">\n<div id=\"fs-id1165137571413\">\n<p id=\"fs-id1165137571414\">[latex]f\\left(x\\right)={x}^{3}+x-2,y=0,1,2[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135336074\">\n<div id=\"fs-id1165135336075\">\n<p id=\"fs-id1165135336076\">[latex]f\\left(x\\right)={x}^{3}+3x-4,y=0,1,2[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135445799\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135445799\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135445799\"]\n<p id=\"fs-id1165135445800\">[latex]\\left(\u20134, 0\\right); \\left(0, 1\\right); \\left(10, 2\\right)[\/latex]<\/p>\n<img src=\"https:\/\/cnx.org\/resources\/5815d06121425c5769bf84de2c8f710222f6adf0\/CNX_Precalc_Figure_03_08_218.jpg\" alt=\"Graph of f(x)= x^3+3x-4.\">[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137768493\">\n<div id=\"fs-id1165137414579\">\n<p id=\"fs-id1165137414580\">[latex]f\\left(x\\right)={x}^{3}+8x-4,y=-1,0,1[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137812553\">\n<div id=\"fs-id1165137812554\">\n<p id=\"fs-id1165137812555\">[latex]f\\left(x\\right)={x}^{4}+5x+1,y=-1,0,1[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137413936\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137413936\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137413936\"]\n<p id=\"fs-id1165137413937\">[latex]\\left(\u20133, -1\\right); \\left(1, 0\\right); \\left(7, 1\\right)[\/latex]<\/p>\n<img src=\"https:\/\/cnx.org\/resources\/0d7ef9b0cb654da615c33f4966eed0ef018910c8\/CNX_Precalc_Figure_03_08_220.jpg\" alt=\"Graph of f(x)= x^4+5x+1.\">[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137833145\" class=\"bc-section section\">\n<h4>Extensions<\/h4>\n<p id=\"fs-id1165137427625\">For the following exercises, find the inverse of the functions with[latex]\\,a,b,c\\,[\/latex]positive real numbers.<\/p>\n\n<div id=\"fs-id1165137601481\">\n<div id=\"fs-id1165137601482\">\n<p id=\"fs-id1165137601483\">[latex]f\\left(x\\right)=a{x}^{3}+b[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165132960779\">\n<div id=\"fs-id1165132960780\">\n<p id=\"fs-id1165137660236\">[latex]f\\left(x\\right)={x}^{2}+bx[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137564896\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137564896\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137564896\"]\n<p id=\"fs-id1165137564897\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+\\frac{{b}^{2}}{4}}-\\frac{b}{2}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135187946\">\n<div id=\"fs-id1165135187947\">\n<p id=\"fs-id1165137548784\">[latex]f\\left(x\\right)=\\sqrt{a{x}^{2}+b}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137590488\">\n<div id=\"fs-id1165137590489\">\n<p id=\"fs-id1165137590490\">[latex]f\\left(x\\right)=\\sqrt[3]{ax+b}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137934451\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137934451\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137934451\"]\n<p id=\"fs-id1165137934452\">[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{3}-b}{a}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137597719\">\n<div id=\"fs-id1165137597720\">\n<p id=\"fs-id1165137597721\">[latex]f\\left(x\\right)=\\frac{ax+b}{x+c}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135347473\" class=\"bc-section section\">\n<h4>Real-World Applications<\/h4>\n<p id=\"fs-id1165137667466\">For the following exercises, determine the function described and then use it to answer the question.<\/p>\n\n<div id=\"fs-id1165137667470\">\n<div id=\"fs-id1165137543382\">\n<p id=\"fs-id1165137543383\">An object dropped from a height of 200 meters has a height,[latex]\\,h\\left(t\\right),\\,[\/latex]in meters after[latex]\\,t\\,[\/latex]seconds have lapsed, such that[latex]\\,h\\left(t\\right)=200-4.9{t}^{2}.\\,[\/latex]Express[latex]\\,t\\,[\/latex]as a function of height,[latex]\\,h,\\,[\/latex]and find the time to reach a height of 50 meters.<\/p>\n\n<\/div>\n<div id=\"fs-id1165133112798\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133112798\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165133112798\"]\n<p id=\"fs-id1165133112799\">[latex]t\\left(h\\right)=\\sqrt{\\frac{200-h}{4.9}},\\,[\/latex]5.53 seconds<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137735080\">\n<div id=\"fs-id1165137735081\">\n<p id=\"fs-id1165137735082\">An object dropped from a height of 600 feet has a height,[latex]\\,h\\left(t\\right),\\,[\/latex]in feet after[latex]\\,t\\,[\/latex]seconds have elapsed, such that[latex]\\,h\\left(t\\right)=600-16{t}^{2}.\\,[\/latex]Express[latex]\\,t\\,[\/latex]\nas a function of height[latex]\\,h,\\,[\/latex]and find the time to reach a height of 400 feet.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137451866\">\n<div id=\"fs-id1165137451867\">\n<p id=\"fs-id1165137451868\">The volume,[latex]\\,V,\\,[\/latex]of a sphere in terms of its radius,[latex]\\,r,\\,[\/latex]is given by[latex]\\,V\\left(r\\right)=\\frac{4}{3}\\pi {r}^{3}.\\,[\/latex]Express[latex]\\,r\\,[\/latex]as a function of[latex]\\,V,\\,[\/latex]and find the radius of a sphere with volume of 200 cubic feet.<\/p>\n\n<\/div>\n<div id=\"fs-id1165137894368\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137894368\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137894368\"]\n<p id=\"fs-id1165137894369\">[latex]r\\left(V\\right)=\\sqrt[3]{\\frac{3V}{4\\pi }},\\,[\/latex]3.63 feet<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137619689\">\n<div id=\"fs-id1165137619690\">\n<p id=\"fs-id1165137446018\">The surface area,[latex]\\,A,\\,[\/latex]of a sphere in terms of its radius,[latex]\\,r,\\,[\/latex]is given by[latex]\\,A\\left(r\\right)=4\\pi {r}^{2}.\\,[\/latex]Express[latex]\\,r\\,[\/latex]as a function of[latex]\\,V,\\,[\/latex]and find the radius of a sphere with a surface area of 1000 square inches.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135199586\">\n<div id=\"fs-id1165137611961\">\n<p id=\"fs-id1165137611962\">A container holds 100 mL of a solution that is 25 mL acid. If[latex]\\,n\\,[\/latex]mL of a solution that is 60% acid is added, the function[latex]\\,C\\left(n\\right)=\\frac{25+.6n}{100+n}\\,[\/latex]gives the concentration,[latex]\\,C,\\,[\/latex]as a function of the number of mL added,[latex]\\,n.\\,[\/latex]Express[latex]\\,n\\,[\/latex]as a function of[latex]\\,C\\,[\/latex]and determine the number of mL that need to be added to have a solution that is 50% acid.<\/p>\n\n<\/div>\n<div id=\"fs-id1165137677899\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137677899\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137677899\"]\n<p id=\"fs-id1165137677900\">[latex]n\\left(C\\right)=\\frac{100C-25}{.6-C},\\,[\/latex]250 mL<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137400716\">\n<div id=\"fs-id1165137400717\">\n<p id=\"fs-id1165137400718\">The period[latex]\\,T,\\,[\/latex]in seconds, of a simple pendulum as a function of its length[latex]\\,l,\\,[\/latex]in feet, is given by[latex]\\,T\\left(l\\right)=2\\pi \\sqrt{\\frac{l}{32.2}}\\,[\/latex] . Express[latex]\\,l\\,[\/latex]as a function of[latex]\\,T\\,[\/latex]and determine the length of a pendulum with period of 2 seconds.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137855335\">\n<div id=\"fs-id1165137855336\">\n<p id=\"fs-id1165137855337\">The volume of a cylinder ,[latex]\\,V,\\,[\/latex]in terms of radius,[latex]\\,r,\\,[\/latex]and height,[latex]\\,h,\\,[\/latex]is given by[latex]\\,V=\\pi {r}^{2}h.\\,[\/latex]If a cylinder has a height of 6 meters, express the radius as a function of[latex]\\,V\\,[\/latex]and find the radius of a cylinder with volume of 300 cubic meters.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135524559\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135524559\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135524559\"]\n<p id=\"fs-id1165135524560\">[latex]r\\left(V\\right)=\\sqrt{\\frac{V}{6\\pi }},\\,[\/latex]3.99 meters<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135543521\">\n<div id=\"fs-id1165135543522\">\n<p id=\"fs-id1165135543524\">The surface area,[latex]\\,A,\\,[\/latex]of a cylinder in terms of its radius,[latex]\\,r,\\,[\/latex]and height,[latex]\\,h,\\,[\/latex]is given by[latex]\\,A=2\\pi {r}^{2}+2\\pi rh.\\,[\/latex]If the height of the cylinder is 4 feet, express the radius as a function of[latex]\\,V\\,[\/latex]and find the radius if the surface area is 200 square feet.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135209080\">\n<div id=\"fs-id1165135209081\">\n<p id=\"fs-id1165135209082\">The volume of a right circular cone,[latex]\\,V,\\,[\/latex]in terms of its radius,[latex]\\,r,\\,[\/latex]and its height,[latex]\\,h,\\,[\/latex]is given by[latex]\\,V=\\frac{1}{3}\\pi {r}^{2}h.\\,[\/latex]Express[latex]\\,r\\,[\/latex]in terms of[latex]\\,V\\,[\/latex]if the height of the cone is 12 feet and find the radius of a cone with volume of 50 cubic inches.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135389876\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135389876\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135389876\"]\n<p id=\"fs-id1165135389877\">[latex]r\\left(V\\right)=\\sqrt{\\frac{V}{4\\pi }},\\,[\/latex]1.99 inches<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137783419\">\n<div id=\"fs-id1165137783420\">\n<p id=\"fs-id1165137783421\">Consider a cone with height of 30 feet. Express the radius,[latex]\\,r,\\,[\/latex]in terms of the volume,[latex]\\,V,\\,[\/latex]and find the radius of a cone with volume of 1000 cubic feet.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165135169260\">\n \t<dt>invertible function<\/dt>\n \t<dd id=\"fs-id1165135169263\">any function that has an inverse function<\/dd>\n<\/dl>\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Find the inverse of an invertible polynomial function.<\/li>\n<li>Restrict the domain to find the inverse of a polynomial function.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137793975\">A mound of gravel is in the shape of a cone with the height equal to twice the radius.<span id=\"fs-id1165137939558\"><\/span><\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/b91ae06dad3cdb0412db8cff73515f372ea917b5\/CNX_Precalc_Figure_03_08_001.jpg\" alt=\"Gravel in the shape of a cone.\" width=\"487\" height=\"410\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 1.<\/strong><\/figcaption><\/figure>\n<p id=\"fs-id1165137411369\">The volume is found using a formula from elementary geometry.<\/p>\n<div id=\"eip-854\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}V& \\hfill =& \\frac{1}{3}\\pi {r}^{2}h\\hfill \\\\ & =& \\frac{1}{3}\\pi {r}^{2}\\left(2r\\right)\\hfill \\\\ & =& \\frac{2}{3}\\pi {r}^{3}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137727278\">We have written the volume[latex]\\,V\\,[\/latex]in terms of the radius[latex]\\,r.\\,[\/latex]However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula<\/p>\n<div id=\"eip-931\" class=\"unnumbered aligncenter\">[latex]r=\\sqrt[3]{\\frac{3V}{2\\pi }}[\/latex]<\/div>\n<p id=\"fs-id1165134129769\">This function is the inverse of the formula for[latex]\\,V\\,[\/latex]in terms of[latex]\\,r.[\/latex]<\/p>\n<p id=\"fs-id1165137656509\">In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process.<\/p>\n<div id=\"fs-id1165135500723\" class=\"bc-section section\">\n<h3>Finding the Inverse of a Polynomial Function<\/h3>\n<p id=\"fs-id1165137439029\">Two functions[latex]\\,f\\,[\/latex]and[latex]\\,g\\,[\/latex]are inverse functions if for every coordinate pair in[latex]\\,f,\\left(a,b\\right),\\,[\/latex]there exists a corresponding coordinate pair in the inverse function,[latex]\\,g,\\left(b,\\,a\\right).\\,[\/latex]In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Only one-to-one functions have inverses. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test.<\/p>\n<p id=\"fs-id1165137448308\">For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_08_002\">(Figure)<\/a>. 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<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/1d361de6ef84f0703aaa4fe3b93dff47afbdb8a9\/CNX_Precalc_Figure_03_08_002.jpg\" alt=\"Diagram of a parabolic trough that is 18\u201d in height, 3\u2019 in length, and 12\u201d in width.\" width=\"487\" height=\"279\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 2.<\/strong><\/figcaption><\/figure>\n<p id=\"fs-id1165137793665\">Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with[latex]\\,x\\,[\/latex]measured horizontally and[latex]\\,y\\,[\/latex]measured vertically, with the origin at the vertex of the parabola. See <a class=\"autogenerated-content\" href=\"#Figure_03_08_003\">(Figure)<\/a>.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/a1157fe5e3ce318e6eb3be33ccbd74ea983d961f\/CNX_Precalc_Figure_03_08_003.jpg\" alt=\"Graph of a parabola.\" width=\"487\" height=\"441\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong><\/figcaption><\/figure>\n<p id=\"fs-id1165137771677\">From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form[latex]\\,y\\left(x\\right)=a{x}^{2}.\\,[\/latex]Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor[latex]\\,a.[\/latex]<\/p>\n<div id=\"eip-893\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill 18& =& a{6}^{2}\\hfill \\\\ \\hfill a& =& \\frac{18}{36}\\hfill \\\\ & =& \\frac{1}{2}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137633973\">Our parabolic cross section has the equation<\/p>\n<div id=\"eip-420\" class=\"unnumbered aligncenter\">[latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/div>\n<p id=\"fs-id1165137770004\">We are interested in the <span class=\"no-emphasis\">surface area<\/span> 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[latex]\\,y,\\,[\/latex]the width will be given by[latex]\\,2x,\\,[\/latex]so we need to solve the equation above for[latex]\\,x\\,[\/latex]and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.<\/p>\n<p id=\"fs-id1165137638570\">To find an inverse, we can restrict our original function to a limited domain on which it <em>is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive[latex]\\,x\\,[\/latex]values. On this domain, we can find an inverse by solving for the input variable:<\/p>\n<div id=\"eip-598\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill y& =& \\frac{1}{2}{x}^{2}\\hfill \\\\ \\hfill 2y& =& {x}^{2}\\hfill \\\\ \\hfill x& =& \u00b1\\sqrt{2y}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137453965\">This is not a function as written. We are limiting ourselves to positive[latex]\\,x\\,[\/latex]values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.<\/p>\n<div class=\"unnumbered\">[latex]y=\\frac{{x}^{2}}{2},\\text{ }x>0[\/latex]<\/div>\n<p id=\"fs-id1165137643958\">Because[latex]\\,x\\,[\/latex]is the distance from the center of the parabola to either side, the entire width of the water at the top will be[latex]\\,2x.\\,[\/latex]The trough is 3 feet (36 inches) long, so the surface area will then be:<\/p>\n<div class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill \\text{Area}& =& l\\cdot w\\hfill \\\\ & =& 36\\cdot 2x\\hfill \\\\ & =& 72x\\hfill \\\\ & =& 72\\sqrt{2y}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137407432\">This example illustrates two important points:<\/p>\n<ol id=\"fs-id1165135545666\" type=\"1\">\n<li>When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.<\/li>\n<li>The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.<\/li>\n<\/ol>\n<p id=\"fs-id1165137618975\">Functions involving roots are often called <span class=\"no-emphasis\">radical functions<\/span>. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation[latex]\\,{f}^{-1}\\left(x\\right).[\/latex]<\/p>\n<p id=\"fs-id1165135185952\">Warning:[latex]\\,{f}^{-1}\\left(x\\right)\\,[\/latex]is not the same as the reciprocal of the function[latex]\\,f\\left(x\\right).\\,[\/latex]This use of \u201c\u20131\u201d is reserved to denote inverse functions. To denote the reciprocal of a function[latex]\\,f\\left(x\\right),\\,[\/latex]we would need to write[latex]\\,{\\left(f\\left(x\\right)\\right)}^{-1}=\\frac{1}{f\\left(x\\right)}.[\/latex]<\/p>\n<p id=\"fs-id1165137561919\">An important relationship between inverse functions is that they \u201cundo\u201d each other. If[latex]\\,{f}^{-1}\\,[\/latex]is the inverse of a function[latex]\\,f,\\,[\/latex]<br \/>\nthen[latex]\\,f\\,[\/latex]is the inverse of the function[latex]\\,{f}^{-1}.\\,[\/latex]In other words, whatever the function[latex]\\,f\\,[\/latex]does to[latex]\\,x,[\/latex]<br \/>\n[latex]\\,{f}^{-1}\\,[\/latex]undoes it\u2014and vice-versa.<\/p>\n<div id=\"eip-519\" class=\"unnumbered aligncenter\">[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x,\\,\\text{for all }x\\text{ in the domain of }f[\/latex]<\/div>\n<p id=\"fs-id1165135503755\">and<\/p>\n<div id=\"eip-590\" class=\"unnumbered aligncenter\">[latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x,\\,\\text{for all }x\\text{ in the domain of }{f}^{-1}[\/latex]<\/div>\n<p id=\"eip-457\">Note that the inverse switches the domain and range of the original function.<\/p>\n<div id=\"fs-id1165137735698\" class=\"textbox key-takeaways\">\n<h3>Verifying Two Functions Are Inverses of One Another<\/h3>\n<p id=\"fs-id1165137852132\">Two functions,[latex]\\,f\\,[\/latex]<br \/>\nand[latex]\\,g,\\,[\/latex]are inverses of one another if for all[latex]\\,x\\,[\/latex]in the domain of[latex]\\,f\\,[\/latex]<br \/>\nand[latex]\\,g.[\/latex]<\/p>\n<div id=\"eip-973\" class=\"unnumbered aligncenter\">[latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137646263\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137501372\"><strong>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.<\/strong><\/p>\n<ol type=\"1\">\n<li>Replace[latex]\\,f\\left(x\\right)\\,[\/latex]with[latex]\\,y.[\/latex]<\/li>\n<li>Interchange[latex]\\,x\\,[\/latex]and[latex]\\,y.[\/latex]<\/li>\n<li>Solve for[latex]\\,y,\\,[\/latex]and rename the function[latex]\\,{f}^{-1}\\left(x\\right).[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_08_01\" class=\"textbox examples\">\n<div id=\"fs-id1165135150650\">\n<div id=\"fs-id1165135620877\">\n<h3>Verifying Inverse Functions<\/h3>\n<p id=\"fs-id1165134148383\">Show that[latex]\\,f\\left(x\\right)=\\frac{1}{x+1}\\,[\/latex]and[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1\\,[\/latex]are inverses, for[latex]\\,x\\ne 0,-1[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165137724950\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137834138\">We must show that[latex]\\,{f}^{-1}\\left(f\\left(x\\right)\\right)=x\\,[\/latex]and[latex]\\,f\\left({f}^{-1}\\left(x\\right)\\right)=x.[\/latex]<\/p>\n<div id=\"eip-id1165131967953\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill {f}^{-1}\\left(f\\left(x\\right)\\right)& =& {f}^{-1}\\left(\\frac{1}{x+1}\\right)\\hfill \\\\ & =& \\frac{1}{\\frac{1}{x+1}}-1\\hfill \\\\ & =& \\left(x+1\\right)-1\\\\ & =& x\\hfill \\\\ \\hfill f\\left({f}^{-1}\\left(x\\right)\\right)& =& f\\left(\\frac{1}{x}-1\\right)\\hfill \\\\ & =& \\frac{1}{\\left(\\frac{1}{x}-1\\right)+1}\\hfill \\\\ & =& \\frac{1}{\\frac{1}{x}}\\hfill \\\\ & =& x\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135168183\">Therefore,[latex]\\,f\\left(x\\right)=\\frac{1}{x+1}\\,[\/latex]<br \/>\nand[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1\\,[\/latex]are inverses.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137594492\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_01\">\n<div id=\"fs-id1165137552094\">\n<p id=\"fs-id1165137426116\">Show that[latex]\\,f\\left(x\\right)=\\frac{x+5}{3}\\,[\/latex]<br \/>\nand[latex]\\,{f}^{-1}\\left(x\\right)=3x-5\\,[\/latex]are inverses.<\/p>\n<\/div>\n<div id=\"fs-id1165135502962\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137696615\">[latex]{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{x+5}{3}\\right)=3\\left(\\frac{x+5}{3}\\right)-5=\\left(x-5\\right)+5=x\\,[\/latex]and[latex]\\,f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(3x-5\\right)=\\frac{\\left(3x-5\\right)+5}{3}=\\frac{3x}{3}=x[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_08_02\" class=\"textbox examples\">\n<div id=\"fs-id1165137600799\">\n<div id=\"fs-id1165135160775\">\n<h3>Finding the Inverse of a Cubic Function<\/h3>\n<p id=\"fs-id1165137569920\">Find the inverse of the function[latex]\\,f\\left(x\\right)=5{x}^{3}+1.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137676384\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135412872\">This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for[latex]\\,x.[\/latex]<\/p>\n<div id=\"eip-id1165133077998\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill y& =& 5{x}^{3}+1\\hfill \\\\ \\hfill x& =& 5{y}^{3}+1\\hfill \\\\ \\hfill x-1& =& 5{y}^{3}\\hfill \\\\ \\hfill \\frac{x-1}{5}& =& {y}^{3}\\hfill \\\\ \\hfill {f}^{-1}\\left(x\\right)& =& \\sqrt[3]{\\frac{x-1}{5}}\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137635322\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165137641602\">Look at the graph of[latex]\\,f\\,[\/latex]and[latex]\\,{f}^{\u20131}.\\,[\/latex]Notice that one graph is the reflection of the other about the line[latex]\\,y=x.\\,[\/latex]This is always the case when graphing a function and its inverse function.<\/p>\n<p id=\"fs-id1165137793468\">Also, since the method involved interchanging[latex]\\,x\\,[\/latex]and[latex]\\,y,\\,[\/latex]notice corresponding points. If[latex]\\,\\left(a,b\\right)\\,[\/latex]is on the graph of[latex]\\,f,[\/latex]then[latex]\\,\\left(b,a\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{\u20131}.\\,[\/latex]Since[latex]\\,\\left(0,1\\right)\\,[\/latex]is on the graph of[latex]\\,f,\\,[\/latex]then[latex]\\,\\left(1,0\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{\u20131}.\\,[\/latex]Similarly, since[latex]\\,\\left(1,6\\right)\\,[\/latex]is on the graph of[latex]\\,f,[\/latex]then[latex]\\,\\left(6,1\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{\u20131}.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Figure_03_08_004\">(Figure)<\/a>.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/90a0ca823116c8647b783db6d540dad88b0b04f0\/CNX_Precalc_Figure_03_08_004.jpg\" alt=\"Graph of f(x)=5x^3+1 and its inverse, f^(-1)(x)=3sqrt((x-1)\/(5)).\" width=\"487\" height=\"554\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 4.<\/strong><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_02\">\n<div id=\"fs-id1165137704564\">\n<p id=\"fs-id1165133047522\">Find the inverse function of[latex]\\,f\\left(x\\right)=\\sqrt[3]{x+4}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137737307\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137529408\">[latex]{f}^{-1}\\left(x\\right)={x}^{3}-4[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137847291\" class=\"bc-section section\">\n<h3>Restricting the Domain to Find the Inverse of a Polynomial Function<\/h3>\n<p id=\"fs-id1165137471808\">So far, we have been able to find the inverse functions of <span class=\"no-emphasis\">cubic functions<\/span> without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an <span class=\"no-emphasis\">inverse function<\/span>. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.<\/p>\n<div id=\"fs-id1165137434585\" class=\"textbox key-takeaways\">\n<h3>Restricting the Domain<\/h3>\n<p id=\"fs-id1165137409777\">If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.<\/p>\n<\/div>\n<div id=\"fs-id1165137431545\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137656706\"><strong>Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.<\/strong><\/p>\n<ol id=\"fs-id1165137532171\" type=\"1\">\n<li>Restrict the domain by determining a domain on which the original function is one-to-one.<\/li>\n<li>Replace[latex]\\,f\\left(x\\right)\\,[\/latex]with[latex]\\,y.[\/latex]<\/li>\n<li>Interchange[latex]\\,x\\,[\/latex]and[latex]\\,y.[\/latex]<\/li>\n<li>Solve for[latex]\\,y,[\/latex]and rename the function or pair of function[latex]\\,{f}^{-1}\\left(x\\right).[\/latex]<\/li>\n<li>Revise the formula for[latex]\\,{f}^{-1}\\left(x\\right)\\,[\/latex]by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_08_03\" class=\"textbox examples\">\n<div>\n<div id=\"fs-id1165137668130\">\n<h3>Restricting the Domain to Find the Inverse of a Polynomial Function<\/h3>\n<p id=\"fs-id1165137482766\">Find the inverse function of[latex]\\,f\\text{:}[\/latex]<\/p>\n<ol type=\"a\">\n<li>[latex]f\\left(x\\right)={\\left(x-4\\right)}^{2}, x\\ge 4[\/latex]<\/li>\n<li>[latex]f\\left(x\\right)={\\left(x-4\\right)}^{2}, x\\le 4[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165137606152\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137506731\">The original function[latex]\\,f\\left(x\\right)={\\left(x-4\\right)}^{2}\\,[\/latex]is not one-to-one, but the function is restricted to a domain of[latex]\\,x\\ge 4\\,[\/latex]or[latex]\\,x\\le 4\\,[\/latex]on which it is one-to-one. See <a class=\"autogenerated-content\" href=\"#Figure_03_08_005\">(Figure)<\/a>.<\/p>\n<div id=\"Figure_03_08_005\" class=\"medium\">\n<figure style=\"width: 731px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/53d5170d583ead5b559b1d76266a29bcaa40c18b\/CNX_Precalc_Figure_03_08_005.jpg\" alt=\"image\" \/><figcaption class=\"wp-caption-text\">=4 and the second is when x&lt;=4.\" width=\"731\" height=\"365\"&gt; <strong>Figure 5.<\/strong><\/figcaption><\/figure>\n<p><span id=\"fs-id1165137532798\"><\/span><\/div>\n<p id=\"fs-id1165137706306\">To find the inverse, start by replacing[latex]\\,f\\left(x\\right)\\,[\/latex]with the simple variable[latex]\\,y.[\/latex]<\/p>\n<div id=\"eip-id1165133259004\" class=\"unnumbered\">[latex]\\begin{array}{cccc}\\hfill y& =& {\\left(x-4\\right)}^{2}\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Interchange}x\\text{and }y.\\hfill \\\\ \\hfill x& =& {\\left(y-4\\right)}^{2}\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Take the square root}.\\hfill \\\\ \\hfill \u00b1\\sqrt{x}& =& y-4\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Add} 4 \\text{to both sides}.\\hfill \\\\ \\hfill 4\u00b1\\sqrt{x}& =& y\\hfill & \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137444285\">This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of[latex]\\,x\\,[\/latex]and[latex]\\,y\\,[\/latex]for the original[latex]\\,f\\left(x\\right),\\,[\/latex]we looked at the domain: the values[latex]\\,x\\,[\/latex]<br \/>\ncould assume. When we reversed the roles of[latex]\\,x\\,[\/latex]and[latex]\\,y,\\,[\/latex]<br \/>\nthis gave us the values[latex]\\,y\\,[\/latex]could assume. For this function,[latex]\\,x\\ge 4,\\,[\/latex]so for the inverse, we should have[latex]\\,y\\ge 4,\\,[\/latex]which is what our inverse function gives.<\/p>\n<ol id=\"fs-id1165137735027\" type=\"a\">\n<li>The domain of the original function was restricted to[latex]\\,x\\ge 4,\\,[\/latex]so the outputs of the inverse need to be the same,[latex]\\,f\\left(x\\right)\\ge 4,\\,[\/latex]and we must use the + case:\n<div id=\"eip-id1165134294825\" class=\"unnumbered\">[latex]{f}^{-1}\\left(x\\right)=4+\\sqrt{x}[\/latex]<\/div>\n<\/li>\n<li>The domain of the original function was restricted to[latex]\\,x\\le 4,\\,[\/latex]so the outputs of the inverse need to be the same,[latex]\\,f\\left(x\\right)\\le 4,\\,[\/latex]and we must use the \u2013 case:\n<div id=\"eip-id1165137482501\" class=\"unnumbered\">[latex]{f}^{-1}\\left(x\\right)=4-\\sqrt{x}[\/latex]<\/details>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div>\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165137534054\">On the graphs in <a class=\"autogenerated-content\" href=\"#Figure_03_08_006\">(Figure)<\/a>, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line[latex]\\,y=x.\\,[\/latex]The coordinate pair[latex]\\,\\left(4, 0\\right)\\,[\/latex]is on the graph of[latex]\\,f\\,[\/latex]and the coordinate pair[latex]\\,\\left(0, 4\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{-1}.\\,[\/latex]For any coordinate pair, if[latex]\\,\\left(a,\\text{ }b\\right)\\,[\/latex]is on the graph of[latex]\\,f,\\,[\/latex]then[latex]\\,\\left(b,\\text{ }a\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{-1}.\\,[\/latex]Finally, observe that the graph of[latex]\\,f\\,[\/latex]intersects the graph of[latex]\\,{f}^{-1}[\/latex]on the line[latex]\\,y=x.\\,[\/latex]Points of intersection for the graphs of[latex]\\,f\\,[\/latex]and[latex]\\,{f}^{-1}\\,[\/latex]will always lie on the line[latex]\\,y=x.[\/latex]<span id=\"fs-id1165137471020\"><\/span><\/p>\n<figure style=\"width: 975px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/6ee202f7074c245ac7cf28a92a363acbf8e6d22a\/CNX_Precalc_Figure_03_08_006.jpg\" alt=\"Two graphs of a parabolic function with half of its inverse.\" width=\"975\" height=\"442\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 6.<\/strong><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_08_04\" class=\"textbox examples\">\n<div id=\"fs-id1165137786597\">\n<div id=\"fs-id1165135481235\">\n<h3>Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified<\/h3>\n<p id=\"fs-id1165137410909\">Restrict the domain and then find the inverse of<\/p>\n<div id=\"eip-id1165133355860\" class=\"unnumbered\">[latex]f\\left(x\\right)={\\left(x-2\\right)}^{2}-3.[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165135394334\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137414415\">We can see this is a parabola with vertex at[latex]\\,\\left(2,\u20133\\right)\\,[\/latex]that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to[latex]\\,x\\ge 2.[\/latex]<\/p>\n<p id=\"fs-id1165137842529\">To find the inverse, we will use the vertex form of the quadratic. We start by replacing[latex]\\,f\\left(x\\right)\\,[\/latex]with a simple variable,[latex]\\,y,\\,[\/latex]then solve for[latex]\\,x.[\/latex]<\/p>\n<div id=\"eip-id1165133305354\" class=\"unnumbered\">[latex]\\begin{array}{cccc}\\hfill y& =& {\\left(x-2\\right)}^{2}-3\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Interchange }x\\text{ and }y.\\hfill \\\\ \\hfill x& =& {\\left(y-2\\right)}^{2}-3\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Add 3 to both sides}.\\hfill \\\\ \\hfill x+3& =& {\\left(y-2\\right)}^{2}\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Take the square root}.\\hfill \\\\ \\hfill \u00b1\\sqrt{x+3}& =& y-2\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Add 2 to both sides}.\\hfill \\\\ \\hfill 2\u00b1\\sqrt{x+3}& =& y\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{\u2003\u2003}\\text{Rename the function}.\\hfill \\\\ \\hfill {f}^{-1}\\left(x\\right)& =& 2\u00b1\\sqrt{x+3}\\hfill & \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137419504\">Now we need to determine which case to use. Because we restricted our original function to a domain of[latex]\\,x\\ge 2,\\,[\/latex]the outputs of the inverse should be the same, telling us to utilize the + case<\/p>\n<div id=\"eip-id1165132193808\" class=\"unnumbered\">[latex]{f}^{-1}\\left(x\\right)=2+\\sqrt{x+3}[\/latex]<\/div>\n<p>If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.<span id=\"fs-id1165135194279\"><\/span><\/details>\n<\/div>\n<div id=\"fs-id1165134362839\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165135259538\">Notice that we arbitrarily decided to restrict the domain on[latex]\\,x\\ge 2.\\,[\/latex]We could just have easily opted to restrict the domain on[latex]\\,x\\le 2,\\,[\/latex]in which case[latex]\\,{f}^{-1}\\left(x\\right)=2-\\sqrt{x+3}.\\,[\/latex]Observe the original function graphed on the same set of axes as its inverse function in <a class=\"autogenerated-content\" href=\"#Figure_03_08_007\">(Figure)<\/a>. Notice that both graphs show symmetry about the line[latex]\\,y=x.\\,[\/latex]The coordinate pair[latex]\\,\\left(2,\\text{ }-3\\right)\\,[\/latex]is on the graph of[latex]\\,f\\,[\/latex]and the coordinate pair[latex]\\,\\left(-3,\\text{ }2\\right)\\,[\/latex]is on the graph of[latex]\\,{f}^{-1}.\\,[\/latex]Observe from the graph of both functions on the same set of axes that<\/p>\n<div id=\"eip-id1165134122215\" class=\"unnumbered\">[latex]\\text{domain of }f=\\text{range of} {f}^{\u20131}=\\left[2,\\infty \\right)[\/latex]<\/div>\n<p id=\"fs-id1165137642128\">and<\/p>\n<div id=\"eip-id1165134279478\" class=\"unnumbered\">[latex]\\text{domain of }{f}^{\u20131}=\\text{range of} f=\\left[\u20133,\\infty \\right).[\/latex]<\/div>\n<p>Finally, observe that the graph of[latex]\\,f\\,[\/latex]intersects the graph of[latex]\\,{f}^{-1}\\,[\/latex]along the line[latex]\\,y=x.[\/latex]<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/93673f4f9c9538c5e8b351f6535b7a9f4886a8f6\/CNX_Precalc_Figure_03_08_007.jpg\" alt=\"Graph of a parabolic function with half of its inverse.\" width=\"487\" height=\"487\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 7.<\/strong><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137419950\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_03\">\n<div id=\"fs-id1165137600895\">\n<p id=\"fs-id1165135596379\">Find the inverse of the function[latex]\\,f\\left(x\\right)={x}^{2}+1,\\,[\/latex]on the domain[latex]\\,x\\ge 0.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137737548\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137737550\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x-1}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137894462\" class=\"bc-section section\">\n<h4>Solving Applications of Radical Functions<\/h4>\n<p id=\"fs-id1165137696560\">Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the <span class=\"no-emphasis\">inverse of a radical function<\/span>, we will need to restrict the domain of the answer because the range of the original function is limited.<\/p>\n<div id=\"fs-id1165137415876\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137455923\"><strong>Given a radical function, find the inverse.<\/strong><\/p>\n<ol id=\"fs-id1165137542989\" type=\"1\">\n<li>Determine the range of the original function.<\/li>\n<li>Replace[latex]\\,f\\left(x\\right)\\,[\/latex] with[latex]\\,y,\\,[\/latex]then solve for[latex]\\,x.[\/latex]<\/li>\n<li>If necessary, restrict the domain of the inverse function to the range of the original function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_08_05\" class=\"textbox examples\">\n<div id=\"fs-id1165135173813\">\n<div id=\"fs-id1165137399685\">\n<h3>Finding the Inverse of a Radical Function<\/h3>\n<p id=\"fs-id1165135570491\">Restrict the domain of the function[latex]\\,f\\left(x\\right)=\\sqrt{x-4}\\,[\/latex]and then find the inverse.<\/p>\n<\/div>\n<div id=\"fs-id1165137766917\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135545767\">Note that the original function has range[latex]\\,f\\left(x\\right)\\ge 0.\\,[\/latex]Replace[latex]\\,f\\left(x\\right)\\,[\/latex]with[latex]\\,y,\\,[\/latex]then solve for[latex]\\,x.[\/latex]<\/p>\n<div id=\"eip-id1165135508330\" class=\"unnumbered\">[latex]\\begin{array}{cccc}\\hfill y& =& \\sqrt{x-4}\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Replace }f\\left(x\\right)\\text{ with }y.\\hfill \\\\ \\hfill x& =& \\sqrt{y-4}\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Interchange }x\\text{ and }y.\\hfill \\\\ \\hfill x& =& \\sqrt{y-4}\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Square each side}.\\hfill \\\\ \\hfill {x}^{2}& =& y-4\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Add 4}.\\hfill \\\\ \\hfill {x}^{2}+4& =& y\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Rename the function }{f}^{-1}\\left(x\\right).\\hfill \\\\ \\hfill {f}^{-1}\\left(x\\right)& =& {x}^{2}+4\\hfill & \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135209570\">Recall that the domain of this function must be limited to the range of the original function.<\/p>\n<div id=\"eip-id1165135452086\" class=\"unnumbered\">[latex]{f}^{-1}\\left(x\\right)={x}^{2}+4,x\\ge 0[\/latex]<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137750058\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165135173239\">Notice in <a class=\"autogenerated-content\" href=\"#Figure_03_08_008\">(Figure)<\/a> that the inverse is a reflection of the original function over the line[latex]\\,y=x.\\,[\/latex]Because the original function has only positive outputs, the inverse function has only positive inputs.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/278b5e9547f105e6c6d64031e7d5c208c55c0996\/CNX_Precalc_Figure_03_08_008.jpg\" alt=\"Graph of f(x)=sqrt(x-4) and its inverse, f^(-1)(x)=x^2+4.\" width=\"487\" height=\"444\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 8.<\/strong><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137662048\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_04\">\n<div id=\"fs-id1165135524557\">\n<p id=\"fs-id1165137784775\">Restrict the domain and then find the inverse of the function[latex]\\,f\\left(x\\right)=\\sqrt{2x+3}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137784778\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137419726\">[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-3}{2},x\\ge 0[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137761571\" class=\"bc-section section\">\n<h4>Solving Applications of Radical Functions<\/h4>\n<p id=\"fs-id1165135435831\">Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section.<\/p>\n<div id=\"Example_03_08_06\" class=\"textbox examples\">\n<div>\n<div id=\"fs-id1165137531120\">\n<h3>Solving an Application with a Cubic Function<\/h3>\n<p id=\"fs-id1165137771982\">A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by<\/p>\n<div id=\"eip-id1165132187568\" class=\"unnumbered\">[latex]V=\\frac{2}{3}\\pi {r}^{3}[\/latex]<\/div>\n<p id=\"fs-id1165135181305\">Find the inverse of the function[latex]\\,V=\\frac{2}{3}\\pi {r}^{3}\\,[\/latex]that determines the volume[latex]\\,V\\,[\/latex]of a cone and is a function of the radius[latex]\\,r.\\,[\/latex]Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use[latex]\\,\\pi =3.14.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137405142\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137405144\">Start with the given function for[latex]\\,V.\\,[\/latex]Notice that the meaningful domain for the function is[latex]\\,r>0\\,[\/latex]since negative radii would not make sense in this context nor would a radius of 0. Also note the range of the function (hence, the domain of the inverse function) is[latex]\\,V>0.\\,[\/latex]Solve for[latex]\\,r\\,[\/latex]in terms of[latex]\\,V,\\,[\/latex]using the method outlined previously. Note that in real-world applications, we do not swap the variables when finding inverses. Instead, we change which variable is considered to be the independent variable.<\/p>\n<div id=\"eip-id1165133447890\" class=\"unnumbered\">[latex]\\begin{array}{cccc}\\hfill V& =& \\frac{2}{3}\\pi {r}^{3}\\hfill & \\\\ \\hfill {r}^{3}& =& \\frac{3V}{2\\pi }\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Solve for }{r}^{3}.\\hfill \\\\ \\hfill r& =& \\sqrt[3]{\\frac{3V}{2\\pi }}\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Solve for }r.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137730324\">This is the result stated in the section opener. Now evaluate this for[latex]\\,V=100\\,[\/latex]and[latex]\\,\\pi =3.14.[\/latex]<\/p>\n<div id=\"eip-id1165135203234\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill r& =& \\sqrt[3]{\\frac{3V}{2\\pi }}\\hfill \\\\ & =& \\sqrt[3]{\\frac{3\\cdot 100}{2\\cdot 3.14}}\\hfill \\\\ & \\approx & \\sqrt[3]{47.7707}\\hfill \\\\ & \\approx \\hfill & 3.63\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137706279\">Therefore, the radius is about 3.63 ft.<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137580023\" class=\"bc-section section\">\n<h4>Determining the Domain of a Radical Function Composed with Other Functions<\/h4>\n<p id=\"fs-id1165134042947\">When radical functions are composed with other functions, determining domain can become more complicated.<\/p>\n<div id=\"Example_03_08_07\" class=\"textbox examples\">\n<div id=\"fs-id1165135378774\">\n<div id=\"fs-id1165135378776\">\n<h3>Finding the Domain of a Radical Function Composed with a Rational Function<\/h3>\n<p id=\"fs-id1165137658778\">Find the domain of the function[latex]\\,f\\left(x\\right)=\\sqrt{\\frac{\\left(x+2\\right)\\left(x-3\\right)}{\\left(x-1\\right)}}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137550072\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137665549\">Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where[latex]\\,\\frac{\\left(x+2\\right)\\left(x-3\\right)}{\\left(x-1\\right)}\\ge 0.\\,[\/latex]The output of a rational function can change signs (change from positive to negative or vice versa) at <em>x<\/em>-intercepts and at vertical asymptotes. For this equation, the graph could change signs at[latex]\\,x=\u20132, 1, \\text{and} 3.[\/latex]<\/p>\n<p id=\"fs-id1165135686721\">To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph as shown in <a class=\"autogenerated-content\" href=\"#Figure_03_08_009\">(Figure)<\/a>.<\/p>\n<div id=\"Figure_03_08_009\" class=\"medium\">\n<figure style=\"width: 731px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140154\/CNX_Precalc_Figure_03_08_009.jpg\" alt=\"Graph of a radical function that shows where the outputs are nonnegative.\" width=\"731\" height=\"439\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 9.<\/strong><\/figcaption><\/figure>\n<\/div>\n<p id=\"fs-id1165137694167\">This function has two <em>x<\/em>-intercepts, both of which exhibit linear behavior near the <em>x<\/em>-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a <em>y<\/em>-intercept at[latex]\\,\\left(0,\\sqrt{6}\\right).[\/latex]<\/p>\n<p id=\"fs-id1165135333589\">From the <em>y<\/em>-intercept and <em>x<\/em>-intercept at[latex]\\,x=-2,\\,[\/latex]we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.<\/p>\n<p id=\"fs-id1165137664081\">From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function[latex]\\,f\\left(x\\right)\\,[\/latex]will be defined.[latex]\\,f\\left(x\\right)\\,[\/latex]has domain[latex]\\,-2\\le x<1\\,\\text{or}\\,x\\ge 3,\\,[\/latex]or in interval notation,[latex]\\,\\left[-2,1\\right)\\cup \\left[3,\\infty \\right).[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\">\n<h4>Finding Inverses of Rational Functions<\/h4>\n<p id=\"fs-id1165135475898\">As with finding inverses of quadratic functions, it is sometimes desirable to find the <span class=\"no-emphasis\">inverse of a rational function<\/span>, particularly of rational functions that are the ratio of linear functions, such as in concentration applications.<\/p>\n<div id=\"Example_03_08_08\" class=\"textbox examples\">\n<div id=\"fs-id1165137642525\">\n<div id=\"fs-id1165137642528\">\n<h3>Finding the Inverse of a Rational Function<\/h3>\n<p id=\"fs-id1165135332364\">The function[latex]\\,C=\\frac{20+0.4n}{100+n}\\,[\/latex]represents the concentration[latex]\\,C\\,[\/latex]of an acid solution after[latex]\\,n\\,[\/latex]mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for[latex]\\,n\\,[\/latex]in terms of[latex]\\,C.\\,[\/latex]Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.<\/p>\n<\/div>\n<div id=\"fs-id1165134223201\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>We first want the inverse of the function in order to determine how many mL we need for a given concentration. We will solve for[latex]\\,n\\,[\/latex]in terms of[latex]\\,C.[\/latex]<\/p>\n<div id=\"eip-id1165132957110\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill C& =& \\frac{20+0.4n}{100+n}\\hfill \\\\ \\hfill C\\left(100+n\\right)& =& 20+0.4n\\hfill \\\\ \\hfill 100C+Cn& =& 20+0.4n\\hfill \\\\ \\hfill 100C-20& =& 0.4n-Cn\\hfill \\\\ \\hfill 100C-20& =& \\left(0.4n-C\\right)n\\hfill \\\\ \\hfill n& =& \\frac{100C-20}{0.4-C}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137637474\">Now evaluate this function at 35%, which is[latex]\\,C=0.35.[\/latex]<\/p>\n<div id=\"eip-id1165131986931\" class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill n& =& \\frac{100\\left(0.35\\right)-20}{0.4-0.35}\\hfill \\\\ & =& \\frac{15}{0.05}\\hfill \\\\ & =& 300\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137706154\">We can conclude that 300 mL of the 40% solution should be added.<\/p>\n<p id=\"fs-id1165137706154\"><\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137697102\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_03_08_05\">\n<div id=\"fs-id1165134042922\">\n<p id=\"fs-id1165134042923\">Find the inverse of the function[latex]\\,f\\left(x\\right)=\\frac{x+3}{x-2}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137731924\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137731925\">[latex]{f}^{-1}\\left(x\\right)=\\frac{2x+3}{x-1}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137551968\" class=\"precalculus media\">\n<p id=\"fs-id1165137761312\">Access these online resources for additional instruction and practice with inverses and radical functions.<\/p>\n<ul id=\"fs-id1165134306731\">\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/graphsquareroot\">Graphing the Basic Square Root Function<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/inversesquare\">Find the Inverse of a Square Root Function<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/inverserational\">Find the Inverse of a Rational Function<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/rationalinverse\">Find the Inverse of a Rational Function and an Inverse Function Value<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/inversefunction\">Inverse Functions<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135192373\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165135528386\">\n<li>The inverse of a quadratic function is a square root function.<\/li>\n<li>If[latex]\\,{f}^{-1}\\,[\/latex] is the inverse of a function[latex]\\,f,\\,[\/latex] then[latex]\\,f\\,[\/latex] is the inverse of the function[latex]\\,{f}^{-1}.\\,[\/latex] See <a class=\"autogenerated-content\" href=\"#Example_03_08_01\">(Figure)<\/a>.<\/li>\n<li>While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible. See <a class=\"autogenerated-content\" href=\"#Example_03_08_02\">(Figure)<\/a>.<\/li>\n<li>To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one. See <a class=\"autogenerated-content\" href=\"#Example_03_08_03\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_03_08_04\">(Figure)<\/a>.<\/li>\n<li>When finding the inverse of a radical function, we need a restriction on the domain of the answer. See <a class=\"autogenerated-content\" href=\"#Example_03_08_05\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_03_08_07\">(Figure)<\/a>.<\/li>\n<li>Inverse and radical and functions can be used to solve application problems. See <a class=\"autogenerated-content\" href=\"#Example_03_08_06\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_03_08_08\">(Figure)<\/a>.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Section Exercises<\/h3>\n<div id=\"fs-id1165137414778\" class=\"bc-section section\">\n<h4>Verbal<\/h4>\n<div id=\"fs-id1165137678194\">\n<div id=\"fs-id1165137452992\">\n<p id=\"fs-id1165137452993\">Explain why we cannot find inverse functions for all polynomial functions.<\/p>\n<\/div>\n<div id=\"fs-id1165137452997\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137723132\">It can be too difficult or impossible to solve for[latex]\\,x\\,[\/latex]in terms of[latex]\\,y.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137648362\">\n<div id=\"fs-id1165137854841\">\n<p>Why must we restrict the domain of a quadratic function when finding its inverse?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137854845\">\n<div id=\"fs-id1165135356600\">\n<p id=\"fs-id1165135356601\">When finding the inverse of a radical function, what restriction will we need to make?<\/p>\n<\/div>\n<div id=\"fs-id1165135356605\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137590162\">We will need a restriction on the domain of the answer.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137590165\">\n<div id=\"fs-id1165137590166\">\n<p id=\"fs-id1165137661780\">The inverse of a quadratic function will always take what form?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137661784\" class=\"bc-section section\">\n<h4>Algebraic<\/h4>\n<p id=\"fs-id1165137936804\">For the following exercises, find the inverse of the function on the given domain.<\/p>\n<div id=\"fs-id1165135193880\">\n<div id=\"fs-id1165135193881\">\n<p id=\"fs-id1165135193882\">[latex]f\\left(x\\right)={\\left(x-4\\right)}^{2}, \\left[4,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137472528\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137472529\">[latex]\\,\\,{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137706126\">\n<div id=\"fs-id1165137706127\">\n<p id=\"fs-id1165137706128\">[latex]f\\left(x\\right)={\\left(x+2\\right)}^{2}, \\left[-2,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137405850\">\n<div id=\"fs-id1165137405851\">\n<p id=\"fs-id1165137405852\">[latex]f\\left(x\\right)={\\left(x+1\\right)}^{2}-3, \\left[-1,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135572106\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135572107\">[latex]\\,\\,{f}^{-1}\\left(x\\right)=\\sqrt{x+3}-1[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135209972\">\n<div id=\"fs-id1165135209973\">\n<p id=\"fs-id1165137425962\">[latex]f\\left(x\\right)=3{x}^{2}+5,\\,\\,\\left(\\infty ,0\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135436476\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135436478\">[latex]{f}^{-1}\\left(x\\right)=-\\sqrt{\\frac{x-5}{3}}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134042512\">\n<div id=\"fs-id1165137696696\">\n<p id=\"fs-id1165137696698\">[latex]f\\left(x\\right)=12-{x}^{2}, \\left[0,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137579624\">\n<div>[latex]f\\left(x\\right)=9-{x}^{2}, \\left[0,\\infty \\right)[\/latex]<\/div>\n<div id=\"fs-id1165135194244\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137862827\">[latex]f\\left(x\\right)=\\sqrt{9-x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134555020\">\n<div id=\"fs-id1165134555021\">\n<p id=\"fs-id1165134555022\">[latex]f\\left(x\\right)=2{x}^{2}+4, \\left[0,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137603367\">For the following exercises, find the inverse of the functions.<\/p>\n<div id=\"fs-id1165137725343\">\n<div id=\"fs-id1165137725344\">\n<p id=\"fs-id1165137725345\">[latex]f\\left(x\\right)={x}^{3}+5[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137885943\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137885944\">[latex]\\,\\,{f}^{-1}\\left(x\\right)=\\sqrt[3]{x-5}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137728086\">\n<div id=\"fs-id1165137728087\">\n<p id=\"fs-id1165137728088\">[latex]f\\left(x\\right)=3{x}^{3}+1[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135530666\">\n<div id=\"fs-id1165133035997\">\n<p id=\"fs-id1165133035998\">[latex]f\\left(x\\right)=4-{x}^{3}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137767092\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137767093\">[latex]\\,{f}^{-1}\\left(x\\right)=\\sqrt[3]{4-x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135500747\">\n<div id=\"fs-id1165135500748\">\n<p id=\"fs-id1165135500749\">[latex]f\\left(x\\right)=4-2{x}^{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135530341\">For the following exercises, find the inverse of the functions.<\/p>\n<div id=\"fs-id1165135530344\">\n<div id=\"fs-id1165135530345\">\n<p id=\"fs-id1165137656262\">[latex]f\\left(x\\right)=\\sqrt{2x+1}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135250674\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135250675\">[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-1}{2},\\,\\,\\left[0,\\infty \\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165137463046\">\n<p id=\"fs-id1165137463047\">[latex]f\\left(x\\right)=\\sqrt{3-4x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137502900\">\n<div id=\"fs-id1165137502901\">\n<p id=\"fs-id1165137502902\">[latex]f\\left(x\\right)=9+\\sqrt{4x-4}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137871173\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137871174\">[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{{\\left(x-9\\right)}^{2}+4}{4},\\,\\,\\left[9,\\infty \\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135319426\">\n<div id=\"fs-id1165137662665\">\n<p id=\"fs-id1165137662666\">[latex]f\\left(x\\right)=\\sqrt{6x-8}+5[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137471620\">\n<div id=\"fs-id1165137471621\">\n<p id=\"fs-id1165137471622\">[latex]f\\left(x\\right)=9+2\\sqrt[3]{x}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137735210\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137735211\">[latex]{f}^{-1}\\left(x\\right)={\\left(\\frac{x-9}{2}\\right)}^{3}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135699114\">\n<div id=\"fs-id1165135699115\">\n<p id=\"fs-id1165137461567\">[latex]f\\left(x\\right)=3-\\sqrt[3]{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137544702\">\n<div id=\"fs-id1165137544703\">\n<p id=\"fs-id1165137660257\">[latex]f\\left(x\\right)=\\frac{2}{x+8}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137737152\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137737153\">[latex]{f}^{-1}\\left(x\\right)={\\frac{2-8x}{x}}^{}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137436932\">\n<div id=\"fs-id1165137436933\">\n<p id=\"fs-id1165137436934\">[latex]f\\left(x\\right)=\\frac{3}{x-4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137442675\">\n<div id=\"fs-id1165137442676\">\n<p id=\"fs-id1165137442677\">[latex]f\\left(x\\right)=\\frac{x+3}{x+7}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137466246\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137466248\">[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{7x-3}{1-x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135701622\">\n<div id=\"fs-id1165135701623\">\n<p id=\"fs-id1165135701624\">[latex]f\\left(x\\right)=\\frac{x-2}{x+7}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135333251\">\n<div id=\"fs-id1165137443976\">\n<p id=\"fs-id1165137443977\">[latex]f\\left(x\\right)=\\frac{3x+4}{5-4x}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137668289\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137668290\">[latex]\\,{f}^{-1}\\left(x\\right)=\\frac{5x-4}{4x+3}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137642936\">\n<div id=\"fs-id1165137642938\">\n<p id=\"fs-id1165137642939\">[latex]f\\left(x\\right)=\\frac{5x+1}{2-5x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137760837\">\n<div id=\"fs-id1165137408974\">\n<p id=\"fs-id1165137408975\">[latex]f\\left(x\\right)={x}^{2}+2x, \\left[-1,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135176553\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135176554\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+1}-1[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137662088\">\n<div id=\"fs-id1165137662089\">\n<p id=\"fs-id1165137436907\">[latex]f\\left(x\\right)={x}^{2}+4x+1, \\left[-2,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133184252\">\n<div id=\"fs-id1165133184253\">\n<p id=\"fs-id1165137675697\">[latex]f\\left(x\\right)={x}^{2}-6x+3, \\left[3,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137760888\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137760889\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+6}+3[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135193360\" class=\"bc-section section\">\n<h4>Graphical<\/h4>\n<p id=\"fs-id1165135193366\">For the following exercises, find the inverse of the function and graph both the function and its inverse.<\/p>\n<div id=\"fs-id1165137715400\">\n<div id=\"fs-id1165137715401\">\n<p id=\"fs-id1165137715402\">[latex]f\\left(x\\right)={x}^{2}+2,\\,x\\ge 0[\/latex]<\/p>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165137640500\">\n<p id=\"fs-id1165137640501\">[latex]f\\left(x\\right)=4-{x}^{2},\\,x\\ge 0[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137502874\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137437197\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{4-x}[\/latex]<\/p>\n<p><span id=\"fs-id1165137462220\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140157\/CNX_Precalc_Figure_03_08_202.jpg\" alt=\"Graph of f(x)=4- x^2 and its inverse, f^(-1)(x)= sqrt(4-x).\" \/><\/span><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135547140\">\n<div id=\"fs-id1165135547141\">\n<p id=\"fs-id1165135547142\">[latex]f\\left(x\\right)={\\left(x+3\\right)}^{2},\\,x\\ge -3[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137452181\">\n<div>\n<p id=\"fs-id1165137452184\">[latex]f\\left(x\\right)={\\left(x-4\\right)}^{2},\\,x\\ge 4[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137459688\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137459689\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]<\/p>\n<p><span id=\"fs-id1165137811172\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140207\/CNX_Precalc_Figure_03_08_204.jpg\" alt=\"Graph of f(x)= (x-4)^2 and its inverse, f^(-1)(x)= sqrt(x)+4.\" \/><\/span><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135187190\">\n<div id=\"fs-id1165135187191\">\n<p id=\"fs-id1165135187192\">[latex]f\\left(x\\right)={x}^{3}+3[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135651499\">\n<div id=\"fs-id1165135651500\">\n<p id=\"fs-id1165135651501\">[latex]f\\left(x\\right)=1-{x}^{3}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137462309\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137462310\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{1-x}[\/latex]<\/p>\n<p><span id=\"fs-id1165135694369\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140209\/CNX_Precalc_Figure_03_08_206.jpg\" alt=\"Graph of f(x)= 1-x^3 and its inverse, f^(-1)(x)= (1-x)^(1\/3).\" \/><\/span><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135368483\">\n<div id=\"fs-id1165135368484\">[latex]f\\left(x\\right)={x}^{2}+4x,\\,x\\ge -2[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137454354\">\n<div id=\"fs-id1165137454356\">\n<p id=\"fs-id1165137454357\">[latex]f\\left(x\\right)={x}^{2}-6x+1,\\,x\\ge 3[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137742528\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+8}+3[\/latex]<span id=\"fs-id1165135678741\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140232\/CNX_Precalc_Figure_03_08_208.jpg\" alt=\"Graph of f(x)= x^2-6x+1 and its inverse, f^(-1)(x)= sqrt(x+8)+3.\" \/><\/span><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135206104\">\n<div id=\"fs-id1165135206105\">\n<p id=\"fs-id1165135206106\">[latex]f\\left(x\\right)=\\frac{2}{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137651509\">\n<div id=\"fs-id1165137651510\">\n<p id=\"fs-id1165137651511\">[latex]f\\left(x\\right)=\\frac{1}{{x}^{2}},\\,x\\ge 0[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134557344\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137634193\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{\\frac{1}{x}}[\/latex]<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/c5037767394b50c474c6620703ea24eb45d7a1c8\/CNX_Precalc_Figure_03_08_210.jpg\" alt=\"Graph of f(x)= 1\/x^2 and its inverse, f^(-1)(x)= sqrt(1\/x).\" \/><\/details>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137451122\">For the following exercises, use a graph to help determine the domain of the functions.<\/p>\n<div id=\"fs-id1165137480291\">\n<div id=\"fs-id1165137480292\">\n<p id=\"fs-id1165137480293\">[latex]f\\left(x\\right)=\\sqrt{\\frac{\\left(x+1\\right)\\left(x-1\\right)}{x}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137422286\">\n<div id=\"fs-id1165137422287\">\n<p id=\"fs-id1165137422288\">[latex]f\\left(x\\right)=\\sqrt{\\frac{\\left(x+2\\right)\\left(x-3\\right)}{x-1}}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137407593\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137407594\">[latex]\\left[-2,1\\right)\\cup \\left[3,\\infty \\right)[\/latex]<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/31d07c963aa8e9a0703c8cd298ead16cbe2e1845\/CNX_Precalc_Figure_03_08_212.jpg\" alt=\"Graph of f(x)= sqrt((x+2)(x-3)\/(x-1)).\" \/><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135173418\">\n<div id=\"fs-id1165135173419\">\n<p id=\"fs-id1165137772203\">[latex]f\\left(x\\right)=\\sqrt{\\frac{x\\left(x+3\\right)}{x-4}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137550583\">\n<div id=\"fs-id1165137550584\">\n<p id=\"fs-id1165137550585\">[latex]f\\left(x\\right)=\\sqrt{\\frac{{x}^{2}-x-20}{x-2}}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137758690\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137539722\">[latex]\\left[-4,2\\right)\\cup \\left[5,\\infty \\right)[\/latex]<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/259c69be03f0d36862a3fceb79891f7db338332b\/CNX_Precalc_Figure_03_08_214.jpg\" alt=\"Graph of f(x)= sqrt((x^2-x-20)\/(x-2)).\" \/><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137898053\">\n<div id=\"fs-id1165137898054\">\n<p id=\"fs-id1165137898056\">[latex]f\\left(x\\right)=\\sqrt{\\frac{9-{x}^{2}}{x+4}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137676077\" class=\"bc-section section\">\n<h4>Technology<\/h4>\n<p id=\"fs-id1165137551626\">For the following exercises, use a calculator to graph the function. Then, using the graph, give three points on the graph of the inverse with <em>y<\/em>-coordinates given.<\/p>\n<div id=\"fs-id1165137855040\">\n<div id=\"fs-id1165137575507\">\n<p id=\"fs-id1165137575508\">[latex]f\\left(x\\right)={x}^{3}-x-2,y=1,2,3[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137724899\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137724900\">[latex]\\left(\u20132, 0\\right); \\left(4, 2\\right); \\left(22, 3\\right)[\/latex]<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/4d856c01c1cc568f1f7ff472813acee8aa8776bf\/CNX_Precalc_Figure_03_08_216.jpg\" alt=\"Graph of f(x)= x^3-x-2.\" \/><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137571412\">\n<div id=\"fs-id1165137571413\">\n<p id=\"fs-id1165137571414\">[latex]f\\left(x\\right)={x}^{3}+x-2,y=0,1,2[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135336074\">\n<div id=\"fs-id1165135336075\">\n<p id=\"fs-id1165135336076\">[latex]f\\left(x\\right)={x}^{3}+3x-4,y=0,1,2[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135445799\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135445800\">[latex]\\left(\u20134, 0\\right); \\left(0, 1\\right); \\left(10, 2\\right)[\/latex]<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/5815d06121425c5769bf84de2c8f710222f6adf0\/CNX_Precalc_Figure_03_08_218.jpg\" alt=\"Graph of f(x)= x^3+3x-4.\" \/><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137768493\">\n<div id=\"fs-id1165137414579\">\n<p id=\"fs-id1165137414580\">[latex]f\\left(x\\right)={x}^{3}+8x-4,y=-1,0,1[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137812553\">\n<div id=\"fs-id1165137812554\">\n<p id=\"fs-id1165137812555\">[latex]f\\left(x\\right)={x}^{4}+5x+1,y=-1,0,1[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137413936\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137413937\">[latex]\\left(\u20133, -1\\right); \\left(1, 0\\right); \\left(7, 1\\right)[\/latex]<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cnx.org\/resources\/0d7ef9b0cb654da615c33f4966eed0ef018910c8\/CNX_Precalc_Figure_03_08_220.jpg\" alt=\"Graph of f(x)= x^4+5x+1.\" \/><\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137833145\" class=\"bc-section section\">\n<h4>Extensions<\/h4>\n<p id=\"fs-id1165137427625\">For the following exercises, find the inverse of the functions with[latex]\\,a,b,c\\,[\/latex]positive real numbers.<\/p>\n<div id=\"fs-id1165137601481\">\n<div id=\"fs-id1165137601482\">\n<p id=\"fs-id1165137601483\">[latex]f\\left(x\\right)=a{x}^{3}+b[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165132960779\">\n<div id=\"fs-id1165132960780\">\n<p id=\"fs-id1165137660236\">[latex]f\\left(x\\right)={x}^{2}+bx[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137564896\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137564897\">[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+\\frac{{b}^{2}}{4}}-\\frac{b}{2}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135187946\">\n<div id=\"fs-id1165135187947\">\n<p id=\"fs-id1165137548784\">[latex]f\\left(x\\right)=\\sqrt{a{x}^{2}+b}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137590488\">\n<div id=\"fs-id1165137590489\">\n<p id=\"fs-id1165137590490\">[latex]f\\left(x\\right)=\\sqrt[3]{ax+b}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137934451\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137934452\">[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{3}-b}{a}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137597719\">\n<div id=\"fs-id1165137597720\">\n<p id=\"fs-id1165137597721\">[latex]f\\left(x\\right)=\\frac{ax+b}{x+c}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135347473\" class=\"bc-section section\">\n<h4>Real-World Applications<\/h4>\n<p id=\"fs-id1165137667466\">For the following exercises, determine the function described and then use it to answer the question.<\/p>\n<div id=\"fs-id1165137667470\">\n<div id=\"fs-id1165137543382\">\n<p id=\"fs-id1165137543383\">An object dropped from a height of 200 meters has a height,[latex]\\,h\\left(t\\right),\\,[\/latex]in meters after[latex]\\,t\\,[\/latex]seconds have lapsed, such that[latex]\\,h\\left(t\\right)=200-4.9{t}^{2}.\\,[\/latex]Express[latex]\\,t\\,[\/latex]as a function of height,[latex]\\,h,\\,[\/latex]and find the time to reach a height of 50 meters.<\/p>\n<\/div>\n<div id=\"fs-id1165133112798\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165133112799\">[latex]t\\left(h\\right)=\\sqrt{\\frac{200-h}{4.9}},\\,[\/latex]5.53 seconds<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137735080\">\n<div id=\"fs-id1165137735081\">\n<p id=\"fs-id1165137735082\">An object dropped from a height of 600 feet has a height,[latex]\\,h\\left(t\\right),\\,[\/latex]in feet after[latex]\\,t\\,[\/latex]seconds have elapsed, such that[latex]\\,h\\left(t\\right)=600-16{t}^{2}.\\,[\/latex]Express[latex]\\,t\\,[\/latex]<br \/>\nas a function of height[latex]\\,h,\\,[\/latex]and find the time to reach a height of 400 feet.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137451866\">\n<div id=\"fs-id1165137451867\">\n<p id=\"fs-id1165137451868\">The volume,[latex]\\,V,\\,[\/latex]of a sphere in terms of its radius,[latex]\\,r,\\,[\/latex]is given by[latex]\\,V\\left(r\\right)=\\frac{4}{3}\\pi {r}^{3}.\\,[\/latex]Express[latex]\\,r\\,[\/latex]as a function of[latex]\\,V,\\,[\/latex]and find the radius of a sphere with volume of 200 cubic feet.<\/p>\n<\/div>\n<div id=\"fs-id1165137894368\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137894369\">[latex]r\\left(V\\right)=\\sqrt[3]{\\frac{3V}{4\\pi }},\\,[\/latex]3.63 feet<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137619689\">\n<div id=\"fs-id1165137619690\">\n<p id=\"fs-id1165137446018\">The surface area,[latex]\\,A,\\,[\/latex]of a sphere in terms of its radius,[latex]\\,r,\\,[\/latex]is given by[latex]\\,A\\left(r\\right)=4\\pi {r}^{2}.\\,[\/latex]Express[latex]\\,r\\,[\/latex]as a function of[latex]\\,V,\\,[\/latex]and find the radius of a sphere with a surface area of 1000 square inches.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135199586\">\n<div id=\"fs-id1165137611961\">\n<p id=\"fs-id1165137611962\">A container holds 100 mL of a solution that is 25 mL acid. If[latex]\\,n\\,[\/latex]mL of a solution that is 60% acid is added, the function[latex]\\,C\\left(n\\right)=\\frac{25+.6n}{100+n}\\,[\/latex]gives the concentration,[latex]\\,C,\\,[\/latex]as a function of the number of mL added,[latex]\\,n.\\,[\/latex]Express[latex]\\,n\\,[\/latex]as a function of[latex]\\,C\\,[\/latex]and determine the number of mL that need to be added to have a solution that is 50% acid.<\/p>\n<\/div>\n<div id=\"fs-id1165137677899\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137677900\">[latex]n\\left(C\\right)=\\frac{100C-25}{.6-C},\\,[\/latex]250 mL<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137400716\">\n<div id=\"fs-id1165137400717\">\n<p id=\"fs-id1165137400718\">The period[latex]\\,T,\\,[\/latex]in seconds, of a simple pendulum as a function of its length[latex]\\,l,\\,[\/latex]in feet, is given by[latex]\\,T\\left(l\\right)=2\\pi \\sqrt{\\frac{l}{32.2}}\\,[\/latex] . Express[latex]\\,l\\,[\/latex]as a function of[latex]\\,T\\,[\/latex]and determine the length of a pendulum with period of 2 seconds.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137855335\">\n<div id=\"fs-id1165137855336\">\n<p id=\"fs-id1165137855337\">The volume of a cylinder ,[latex]\\,V,\\,[\/latex]in terms of radius,[latex]\\,r,\\,[\/latex]and height,[latex]\\,h,\\,[\/latex]is given by[latex]\\,V=\\pi {r}^{2}h.\\,[\/latex]If a cylinder has a height of 6 meters, express the radius as a function of[latex]\\,V\\,[\/latex]and find the radius of a cylinder with volume of 300 cubic meters.<\/p>\n<\/div>\n<div id=\"fs-id1165135524559\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135524560\">[latex]r\\left(V\\right)=\\sqrt{\\frac{V}{6\\pi }},\\,[\/latex]3.99 meters<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135543521\">\n<div id=\"fs-id1165135543522\">\n<p id=\"fs-id1165135543524\">The surface area,[latex]\\,A,\\,[\/latex]of a cylinder in terms of its radius,[latex]\\,r,\\,[\/latex]and height,[latex]\\,h,\\,[\/latex]is given by[latex]\\,A=2\\pi {r}^{2}+2\\pi rh.\\,[\/latex]If the height of the cylinder is 4 feet, express the radius as a function of[latex]\\,V\\,[\/latex]and find the radius if the surface area is 200 square feet.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135209080\">\n<div id=\"fs-id1165135209081\">\n<p id=\"fs-id1165135209082\">The volume of a right circular cone,[latex]\\,V,\\,[\/latex]in terms of its radius,[latex]\\,r,\\,[\/latex]and its height,[latex]\\,h,\\,[\/latex]is given by[latex]\\,V=\\frac{1}{3}\\pi {r}^{2}h.\\,[\/latex]Express[latex]\\,r\\,[\/latex]in terms of[latex]\\,V\\,[\/latex]if the height of the cone is 12 feet and find the radius of a cone with volume of 50 cubic inches.<\/p>\n<\/div>\n<div id=\"fs-id1165135389876\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135389877\">[latex]r\\left(V\\right)=\\sqrt{\\frac{V}{4\\pi }},\\,[\/latex]1.99 inches<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137783419\">\n<div id=\"fs-id1165137783420\">\n<p id=\"fs-id1165137783421\">Consider a cone with height of 30 feet. Express the radius,[latex]\\,r,\\,[\/latex]in terms of the volume,[latex]\\,V,\\,[\/latex]and find the radius of a cone with volume of 1000 cubic feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165135169260\">\n<dt>invertible function<\/dt>\n<dd id=\"fs-id1165135169263\">any function that has an inverse function<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":291,"menu_order":8,"template":"","meta":{"pb_show_title":null,"pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-91","chapter","type-chapter","status-publish","hentry"],"part":76,"_links":{"self":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/91","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/users\/291"}],"version-history":[{"count":1,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/91\/revisions"}],"predecessor-version":[{"id":92,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/91\/revisions\/92"}],"part":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/parts\/76"}],"metadata":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/91\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/media?parent=91"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapter-type?post=91"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/contributor?post=91"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/license?post=91"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}