{"id":1043,"date":"2025-07-21T23:26:15","date_gmt":"2025-07-21T23:26:15","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1043"},"modified":"2026-01-14T19:24:56","modified_gmt":"2026-01-14T19:24:56","slug":"applications-of-rational-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/applications-of-rational-functions-learn-it-3\/","title":{"raw":"Applications of Rational Functions: Learn It 3","rendered":"Applications of Rational Functions: Learn It 3"},"content":{"raw":"<h2 data-start=\"189\" data-end=\"221\">Solving a Rational Inequality<\/h2>\r\n<p data-start=\"223\" data-end=\"472\">A rational inequality is similar to a rational equation\u2014it contains one or more rational expressions\u2014but instead of an equal sign, it includes an inequality symbol like [latex]&lt;[\/latex], [latex]\\leq[\/latex], [latex]&gt;[\/latex], or [latex]\\geq[\/latex].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>rational inequality<\/h3>\r\nA rational inequality is an inequality that contains at least one rational expression where the variable appears in a denominator.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the rational inequality:[latex]\\dfrac{x - 4}{x + 2} &gt; 0[\/latex][reveal-answer q=\"56901\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"56901\"]Step 1: Identify critical values The numerator is zero when [latex]x = 4[\/latex] The denominator is zero when [latex]x = -2[\/latex] (excluded from the solution) These values divide the number line into intervals:\r\n<ul>\r\n \t<li>Interval 1: [latex]x &lt; -2[\/latex]<\/li>\r\n \t<li>Interval 2: [latex]-2 &lt; x &lt; 4[\/latex]<\/li>\r\n \t<li>Interval 3: [latex]x &gt; 4[\/latex]<\/li>\r\n<\/ul>\r\nStep 2: Test a point from each interval\r\n<table style=\"border-collapse: collapse; width: 100%; height: 66px;\">\r\n<tbody>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"width: 30.7176%; height: 22px;\">Interval<\/td>\r\n<td style=\"width: 26.0898%; height: 22px;\">Test Point<\/td>\r\n<td style=\"width: 43.1925%; height: 22px;\">Sign of [latex]\\frac{x - 4}{x + 2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"width: 30.7176%; height: 22px;\">\u00a0[latex]x &lt; -2[\/latex]<\/td>\r\n<td style=\"width: 26.0898%; height: 22px;\">\u00a0[latex]-3[\/latex]<\/td>\r\n<td style=\"width: 43.1925%; height: 22px;\">[latex]\\frac{-7}{-1} = 7[\/latex] \u2192 positive<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"width: 30.7176%; height: 22px;\">[latex]-2 &lt; x &lt; 4[\/latex]<\/td>\r\n<td style=\"width: 26.0898%; height: 22px;\">[latex]0[\/latex]<\/td>\r\n<td style=\"width: 43.1925%; height: 22px;\">[latex]\\frac{-4}{2} = -2[\/latex] \u2192 negative<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 30.7176%;\">[latex]x &gt; 4[\/latex]<\/td>\r\n<td style=\"width: 26.0898%;\">\u00a0[latex]5[\/latex]<\/td>\r\n<td style=\"width: 43.1925%;\">[latex]\\frac{1}{7}[\/latex] \u2192 positive<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe are looking for where the expression is greater than 0, so we want the intervals where the expression is positive.\r\n\r\nStep 3: Write the solution The solution is: [latex](-\\infty, -2) \\cup (4, \\infty)[\/latex]\r\n\r\nNote: We use open intervals because the expression is strictly greater than 0 [latex]x = -2[\/latex] makes the expression undefined [latex]x = 4[\/latex] makes the numerator zero, which gives 0\u2014not greater than 0[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>To solve a rational inequality:<\/strong>\r\n<ol>\r\n \t<li>Write the inequality so that one side is zero and the other side is a single rational expression<\/li>\r\n \t<li>Identify critical values (zeros of the numerator and undefined points from the denominator)<\/li>\r\n \t<li>Use a sign chart to test the sign of the expression in each interval<\/li>\r\n \t<li>Determine the interval(s) where the expression satisfies the inequality<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p data-start=\"2752\" data-end=\"2806\"><strong>Common Mistakes When Solving Rational Inequalities<\/strong><\/p>\r\n\r\n<ul data-start=\"2808\" data-end=\"3381\">\r\n \t<li data-start=\"2808\" data-end=\"2927\">\r\n<p data-start=\"2810\" data-end=\"2927\">Including excluded values in the solution:\r\nAlways identify and exclude values that make the denominator zero.<\/p>\r\n<\/li>\r\n \t<li data-start=\"2928\" data-end=\"3058\">\r\n<p data-start=\"2930\" data-end=\"3058\">Not setting one side to zero before testing signs:\r\nThe sign chart method only works when the inequality is compared to zero.<\/p>\r\n<\/li>\r\n \t<li data-start=\"3059\" data-end=\"3381\">\r\n<p data-start=\"3061\" data-end=\"3381\">Forgetting open vs. closed intervals:\r\nIf the inequality is strict ([latex]&lt;[\/latex] or [latex]&gt;[\/latex]), use open intervals for all values. If it includes equality ([latex]\\leq[\/latex] or [latex]\\geq[\/latex]), to include values that make the numerator zero (but still exclude points that make the denominator zero).<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p data-start=\"3921\" data-end=\"3955\">Solve the rational inequality:<\/p>\r\n<p data-start=\"3957\" data-end=\"3999\">[latex]\\dfrac{x + 3}{x - 5} \\leq 0[\/latex]<\/p>\r\n[reveal-answer q=\"218664\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"218664\"]\r\n\r\nStep 1: Identify critical values\r\n\r\nNumerator: [latex]x + 3 = 0 \\Rightarrow x = -3[\/latex]\r\n\r\nDenominator: [latex]x - 5 = 0 \\Rightarrow x = 5[\/latex] (excluded)\r\n\r\nIntervals to test:\r\n<ul>\r\n \t<li>[latex](-\\infty, -3)[\/latex]<\/li>\r\n \t<li>[latex](-3, 5)[\/latex]<\/li>\r\n \t<li>[latex](5, \\infty)[\/latex]<\/li>\r\n<\/ul>\r\nStep 2: Sign test\r\n<table style=\"border-collapse: collapse; width: 100%; height: 66px;\">\r\n<tbody>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"width: 30.7176%; height: 22px;\">Interval<\/td>\r\n<td style=\"width: 26.0898%; height: 22px;\">Test Point<\/td>\r\n<td style=\"width: 43.1925%; height: 22px;\">Sign of [latex]\\frac{x +3}{x -5}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"width: 30.7176%; height: 22px;\">\u00a0[latex]x &lt; -3[\/latex]<\/td>\r\n<td style=\"width: 26.0898%; height: 22px;\">\u00a0[latex]-4[\/latex]<\/td>\r\n<td style=\"width: 43.1925%; height: 22px;\">[latex]\\frac{-1}{-9}[\/latex] \u2192 positive<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"width: 30.7176%; height: 22px;\">[latex]-3 &lt; x &lt; 5*[\/latex]<\/td>\r\n<td style=\"width: 26.0898%; height: 22px;\">[latex]0[\/latex]<\/td>\r\n<td style=\"width: 43.1925%; height: 22px;\">[latex]\\frac{3}{-5} [\/latex] \u2192 negative<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 30.7176%;\">[latex]x &gt; 5*[\/latex]<\/td>\r\n<td style=\"width: 26.0898%;\">\u00a0[latex]6[\/latex]<\/td>\r\n<td style=\"width: 43.1925%;\">[latex]\\frac{9}{1}[\/latex] \u2192 positive<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n*This is an asymptote, the rational function is undefined so we won't use [ ] for this value.\r\n\r\nWe are looking for [latex]\\leq 0[\/latex] so we want the intervals where the expression is negative.\r\n\r\nStep 3: Solution [latex][-3, 5)[\/latex][\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]317827[\/ohm_question]<\/section><section aria-label=\"Try It\"><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]317828[\/ohm_question]<\/section><\/section><section aria-label=\"Try It\"><\/section>","rendered":"<h2 data-start=\"189\" data-end=\"221\">Solving a Rational Inequality<\/h2>\n<p data-start=\"223\" data-end=\"472\">A rational inequality is similar to a rational equation\u2014it contains one or more rational expressions\u2014but instead of an equal sign, it includes an inequality symbol like [latex]<[\/latex], [latex]\\leq[\/latex], [latex]>[\/latex], or [latex]\\geq[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>rational inequality<\/h3>\n<p>A rational inequality is an inequality that contains at least one rational expression where the variable appears in a denominator.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the rational inequality:[latex]\\dfrac{x - 4}{x + 2} > 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q56901\">Show Answer<\/button><\/p>\n<div id=\"q56901\" class=\"hidden-answer\" style=\"display: none\">Step 1: Identify critical values The numerator is zero when [latex]x = 4[\/latex] The denominator is zero when [latex]x = -2[\/latex] (excluded from the solution) These values divide the number line into intervals:<\/p>\n<ul>\n<li>Interval 1: [latex]x < -2[\/latex]<\/li>\n<li>Interval 2: [latex]-2 < x < 4[\/latex]<\/li>\n<li>Interval 3: [latex]x > 4[\/latex]<\/li>\n<\/ul>\n<p>Step 2: Test a point from each interval<\/p>\n<table style=\"border-collapse: collapse; width: 100%; height: 66px;\">\n<tbody>\n<tr style=\"height: 22px;\">\n<td style=\"width: 30.7176%; height: 22px;\">Interval<\/td>\n<td style=\"width: 26.0898%; height: 22px;\">Test Point<\/td>\n<td style=\"width: 43.1925%; height: 22px;\">Sign of [latex]\\frac{x - 4}{x + 2}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"width: 30.7176%; height: 22px;\">\u00a0[latex]x < -2[\/latex]<\/td>\n<td style=\"width: 26.0898%; height: 22px;\">\u00a0[latex]-3[\/latex]<\/td>\n<td style=\"width: 43.1925%; height: 22px;\">[latex]\\frac{-7}{-1} = 7[\/latex] \u2192 positive<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"width: 30.7176%; height: 22px;\">[latex]-2 < x < 4[\/latex]<\/td>\n<td style=\"width: 26.0898%; height: 22px;\">[latex]0[\/latex]<\/td>\n<td style=\"width: 43.1925%; height: 22px;\">[latex]\\frac{-4}{2} = -2[\/latex] \u2192 negative<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 30.7176%;\">[latex]x > 4[\/latex]<\/td>\n<td style=\"width: 26.0898%;\">\u00a0[latex]5[\/latex]<\/td>\n<td style=\"width: 43.1925%;\">[latex]\\frac{1}{7}[\/latex] \u2192 positive<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We are looking for where the expression is greater than 0, so we want the intervals where the expression is positive.<\/p>\n<p>Step 3: Write the solution The solution is: [latex](-\\infty, -2) \\cup (4, \\infty)[\/latex]<\/p>\n<p>Note: We use open intervals because the expression is strictly greater than 0 [latex]x = -2[\/latex] makes the expression undefined [latex]x = 4[\/latex] makes the numerator zero, which gives 0\u2014not greater than 0<\/p><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>To solve a rational inequality:<\/strong><\/p>\n<ol>\n<li>Write the inequality so that one side is zero and the other side is a single rational expression<\/li>\n<li>Identify critical values (zeros of the numerator and undefined points from the denominator)<\/li>\n<li>Use a sign chart to test the sign of the expression in each interval<\/li>\n<li>Determine the interval(s) where the expression satisfies the inequality<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p data-start=\"2752\" data-end=\"2806\"><strong>Common Mistakes When Solving Rational Inequalities<\/strong><\/p>\n<ul data-start=\"2808\" data-end=\"3381\">\n<li data-start=\"2808\" data-end=\"2927\">\n<p data-start=\"2810\" data-end=\"2927\">Including excluded values in the solution:<br \/>\nAlways identify and exclude values that make the denominator zero.<\/p>\n<\/li>\n<li data-start=\"2928\" data-end=\"3058\">\n<p data-start=\"2930\" data-end=\"3058\">Not setting one side to zero before testing signs:<br \/>\nThe sign chart method only works when the inequality is compared to zero.<\/p>\n<\/li>\n<li data-start=\"3059\" data-end=\"3381\">\n<p data-start=\"3061\" data-end=\"3381\">Forgetting open vs. closed intervals:<br \/>\nIf the inequality is strict ([latex]<[\/latex] or [latex]>[\/latex]), use open intervals for all values. If it includes equality ([latex]\\leq[\/latex] or [latex]\\geq[\/latex]), to include values that make the numerator zero (but still exclude points that make the denominator zero).<\/p>\n<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p data-start=\"3921\" data-end=\"3955\">Solve the rational inequality:<\/p>\n<p data-start=\"3957\" data-end=\"3999\">[latex]\\dfrac{x + 3}{x - 5} \\leq 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q218664\">Show Answer<\/button><\/p>\n<div id=\"q218664\" class=\"hidden-answer\" style=\"display: none\">\n<p>Step 1: Identify critical values<\/p>\n<p>Numerator: [latex]x + 3 = 0 \\Rightarrow x = -3[\/latex]<\/p>\n<p>Denominator: [latex]x - 5 = 0 \\Rightarrow x = 5[\/latex] (excluded)<\/p>\n<p>Intervals to test:<\/p>\n<ul>\n<li>[latex](-\\infty, -3)[\/latex]<\/li>\n<li>[latex](-3, 5)[\/latex]<\/li>\n<li>[latex](5, \\infty)[\/latex]<\/li>\n<\/ul>\n<p>Step 2: Sign test<\/p>\n<table style=\"border-collapse: collapse; width: 100%; height: 66px;\">\n<tbody>\n<tr style=\"height: 22px;\">\n<td style=\"width: 30.7176%; height: 22px;\">Interval<\/td>\n<td style=\"width: 26.0898%; height: 22px;\">Test Point<\/td>\n<td style=\"width: 43.1925%; height: 22px;\">Sign of [latex]\\frac{x +3}{x -5}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"width: 30.7176%; height: 22px;\">\u00a0[latex]x < -3[\/latex]<\/td>\n<td style=\"width: 26.0898%; height: 22px;\">\u00a0[latex]-4[\/latex]<\/td>\n<td style=\"width: 43.1925%; height: 22px;\">[latex]\\frac{-1}{-9}[\/latex] \u2192 positive<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"width: 30.7176%; height: 22px;\">[latex]-3 < x < 5*[\/latex]<\/td>\n<td style=\"width: 26.0898%; height: 22px;\">[latex]0[\/latex]<\/td>\n<td style=\"width: 43.1925%; height: 22px;\">[latex]\\frac{3}{-5}[\/latex] \u2192 negative<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 30.7176%;\">[latex]x > 5*[\/latex]<\/td>\n<td style=\"width: 26.0898%;\">\u00a0[latex]6[\/latex]<\/td>\n<td style=\"width: 43.1925%;\">[latex]\\frac{9}{1}[\/latex] \u2192 positive<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>*This is an asymptote, the rational function is undefined so we won&#8217;t use [ ] for this value.<\/p>\n<p>We are looking for [latex]\\leq 0[\/latex] so we want the intervals where the expression is negative.<\/p>\n<p>Step 3: Solution [latex][-3, 5)[\/latex]<\/p><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm317827\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317827&theme=lumen&iframe_resize_id=ohm317827&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section aria-label=\"Try It\">\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm317828\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317828&theme=lumen&iframe_resize_id=ohm317828&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/section>\n<section aria-label=\"Try It\"><\/section>\n","protected":false},"author":13,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":508,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1043"}],"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\/13"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1043\/revisions"}],"predecessor-version":[{"id":5360,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1043\/revisions\/5360"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/508"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1043\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1043"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1043"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1043"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1043"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}