{"id":650,"date":"2025-10-09T15:28:08","date_gmt":"2025-10-09T15:28:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=650"},"modified":"2026-05-26T14:07:33","modified_gmt":"2026-05-26T14:07:33","slug":"optimization-and-applications-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/optimization-and-applications-background-youll-need-2\/","title":{"raw":"Optimization and Applications: Background You'll Need 2","rendered":"Optimization and Applications: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Apply the second derivative to determine concavity and locate inflection points<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 class=\"font-claude-response-heading text-text-100 mt-1 -mb-0.5\">Concavity and the Second Derivative<\/h2>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The second derivative tells you about the <strong>concavity<\/strong> of a function\u2014whether the graph curves upward or downward.<\/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\">When [latex]f[\/latex] is <strong>concave up<\/strong>, the slope [latex]f'[\/latex] is increasing, so [latex]f''(x) &gt; 0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">When [latex]f[\/latex] is <strong>concave down<\/strong>, the slope [latex]f'[\/latex] is decreasing, so [latex]f''(x) &lt; 0[\/latex]<\/li>\r\n<\/ul>\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>test for concavity<\/h3>\r\nLet [latex]f[\/latex] be twice differentiable over an interval [latex]I[\/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''(x) &gt; 0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f[\/latex] is <strong>concave up<\/strong> on [latex]I[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]f''(x) &lt; 0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f[\/latex] is <strong>concave down<\/strong> on [latex]I[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>A function can switch concavity, but only at points where [latex]f''(x) = 0[\/latex] or [latex]f''(x)[\/latex] is undefined. When [latex]f[\/latex] is continuous at such a point and actually changes concavity there, we call it an <strong>inflection point<\/strong>.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>inflection point<\/h3>\r\nIf [latex]f[\/latex] is continuous at [latex]a[\/latex] and [latex]f[\/latex] changes concavity at [latex]a[\/latex], then the point [latex](a, f(a))[\/latex] is an inflection point of [latex]f[\/latex].\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p id=\"fs-id1165043163933\"><strong>Quick Reference: What Derivatives Tell You<\/strong><\/p>\r\n\r\n<table id=\"fs-id1165043353366\" summary=\"This table has five rows and four columns. The first row is a header row, and it reads from left to right Sign of f\u2019, Sign of f\u2019\u2019, Is f increasing or decreasing?, and Concavity. Below the header, the first column reads Positive, Positive, Negative, and Negative. The second column reads Positive, Negative, Positive, and Negative. The third column reads Increasing, Increasing, Decreasing, and Decreasing. The fourth column reads Concave up, Concave down, Concave up, and Concave down.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Sign of [latex]f^{\\prime}[\/latex]<\/th>\r\n<th>Sign of [latex]f^{\\prime \\prime}[\/latex]<\/th>\r\n<th>Is [latex]f[\/latex] increasing or decreasing?<\/th>\r\n<th>Concavity<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>Positive<\/td>\r\n<td>Positive<\/td>\r\n<td>Increasing<\/td>\r\n<td>Concave up<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Positive<\/td>\r\n<td>Negative<\/td>\r\n<td>Increasing<\/td>\r\n<td>Concave down<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Negative<\/td>\r\n<td>Positive<\/td>\r\n<td>Decreasing<\/td>\r\n<td>Concave up<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Negative<\/td>\r\n<td>Negative<\/td>\r\n<td>Decreasing<\/td>\r\n<td>Concave down<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section>\r\n<h3 class=\"font-claude-response-subheading text-text-100 mt-1 -mb-1.5\">The Second Derivative Test<\/h3>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The second derivative provides an alternative method for classifying critical points. At a critical point [latex]c[\/latex] where [latex]f'(c) = 0[\/latex], the concavity tells you whether you have a maximum or minimum.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>second derivative test<\/h3>\r\nSuppose [latex]f'(c) = 0[\/latex] and [latex]f''[\/latex] is continuous near [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''(c) &gt; 0[\/latex], then [latex]f[\/latex] has a <strong>local minimum<\/strong> at [latex]c[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]f''(c) &lt; 0[\/latex], then [latex]f[\/latex] has a <strong>local maximum<\/strong> at [latex]c[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]f''(c) = 0[\/latex], the test is <strong>inconclusive<\/strong> (use the first derivative test instead)<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the second derivative to find the location of all local extrema for [latex]f(x)=x^5-5x^3[\/latex].[reveal-answer q=\"fs-id1165042320876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042320876\"]\r\n<p id=\"fs-id1165042320876\">To apply the second derivative test, we first need to find critical points [latex]c[\/latex] where [latex]f^{\\prime}(c)=0[\/latex].<\/p>\r\nThe derivative is [latex]f^{\\prime}(x)=5x^4-15x^2[\/latex]. Therefore, [latex]f^{\\prime}(x)=5x^4-15x^2=5x^2(x^2-3)=0[\/latex] when [latex]x=0,\\pm \\sqrt{3}[\/latex].\r\n<p id=\"fs-id1165043431475\">To determine whether [latex]f[\/latex] has a local extrema at any of these points, we need to evaluate the sign of [latex]f^{\\prime \\prime}[\/latex] at these points. The second derivative is<\/p>\r\n\r\n<div id=\"fs-id1165042318723\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=20x^3-30x=10x(2x^2-3)[\/latex].<\/div>\r\n<p id=\"fs-id1165042373916\">In the following table, we evaluate the second derivative at each of the critical points and use the second derivative test to determine whether [latex]f[\/latex] has a local maximum or local minimum at any of these points.<\/p>\r\n\r\n<table id=\"fs-id1165042705774\" class=\"unnumbered\" summary=\"This table has four rows and three columns. The first row is a header row, and it reads x, f\u2019\u2019(x), and Conclusion. After the header, the first column reads negative square root of 3, 0, and square root of 3. The second column reads negative 30 times the square root of 3, 0, and 30 times the square root of 3. The third column reads Local maxiumum, Second derivative test is inconclusive, and Local minimum.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]f^{\\prime \\prime}(x)[\/latex]<\/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\\sqrt{3}[\/latex]<\/td>\r\n<td>[latex]-30\\sqrt{3}[\/latex]<\/td>\r\n<td>Local maximum<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>Second derivative test is inconclusive<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>[latex]30\\sqrt{3}[\/latex]<\/td>\r\n<td>Local minimum<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042476082\">By the second derivative test, we conclude that [latex]f[\/latex] has a local maximum at [latex]x=\u2212\\sqrt{3}[\/latex] and [latex]f[\/latex] has a local minimum at [latex]x=\\sqrt{3}[\/latex]. The second derivative test is inconclusive at [latex]x=0[\/latex].<\/p>\r\nTo determine whether [latex]f[\/latex] has a local extrema at [latex]x=0[\/latex], we apply the first derivative test.\r\n\r\nTo evaluate the sign of [latex]f^{\\prime}(x)=5x^2(x^2-3)[\/latex] for [latex]x \\in (\u2212\\sqrt{3},0)[\/latex] and [latex]x \\in (0,\\sqrt{3})[\/latex], let [latex]x=-1[\/latex] and [latex]x=1[\/latex] be the two test points. Since [latex]f^{\\prime}(-1)&lt;0[\/latex] and [latex]f^{\\prime}(1)&lt;0[\/latex], we conclude that [latex]f[\/latex] is decreasing on both intervals and, therefore, [latex]f[\/latex] does not have a local extrema at [latex]x=0[\/latex] as shown in the following graph.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"801\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210942\/CNX_Calc_Figure_04_05_007.jpg\" alt=\"The function f(x) = x5 \u2013 5x3 is graphed. The function increases to (negative square root of 3, 10), then decreases to an inflection point at 0, continues decreasing to (square root of 3, \u221210), and then increases.\" width=\"801\" height=\"408\" \/> Figure 10. The function [latex]f[\/latex] has a local maximum at [latex]x=\u2212\\sqrt{3}[\/latex] and a local minimum at [latex]x=\\sqrt{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=789&amp;end=1016&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 style=\"font-size: 1rem; orphans: 1; text-align: initial;\" href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.5DerivativesAndTheShapeOfAGraph789to1016_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_numbers=1]325816[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Apply the second derivative to determine concavity and locate inflection points<\/span><\/li>\n<\/ul>\n<\/section>\n<h2 class=\"font-claude-response-heading text-text-100 mt-1 -mb-0.5\">Concavity and the Second Derivative<\/h2>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The second derivative tells you about the <strong>concavity<\/strong> of a function\u2014whether the graph curves upward or downward.<\/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\">When [latex]f[\/latex] is <strong>concave up<\/strong>, the slope [latex]f'[\/latex] is increasing, so [latex]f''(x) > 0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">When [latex]f[\/latex] is <strong>concave down<\/strong>, the slope [latex]f'[\/latex] is decreasing, so [latex]f''(x) < 0[\/latex]<\/li>\n<\/ul>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>test for concavity<\/h3>\n<p>Let [latex]f[\/latex] be twice differentiable over an interval [latex]I[\/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''(x) > 0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f[\/latex] is <strong>concave up<\/strong> on [latex]I[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]f''(x) < 0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f[\/latex] is <strong>concave down<\/strong> on [latex]I[\/latex]<\/li>\n<\/ul>\n<\/section>\n<p>A function can switch concavity, but only at points where [latex]f''(x) = 0[\/latex] or [latex]f''(x)[\/latex] is undefined. When [latex]f[\/latex] is continuous at such a point and actually changes concavity there, we call it an <strong>inflection point<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>inflection point<\/h3>\n<p>If [latex]f[\/latex] is continuous at [latex]a[\/latex] and [latex]f[\/latex] changes concavity at [latex]a[\/latex], then the point [latex](a, f(a))[\/latex] is an inflection point of [latex]f[\/latex].<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p id=\"fs-id1165043163933\"><strong>Quick Reference: What Derivatives Tell You<\/strong><\/p>\n<table id=\"fs-id1165043353366\" summary=\"This table has five rows and four columns. The first row is a header row, and it reads from left to right Sign of f\u2019, Sign of f\u2019\u2019, Is f increasing or decreasing?, and Concavity. Below the header, the first column reads Positive, Positive, Negative, and Negative. The second column reads Positive, Negative, Positive, and Negative. The third column reads Increasing, Increasing, Decreasing, and Decreasing. The fourth column reads Concave up, Concave down, Concave up, and Concave down.\">\n<thead>\n<tr valign=\"top\">\n<th>Sign of [latex]f^{\\prime}[\/latex]<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}[\/latex]<\/th>\n<th>Is [latex]f[\/latex] increasing or decreasing?<\/th>\n<th>Concavity<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>Positive<\/td>\n<td>Positive<\/td>\n<td>Increasing<\/td>\n<td>Concave up<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Positive<\/td>\n<td>Negative<\/td>\n<td>Increasing<\/td>\n<td>Concave down<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Negative<\/td>\n<td>Positive<\/td>\n<td>Decreasing<\/td>\n<td>Concave up<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Negative<\/td>\n<td>Negative<\/td>\n<td>Decreasing<\/td>\n<td>Concave down<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<h3 class=\"font-claude-response-subheading text-text-100 mt-1 -mb-1.5\">The Second Derivative Test<\/h3>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The second derivative provides an alternative method for classifying critical points. At a critical point [latex]c[\/latex] where [latex]f'(c) = 0[\/latex], the concavity tells you whether you have a maximum or minimum.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>second derivative test<\/h3>\n<p>Suppose [latex]f'(c) = 0[\/latex] and [latex]f''[\/latex] is continuous near [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''(c) > 0[\/latex], then [latex]f[\/latex] has a <strong>local minimum<\/strong> at [latex]c[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]f''(c) < 0[\/latex], then [latex]f[\/latex] has a <strong>local maximum<\/strong> at [latex]c[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]f''(c) = 0[\/latex], the test is <strong>inconclusive<\/strong> (use the first derivative test instead)<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the second derivative to find the location of all local extrema for [latex]f(x)=x^5-5x^3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042320876\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042320876\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042320876\">To apply the second derivative test, we first need to find critical points [latex]c[\/latex] where [latex]f^{\\prime}(c)=0[\/latex].<\/p>\n<p>The derivative is [latex]f^{\\prime}(x)=5x^4-15x^2[\/latex]. Therefore, [latex]f^{\\prime}(x)=5x^4-15x^2=5x^2(x^2-3)=0[\/latex] when [latex]x=0,\\pm \\sqrt{3}[\/latex].<\/p>\n<p id=\"fs-id1165043431475\">To determine whether [latex]f[\/latex] has a local extrema at any of these points, we need to evaluate the sign of [latex]f^{\\prime \\prime}[\/latex] at these points. The second derivative is<\/p>\n<div id=\"fs-id1165042318723\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=20x^3-30x=10x(2x^2-3)[\/latex].<\/div>\n<p id=\"fs-id1165042373916\">In the following table, we evaluate the second derivative at each of the critical points and use the second derivative test to determine whether [latex]f[\/latex] has a local maximum or local minimum at any of these points.<\/p>\n<table id=\"fs-id1165042705774\" class=\"unnumbered\" summary=\"This table has four rows and three columns. The first row is a header row, and it reads x, f\u2019\u2019(x), and Conclusion. After the header, the first column reads negative square root of 3, 0, and square root of 3. The second column reads negative 30 times the square root of 3, 0, and 30 times the square root of 3. The third column reads Local maxiumum, Second derivative test is inconclusive, and Local minimum.\">\n<thead>\n<tr valign=\"top\">\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]f^{\\prime \\prime}(x)[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]\u2212\\sqrt{3}[\/latex]<\/td>\n<td>[latex]-30\\sqrt{3}[\/latex]<\/td>\n<td>Local maximum<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>Second derivative test is inconclusive<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>[latex]30\\sqrt{3}[\/latex]<\/td>\n<td>Local minimum<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042476082\">By the second derivative test, we conclude that [latex]f[\/latex] has a local maximum at [latex]x=\u2212\\sqrt{3}[\/latex] and [latex]f[\/latex] has a local minimum at [latex]x=\\sqrt{3}[\/latex]. The second derivative test is inconclusive at [latex]x=0[\/latex].<\/p>\n<p>To determine whether [latex]f[\/latex] has a local extrema at [latex]x=0[\/latex], we apply the first derivative test.<\/p>\n<p>To evaluate the sign of [latex]f^{\\prime}(x)=5x^2(x^2-3)[\/latex] for [latex]x \\in (\u2212\\sqrt{3},0)[\/latex] and [latex]x \\in (0,\\sqrt{3})[\/latex], let [latex]x=-1[\/latex] and [latex]x=1[\/latex] be the two test points. Since [latex]f^{\\prime}(-1)<0[\/latex] and [latex]f^{\\prime}(1)<0[\/latex], we conclude that [latex]f[\/latex] is decreasing on both intervals and, therefore, [latex]f[\/latex] does not have a local extrema at [latex]x=0[\/latex] as shown in the following graph.\n\n\n\n<figure style=\"width: 801px\" 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\/11210942\/CNX_Calc_Figure_04_05_007.jpg\" alt=\"The function f(x) = x5 \u2013 5x3 is graphed. The function increases to (negative square root of 3, 10), then decreases to an inflection point at 0, continues decreasing to (square root of 3, \u221210), and then increases.\" width=\"801\" height=\"408\" \/><figcaption class=\"wp-caption-text\">Figure 10. The function [latex]f[\/latex] has a local maximum at [latex]x=\u2212\\sqrt{3}[\/latex] and a local minimum at [latex]x=\\sqrt{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=789&amp;end=1016&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 style=\"font-size: 1rem; orphans: 1; text-align: initial;\" href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.5DerivativesAndTheShapeOfAGraph789to1016_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=\"ohm325816\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=325816&theme=lumen&iframe_resize_id=ohm325816&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":3,"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-650","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\/650","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\/650\/revisions"}],"predecessor-version":[{"id":2872,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/650\/revisions\/2872"}],"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\/650\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=650"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=650"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=650"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=650"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}