{"id":1418,"date":"2025-07-25T01:11:38","date_gmt":"2025-07-25T01:11:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1418"},"modified":"2026-03-11T09:02:23","modified_gmt":"2026-03-11T09:02:23","slug":"applications-of-rational-functions-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/applications-of-rational-functions-fresh-take\/","title":{"raw":"Applications of Rational Functions: Fresh Take","rendered":"Applications of Rational Functions: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Solve applied problems involving rational functions.<\/li>\r\n \t<li>Solve rational inequalities<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solving Applied Problems Involving Rational Functions<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea<\/strong>\r\n\r\nRational functions show up in real-world situations involving rates, concentrations, and ratios. A <strong>rational function<\/strong> is simply one polynomial divided by another\u2014it's a fraction where both the top and bottom are polynomials.\r\n\r\nCommon applications include:\r\n<ul>\r\n \t<li>Mixture problems (concentration of solutions)<\/li>\r\n \t<li>Rate problems (speed, work rates)<\/li>\r\n \t<li>Average cost in business<\/li>\r\n \t<li>Population density<\/li>\r\n<\/ul>\r\nThe key is setting up the ratio correctly and understanding what happens as time passes or quantities change.\r\n\r\n<\/div>\r\n<h3>Understanding Asymptotes in Context<\/h3>\r\n<div class=\"textbox shaded\"><strong>The Main Idea\r\n<\/strong>\r\nAsymptotes tell us about the long-term behavior of rational functions\u2014what happens as values get very large or approach restricted values.<strong>Horizontal Asymptote:<\/strong> A horizontal line [latex]y = c[\/latex] that shows what value the function approaches as [latex]x[\/latex] gets very large (positively or negatively). This represents the long-run behavior or end behavior.<strong>Vertical Asymptote:<\/strong> A vertical line [latex]x = a[\/latex] where the function is undefined (the denominator equals zero). The graph \"bends around\" this line, shooting up or down toward infinity.<\/div>\r\n<div class=\"textbox recall\"><strong>Recall:<\/strong> To find a horizontal asymptote, look at the leading terms of the numerator and denominator. If both have the same degree, divide their coefficients. For vertical asymptotes, set the denominator equal to zero and solve.<\/div>\r\n<section class=\"textbox example\">A large mixing tank currently contains [latex]100[\/latex] gallons of water into which [latex]5[\/latex] pounds of sugar have been mixed. A tap opens, pouring [latex]10[\/latex] gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of [latex]1[\/latex] pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after [latex]12[\/latex] minutes. Is that a greater concentration than at the beginning?\r\n[reveal-answer q=\"rat1\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"rat1\"]\r\nLet [latex]t[\/latex] = number of minutes since the tap opened.\r\n<strong>\r\nSet up equations for water and sugar:<\/strong>\r\n<ul>\r\n \t<li>Water: [latex]W(t) = 100 + 10t[\/latex] gallons<\/li>\r\n \t<li>Sugar: [latex]S(t) = 5 + t[\/latex] pounds<\/li>\r\n<\/ul>\r\n<strong>Concentration is the ratio:<\/strong>\r\n[latex]C(t) = \\frac{S(t)}{W(t)} = \\frac{5 + t}{100 + 10t}[\/latex]\r\n<strong>\r\nAfter 12 minutes:<\/strong>\r\n[latex]\\begin{align}\r\nC(12) &amp;= \\frac{5 + 12}{100 + 10(12)} \\\r\n&amp;= \\frac{17}{220} \\\r\n&amp;\\approx 0.077 \\text{ pounds per gallon}\r\n\\end{align}[\/latex]\r\n<strong>\r\nAt the beginning:<\/strong>\r\n[latex]\\begin{align}\r\nC(0) &amp;= \\frac{5 + 0}{100 + 10(0)} \\\r\n&amp;= \\frac{5}{100} \\\r\n&amp;= \\frac{1}{20} = 0.05 \\text{ pounds per gallon}\r\n\\end{align}[\/latex]\r\nSince [latex]0.077 &gt; 0.05[\/latex], the concentration is greater after 12 minutes.\r\n<strong>\r\nLong-term behavior:<\/strong>\r\nThe horizontal asymptote is [latex]y = \\frac{1}{10} = 0.1[\/latex] (divide leading coefficients). This means the concentration will approach [latex]0.1[\/latex] pounds per gallon in the long run.\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<div class=\"textbox proTip\">When working with mixture problems, always identify what's being added and at what rate. The concentration is always amount of substance divided by total volume.<\/div>\r\n<div><section class=\"textbox watchIt\" aria-label=\"Watch It\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api \"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-feecgdda-h0bPyfXK6FY\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/h0bPyfXK6FY?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-feecgdda-h0bPyfXK6FY\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=14660223&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-feecgdda-h0bPyfXK6FY&vembed=0&video_id=h0bPyfXK6FY&video_target=tpm-plugin-feecgdda-h0bPyfXK6FY'><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Applications+of+Rational+Equations+I_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cApplications of Rational Equations I\u201d here (opens in new window).<\/a>\r\n\r\n<\/section><\/div>\r\n<h2>Solving Rational Inequalities<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea<\/strong>\r\n\r\nA rational inequality is like a rational equation, but instead of an equal sign, it uses an inequality symbol ([latex]&lt;[\/latex], [latex]\\leq[\/latex], [latex]&gt;[\/latex], or [latex]\\geq[\/latex]). These inequalities ask: \"For which values of [latex]x[\/latex] is this rational expression positive (or negative)?\"\r\n\r\nThe key difference from polynomial inequalities: we must be extra careful about values that make the denominator zero\u2014these create vertical asymptotes and are always excluded from solutions.\r\n\r\n<\/div>\r\n<div class=\"textbox questionHelp\">\r\n\r\n<strong>Solving Rational Inequalities<\/strong>\r\n<ol>\r\n \t<li>Rewrite the inequality so one side is zero and the other side is a single rational expression.<\/li>\r\n \t<li>Identify <strong>critical values<\/strong>:\r\n<ul>\r\n \t<li>Zeros of the numerator (where the expression equals zero)<\/li>\r\n \t<li>Zeros of the denominator (where the expression is undefined\u2014these are ALWAYS excluded)<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Plot critical values on a number line to create intervals.<\/li>\r\n \t<li>Test a value from each interval to determine if the expression is positive or negative there.<\/li>\r\n \t<li>Select intervals that satisfy the inequality.<\/li>\r\n \t<li>Write your answer in interval notation:\r\n<ul>\r\n \t<li>Use parentheses [latex]( )[\/latex] for excluded values<\/li>\r\n \t<li>Use brackets [latex][ ][\/latex] only for numerator zeros when the inequality includes \"or equal to\"<\/li>\r\n \t<li>NEVER use brackets for denominator zeros\u2014they're always excluded<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api \"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-fcdgfhaf-UhZJiPSmYps\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/UhZJiPSmYps?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-fcdgfhaf-UhZJiPSmYps\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=14660224&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-fcdgfhaf-UhZJiPSmYps&vembed=0&video_id=UhZJiPSmYps&video_target=tpm-plugin-fcdgfhaf-UhZJiPSmYps'><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Easy+Steps+to+Solving+a+Rational+Inequality+and+WHY+it+works_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEasy Steps to Solving a Rational Inequality and WHY it works\u201d here (opens in new window).<\/a>\r\n\r\n<\/section><section class=\"textbox example\">Solve [latex]\\frac{x - 4}{x + 2} &gt; 0[\/latex]\r\n[reveal-answer q=\"rat2\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"rat2\"]\r\n<strong>\r\nIdentify critical values<\/strong>\r\n<ul>\r\n \t<li>Numerator zero: [latex]x - 4 = 0 \\Rightarrow x = 4[\/latex]<\/li>\r\n \t<li>Denominator zero: [latex]x + 2 = 0 \\Rightarrow x = -2[\/latex] (excluded!)<\/li>\r\n<\/ul>\r\nThese divide the number line into three intervals:\r\n<ul>\r\n \t<li>[latex]x &lt; -2[\/latex]<\/li>\r\n \t<li>[latex]-2 &lt; x &lt; 4[\/latex]<\/li>\r\n \t<li>[latex]x &gt; 4[\/latex]<\/li>\r\n<\/ul>\r\n<strong>\r\nTest each interval<\/strong>\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]\\frac{x - 4}{x + 2}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]x &lt; -2[\/latex]<\/td>\r\n<td>[latex]x = -3[\/latex]<\/td>\r\n<td>[latex]\\frac{-7}{-1} = 7[\/latex] \u2192 positive \u2713<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2 &lt; x &lt; 4[\/latex]<\/td>\r\n<td>[latex]x = 0[\/latex]<\/td>\r\n<td>[latex]\\frac{-4}{2} = -2[\/latex] \u2192 negative<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x &gt; 4[\/latex]<\/td>\r\n<td>[latex]x = 5[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{7}[\/latex] \u2192 positive \u2713<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<strong>\r\nWrite the solution<\/strong>\r\n\r\nWe want where the expression is greater than 0 (positive), so we select the intervals where our tests were positive.\r\nSolution: [latex](-\\infty, -2) \\cup (4, \\infty)[\/latex]\r\nNote: We use parentheses at [latex]-2[\/latex] because it makes the denominator zero (undefined), and at [latex]4[\/latex] because the inequality is strictly greater than zero (not \"or equal to\").\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<div class=\"textbox recall\">Denominator zeros are ALWAYS excluded from the solution\u2014they make the expression undefined. Even if the inequality includes \"or equal to,\" you cannot include values that make the denominator zero.<\/div>\r\n<section class=\"textbox example\">Solve [latex]\\frac{x + 3}{x - 5} \\leq 0[\/latex]\r\n[reveal-answer q=\"rat3\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"rat3\"]\r\n<strong>Identify critical values<\/strong>\r\n<ul>\r\n \t<li>Numerator: [latex]x + 3 = 0 \\Rightarrow x = -3[\/latex]<\/li>\r\n \t<li>Denominator: [latex]x - 5 = 0 \\Rightarrow x = 5[\/latex] (excluded)<\/li>\r\n<\/ul>\r\n<strong>Test intervals<\/strong>\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]\\frac{x + 3}{x - 5}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]x &lt; -3[\/latex]<\/td>\r\n<td>[latex]x = -4[\/latex]<\/td>\r\n<td>[latex]\\frac{-1}{-9}[\/latex] \u2192 positive<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3 &lt; x &lt; 5[\/latex]<\/td>\r\n<td>[latex]x = 0[\/latex]<\/td>\r\n<td>[latex]\\frac{3}{-5}[\/latex] \u2192 negative \u2713<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x &gt; 5[\/latex]<\/td>\r\n<td>[latex]x = 6[\/latex]<\/td>\r\n<td>[latex]\\frac{9}{1}[\/latex] \u2192 positive<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<strong>Write the solution<\/strong>\r\nWe want [latex]\\leq 0[\/latex], which means negative or zero. The expression is negative on [latex](-3, 5)[\/latex] and equals zero at [latex]x = -3[\/latex].\r\nSolution: [latex][-3, 5)[\/latex]\r\n\r\nNote: We use a bracket at [latex]-3[\/latex] because the inequality includes \"or equal to\" and [latex]-3[\/latex] makes the numerator zero (which gives 0). We use a parenthesis at [latex]5[\/latex] because it makes the denominator zero (always excluded).\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<div class=\"textbox proTip\">Create a sign chart! Draw a number line, mark your critical values, test each interval, and label each region as positive or negative. This visual representation makes it much easier to see which intervals satisfy your inequality.<\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Solve applied problems involving rational functions.<\/li>\n<li>Solve rational inequalities<\/li>\n<\/ul>\n<\/section>\n<h2>Solving Applied Problems Involving Rational Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<\/strong><\/p>\n<p>Rational functions show up in real-world situations involving rates, concentrations, and ratios. A <strong>rational function<\/strong> is simply one polynomial divided by another\u2014it&#8217;s a fraction where both the top and bottom are polynomials.<\/p>\n<p>Common applications include:<\/p>\n<ul>\n<li>Mixture problems (concentration of solutions)<\/li>\n<li>Rate problems (speed, work rates)<\/li>\n<li>Average cost in business<\/li>\n<li>Population density<\/li>\n<\/ul>\n<p>The key is setting up the ratio correctly and understanding what happens as time passes or quantities change.<\/p>\n<\/div>\n<h3>Understanding Asymptotes in Context<\/h3>\n<div class=\"textbox shaded\"><strong>The Main Idea<br \/>\n<\/strong><br \/>\nAsymptotes tell us about the long-term behavior of rational functions\u2014what happens as values get very large or approach restricted values.<strong>Horizontal Asymptote:<\/strong> A horizontal line [latex]y = c[\/latex] that shows what value the function approaches as [latex]x[\/latex] gets very large (positively or negatively). This represents the long-run behavior or end behavior.<strong>Vertical Asymptote:<\/strong> A vertical line [latex]x = a[\/latex] where the function is undefined (the denominator equals zero). The graph &#8220;bends around&#8221; this line, shooting up or down toward infinity.<\/div>\n<div class=\"textbox recall\"><strong>Recall:<\/strong> To find a horizontal asymptote, look at the leading terms of the numerator and denominator. If both have the same degree, divide their coefficients. For vertical asymptotes, set the denominator equal to zero and solve.<\/div>\n<section class=\"textbox example\">A large mixing tank currently contains [latex]100[\/latex] gallons of water into which [latex]5[\/latex] pounds of sugar have been mixed. A tap opens, pouring [latex]10[\/latex] gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of [latex]1[\/latex] pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after [latex]12[\/latex] minutes. Is that a greater concentration than at the beginning?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qrat1\">Show Solution<\/button><\/p>\n<div id=\"qrat1\" class=\"hidden-answer\" style=\"display: none\">\nLet [latex]t[\/latex] = number of minutes since the tap opened.<br \/>\n<strong><br \/>\nSet up equations for water and sugar:<\/strong><\/p>\n<ul>\n<li>Water: [latex]W(t) = 100 + 10t[\/latex] gallons<\/li>\n<li>Sugar: [latex]S(t) = 5 + t[\/latex] pounds<\/li>\n<\/ul>\n<p><strong>Concentration is the ratio:<\/strong><br \/>\n[latex]C(t) = \\frac{S(t)}{W(t)} = \\frac{5 + t}{100 + 10t}[\/latex]<br \/>\n<strong><br \/>\nAfter 12 minutes:<\/strong><br \/>\n[latex]\\begin{align}  C(12) &= \\frac{5 + 12}{100 + 10(12)} \\  &= \\frac{17}{220} \\  &\\approx 0.077 \\text{ pounds per gallon}  \\end{align}[\/latex]<br \/>\n<strong><br \/>\nAt the beginning:<\/strong><br \/>\n[latex]\\begin{align}  C(0) &= \\frac{5 + 0}{100 + 10(0)} \\  &= \\frac{5}{100} \\  &= \\frac{1}{20} = 0.05 \\text{ pounds per gallon}  \\end{align}[\/latex]<br \/>\nSince [latex]0.077 > 0.05[\/latex], the concentration is greater after 12 minutes.<br \/>\n<strong><br \/>\nLong-term behavior:<\/strong><br \/>\nThe horizontal asymptote is [latex]y = \\frac{1}{10} = 0.1[\/latex] (divide leading coefficients). This means the concentration will approach [latex]0.1[\/latex] pounds per gallon in the long run.\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox proTip\">When working with mixture problems, always identify what&#8217;s being added and at what rate. The concentration is always amount of substance divided by total volume.<\/div>\n<div>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-feecgdda-h0bPyfXK6FY\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/h0bPyfXK6FY?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-feecgdda-h0bPyfXK6FY\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script type=\"text\/javascript\" src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&#38;cc_minimizable=1&#38;cc_minimize_on_load=0&#38;cc_multi_text_track=0&#38;cc_overlay=1&#38;cc_searchable=0&#38;embed=ajax&#38;mf=14660223&#38;p3sdk_version=1.11.7&#38;p=20361&#38;player_type=youtube&#38;plugin_skin=dark&#38;target=3p-plugin-target-feecgdda-h0bPyfXK6FY&#38;vembed=0&#38;video_id=h0bPyfXK6FY&#38;video_target=tpm-plugin-feecgdda-h0bPyfXK6FY\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Applications+of+Rational+Equations+I_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cApplications of Rational Equations I\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<\/div>\n<h2>Solving Rational Inequalities<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<\/strong><\/p>\n<p>A rational inequality is like a rational equation, but instead of an equal sign, it uses an inequality symbol ([latex]<[\/latex], [latex]\\leq[\/latex], [latex]>[\/latex], or [latex]\\geq[\/latex]). These inequalities ask: &#8220;For which values of [latex]x[\/latex] is this rational expression positive (or negative)?&#8221;<\/p>\n<p>The key difference from polynomial inequalities: we must be extra careful about values that make the denominator zero\u2014these create vertical asymptotes and are always excluded from solutions.<\/p>\n<\/div>\n<div class=\"textbox questionHelp\">\n<p><strong>Solving Rational Inequalities<\/strong><\/p>\n<ol>\n<li>Rewrite the inequality so one side is zero and the other side is a single rational expression.<\/li>\n<li>Identify <strong>critical values<\/strong>:\n<ul>\n<li>Zeros of the numerator (where the expression equals zero)<\/li>\n<li>Zeros of the denominator (where the expression is undefined\u2014these are ALWAYS excluded)<\/li>\n<\/ul>\n<\/li>\n<li>Plot critical values on a number line to create intervals.<\/li>\n<li>Test a value from each interval to determine if the expression is positive or negative there.<\/li>\n<li>Select intervals that satisfy the inequality.<\/li>\n<li>Write your answer in interval notation:\n<ul>\n<li>Use parentheses [latex]( )[\/latex] for excluded values<\/li>\n<li>Use brackets [latex][ ][\/latex] only for numerator zeros when the inequality includes &#8220;or equal to&#8221;<\/li>\n<li>NEVER use brackets for denominator zeros\u2014they&#8217;re always excluded<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-fcdgfhaf-UhZJiPSmYps\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/UhZJiPSmYps?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-fcdgfhaf-UhZJiPSmYps\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script type=\"text\/javascript\" src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&#38;cc_minimizable=1&#38;cc_minimize_on_load=0&#38;cc_multi_text_track=0&#38;cc_overlay=1&#38;cc_searchable=0&#38;embed=ajax&#38;mf=14660224&#38;p3sdk_version=1.11.7&#38;p=20361&#38;player_type=youtube&#38;plugin_skin=dark&#38;target=3p-plugin-target-fcdgfhaf-UhZJiPSmYps&#38;vembed=0&#38;video_id=UhZJiPSmYps&#38;video_target=tpm-plugin-fcdgfhaf-UhZJiPSmYps\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Easy+Steps+to+Solving+a+Rational+Inequality+and+WHY+it+works_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEasy Steps to Solving a Rational Inequality and WHY it works\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<section class=\"textbox example\">Solve [latex]\\frac{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=\"qrat2\">Show Solution<\/button><\/p>\n<div id=\"qrat2\" class=\"hidden-answer\" style=\"display: none\">\n<strong><br \/>\nIdentify critical values<\/strong><\/p>\n<ul>\n<li>Numerator zero: [latex]x - 4 = 0 \\Rightarrow x = 4[\/latex]<\/li>\n<li>Denominator zero: [latex]x + 2 = 0 \\Rightarrow x = -2[\/latex] (excluded!)<\/li>\n<\/ul>\n<p>These divide the number line into three intervals:<\/p>\n<ul>\n<li>[latex]x < -2[\/latex]<\/li>\n<li>[latex]-2 < x < 4[\/latex]<\/li>\n<li>[latex]x > 4[\/latex]<\/li>\n<\/ul>\n<p><strong><br \/>\nTest each interval<\/strong><\/p>\n<table>\n<thead>\n<tr>\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]\\frac{x - 4}{x + 2}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]x < -2[\/latex]<\/td>\n<td>[latex]x = -3[\/latex]<\/td>\n<td>[latex]\\frac{-7}{-1} = 7[\/latex] \u2192 positive \u2713<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2 < x < 4[\/latex]<\/td>\n<td>[latex]x = 0[\/latex]<\/td>\n<td>[latex]\\frac{-4}{2} = -2[\/latex] \u2192 negative<\/td>\n<\/tr>\n<tr>\n<td>[latex]x > 4[\/latex]<\/td>\n<td>[latex]x = 5[\/latex]<\/td>\n<td>[latex]\\frac{1}{7}[\/latex] \u2192 positive \u2713<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong><br \/>\nWrite the solution<\/strong><\/p>\n<p>We want where the expression is greater than 0 (positive), so we select the intervals where our tests were positive.<br \/>\nSolution: [latex](-\\infty, -2) \\cup (4, \\infty)[\/latex]<br \/>\nNote: We use parentheses at [latex]-2[\/latex] because it makes the denominator zero (undefined), and at [latex]4[\/latex] because the inequality is strictly greater than zero (not &#8220;or equal to&#8221;).\n<\/p><\/div>\n<\/div>\n<\/section>\n<div class=\"textbox recall\">Denominator zeros are ALWAYS excluded from the solution\u2014they make the expression undefined. Even if the inequality includes &#8220;or equal to,&#8221; you cannot include values that make the denominator zero.<\/div>\n<section class=\"textbox example\">Solve [latex]\\frac{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=\"qrat3\">Show Solution<\/button><\/p>\n<div id=\"qrat3\" class=\"hidden-answer\" style=\"display: none\">\n<strong>Identify critical values<\/strong><\/p>\n<ul>\n<li>Numerator: [latex]x + 3 = 0 \\Rightarrow x = -3[\/latex]<\/li>\n<li>Denominator: [latex]x - 5 = 0 \\Rightarrow x = 5[\/latex] (excluded)<\/li>\n<\/ul>\n<p><strong>Test intervals<\/strong><\/p>\n<table>\n<thead>\n<tr>\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]\\frac{x + 3}{x - 5}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]x < -3[\/latex]<\/td>\n<td>[latex]x = -4[\/latex]<\/td>\n<td>[latex]\\frac{-1}{-9}[\/latex] \u2192 positive<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3 < x < 5[\/latex]<\/td>\n<td>[latex]x = 0[\/latex]<\/td>\n<td>[latex]\\frac{3}{-5}[\/latex] \u2192 negative \u2713<\/td>\n<\/tr>\n<tr>\n<td>[latex]x > 5[\/latex]<\/td>\n<td>[latex]x = 6[\/latex]<\/td>\n<td>[latex]\\frac{9}{1}[\/latex] \u2192 positive<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Write the solution<\/strong><br \/>\nWe want [latex]\\leq 0[\/latex], which means negative or zero. The expression is negative on [latex](-3, 5)[\/latex] and equals zero at [latex]x = -3[\/latex].<br \/>\nSolution: [latex][-3, 5)[\/latex]<\/p>\n<p>Note: We use a bracket at [latex]-3[\/latex] because the inequality includes &#8220;or equal to&#8221; and [latex]-3[\/latex] makes the numerator zero (which gives 0). We use a parenthesis at [latex]5[\/latex] because it makes the denominator zero (always excluded).\n<\/p><\/div>\n<\/div>\n<\/section>\n<div class=\"textbox proTip\">Create a sign chart! Draw a number line, mark your critical values, test each interval, and label each region as positive or negative. This visual representation makes it much easier to see which intervals satisfy your inequality.<\/div>\n","protected":false},"author":67,"menu_order":19,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Applications of Rational Equations I\",\"author\":\"\",\"organization\":\"Mathispower4u\",\"url\":\"https:\/\/youtu.be\/h0bPyfXK6FY\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"copyrighted_video\",\"description\":\"Easy Steps to Solving a Rational Inequality and WHY it works\",\"author\":\"\",\"organization\":\"Midnight Math Tutor\",\"url\":\"https:\/\/youtu.be\/UhZJiPSmYps\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":508,"module-header":"fresh_take","content_attributions":[{"type":"copyrighted_video","description":"Applications of Rational Equations I","author":"","organization":"Mathispower4u","url":"https:\/\/youtu.be\/h0bPyfXK6FY","project":"","license":"arr","license_terms":"Standard YouTube License"},{"type":"copyrighted_video","description":"Easy Steps to Solving a Rational Inequality and WHY it works","author":"","organization":"Midnight Math Tutor","url":"https:\/\/youtu.be\/UhZJiPSmYps","project":"","license":"arr","license_terms":"Standard YouTube License"}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"<script type='text\/javascript' src='https:\/\/www.youtube.com\/iframe_api'><\/script><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=14660223&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-feecgdda-h0bPyfXK6FY&vembed=0&video_id=h0bPyfXK6FY&video_target=tpm-plugin-feecgdda-h0bPyfXK6FY'><\/script>\n<script type='text\/javascript' src='https:\/\/www.youtube.com\/iframe_api'><\/script><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=14660224&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-fcdgfhaf-UhZJiPSmYps&vembed=0&video_id=UhZJiPSmYps&video_target=tpm-plugin-fcdgfhaf-UhZJiPSmYps'><\/script>\n","media_targets":["tpm-plugin-feecgdda-h0bPyfXK6FY","tpm-plugin-fcdgfhaf-UhZJiPSmYps"]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1418"}],"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\/67"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1418\/revisions"}],"predecessor-version":[{"id":5755,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1418\/revisions\/5755"}],"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\/1418\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1418"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1418"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1418"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1418"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}