{"id":85,"date":"2025-02-13T22:43:35","date_gmt":"2025-02-13T22:43:35","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/inverse-functions-2\/"},"modified":"2026-01-13T20:59:12","modified_gmt":"2026-01-13T20:59:12","slug":"inverse-functions-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/inverse-functions-2\/","title":{"raw":"Polynomial Equations and Inequalities: Learn It 1","rendered":"Polynomial Equations and Inequalities: Learn It 1"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\"><section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Solve polynomial inequalities.<\/li>\r\n \t<li>Find the inverse of invertible polynomial functions<\/li>\r\n \t<li>Restrict the domain to find the inverse of a polynomial function<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solving Polynomial Inequalities<\/h2>\r\nOne application of our ability to find intercepts and sketch a graph of polynomials is the ability to solve polynomial inequalities. It is a very common question to ask when a function will be positive and negative. We can solve polynomial inequalities by either utilizing the graph, or by using test values.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Solve [latex]\\left(x+3\\right){\\left(x+1\\right)}^{2}\\left(x-4\\right)&gt; 0[\/latex][reveal-answer q=\"494744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"494744\"]As with all inequalities, we start by solving the equality [latex]\\left(x+3\\right){\\left(x+1\\right)}^{2}\\left(x-4\\right)= 0[\/latex], which has solutions at [latex]x = -3, -1[\/latex], and [latex]4[\/latex]. We know the function can only change from positive to negative at these values, so these divide the inputs into four intervals.\r\nWe could choose a test value in each interval and evaluate the function [latex]f\\left(x\\right) = \\left(x+3\\right){\\left(x+1\\right)}^{2}\\left(x-4\\right)[\/latex] at each test value to determine if the function is positive or negative in that interval\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Interval<\/td>\r\n<td>Test x in interval<\/td>\r\n<td>f(test value)<\/td>\r\n<td>&gt; 0 or &lt; 0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>x &lt; -3<\/td>\r\n<td>-4<\/td>\r\n<td>72<\/td>\r\n<td>&gt; 0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>-3 &lt; x &lt; -1<\/td>\r\n<td>-2<\/td>\r\n<td>-6<\/td>\r\n<td>&lt; 0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>-1 &lt;\u00a0 x &lt; 4<\/td>\r\n<td>0<\/td>\r\n<td>-12<\/td>\r\n<td>&lt; 0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>x &gt; 4<\/td>\r\n<td>5<\/td>\r\n<td>288<\/td>\r\n<td>\u00a0&gt; 0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nOn a number line this would look like:\r\n\r\n<img class=\"aligncenter wp-image-13403 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/139\/2016\/04\/28183410\/1.png\" alt=\"Number line with values from -6 to 6 double headed arrows from -6 to -3 read positive, from -3 to -1 read negative, from -1 to positive 4 read negative and from 4 to 6 read positive.\" width=\"630\" height=\"110\" \/>\r\nFrom our test values, we can determine this function is positive when <em>x<\/em> &lt; -3 or <em>x<\/em> &gt; 4, or in interval notation, [latex]\\left(-\\infty, -3\\right)\\cup\\left(4,\\infty\\right)[\/latex]. We could have also determined on which intervals the function was positive by sketching a graph of the function. We illustrate that technique in the next example.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>We can also use polynomial inequalities to find the domain of more complex radical functions.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the domain of the function [latex]v\\left(t\\right)=\\sqrt{6-5t-{t}^{2}}[\/latex][reveal-answer q=\"359527\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"359527\"]A square root is only defined when the quantity we are taking the square root of, the quantity inside the square root, is zero or greater. Thus, the domain of this function will be when [latex]6 - 5t - {t}^{2}\\ge 0[\/latex]. Again we start by solving the equality [latex]6 - 5t - {t}^{2}= 0[\/latex]. While we could use the quadratic formula, this equation factors nicely to [latex]\\left(6 + t\\right)\\left(1-t\\right)=0[\/latex], giving horizontal intercepts\r\nt = 1 and t = -6.\r\nSketching a graph of this quadratic will allow us to determine when it is positive.<img class=\"aligncenter size-full wp-image-13404\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/139\/2016\/04\/28183442\/Screen-Shot2.png\" alt=\"Graph of upside down parabola on cartesian coordinate axes passing through (-6,0) and (1,0)\" width=\"278\" height=\"204\" \/>\r\nFrom the graph we can see this function is positive for inputs between the intercepts. So [latex]6 - 5t - {t}^{2}\\ge 0[\/latex] is positive for [latex]-6 \\le t\\le 1[\/latex], and this will be the domain of the v(t) function.[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the inequality [latex]{x}^{4} - 2{x}^{3} - 3{x}^{2} \\gt 0[\/latex][reveal-answer q=\"273617\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"273617\"]In our other examples, we were given polynomials that were already in factored form, here we have an additional step to finding the intervals on which solutions to the given inequality lie. Again, we will start by solving the equality [latex]{x}^{4} - 2{x}^{3} - 3{x}^{2} = 0[\/latex]\r\n<p style=\"text-align: left;\">Notice that there is a common factor of [latex]{x}^{2}[\/latex] in each term of this polynomial. We can use factoring to simplify in the following way:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{x}^{4} - 2{x}^{3} - 3{x}^{2} &amp;= 0&amp;\\\\{x}^{2}\\left({x}^{2} - 2{x} - 3\\right) &amp;= 0\\\\ {x}^{2}\\left(x - 3\\right)\\left(x + 1 \\right)&amp;= 0\\end{align}[\/latex]<\/p>\r\nNow we can set each factor equal to zero to find the solution to the equality.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc} {x}^{2} = 0 &amp; \\left(x - 3\\right) = 0 &amp;\\left(x+1\\right) = 0\\\\ {x} = 0 &amp; x = 3 &amp; x = -1\\\\ \\end{array}[\/latex].<\/p>\r\nNote that x = 0 has multiplicity of two, but since our inequality is strictly greater than, we don't need to include it in our solutions.\r\nWe can choose a test value in each interval and evaluate the function\r\n<p style=\"text-align: center;\">[latex]{x}^{4} - 2{x}^{3} - 3{x}^{2} = 0[\/latex]<\/p>\r\n<p style=\"text-align: left;\">at each test value to determine if the function is positive or negative in that interval<\/p>\r\n\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Interval<\/td>\r\n<td>Test x in interval<\/td>\r\n<td>&gt; 0,\u00a0 &lt; 0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>x &lt; -1<\/td>\r\n<td>-2<\/td>\r\n<td>x &gt; 0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>-1 &lt; x &lt; 0<\/td>\r\n<td>-1\/2<\/td>\r\n<td>\u00a0x &lt;\u00a0 0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0 &lt; x &lt; 3<\/td>\r\n<td>1<\/td>\r\n<td>x &lt; 0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>x &gt; 3<\/td>\r\n<td>5<\/td>\r\n<td>x &gt; 0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe want to have the set of x values that will give us the intervals where the polynomial is greater than zero. Our answer will be [latex]\\left(-\\infty, -1\\right]\\cup\\left[3,\\infty\\right)[\/latex].\r\n\r\n&nbsp;\r\n\r\nThe graph of the function gives us additional confirmation of our solution.\r\n\r\n<img class=\"aligncenter wp-image-13406 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/139\/2016\/04\/28183505\/Screen-Shot3.png\" alt=\"Line dips down, dips slightly up, dips very far down, then sharply goes up\" width=\"302\" height=\"445\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question]318904[\/ohm_question]<\/section><\/div>\r\n<dl id=\"fs-id1165135169260\" class=\"definition\">\r\n \t<dd id=\"fs-id1165135169263\"><\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Solve polynomial inequalities.<\/li>\n<li>Find the inverse of invertible polynomial functions<\/li>\n<li>Restrict the domain to find the inverse of a polynomial function<\/li>\n<\/ul>\n<\/section>\n<h2>Solving Polynomial Inequalities<\/h2>\n<p>One application of our ability to find intercepts and sketch a graph of polynomials is the ability to solve polynomial inequalities. It is a very common question to ask when a function will be positive and negative. We can solve polynomial inequalities by either utilizing the graph, or by using test values.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Solve [latex]\\left(x+3\\right){\\left(x+1\\right)}^{2}\\left(x-4\\right)> 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q494744\">Show Solution<\/button><\/p>\n<div id=\"q494744\" class=\"hidden-answer\" style=\"display: none\">As with all inequalities, we start by solving the equality [latex]\\left(x+3\\right){\\left(x+1\\right)}^{2}\\left(x-4\\right)= 0[\/latex], which has solutions at [latex]x = -3, -1[\/latex], and [latex]4[\/latex]. We know the function can only change from positive to negative at these values, so these divide the inputs into four intervals.<br \/>\nWe could choose a test value in each interval and evaluate the function [latex]f\\left(x\\right) = \\left(x+3\\right){\\left(x+1\\right)}^{2}\\left(x-4\\right)[\/latex] at each test value to determine if the function is positive or negative in that interval<\/p>\n<table>\n<tbody>\n<tr>\n<td>Interval<\/td>\n<td>Test x in interval<\/td>\n<td>f(test value)<\/td>\n<td>&gt; 0 or &lt; 0<\/td>\n<\/tr>\n<tr>\n<td>x &lt; -3<\/td>\n<td>-4<\/td>\n<td>72<\/td>\n<td>&gt; 0<\/td>\n<\/tr>\n<tr>\n<td>-3 &lt; x &lt; -1<\/td>\n<td>-2<\/td>\n<td>-6<\/td>\n<td>&lt; 0<\/td>\n<\/tr>\n<tr>\n<td>-1 &lt;\u00a0 x &lt; 4<\/td>\n<td>0<\/td>\n<td>-12<\/td>\n<td>&lt; 0<\/td>\n<\/tr>\n<tr>\n<td>x &gt; 4<\/td>\n<td>5<\/td>\n<td>288<\/td>\n<td>\u00a0&gt; 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>On a number line this would look like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-13403 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/139\/2016\/04\/28183410\/1.png\" alt=\"Number line with values from -6 to 6 double headed arrows from -6 to -3 read positive, from -3 to -1 read negative, from -1 to positive 4 read negative and from 4 to 6 read positive.\" width=\"630\" height=\"110\" \/><br \/>\nFrom our test values, we can determine this function is positive when <em>x<\/em> &lt; -3 or <em>x<\/em> &gt; 4, or in interval notation, [latex]\\left(-\\infty, -3\\right)\\cup\\left(4,\\infty\\right)[\/latex]. We could have also determined on which intervals the function was positive by sketching a graph of the function. We illustrate that technique in the next example.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>We can also use polynomial inequalities to find the domain of more complex radical functions.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Find the domain of the function [latex]v\\left(t\\right)=\\sqrt{6-5t-{t}^{2}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q359527\">Show Solution<\/button><\/p>\n<div id=\"q359527\" class=\"hidden-answer\" style=\"display: none\">A square root is only defined when the quantity we are taking the square root of, the quantity inside the square root, is zero or greater. Thus, the domain of this function will be when [latex]6 - 5t - {t}^{2}\\ge 0[\/latex]. Again we start by solving the equality [latex]6 - 5t - {t}^{2}= 0[\/latex]. While we could use the quadratic formula, this equation factors nicely to [latex]\\left(6 + t\\right)\\left(1-t\\right)=0[\/latex], giving horizontal intercepts<br \/>\nt = 1 and t = -6.<br \/>\nSketching a graph of this quadratic will allow us to determine when it is positive.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-13404\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/139\/2016\/04\/28183442\/Screen-Shot2.png\" alt=\"Graph of upside down parabola on cartesian coordinate axes passing through (-6,0) and (1,0)\" width=\"278\" height=\"204\" \/><br \/>\nFrom the graph we can see this function is positive for inputs between the intercepts. So [latex]6 - 5t - {t}^{2}\\ge 0[\/latex] is positive for [latex]-6 \\le t\\le 1[\/latex], and this will be the domain of the v(t) function.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the inequality [latex]{x}^{4} - 2{x}^{3} - 3{x}^{2} \\gt 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q273617\">Show Solution<\/button><\/p>\n<div id=\"q273617\" class=\"hidden-answer\" style=\"display: none\">In our other examples, we were given polynomials that were already in factored form, here we have an additional step to finding the intervals on which solutions to the given inequality lie. Again, we will start by solving the equality [latex]{x}^{4} - 2{x}^{3} - 3{x}^{2} = 0[\/latex]<\/p>\n<p style=\"text-align: left;\">Notice that there is a common factor of [latex]{x}^{2}[\/latex] in each term of this polynomial. We can use factoring to simplify in the following way:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{x}^{4} - 2{x}^{3} - 3{x}^{2} &= 0&\\\\{x}^{2}\\left({x}^{2} - 2{x} - 3\\right) &= 0\\\\ {x}^{2}\\left(x - 3\\right)\\left(x + 1 \\right)&= 0\\end{align}[\/latex]<\/p>\n<p>Now we can set each factor equal to zero to find the solution to the equality.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc} {x}^{2} = 0 & \\left(x - 3\\right) = 0 &\\left(x+1\\right) = 0\\\\ {x} = 0 & x = 3 & x = -1\\\\ \\end{array}[\/latex].<\/p>\n<p>Note that x = 0 has multiplicity of two, but since our inequality is strictly greater than, we don&#8217;t need to include it in our solutions.<br \/>\nWe can choose a test value in each interval and evaluate the function<\/p>\n<p style=\"text-align: center;\">[latex]{x}^{4} - 2{x}^{3} - 3{x}^{2} = 0[\/latex]<\/p>\n<p style=\"text-align: left;\">at each test value to determine if the function is positive or negative in that interval<\/p>\n<table>\n<tbody>\n<tr>\n<td>Interval<\/td>\n<td>Test x in interval<\/td>\n<td>&gt; 0,\u00a0 &lt; 0<\/td>\n<\/tr>\n<tr>\n<td>x &lt; -1<\/td>\n<td>-2<\/td>\n<td>x &gt; 0<\/td>\n<\/tr>\n<tr>\n<td>-1 &lt; x &lt; 0<\/td>\n<td>-1\/2<\/td>\n<td>\u00a0x &lt;\u00a0 0<\/td>\n<\/tr>\n<tr>\n<td>0 &lt; x &lt; 3<\/td>\n<td>1<\/td>\n<td>x &lt; 0<\/td>\n<\/tr>\n<tr>\n<td>x &gt; 3<\/td>\n<td>5<\/td>\n<td>x &gt; 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We want to have the set of x values that will give us the intervals where the polynomial is greater than zero. Our answer will be [latex]\\left(-\\infty, -1\\right]\\cup\\left[3,\\infty\\right)[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p>The graph of the function gives us additional confirmation of our solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-13406 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/139\/2016\/04\/28183505\/Screen-Shot3.png\" alt=\"Line dips down, dips slightly up, dips very far down, then sharply goes up\" width=\"302\" height=\"445\" \/><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm318904\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318904&theme=lumen&iframe_resize_id=ohm318904&source=tnh&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<dl id=\"fs-id1165135169260\" class=\"definition\">\n<dd id=\"fs-id1165135169263\"><\/dd>\n<\/dl>\n","protected":false},"author":6,"menu_order":23,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":506,"module-header":"learn_it","content_attributions":[{"type":"cc-attribution","description":"Precalculus","author":"OpenStax College","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface","project":"","license":"cc-by","license_terms":""}],"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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/85"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/85\/revisions"}],"predecessor-version":[{"id":5340,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/85\/revisions\/5340"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/506"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/85\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=85"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=85"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=85"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=85"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}