{"id":625,"date":"2025-07-14T17:57:34","date_gmt":"2025-07-14T17:57:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=625"},"modified":"2026-01-29T18:41:13","modified_gmt":"2026-01-29T18:41:13","slug":"composition-of-functions-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/composition-of-functions-learn-it-4\/","title":{"raw":"Composition of Functions: Learn It 5","rendered":"Composition of Functions: Learn It 5"},"content":{"raw":"<h2>Finding the Domain of a Composite Function<\/h2>\r\n<p id=\"fs-id1165135519324\">The <strong>domain of a composite function<\/strong> such as [latex]f\\circ g[\/latex] is dependent on the domain of [latex]g[\/latex] and the domain of [latex]f[\/latex]. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as [latex]f\\circ g[\/latex]. Let us assume we know the domains of the functions [latex]f[\/latex] and [latex]g[\/latex] separately. If we write the composite function for an input [latex]x[\/latex] as [latex]f\\left(g\\left(x\\right)\\right)[\/latex], we can see right away that [latex]x[\/latex] must be a member of the domain of [latex]g[\/latex] in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that [latex]g\\left(x\\right)[\/latex] must be a member of the domain of [latex]f[\/latex], otherwise the second function evaluation in [latex]f\\left(g\\left(x\\right)\\right)[\/latex] cannot be completed, and the expression is still undefined. Thus the domain of [latex]f\\circ g[\/latex] consists of only those inputs in the domain of [latex]g[\/latex] that produce outputs from [latex]g[\/latex] belonging to the domain of [latex]f[\/latex]. Note that the domain of [latex]f[\/latex] composed with [latex]g[\/latex] is the set of all [latex]x[\/latex] such that [latex]x[\/latex] is in the domain of [latex]g[\/latex] and [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f[\/latex].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>domain of a composite function<\/h3>\r\nThe domain of a composite function [latex]f\\left(g\\left(x\\right)\\right)[\/latex] is the set of those inputs [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f[\/latex].\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p id=\"fs-id1165135203267\"><strong>How To: Given a function composition [latex]f\\left(g\\left(x\\right)\\right)[\/latex], determine its domain.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137714200\">\r\n \t<li>Find the domain of [latex]g[\/latex].<\/li>\r\n \t<li>Find the domain of [latex]f[\/latex].<\/li>\r\n \t<li>Find those inputs, [latex]x[\/latex], in the domain of [latex]g[\/latex] for which [latex]g(x)[\/latex] is in the domain of [latex]f[\/latex]. That is, exclude those inputs, [latex]x[\/latex], from the domain of [latex]g[\/latex] for which [latex]g(x)[\/latex] is not in the domain of [latex]f[\/latex]. The resulting set is the domain of [latex]f\\circ g[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the domain of\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\frac{5}{x - 1}\\text{ and }g\\left(x\\right)=\\frac{4}{3x - 2}[\/latex]<\/p>\r\n[reveal-answer q=\"262512\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"262512\"]\r\n<p id=\"fs-id1165137627922\">The domain of [latex]g\\left(x\\right)[\/latex] consists of all real numbers except [latex]x=\\frac{2}{3}[\/latex], since that input value would cause us to divide by 0. Likewise, the domain of [latex]f[\/latex] consists of all real numbers except 1. So we need to exclude from the domain of [latex]g\\left(x\\right)[\/latex] that value of [latex]x[\/latex] for which [latex]g\\left(x\\right)=1[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{4}{3x - 2}&amp;=1 \\\\ 4&amp;=3x - 2 &amp;&amp;\\text{Multiply by } 3x-2 \\\\ 6&amp;=3x &amp;&amp;\\text{Add }2 \\\\ x&amp;=2 &amp;&amp; \\text{Divide by }3 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137891240\">So the domain of [latex]f\\circ g[\/latex] is the set of all real numbers except [latex]\\frac{2}{3}[\/latex] and [latex]2[\/latex]. This means that<\/p>\r\n<p style=\"text-align: center;\">[latex]x\\ne \\frac{2}{3}\\text{or}x\\ne 2[\/latex]<\/p>\r\n<p id=\"fs-id1165135152098\">We can write this in interval notation as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(-\\infty ,\\frac{2}{3}\\right)\\cup \\left(\\frac{2}{3},2\\right)\\cup \\left(2,\\infty \\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1165135547426\">Find the domain of<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\sqrt{x+2}\\text{ and }g\\left(x\\right)=\\sqrt{3-x}[\/latex]<\/p>\r\n[reveal-answer q=\"959136\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"959136\"]\r\n<p id=\"fs-id1165137812353\">Because we cannot take the square root of a negative number, the domain of [latex]g[\/latex] is [latex]\\left(-\\infty ,3\\right][\/latex]. Now we check the domain of the composite function<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{3-x+2}\\text{ or }\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{5-x}[\/latex]<\/p>\r\n<p id=\"fs-id1165137871761\">The domain of this function is [latex]\\left(-\\infty ,5\\right][\/latex]. To find the domain of [latex]f\\circ g[\/latex], we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since [latex]\\left(-\\infty ,3\\right][\/latex] is a proper subset of the domain of [latex]f\\circ g[\/latex]. This means the domain of [latex]f\\circ g[\/latex] is the same as the domain of [latex]g[\/latex], namely, [latex]\\left(-\\infty ,3\\right][\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135241313\">This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of [latex]f\\circ g[\/latex] can contain values that are not in the domain of [latex]f[\/latex], though they must be in the domain of [latex]g[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\n[ohm_question hide_question_numbers=1]317557[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]317906[\/ohm_question]<\/section>\r\n<div id=\"fs-id1165135435597\" class=\"note precalculus try\">\r\n<div id=\"ti_01_04_07\" class=\"exercise\">\r\n<div id=\"fs-id1165137937116\" class=\"problem\"><section id=\"fs-id1165135388428\" class=\"key-equations\">\r\n<h2>Decomposing a Composite Function<\/h2>\r\nSometimes a function is easier to work with if we break it into simpler pieces. That means writing the function as a composition of two simpler functions. Because there can be more than one way to do this, we usually choose the decomposition that makes the problem easiest to handle.\r\n\r\n<section class=\"textbox example\"><strong>Write [latex]f(x)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions. <\/strong>We are looking for two functions, [latex]g[\/latex] and [latex]h[\/latex], so [latex]f\\left(x\\right)=g\\left(h\\left(x\\right)\\right)[\/latex]. To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right)[\/latex].There are multiple ways to express [latex]f(x)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions.[reveal-answer q=\"53139\"]Option 1[\/reveal-answer]\r\n[hidden-answer a=\"53139\"]\r\n<ul>\r\n \t<li>[latex]h(x) = 5 - x^2[\/latex]<\/li>\r\n \t<li>[latex]g(x) = \\sqrt{x}[\/latex]<\/li>\r\n<\/ul>\r\nThus:\r\n\r\n<center>[latex]\\begin{align*} f(x) = g(h(x)) &amp;= g(5 - x^2) \\\\ \\text{Since } g(x) &amp;= \\sqrt{x}, \\text{ we have:} \\\\ g(5 - x^2) &amp;= \\sqrt{5 - x^2} \\end{align*}[\/latex]<\/center>[\/hidden-answer]\r\n\r\n[reveal-answer q=\"80338\"]Option 2[\/reveal-answer]\r\n[hidden-answer a=\"80338\"]\r\n<ul>\r\n \t<li>[latex]h(x) = x^2[\/latex]<\/li>\r\n \t<li>[latex]g(x) = \\sqrt{5 - x}[\/latex]<\/li>\r\n<\/ul>\r\nThus:\r\n\r\n<center>[latex]\\begin{align*} f(x) = g(h(x)) &amp;= g(x^2) \\\\ \\text{Since } g(x) &amp;= \\sqrt{5 - x}, \\text{ we have:} \\\\ g(x^2) &amp;= \\sqrt{5 - x^2} \\end{align*}[\/latex]<\/center>[\/hidden-answer]\r\n\r\n[reveal-answer q=\"399617\"]Option 3[\/reveal-answer]\r\n[hidden-answer a=\"399617\"]\r\n<ul>\r\n \t<li>[latex]h(x) = x[\/latex]<\/li>\r\n \t<li>[latex]g(x) = \\sqrt{5 - x^2}[\/latex]<\/li>\r\n<\/ul>\r\nThus:\r\n\r\n<center>[latex]\\begin{align*} f(x) = g(h(x)) &amp;= g(x) \\\\ \\text{Since } g(x) &amp;= \\sqrt{5 - x^2}, \\text{ we have:} \\\\ g(x) &amp;= \\sqrt{5 - x^2} \\end{align*}[\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]317907[\/ohm_question]<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]317908[\/ohm_question]<\/section><\/section><\/div>\r\n<\/div>\r\n<\/div>","rendered":"<h2>Finding the Domain of a Composite Function<\/h2>\n<p id=\"fs-id1165135519324\">The <strong>domain of a composite function<\/strong> such as [latex]f\\circ g[\/latex] is dependent on the domain of [latex]g[\/latex] and the domain of [latex]f[\/latex]. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as [latex]f\\circ g[\/latex]. Let us assume we know the domains of the functions [latex]f[\/latex] and [latex]g[\/latex] separately. If we write the composite function for an input [latex]x[\/latex] as [latex]f\\left(g\\left(x\\right)\\right)[\/latex], we can see right away that [latex]x[\/latex] must be a member of the domain of [latex]g[\/latex] in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that [latex]g\\left(x\\right)[\/latex] must be a member of the domain of [latex]f[\/latex], otherwise the second function evaluation in [latex]f\\left(g\\left(x\\right)\\right)[\/latex] cannot be completed, and the expression is still undefined. Thus the domain of [latex]f\\circ g[\/latex] consists of only those inputs in the domain of [latex]g[\/latex] that produce outputs from [latex]g[\/latex] belonging to the domain of [latex]f[\/latex]. Note that the domain of [latex]f[\/latex] composed with [latex]g[\/latex] is the set of all [latex]x[\/latex] such that [latex]x[\/latex] is in the domain of [latex]g[\/latex] and [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>domain of a composite function<\/h3>\n<p>The domain of a composite function [latex]f\\left(g\\left(x\\right)\\right)[\/latex] is the set of those inputs [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f[\/latex].<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p id=\"fs-id1165135203267\"><strong>How To: Given a function composition [latex]f\\left(g\\left(x\\right)\\right)[\/latex], determine its domain.<\/strong><\/p>\n<ol id=\"fs-id1165137714200\">\n<li>Find the domain of [latex]g[\/latex].<\/li>\n<li>Find the domain of [latex]f[\/latex].<\/li>\n<li>Find those inputs, [latex]x[\/latex], in the domain of [latex]g[\/latex] for which [latex]g(x)[\/latex] is in the domain of [latex]f[\/latex]. That is, exclude those inputs, [latex]x[\/latex], from the domain of [latex]g[\/latex] for which [latex]g(x)[\/latex] is not in the domain of [latex]f[\/latex]. The resulting set is the domain of [latex]f\\circ g[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the domain of<\/p>\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\frac{5}{x - 1}\\text{ and }g\\left(x\\right)=\\frac{4}{3x - 2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q262512\">Show Solution<\/button><\/p>\n<div id=\"q262512\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137627922\">The domain of [latex]g\\left(x\\right)[\/latex] consists of all real numbers except [latex]x=\\frac{2}{3}[\/latex], since that input value would cause us to divide by 0. Likewise, the domain of [latex]f[\/latex] consists of all real numbers except 1. So we need to exclude from the domain of [latex]g\\left(x\\right)[\/latex] that value of [latex]x[\/latex] for which [latex]g\\left(x\\right)=1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{4}{3x - 2}&=1 \\\\ 4&=3x - 2 &&\\text{Multiply by } 3x-2 \\\\ 6&=3x &&\\text{Add }2 \\\\ x&=2 && \\text{Divide by }3 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137891240\">So the domain of [latex]f\\circ g[\/latex] is the set of all real numbers except [latex]\\frac{2}{3}[\/latex] and [latex]2[\/latex]. This means that<\/p>\n<p style=\"text-align: center;\">[latex]x\\ne \\frac{2}{3}\\text{or}x\\ne 2[\/latex]<\/p>\n<p id=\"fs-id1165135152098\">We can write this in interval notation as<\/p>\n<p style=\"text-align: center;\">[latex]\\left(-\\infty ,\\frac{2}{3}\\right)\\cup \\left(\\frac{2}{3},2\\right)\\cup \\left(2,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1165135547426\">Find the domain of<\/p>\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\sqrt{x+2}\\text{ and }g\\left(x\\right)=\\sqrt{3-x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q959136\">Show Solution<\/button><\/p>\n<div id=\"q959136\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137812353\">Because we cannot take the square root of a negative number, the domain of [latex]g[\/latex] is [latex]\\left(-\\infty ,3\\right][\/latex]. Now we check the domain of the composite function<\/p>\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{3-x+2}\\text{ or }\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{5-x}[\/latex]<\/p>\n<p id=\"fs-id1165137871761\">The domain of this function is [latex]\\left(-\\infty ,5\\right][\/latex]. To find the domain of [latex]f\\circ g[\/latex], we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since [latex]\\left(-\\infty ,3\\right][\/latex] is a proper subset of the domain of [latex]f\\circ g[\/latex]. This means the domain of [latex]f\\circ g[\/latex] is the same as the domain of [latex]g[\/latex], namely, [latex]\\left(-\\infty ,3\\right][\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135241313\">This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of [latex]f\\circ g[\/latex] can contain values that are not in the domain of [latex]f[\/latex], though they must be in the domain of [latex]g[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p><iframe loading=\"lazy\" id=\"ohm317557\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317557&theme=lumen&iframe_resize_id=ohm317557&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm317906\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317906&theme=lumen&iframe_resize_id=ohm317906&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<div id=\"fs-id1165135435597\" class=\"note precalculus try\">\n<div id=\"ti_01_04_07\" class=\"exercise\">\n<div id=\"fs-id1165137937116\" class=\"problem\">\n<section id=\"fs-id1165135388428\" class=\"key-equations\">\n<h2>Decomposing a Composite Function<\/h2>\n<p>Sometimes a function is easier to work with if we break it into simpler pieces. That means writing the function as a composition of two simpler functions. Because there can be more than one way to do this, we usually choose the decomposition that makes the problem easiest to handle.<\/p>\n<section class=\"textbox example\"><strong>Write [latex]f(x)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions. <\/strong>We are looking for two functions, [latex]g[\/latex] and [latex]h[\/latex], so [latex]f\\left(x\\right)=g\\left(h\\left(x\\right)\\right)[\/latex]. To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right)[\/latex].There are multiple ways to express [latex]f(x)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q53139\">Option 1<\/button><\/p>\n<div id=\"q53139\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>[latex]h(x) = 5 - x^2[\/latex]<\/li>\n<li>[latex]g(x) = \\sqrt{x}[\/latex]<\/li>\n<\/ul>\n<p>Thus:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*} f(x) = g(h(x)) &= g(5 - x^2) \\\\ \\text{Since } g(x) &= \\sqrt{x}, \\text{ we have:} \\\\ g(5 - x^2) &= \\sqrt{5 - x^2} \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q80338\">Option 2<\/button><\/p>\n<div id=\"q80338\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>[latex]h(x) = x^2[\/latex]<\/li>\n<li>[latex]g(x) = \\sqrt{5 - x}[\/latex]<\/li>\n<\/ul>\n<p>Thus:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*} f(x) = g(h(x)) &= g(x^2) \\\\ \\text{Since } g(x) &= \\sqrt{5 - x}, \\text{ we have:} \\\\ g(x^2) &= \\sqrt{5 - x^2} \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q399617\">Option 3<\/button><\/p>\n<div id=\"q399617\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>[latex]h(x) = x[\/latex]<\/li>\n<li>[latex]g(x) = \\sqrt{5 - x^2}[\/latex]<\/li>\n<\/ul>\n<p>Thus:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*} f(x) = g(h(x)) &= g(x) \\\\ \\text{Since } g(x) &= \\sqrt{5 - x^2}, \\text{ we have:} \\\\ g(x) &= \\sqrt{5 - x^2} \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm317907\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317907&theme=lumen&iframe_resize_id=ohm317907&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm317908\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=317908&theme=lumen&iframe_resize_id=ohm317908&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":13,"menu_order":10,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":498,"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\/625"}],"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":11,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/625\/revisions"}],"predecessor-version":[{"id":5443,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/625\/revisions\/5443"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/498"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/625\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=625"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=625"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=625"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=625"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}