{"id":567,"date":"2025-07-10T19:44:45","date_gmt":"2025-07-10T19:44:45","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=567"},"modified":"2026-01-27T17:55:39","modified_gmt":"2026-01-27T17:55:39","slug":"domain-and-range-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/domain-and-range-learn-it-2\/","title":{"raw":"Domain and Range: Learn It 2","rendered":"Domain and Range: Learn It 2"},"content":{"raw":"<h2>Domain and Range from Function Equations<\/h2>\r\nLet\u2019s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms.\r\n<ul>\r\n \t<li>First, if the function has no denominator or an even root, consider whether the domain could be all real numbers.<\/li>\r\n \t<li>Second, if there is a denominator in the function\u2019s equation, exclude values in the domain that force the denominator to be zero.<\/li>\r\n \t<li>Third, if there is an even root, consider excluding values that would make the radicand negative.<\/li>\r\n<\/ul>\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a function written in equation form, find the domain.<\/strong>\r\n<ol>\r\n \t<li>Identify the input values.<\/li>\r\n \t<li>Identify any restrictions on the input and exclude those values from the domain.<\/li>\r\n \t<li>Write the domain in interval form, if possible.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the domain of the following function:<center>[latex]f\\left(x\\right)={x}^{2}-1[\/latex]<\/center>[reveal-answer q=\"100687\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"100687\"]The input value, shown by the variable [latex]x[\/latex] in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers.In interval form the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].[\/hidden-answer]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]293801[\/ohm_question]<\/section><section aria-label=\"Try It\"><section class=\"textbox recall\" aria-label=\"Recall\">While zero divided by any number equals zero, division by zero results in an undefined ratio.<center>[latex]\\dfrac{0}{a} = 0 \\quad \\text{but } \\quad \\dfrac{b}{0} = \\text{undefined} [\/latex]<\/center>An even root of a negative number does not exist in the real numbers.<center>[latex]\\sqrt{-1} = i[\/latex]<\/center><\/section>Since the domain of any function defined in the real plane is the set of all real input into the function, we must exclude any values of the input variable that create undefined expressions or even roots of a negative.\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To:\u00a0Given a function written in an equation form that includes a fraction, find the domain.<\/strong>\r\n<ol>\r\n \t<li>Identify the input values.<\/li>\r\n \t<li>Identify any restrictions on the input. If there is a denominator in the function\u2019s formula, set the denominator equal to zero and solve for [latex]x[\/latex] . These are the values that cannot be inputs in the function.<\/li>\r\n \t<li>Write the domain in interval form, making sure to exclude any restricted values from the domain.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the domain of the following function:<center>[latex]f\\left(x\\right)=\\dfrac{x+1}{2-x}[\/latex]<\/center>[reveal-answer q=\"759017\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"759017\"]When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to [latex]0[\/latex] and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2-x&amp;=0 \\\\ -x&amp;=-2 \\\\ x&amp;=2 \\end{align}[\/latex]<\/p>\r\nNow, we will exclude [latex]2[\/latex] from the domain. The answers are all real numbers where [latex]x&lt;2[\/latex] or [latex]x&gt;2[\/latex]. We can use a symbol known as the union, [latex]\\cup [\/latex], to combine the two sets. In interval notation, we write the solution: [latex]\\left(\\mathrm{-\\infty },2\\right)\\cup \\left(2,\\infty \\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18193532\/CNX_Precalc_Figure_01_02_028n2.jpg\" alt=\"Line graph of x=!2.\" width=\"487\" height=\"164\" \/>\r\n\r\nIn interval form, the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,2\\right)\\cup \\left(2,\\infty \\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]293803[\/ohm_question]<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a function written in equation form including an even root, find the domain.<\/strong>\r\n<ol>\r\n \t<li>Identify the input values.<\/li>\r\n \t<li>Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for [latex]x[\/latex].<\/li>\r\n \t<li>The solution(s) are the domain of the function. If possible, write the answer in interval form.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the domain of the following function:<center>[latex]f\\left(x\\right)=\\sqrt{7-x}[\/latex]<\/center>[reveal-answer q=\"722013\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"722013\"]When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}7-x&amp;\\ge 0 \\\\ -x&amp;\\ge -7 \\\\ x&amp;\\le 7 \\end{align}[\/latex]<\/p>\r\nNow, we will exclude any number greater than [latex]7[\/latex] from the domain. The answers are all real numbers less than or equal to [latex]7[\/latex], or [latex]\\left(-\\infty ,7\\right][\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]293804[\/ohm_question]<\/section>\r\n<h3 aria-label=\"Try It\">Rational Functions with Even Roots<\/h3>\r\n<section aria-label=\"Try It\"><section class=\"textbox example\" aria-label=\"Example\">Find the domain of [latex]\\frac{\\sqrt{x-2}}{x+1}[\/latex]\r\n\r\n\r\n[reveal-answer q=\"720576\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"720576\"]When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand.\r\n\r\nSet the radicand greater than or equal to zero and solve for [latex]x[\/latex].\r\n\r\nThe domain restriction for the even root is [latex]x-2\\geq 0[\/latex] which is solved as [latex]x\\geq 2[\/latex].\r\n\r\nWhen there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to [latex]0[\/latex] and solve for [latex]x[\/latex].\r\n\r\nThe domain restriction for the denominator is [latex]x+1 \\neq 0[\/latex] which gives us [latex]x\\neq -1[\/latex].\r\n\r\nSince numbers greater than or equal to [latex]2[\/latex] do not include [latex]-1[\/latex] the domain of this function is [latex]x\\geq 2[\/latex] or [latex][2,\\infty)[\/latex] in interval notation.[\/hidden-answer]<\/section><section aria-label=\"Example\"><section class=\"textbox questionHelp\" aria-label=\"Question Help\">To find the domain of a function with both an even root and a denominator:\r\n<ul>\r\n \t<li>Find the restriction for the root<\/li>\r\n \t<li>Find the restriction for the denominator<\/li>\r\n \t<li>Determine where the restrictions overlap and combine them<\/li>\r\n<\/ul>\r\n<\/section><\/section><section aria-label=\"Example\"><section class=\"textbox example\" aria-label=\"Example\">\r\n<p data-start=\"154\" data-end=\"212\">(a) Find the domain of [latex]\\frac{1}{\\sqrt{x^2 - 4}}[\/latex]<\/p>\r\n<p data-start=\"154\" data-end=\"212\">[reveal-answer q=\"486975\"]Show Answer[\/reveal-answer][hidden-answer a=\"486975\"]<\/p>\r\n<p data-start=\"214\" data-end=\"418\"><strong data-start=\"214\" data-end=\"225\">Step 1:<\/strong> The square root tells us that [latex]x^2 - 4 \\geq 0[\/latex]. Which factors to [latex](x - 2)(x + 2) \\geq 0[\/latex] so [latex]x\\geq 2[\/latex] or [latex]x\\leq 2[\/latex]<\/p>\r\n<p data-start=\"568\" data-end=\"733\"><strong data-start=\"568\" data-end=\"579\">Step 2:<\/strong> The domain restriction for the denominator is that [latex]x^2-4\\neq 0[\/latex] which factors to [latex](x-2)(x+2)\\neq 0[\/latex]. So [latex]x\\neq 2[\/latex] and [latex]x\\neq -2[\/latex].<\/p>\r\n<p data-start=\"568\" data-end=\"733\"><strong>Step 3:<\/strong> Combining the restrictions tells us that [latex]x&gt;2[\/latex] or [latex]x&lt;-2[\/latex] which is [latex](-\\infty,-2)\\cup (2,\\infty)[\/latex].<\/p>\r\n<p data-start=\"154\" data-end=\"212\">[\/hidden-answer]<\/p>\r\n<p data-start=\"154\" data-end=\"212\">(b) Find the domain of [latex]\\frac{\\sqrt{x - 1}}{\\sqrt{x^2 - 9}}[\/latex]<\/p>\r\n<p data-start=\"154\" data-end=\"212\">[reveal-answer q=\"870280\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"870280\"]<\/p>\r\n<p data-start=\"831\" data-end=\"969\">The numerator contains a square root, so we require [latex]x - 1 \\geq 0[\/latex] which is solved as [latex]x \\geq 1[\/latex]<\/p>\r\n<p data-start=\"971\" data-end=\"1222\">The denominator also contains a square root, but because it is in the denominator, it also can't equal 0. Therefore we must have [latex]x^2 - 9 &gt; 0[\/latex]<\/p>\r\n<p data-start=\"971\" data-end=\"1222\">Factor and solve:<br data-start=\"1127\" data-end=\"1130\" \/>[latex](x - 3)(x + 3) &gt; 0[\/latex]<br data-start=\"1163\" data-end=\"1166\" \/>This gives [latex]x &lt; -3[\/latex] or [latex]x &gt; 3[\/latex]<\/p>\r\n<p data-start=\"1224\" data-end=\"1382\">To satisfy <em data-start=\"1247\" data-end=\"1253\">both<\/em> conditions, we take the overlap of the intervals [latex]x \\geq 1[\/latex] and [latex]x &gt; 3[\/latex], which is [latex]x &gt; 3[\/latex]<\/p>\r\n<p data-start=\"1384\" data-end=\"1499\">The domain of this function is [latex]x &gt; 3[\/latex], or [latex](3, \\infty)[\/latex] in interval notation<\/p>\r\n<p data-start=\"154\" data-end=\"212\">[\/hidden-answer]<\/p>\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\"><span class=\"katex-html\" aria-hidden=\"true\">[ohm_question hide_question_numbers=1]317469[\/ohm_question]<span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/section><\/section><\/section>","rendered":"<h2>Domain and Range from Function Equations<\/h2>\n<p>Let\u2019s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms.<\/p>\n<ul>\n<li>First, if the function has no denominator or an even root, consider whether the domain could be all real numbers.<\/li>\n<li>Second, if there is a denominator in the function\u2019s equation, exclude values in the domain that force the denominator to be zero.<\/li>\n<li>Third, if there is an even root, consider excluding values that would make the radicand negative.<\/li>\n<\/ul>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a function written in equation form, find the domain.<\/strong><\/p>\n<ol>\n<li>Identify the input values.<\/li>\n<li>Identify any restrictions on the input and exclude those values from the domain.<\/li>\n<li>Write the domain in interval form, if possible.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the domain of the following function:<\/p>\n<div style=\"text-align: center;\">[latex]f\\left(x\\right)={x}^{2}-1[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q100687\">Show Solution<\/button><\/p>\n<div id=\"q100687\" class=\"hidden-answer\" style=\"display: none\">The input value, shown by the variable [latex]x[\/latex] in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers.In interval form the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm293801\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=293801&theme=lumen&iframe_resize_id=ohm293801&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section aria-label=\"Try It\">\n<section class=\"textbox recall\" aria-label=\"Recall\">While zero divided by any number equals zero, division by zero results in an undefined ratio.<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{0}{a} = 0 \\quad \\text{but } \\quad \\dfrac{b}{0} = \\text{undefined}[\/latex]<\/div>\n<p>An even root of a negative number does not exist in the real numbers.<\/p>\n<div style=\"text-align: center;\">[latex]\\sqrt{-1} = i[\/latex]<\/div>\n<\/section>\n<p>Since the domain of any function defined in the real plane is the set of all real input into the function, we must exclude any values of the input variable that create undefined expressions or even roots of a negative.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To:\u00a0Given a function written in an equation form that includes a fraction, find the domain.<\/strong><\/p>\n<ol>\n<li>Identify the input values.<\/li>\n<li>Identify any restrictions on the input. If there is a denominator in the function\u2019s formula, set the denominator equal to zero and solve for [latex]x[\/latex] . These are the values that cannot be inputs in the function.<\/li>\n<li>Write the domain in interval form, making sure to exclude any restricted values from the domain.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the domain of the following function:<\/p>\n<div style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{x+1}{2-x}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q759017\">Show Solution<\/button><\/p>\n<div id=\"q759017\" class=\"hidden-answer\" style=\"display: none\">When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to [latex]0[\/latex] and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2-x&=0 \\\\ -x&=-2 \\\\ x&=2 \\end{align}[\/latex]<\/p>\n<p>Now, we will exclude [latex]2[\/latex] from the domain. The answers are all real numbers where [latex]x<2[\/latex] or [latex]x>2[\/latex]. We can use a symbol known as the union, [latex]\\cup[\/latex], to combine the two sets. In interval notation, we write the solution: [latex]\\left(\\mathrm{-\\infty },2\\right)\\cup \\left(2,\\infty \\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18193532\/CNX_Precalc_Figure_01_02_028n2.jpg\" alt=\"Line graph of x=!2.\" width=\"487\" height=\"164\" \/><\/p>\n<p>In interval form, the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,2\\right)\\cup \\left(2,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm293803\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=293803&theme=lumen&iframe_resize_id=ohm293803&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a function written in equation form including an even root, find the domain.<\/strong><\/p>\n<ol>\n<li>Identify the input values.<\/li>\n<li>Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for [latex]x[\/latex].<\/li>\n<li>The solution(s) are the domain of the function. If possible, write the answer in interval form.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the domain of the following function:<\/p>\n<div style=\"text-align: center;\">[latex]f\\left(x\\right)=\\sqrt{7-x}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q722013\">Show Solution<\/button><\/p>\n<div id=\"q722013\" class=\"hidden-answer\" style=\"display: none\">When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}7-x&\\ge 0 \\\\ -x&\\ge -7 \\\\ x&\\le 7 \\end{align}[\/latex]<\/p>\n<p>Now, we will exclude any number greater than [latex]7[\/latex] from the domain. The answers are all real numbers less than or equal to [latex]7[\/latex], or [latex]\\left(-\\infty ,7\\right][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm293804\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=293804&theme=lumen&iframe_resize_id=ohm293804&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3 aria-label=\"Try It\">Rational Functions with Even Roots<\/h3>\n<section aria-label=\"Try It\">\n<section class=\"textbox example\" aria-label=\"Example\">Find the domain of [latex]\\frac{\\sqrt{x-2}}{x+1}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q720576\">Show Solution<\/button><\/p>\n<div id=\"q720576\" class=\"hidden-answer\" style=\"display: none\">When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand.<\/p>\n<p>Set the radicand greater than or equal to zero and solve for [latex]x[\/latex].<\/p>\n<p>The domain restriction for the even root is [latex]x-2\\geq 0[\/latex] which is solved as [latex]x\\geq 2[\/latex].<\/p>\n<p>When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to [latex]0[\/latex] and solve for [latex]x[\/latex].<\/p>\n<p>The domain restriction for the denominator is [latex]x+1 \\neq 0[\/latex] which gives us [latex]x\\neq -1[\/latex].<\/p>\n<p>Since numbers greater than or equal to [latex]2[\/latex] do not include [latex]-1[\/latex] the domain of this function is [latex]x\\geq 2[\/latex] or [latex][2,\\infty)[\/latex] in interval notation.<\/p><\/div>\n<\/div>\n<\/section>\n<section aria-label=\"Example\">\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">To find the domain of a function with both an even root and a denominator:<\/p>\n<ul>\n<li>Find the restriction for the root<\/li>\n<li>Find the restriction for the denominator<\/li>\n<li>Determine where the restrictions overlap and combine them<\/li>\n<\/ul>\n<\/section>\n<\/section>\n<section aria-label=\"Example\">\n<section class=\"textbox example\" aria-label=\"Example\">\n<p data-start=\"154\" data-end=\"212\">(a) Find the domain of [latex]\\frac{1}{\\sqrt{x^2 - 4}}[\/latex]<\/p>\n<p data-start=\"154\" data-end=\"212\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q486975\">Show Answer<\/button><\/p>\n<div id=\"q486975\" class=\"hidden-answer\" style=\"display: none\">\n<p data-start=\"214\" data-end=\"418\"><strong data-start=\"214\" data-end=\"225\">Step 1:<\/strong> The square root tells us that [latex]x^2 - 4 \\geq 0[\/latex]. Which factors to [latex](x - 2)(x + 2) \\geq 0[\/latex] so [latex]x\\geq 2[\/latex] or [latex]x\\leq 2[\/latex]<\/p>\n<p data-start=\"568\" data-end=\"733\"><strong data-start=\"568\" data-end=\"579\">Step 2:<\/strong> The domain restriction for the denominator is that [latex]x^2-4\\neq 0[\/latex] which factors to [latex](x-2)(x+2)\\neq 0[\/latex]. So [latex]x\\neq 2[\/latex] and [latex]x\\neq -2[\/latex].<\/p>\n<p data-start=\"568\" data-end=\"733\"><strong>Step 3:<\/strong> Combining the restrictions tells us that [latex]x>2[\/latex] or [latex]x<-2[\/latex] which is [latex](-\\infty,-2)\\cup (2,\\infty)[\/latex].<\/p>\n<p data-start=\"154\" data-end=\"212\"><\/div>\n<\/div>\n<p data-start=\"154\" data-end=\"212\">(b) Find the domain of [latex]\\frac{\\sqrt{x - 1}}{\\sqrt{x^2 - 9}}[\/latex]<\/p>\n<p data-start=\"154\" data-end=\"212\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q870280\">Show Answer<\/button><\/p>\n<div id=\"q870280\" class=\"hidden-answer\" style=\"display: none\">\n<p data-start=\"831\" data-end=\"969\">The numerator contains a square root, so we require [latex]x - 1 \\geq 0[\/latex] which is solved as [latex]x \\geq 1[\/latex]<\/p>\n<p data-start=\"971\" data-end=\"1222\">The denominator also contains a square root, but because it is in the denominator, it also can&#8217;t equal 0. Therefore we must have [latex]x^2 - 9 > 0[\/latex]<\/p>\n<p data-start=\"971\" data-end=\"1222\">Factor and solve:<br data-start=\"1127\" data-end=\"1130\" \/>[latex](x - 3)(x + 3) > 0[\/latex]<br data-start=\"1163\" data-end=\"1166\" \/>This gives [latex]x < -3[\/latex] or [latex]x > 3[\/latex]<\/p>\n<p data-start=\"1224\" data-end=\"1382\">To satisfy <em data-start=\"1247\" data-end=\"1253\">both<\/em> conditions, we take the overlap of the intervals [latex]x \\geq 1[\/latex] and [latex]x > 3[\/latex], which is [latex]x > 3[\/latex]<\/p>\n<p data-start=\"1384\" data-end=\"1499\">The domain of this function is [latex]x > 3[\/latex], or [latex](3, \\infty)[\/latex] in interval notation<\/p>\n<p data-start=\"154\" data-end=\"212\"><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><span class=\"katex-html\" aria-hidden=\"true\"><iframe loading=\"lazy\" id=\"ohm317469\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317469&theme=lumen&iframe_resize_id=ohm317469&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/section>\n<\/section>\n<\/section>\n","protected":false},"author":13,"menu_order":14,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":36,"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\/567"}],"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":15,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/567\/revisions"}],"predecessor-version":[{"id":5436,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/567\/revisions\/5436"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/36"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/567\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=567"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=567"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=567"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=567"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}