{"id":649,"date":"2025-10-09T15:28:06","date_gmt":"2025-10-09T15:28:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=649"},"modified":"2025-11-20T16:12:50","modified_gmt":"2025-11-20T16:12:50","slug":"optimization-and-applications-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/optimization-and-applications-background-youll-need-1\/","title":{"raw":"Optimization and Applications: Background You'll Need 1","rendered":"Optimization and Applications: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Use the first derivative to determine where a function is increasing or decreasing and identify local extrema<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The First Derivative and Function Behavior<\/h2>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The first derivative tells you whether a function is increasing or decreasing. This comes directly from the Mean Value Theorem:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">If [latex]f'(x) &gt; 0[\/latex] on an interval [latex]I[\/latex], then [latex]f[\/latex] is <strong>increasing<\/strong> on [latex]I[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]f'(x) &lt; 0[\/latex] on an interval [latex]I[\/latex], then [latex]f[\/latex] is <strong>decreasing<\/strong> on [latex]I[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">A continuous function [latex]f[\/latex] has a <strong>local maximum<\/strong> at [latex]c[\/latex] when [latex]f[\/latex] switches from increasing to decreasing at that point. Similarly, [latex]f[\/latex] has a <strong>local minimum<\/strong> at [latex]c[\/latex] when it switches from decreasing to increasing.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a continuous function [latex]f[\/latex] to switch from increasing to decreasing (or vice versa) at point [latex]c[\/latex], the derivative must change sign. This can only happen when [latex]f'(c) = 0[\/latex] or [latex]f'(c)[\/latex] is undefined\u2014these are your <strong>critical points<\/strong>.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">A critical point of [latex]f[\/latex] occurs at [latex]x = c[\/latex] when [latex]f'(c) = 0[\/latex] or [latex]f'(c)[\/latex] is undefined. Critical points are candidates for local extrema, but not every critical point is a local extremum.<\/section><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>first derivative test<\/h3>\r\nLet [latex]f[\/latex] be continuous over an interval [latex]I[\/latex] containing a critical point [latex]c[\/latex]:\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">If [latex]f'[\/latex] changes from <strong>positive to negative<\/strong> at [latex]c[\/latex], then [latex]f(c)[\/latex] is a <strong>local maximum<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]f'[\/latex] changes from <strong>negative to positive<\/strong> at [latex]c[\/latex], then [latex]f(c)[\/latex] is a <strong>local minimum<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]f'[\/latex] does <strong>not change sign<\/strong> at [latex]c[\/latex], then [latex]f(c)[\/latex] is <strong>neither<\/strong> a maximum nor minimum<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Apply the First Derivative Test<\/strong>\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find all critical points<\/strong> of [latex]f(x)[\/latex] on the interval (where [latex]f'(x) = 0[\/latex] or [latex]f'(x)[\/latex] is undefined)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Divide the interval<\/strong> into subintervals using the critical points as boundaries<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Test the sign of [latex]f'(x)[\/latex]<\/strong> in each subinterval by choosing a test point<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Determine extrema<\/strong> by checking if [latex]f'[\/latex] changes sign at each critical point<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the first derivative test to find the location of all local extrema for [latex]f(x)=x^3-3x^2-9x-1[\/latex]. Use a graphing utility to confirm your results.[reveal-answer q=\"fs-id1165043253635\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043253635\"]\r\n<p id=\"fs-id1165043253635\"><strong>Step 1.<\/strong> The derivative is [latex]f^{\\prime}(x)=3x^2-6x-9[\/latex]. To find the critical points, we need to find where [latex]f^{\\prime}(x)=0[\/latex]. Factoring the polynomial, we conclude that the critical points must satisfy:<\/p>\r\n\r\n<div id=\"fs-id1165043318988\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]3(x^2-2x-3)=3(x-3)(x+1)=0[\/latex]<\/div>\r\n<p id=\"fs-id1165043041063\">Therefore, the critical points are [latex]x=3,-1[\/latex]. Now divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the smaller intervals [latex](\u2212\\infty ,-1), \\, (-1,3)[\/latex], and [latex](3,\\infty)[\/latex].<\/p>\r\n<p id=\"fs-id1165043097583\"><strong>Step 2<\/strong>. Since [latex]f^{\\prime}[\/latex] is a continuous function, to determine the sign of [latex]f^{\\prime}(x)[\/latex] over each subinterval, it suffices to choose a point over each of the intervals [latex](\u2212\\infty ,-1), \\, (-1,3)[\/latex], and [latex](3,\\infty)[\/latex] and determine the sign of [latex]f^{\\prime}[\/latex] at each of these points.<\/p>\r\nFor example, let\u2019s choose [latex]x=-2, \\, x=0[\/latex], and [latex]x=4[\/latex] as test points.\r\n<table id=\"fs-id1165042978447\" class=\"unnumbered\" summary=\"This table has four rows and four columns. The first row is a header row, and it reads from left to right Interval, Test Point, Sign of f\u2019(x) = 3(x \u22123)(x + 1) at Test Point, and Conclusion. Below the header, the first column reads (\u2212\u221e, \u22121), (\u22121, 3), and (3, \u221e). The second column reads x = \u22122, x = 0, and x = 4. The third column reads (+)(\u2212)(\u2212) = +, (+)(\u2212)(+) = \u2212, and (+)(+)(+) = +. The fourth column reads f is increasing, f is decreasing, and f is increasing.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime}(x)=3(x-3)(x+1)[\/latex] at Test Point<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\r\n<td>[latex]x=-2[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)(\u2212)=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](-1,3)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)(+)=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](3,\\infty)[\/latex]<\/td>\r\n<td>[latex]x=4[\/latex]<\/td>\r\n<td>[latex](+)(+)(+)=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042931764\"><strong>Step 3.<\/strong> Since [latex]f^{\\prime}[\/latex] switches sign from positive to negative as [latex]x[\/latex] increases through [latex]1, \\, f[\/latex] has a local maximum at [latex]x=-1[\/latex].<\/p>\r\nSince [latex]f^{\\prime}[\/latex] switches sign from negative to positive as [latex]x[\/latex] increases through [latex]3, \\, f[\/latex] has a local minimum at [latex]x=3[\/latex].\r\n\r\nThese analytical results agree with the following graph.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"455\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210918\/CNX_Calc_Figure_04_05_008.jpg\" alt=\"The function f(x) = x3 \u2013 3x2 \u2013 9x \u2013 1 is graphed. It has a maximum at x = \u22121 and a minimum at x = 3. The function is increasing before x = \u22121, decreasing until x = 3, and then increasing after that.\" width=\"455\" height=\"764\" \/> Figure 3. The function [latex]f[\/latex] has a maximum at [latex]x=-1[\/latex] and a minimum at [latex]x=3[\/latex][\/caption]Watch the following video to see the worked solution to this example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/PBeo4ZJ-FGY?controls=0&amp;start=164&amp;end=313&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.5DerivativesAndTheShapeOfAGraph164to313_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.5 Derivatives and the Shape of a Graph\" here (opens in new window)<\/a>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_number=1]288395[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Use the first derivative to determine where a function is increasing or decreasing and identify local extrema<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>The First Derivative and Function Behavior<\/h2>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The first derivative tells you whether a function is increasing or decreasing. This comes directly from the Mean Value Theorem:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">If [latex]f'(x) > 0[\/latex] on an interval [latex]I[\/latex], then [latex]f[\/latex] is <strong>increasing<\/strong> on [latex]I[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]f'(x) < 0[\/latex] on an interval [latex]I[\/latex], then [latex]f[\/latex] is <strong>decreasing<\/strong> on [latex]I[\/latex]<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\">A continuous function [latex]f[\/latex] has a <strong>local maximum<\/strong> at [latex]c[\/latex] when [latex]f[\/latex] switches from increasing to decreasing at that point. Similarly, [latex]f[\/latex] has a <strong>local minimum<\/strong> at [latex]c[\/latex] when it switches from decreasing to increasing.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For a continuous function [latex]f[\/latex] to switch from increasing to decreasing (or vice versa) at point [latex]c[\/latex], the derivative must change sign. This can only happen when [latex]f'(c) = 0[\/latex] or [latex]f'(c)[\/latex] is undefined\u2014these are your <strong>critical points<\/strong>.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">A critical point of [latex]f[\/latex] occurs at [latex]x = c[\/latex] when [latex]f'(c) = 0[\/latex] or [latex]f'(c)[\/latex] is undefined. Critical points are candidates for local extrema, but not every critical point is a local extremum.<\/section>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>first derivative test<\/h3>\n<p>Let [latex]f[\/latex] be continuous over an interval [latex]I[\/latex] containing a critical point [latex]c[\/latex]:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">If [latex]f'[\/latex] changes from <strong>positive to negative<\/strong> at [latex]c[\/latex], then [latex]f(c)[\/latex] is a <strong>local maximum<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">If [latex]f'[\/latex] changes from <strong>negative to positive<\/strong> at [latex]c[\/latex], then [latex]f(c)[\/latex] is a <strong>local minimum<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">If [latex]f'[\/latex] does <strong>not change sign<\/strong> at [latex]c[\/latex], then [latex]f(c)[\/latex] is <strong>neither<\/strong> a maximum nor minimum<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Apply the First Derivative Test<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Find all critical points<\/strong> of [latex]f(x)[\/latex] on the interval (where [latex]f'(x) = 0[\/latex] or [latex]f'(x)[\/latex] is undefined)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Divide the interval<\/strong> into subintervals using the critical points as boundaries<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Test the sign of [latex]f'(x)[\/latex]<\/strong> in each subinterval by choosing a test point<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Determine extrema<\/strong> by checking if [latex]f'[\/latex] changes sign at each critical point<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the first derivative test to find the location of all local extrema for [latex]f(x)=x^3-3x^2-9x-1[\/latex]. Use a graphing utility to confirm your results.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043253635\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043253635\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043253635\"><strong>Step 1.<\/strong> The derivative is [latex]f^{\\prime}(x)=3x^2-6x-9[\/latex]. To find the critical points, we need to find where [latex]f^{\\prime}(x)=0[\/latex]. Factoring the polynomial, we conclude that the critical points must satisfy:<\/p>\n<div id=\"fs-id1165043318988\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]3(x^2-2x-3)=3(x-3)(x+1)=0[\/latex]<\/div>\n<p id=\"fs-id1165043041063\">Therefore, the critical points are [latex]x=3,-1[\/latex]. Now divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the smaller intervals [latex](\u2212\\infty ,-1), \\, (-1,3)[\/latex], and [latex](3,\\infty)[\/latex].<\/p>\n<p id=\"fs-id1165043097583\"><strong>Step 2<\/strong>. Since [latex]f^{\\prime}[\/latex] is a continuous function, to determine the sign of [latex]f^{\\prime}(x)[\/latex] over each subinterval, it suffices to choose a point over each of the intervals [latex](\u2212\\infty ,-1), \\, (-1,3)[\/latex], and [latex](3,\\infty)[\/latex] and determine the sign of [latex]f^{\\prime}[\/latex] at each of these points.<\/p>\n<p>For example, let\u2019s choose [latex]x=-2, \\, x=0[\/latex], and [latex]x=4[\/latex] as test points.<\/p>\n<table id=\"fs-id1165042978447\" class=\"unnumbered\" summary=\"This table has four rows and four columns. The first row is a header row, and it reads from left to right Interval, Test Point, Sign of f\u2019(x) = 3(x \u22123)(x + 1) at Test Point, and Conclusion. Below the header, the first column reads (\u2212\u221e, \u22121), (\u22121, 3), and (3, \u221e). The second column reads x = \u22122, x = 0, and x = 4. The third column reads (+)(\u2212)(\u2212) = +, (+)(\u2212)(+) = \u2212, and (+)(+)(+) = +. The fourth column reads f is increasing, f is decreasing, and f is increasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime}(x)=3(x-3)(x+1)[\/latex] at Test Point<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\n<td>[latex]x=-2[\/latex]<\/td>\n<td>[latex](+)(\u2212)(\u2212)=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](-1,3)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex](+)(\u2212)(+)=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](3,\\infty)[\/latex]<\/td>\n<td>[latex]x=4[\/latex]<\/td>\n<td>[latex](+)(+)(+)=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042931764\"><strong>Step 3.<\/strong> Since [latex]f^{\\prime}[\/latex] switches sign from positive to negative as [latex]x[\/latex] increases through [latex]1, \\, f[\/latex] has a local maximum at [latex]x=-1[\/latex].<\/p>\n<p>Since [latex]f^{\\prime}[\/latex] switches sign from negative to positive as [latex]x[\/latex] increases through [latex]3, \\, f[\/latex] has a local minimum at [latex]x=3[\/latex].<\/p>\n<p>These analytical results agree with the following graph.<\/p>\n<figure style=\"width: 455px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210918\/CNX_Calc_Figure_04_05_008.jpg\" alt=\"The function f(x) = x3 \u2013 3x2 \u2013 9x \u2013 1 is graphed. It has a maximum at x = \u22121 and a minimum at x = 3. The function is increasing before x = \u22121, decreasing until x = 3, and then increasing after that.\" width=\"455\" height=\"764\" \/><figcaption class=\"wp-caption-text\">Figure 3. The function [latex]f[\/latex] has a maximum at [latex]x=-1[\/latex] and a minimum at [latex]x=3[\/latex]<\/figcaption><\/figure>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/PBeo4ZJ-FGY?controls=0&amp;start=164&amp;end=313&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.5DerivativesAndTheShapeOfAGraph164to313_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.5 Derivatives and the Shape of a Graph&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm288395\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288395&theme=lumen&iframe_resize_id=ohm288395&source=tnh&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-649","chapter","type-chapter","status-publish","hentry"],"part":430,"module-header":"background_you_need","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\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/649","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/649\/revisions"}],"predecessor-version":[{"id":2651,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/649\/revisions\/2651"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/430"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/649\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=649"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=649"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=649"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=649"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}