{"id":2040,"date":"2024-07-02T23:44:57","date_gmt":"2024-07-02T23:44:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=2040"},"modified":"2025-08-15T13:59:59","modified_gmt":"2025-08-15T13:59:59","slug":"radical-functions-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/radical-functions-apply-it-1\/","title":{"raw":"Inverses and Radical Functions: Apply It 1","rendered":"Inverses and Radical Functions: Apply It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Learn how to find the inverse (or \"reverse\") of a polynomial function when it's possible<\/li>\r\n \t<li>Figure out how to limit the domain of a polynomial function so you can find its inverse<\/li>\r\n \t<li>Use radical functions to solve real-world problems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solve Applied Problems Involving Radical Functions<\/h2>\r\nRadical functions are frequently used in real-world applications, especially in physical models where they can describe various natural phenomena. Radical functions often appear in physics, engineering, and biology. They help model situations where relationships between quantities involve roots.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Suppose a water runoff collector is built in the shape of a parabolic trough as shown below. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221704\/CNX_Precalc_Figure_03_08_0022.jpg\" alt=\"Diagram of a parabolic trough that is 18\" width=\"487\" height=\"279\" \/> Diagram of a parabolic trough[\/caption]\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"300\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221706\/CNX_Precalc_Figure_03_08_0032.jpg\" alt=\"Graph of a parabola.\" width=\"300\" height=\"272\" \/> Graph of the parabola on the trough[\/caption]\r\n\r\nBecause it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with [latex]x[\/latex]\u00a0measured horizontally and [latex]y[\/latex]\u00a0measured vertically, with the origin at the vertex of the parabola.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 [latex](6, 18)[\/latex], from which we can solve for the stretch factor [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 18&amp;=a{6}^{2} \\\\[1mm] a&amp;=\\frac{18}{36} \\\\[1mm] a&amp;=\\frac{1}{2} \\end{align}[\/latex]<\/p>\r\nOur parabolic cross section has the equation\r\n<p style=\"text-align: center;\">[latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/p>\r\nWe are interested in the <strong>surface area<\/strong> of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth [latex]y[\/latex]\u00a0the width will be given by [latex]2x[\/latex], so we need to solve the equation above for [latex]x[\/latex]\u00a0and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.\r\n\r\nTo find an inverse, we can restrict our original function to a limited domain on which it <em>is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive [latex]x[\/latex]\u00a0values. On this domain, we can find an inverse by solving for the input variable:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}y&amp;=\\frac{1}{2}{x}^{2} \\\\[1mm] 2y&amp;={x}^{2} \\\\[1mm] x&amp;=\\pm \\sqrt{2y} \\end{align}[\/latex]<\/p>\r\nThis is not a function as written. We are limiting ourselves to positive [latex]x[\/latex]\u00a0values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.\r\n<p style=\"text-align: center;\">[latex]y=\\dfrac{{x}^{2}}{2},\\text{ }x&gt;0[\/latex]<\/p>\r\nBecause [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 [latex]3[\/latex] feet ([latex]36[\/latex] inches) long, so the surface area will then be:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\text{Area} &amp;=l\\cdot w \\\\[1mm] &amp;=36\\cdot 2x \\\\[1mm] &amp;=72x \\\\[1mm] &amp;=72\\sqrt{2y} \\end{align}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">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\r\n<p style=\"text-align: center;\">[latex]V=\\frac{2}{3}\\pi {r}^{3}[\/latex]<\/p>\r\nFind 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 [latex]100 [\/latex]cubic feet. Use [latex]\\pi =3.14[\/latex].\r\n\r\n[reveal-answer q=\"158008\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"158008\"]\r\n\r\nStart with the given function for [latex]V[\/latex]. Notice that the meaningful domain for the function is [latex]r\\ge 0[\/latex] since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is [latex]V\\ge 0[\/latex]. Solve for [latex]r[\/latex]\u00a0in terms of [latex]V[\/latex], using the method outlined previously.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}V&amp;=\\frac{2}{3}\\pi {r}^{3} \\\\[1mm] {r}^{3}&amp;=\\dfrac{3V}{2\\pi } &amp;&amp; \\text{Solve for }{r}^{3}. \\\\[1mm] r&amp;=\\sqrt[3]{\\frac{3V}{2\\pi }} &amp;&amp; \\text{Solve for }r. \\end{align}[\/latex]<\/p>\r\nThis is the result stated in the section opener. Now evaluate this for [latex]V=100[\/latex]\u00a0and [latex]\\pi =3.14[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}r&amp;=\\sqrt[3]{\\dfrac{3V}{2\\pi }} \\\\[1mm] &amp;=\\sqrt[3]{\\dfrac{3\\cdot 100}{2\\cdot 3.14}} \\\\[1mm] &amp;\\approx \\sqrt[3]{47.7707} \\\\ &amp; \\approx 3.63 \\end{align}[\/latex]<\/p>\r\nTherefore, the radius is about [latex]3.63[\/latex] ft.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24674[\/ohm2_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24675[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Learn how to find the inverse (or &#8220;reverse&#8221;) of a polynomial function when it&#8217;s possible<\/li>\n<li>Figure out how to limit the domain of a polynomial function so you can find its inverse<\/li>\n<li>Use radical functions to solve real-world problems<\/li>\n<\/ul>\n<\/section>\n<h2>Solve Applied Problems Involving Radical Functions<\/h2>\n<p>Radical functions are frequently used in real-world applications, especially in physical models where they can describe various natural phenomena. Radical functions often appear in physics, engineering, and biology. They help model situations where relationships between quantities involve roots.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Suppose a water runoff collector is built in the shape of a parabolic trough as shown below. 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:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221704\/CNX_Precalc_Figure_03_08_0022.jpg\" alt=\"Diagram of a parabolic trough that is 18\" width=\"487\" height=\"279\" \/><figcaption class=\"wp-caption-text\">Diagram of a parabolic trough<\/figcaption><\/figure>\n<figure style=\"width: 300px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02221706\/CNX_Precalc_Figure_03_08_0032.jpg\" alt=\"Graph of a parabola.\" width=\"300\" height=\"272\" \/><figcaption class=\"wp-caption-text\">Graph of the parabola on the trough<\/figcaption><\/figure>\n<p>Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with [latex]x[\/latex]\u00a0measured horizontally and [latex]y[\/latex]\u00a0measured vertically, with the origin at the vertex of the parabola.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 [latex](6, 18)[\/latex], from which we can solve for the stretch factor [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 18&=a{6}^{2} \\\\[1mm] a&=\\frac{18}{36} \\\\[1mm] a&=\\frac{1}{2} \\end{align}[\/latex]<\/p>\n<p>Our parabolic cross section has the equation<\/p>\n<p style=\"text-align: center;\">[latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/p>\n<p>We are interested in the <strong>surface area<\/strong> of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth [latex]y[\/latex]\u00a0the width will be given by [latex]2x[\/latex], so we need to solve the equation above for [latex]x[\/latex]\u00a0and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.<\/p>\n<p>To find an inverse, we can restrict our original function to a limited domain on which it <em>is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive [latex]x[\/latex]\u00a0values. On this domain, we can find an inverse by solving for the input variable:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}y&=\\frac{1}{2}{x}^{2} \\\\[1mm] 2y&={x}^{2} \\\\[1mm] x&=\\pm \\sqrt{2y} \\end{align}[\/latex]<\/p>\n<p>This is not a function as written. We are limiting ourselves to positive [latex]x[\/latex]\u00a0values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.<\/p>\n<p style=\"text-align: center;\">[latex]y=\\dfrac{{x}^{2}}{2},\\text{ }x>0[\/latex]<\/p>\n<p>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 [latex]3[\/latex] feet ([latex]36[\/latex] inches) long, so the surface area will then be:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\text{Area} &=l\\cdot w \\\\[1mm] &=36\\cdot 2x \\\\[1mm] &=72x \\\\[1mm] &=72\\sqrt{2y} \\end{align}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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<p style=\"text-align: center;\">[latex]V=\\frac{2}{3}\\pi {r}^{3}[\/latex]<\/p>\n<p>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 [latex]100[\/latex]cubic feet. Use [latex]\\pi =3.14[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q158008\">Show Solution<\/button><\/p>\n<div id=\"q158008\" class=\"hidden-answer\" style=\"display: none\">\n<p>Start with the given function for [latex]V[\/latex]. Notice that the meaningful domain for the function is [latex]r\\ge 0[\/latex] since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is [latex]V\\ge 0[\/latex]. Solve for [latex]r[\/latex]\u00a0in terms of [latex]V[\/latex], using the method outlined previously.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}V&=\\frac{2}{3}\\pi {r}^{3} \\\\[1mm] {r}^{3}&=\\dfrac{3V}{2\\pi } && \\text{Solve for }{r}^{3}. \\\\[1mm] r&=\\sqrt[3]{\\frac{3V}{2\\pi }} && \\text{Solve for }r. \\end{align}[\/latex]<\/p>\n<p>This is the result stated in the section opener. Now evaluate this for [latex]V=100[\/latex]\u00a0and [latex]\\pi =3.14[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}r&=\\sqrt[3]{\\dfrac{3V}{2\\pi }} \\\\[1mm] &=\\sqrt[3]{\\dfrac{3\\cdot 100}{2\\cdot 3.14}} \\\\[1mm] &\\approx \\sqrt[3]{47.7707} \\\\ & \\approx 3.63 \\end{align}[\/latex]<\/p>\n<p>Therefore, the radius is about [latex]3.63[\/latex] ft.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24674\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24674&theme=lumen&iframe_resize_id=ohm24674&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24675\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24675&theme=lumen&iframe_resize_id=ohm24675&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":15,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":232,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2040"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":11,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2040\/revisions"}],"predecessor-version":[{"id":7797,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2040\/revisions\/7797"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/232"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2040\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=2040"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2040"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=2040"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=2040"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}