{"id":308,"date":"2023-09-20T22:48:58","date_gmt":"2023-09-20T22:48:58","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-first-derivative-test-and-concavity\/"},"modified":"2024-08-05T13:07:20","modified_gmt":"2024-08-05T13:07:20","slug":"derivatives-and-the-shape-of-a-graph-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-and-the-shape-of-a-graph-learn-it-1\/","title":{"raw":"Derivatives and the Shape of a Graph: Learn It 1","rendered":"Derivatives and the Shape of a Graph: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Use the first derivative to determine where a function is going up or down, and identify points that might be local highs or lows<\/li>\r\n\t<li>Apply the second derivative to find out where a function curves upward or downward and locate points where this curvature changes<\/li>\r\n\t<li>Use the second derivative test to determine if a point on a graph is the highest or lowest within specific sections<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The First Derivative Test<\/h2>\r\n<p id=\"fs-id1165043093996\">Corollary 3 of the Mean Value Theorem showed that if the derivative of a function is positive over an interval [latex]I[\/latex] then the function is increasing over [latex]I[\/latex]. On the other hand, if the derivative of the function is negative over an interval [latex]I[\/latex], then the function is decreasing over [latex]I[\/latex] as shown in the following figure.<\/p>\r\n\r\n[caption id=\"attachment_3274\" align=\"alignnone\" width=\"487\"]<img class=\"wp-image-3274 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/18170255\/9f1308c124662c7cc7ef9debf1112c9c47e96df6.png\" alt=\"This figure is broken into four figures labeled a, b, c, and d. Figure a shows a function increasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f\u2019 &gt; 0. In other words, f is increasing. Figure b shows a function increasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f\u2019 &gt; 0. In other words, f is increasing. Figure c shows a function decreasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f\u2019 &lt; 0. In other words, f is decreasing. Figure d shows a function decreasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f\u2019 &lt; 0. In other words, f is decreasing.\" width=\"487\" height=\"548\" \/> Figure 1. Both functions are increasing over the interval [latex](a,b)[\/latex]. At each point [latex]x[\/latex], the derivative [latex]f^{\\prime}(x)&gt;0[\/latex]. Both functions are decreasing over the interval [latex](a,b)[\/latex]. At each point [latex]x[\/latex], the derivative [latex]f^{\\prime}(x)&lt;0[\/latex].[\/caption]\r\n\r\n<p id=\"fs-id1165042319627\">A continuous function [latex]f[\/latex] has a local maximum at point [latex]c[\/latex] if and only if [latex]f[\/latex] switches from increasing to decreasing at point [latex]c[\/latex]. Similarly, [latex]f[\/latex] has a local minimum at [latex]c[\/latex] if and only if [latex]f[\/latex] switches from decreasing to increasing at [latex]c[\/latex].<\/p>\r\n<p>If [latex]f[\/latex] is a continuous function over an interval [latex]I[\/latex] containing [latex]c[\/latex] and differentiable over [latex]I[\/latex], except possibly at [latex]c[\/latex], the only way [latex]f[\/latex] can switch from increasing to decreasing (or vice versa) at point [latex]c[\/latex] is if [latex]{f}^{\\prime }[\/latex] changes sign as [latex]x[\/latex] increases through [latex]c.[\/latex]<\/p>\r\n<p>If [latex]f[\/latex] is differentiable at [latex]c,[\/latex] the only way that [latex]{f}^{\\prime }.[\/latex] can change sign as [latex]x[\/latex] increases through [latex]c[\/latex] is if [latex]f^{\\prime}(c)=0[\/latex].<\/p>\r\n<p>Therefore, for a function [latex]f[\/latex] that is continuous over an interval [latex]I[\/latex] containing [latex]c[\/latex] and differentiable over [latex]I[\/latex], except possibly at [latex]c[\/latex], the only way [latex]f[\/latex] can switch from increasing to decreasing (or vice versa) is if [latex]f^{\\prime}(c)=0[\/latex] or [latex]f^{\\prime}(c)[\/latex] is undefined.<\/p>\r\n<p>Consequently, to locate local extrema for a function [latex]f[\/latex], we look for points [latex]c[\/latex] in the domain of [latex]f[\/latex] such that [latex]f^{\\prime}(c)=0[\/latex] or [latex]f^{\\prime}(c)[\/latex] is undefined. Recall that such points are called critical points of [latex]f[\/latex]. Note that [latex]f[\/latex] need not have a local extrema at a critical point. The critical points are candidates for local extrema only.<\/p>\r\n<p>In Figure 2, we show that if a continuous function [latex]f[\/latex] has a local extremum, it must occur at a critical point, but a function may not have a local extremum at a critical point. We show that if [latex]f[\/latex] has a local extremum at a critical point, then the sign of [latex]f^{\\prime}[\/latex] switches as [latex]x[\/latex] increases through that point.<\/p>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"867\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210914\/CNX_Calc_Figure_04_05_002.jpg\" alt=\"A function f(x) is graphed. It starts in the second quadrant and increases to x = a, which is too sharp and hence f\u2019(a) is undefined. In this section f\u2019 &gt; 0. Then, f decreases from x = a to x = b (so f\u2019 &lt; 0 here), before increasing at x = b. It is noted that f\u2019(b) = 0. While increasing from x = b to x = c, f\u2019 &gt; 0. The function has an inversion point at c, and it is marked f\u2019(c) = 0. The function increases some more to d (so f\u2019 &gt; 0), which is the global maximum. It is marked that f\u2019(d) = 0. Then the function decreases and it is marked that f\u2019 &gt; 0.\" width=\"867\" height=\"429\" \/> Figure 2. The function [latex]f[\/latex] has four critical points: [latex]a,b,c[\/latex], and [latex]d[\/latex]. The function [latex]f[\/latex] has local maxima at [latex]a[\/latex] and [latex]d[\/latex], and a local minimum at [latex]b[\/latex]. The function [latex]f[\/latex] does not have a local extremum at [latex]c[\/latex]. The sign of [latex]f^{\\prime}[\/latex] changes at all local extrema.[\/caption]\r\n\r\n<p id=\"fs-id1165043095850\">Using Figure 2, we summarize the main results regarding local extrema.<\/p>\r\n<ul id=\"fs-id1165043257290\">\r\n\t<li>If a continuous function [latex]f[\/latex] has a local extremum, it must occur at a critical point [latex]c[\/latex].<\/li>\r\n\t<li>The function has a local extremum at the critical point [latex]c[\/latex] if and only if the derivative [latex]f^{\\prime}[\/latex] switches sign as [latex]x[\/latex] increases through [latex]c[\/latex].<\/li>\r\n\t<li>Therefore, to test whether a function has a local extremum at a critical point [latex]c[\/latex], we must determine the sign of [latex]f^{\\prime}(x)[\/latex] to the left and right of [latex]c[\/latex].<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165043262520\">This result is known as the <strong>first derivative test<\/strong>.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">first derivative test<\/h3>\r\n<p id=\"fs-id1165043067802\">Suppose that [latex]f[\/latex] is a continuous function over an interval [latex]I[\/latex] containing a critical point [latex]c[\/latex]. If [latex]f[\/latex] is differentiable over [latex]I[\/latex], except possibly at point [latex]c[\/latex], then [latex]f(c)[\/latex] satisfies one of the following descriptions:<\/p>\r\n<ol id=\"fs-id1165043068012\" style=\"list-style-type: lower-roman;\">\r\n\t<li>If [latex]f^{\\prime}[\/latex] changes sign from positive when [latex]x &lt; c[\/latex] to negative when [latex]x &gt; c[\/latex], then [latex]f(c)[\/latex] is a local maximum of [latex]f[\/latex].<\/li>\r\n\t<li>If [latex]f^{\\prime}[\/latex] changes sign from negative when [latex]x &lt; c[\/latex] to positive when [latex]x &gt; [\/latex], then [latex]f(c)[\/latex] is a local minimum of [latex]f[\/latex].<\/li>\r\n\t<li>If [latex]f^{\\prime}[\/latex] has the same sign for [latex]x &lt; c[\/latex] and [latex]x&gt;c[\/latex], then [latex]f(c)[\/latex] is neither a local maximum nor a local minimum of [latex]f[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>\r\n<p id=\"fs-id1165043354060\">We can summarize the first derivative test as a strategy for locating local extrema.<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How to: Use the First Derivative Test<\/strong><\/p>\r\n<p id=\"fs-id1165042349982\">To determine local extrema of a function [latex]f(x)[\/latex] that is continuous on an interval [latex][a,b][\/latex], follow these steps:<\/p>\r\n<ol>\r\n\t<li style=\"list-style-type: none;\">\r\n<ol>\r\n\t<li><strong>Identify Critical Points:<\/strong> Find all critical points of [latex]f(x)[\/latex] within the interval and note these points along with the endpoints [latex]a[\/latex] and [latex]b[\/latex].<\/li>\r\n\t<li><strong>Divide the Interval:<\/strong> Segment the interval [latex][a,b][\/latex] using the critical points as boundaries, creating smaller subintervals.<\/li>\r\n\t<li><strong>Analyze the Sign of [latex]f\u2032(x)[\/latex]<\/strong>: Evaluate the derivative [latex]f\u2032(x)[\/latex] within each subinterval. A change in the sign of [latex]f\u2032(x) [\/latex] across a critical point indicates a local extremum at that point:\r\n\r\n<ul>\r\n\t<li>If [latex]f\u2032(x)[\/latex] changes from positive to negative, [latex]f(x)[\/latex] has a local maximum at the critical point.<\/li>\r\n\t<li>If [latex]f\u2032(x)[\/latex] changes from negative to positive, [latex]f(x)[\/latex] has a local minimum at the critical point.<\/li>\r\n\t<li>If [latex]f\u2032(x)[\/latex] does not change sign, [latex]f(x)[\/latex] does not have a local extremum at that critical point.<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li><strong>Compare Values:<\/strong> Evaluate [latex]f(x)[\/latex] at the critical points and endpoints to identify the absolute maximum and minimum over the interval.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox recall\">\r\n<p>Recall that when talking about local extrema, <em>the value<\/em> of the extremum is the [latex]y[\/latex] value and <em>the location<\/em> of the extremum is the [latex]x[\/latex] value.<\/p>\r\n<\/section>\r\n<p>Now let\u2019s look at how to use this strategy to locate all local extrema for particular functions.<\/p>\r\n<section class=\"textbox example\">\r\n<p>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>\r\n\r\n[reveal-answer q=\"fs-id1165043253635\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043253635\"]\r\n\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<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\n<p>For example, let\u2019s choose [latex]x=-2, \\, x=0[\/latex], and [latex]x=4[\/latex] as test points.<\/p>\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\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>\r\n<p>These analytical results agree with the following graph.<\/p>\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]\r\n\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<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>\r\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>\r\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 \"4.5 Derivatives and the Shape of a Graph\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Use the first derivative test to find the location of all local extrema for [latex]f(x)=5x^{\\frac{1}{3}}-x^{\\frac{5}{3}}[\/latex]. Use a graphing utility to confirm your results.<\/p>\r\n\r\n[reveal-answer q=\"fs-id1165043109175\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043109175\"]\r\n\r\n<p id=\"fs-id1165043109175\"><strong>Step 1<\/strong>. The derivative is<\/p>\r\n<div id=\"fs-id1165043066545\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{5}{3}x^{-2\/3}-\\frac{5}{3}x^{2\/3}=\\frac{5}{3x^{2\/3}}-\\frac{5x^{2\/3}}{3}=\\frac{5-5x^{4\/3}}{3x^{2\/3}}=\\frac{5(1-x^{4\/3})}{3x^{2\/3}}[\/latex].<\/div>\r\n<p>The derivative [latex]f^{\\prime}(x)=0[\/latex] when [latex]1-x^{4\/3}=0[\/latex]. Therefore, [latex]f^{\\prime}(x)=0[\/latex] at [latex]x=\\pm 1[\/latex].<\/p>\r\n<p>The derivative [latex]f^{\\prime}(x)[\/latex] is undefined at [latex]x=0[\/latex]. Therefore, we have three critical points: [latex]x=0[\/latex], [latex]x=1[\/latex], and [latex]x=-1[\/latex].<\/p>\r\n<p>Consequently, divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the smaller intervals [latex](\u2212\\infty ,-1), \\, (-1,0), \\, (0,1)[\/latex], and [latex](1,\\infty )[\/latex].<\/p>\r\n<p id=\"fs-id1165042354821\"><strong>Step 2<\/strong>: Since [latex]f^{\\prime}[\/latex] is continuous over each subinterval, it suffices to choose a test point [latex]x[\/latex] in each of the intervals from step 1 and determine the sign of [latex]f^{\\prime}[\/latex] at each of these points. The points [latex]x=-2, \\, x=-\\frac{1}{2}, \\, x=\\frac{1}{2}[\/latex], and [latex]x=2[\/latex] are test points for these intervals.<\/p>\r\n<table id=\"fs-id1165043032734\" class=\"unnumbered\" summary=\"This table has five 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) = 5(1 \u2013 x4\/3)\/(3x2\/3) at Test Point, and Conclusion. Below the header, the first column reads (\u2212\u221e, \u22121), (\u22121, 0), (0, 1), and (1, \u221e). The second column reads x = \u22122, x = \u22121\/2, x = 1\/2, and x = 2. The third column reads (+)(\u2212)\/(+) = \u2212, (+)(+)\/(+) = +, (+)(+)\/(+) = +, and (+)(\u2212)\/(+) = \u2212. The fourth column reads f is decreasing, f is increasing, f is increasing, and f is decreasing.\">\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)=\\frac{5(1-x^{4\/3})}{3x^{2\/3}}[\/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]\\frac{(+)(\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](-1,0)[\/latex]<\/td>\r\n<td>[latex]x=-\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{(+)(+)}{+}=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](0,1)[\/latex]<\/td>\r\n<td>[latex]x=\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{(+)(+)}{+}=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]\\frac{(+)(\u2212)}{+}=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p><strong>Step 3<\/strong>: Since [latex]f[\/latex] is decreasing over the interval [latex](\u2212\\infty ,-1)[\/latex] and increasing over the interval [latex](-1,0)[\/latex], [latex]f[\/latex] has a local minimum at [latex]x=-1[\/latex].<\/p>\r\n<p>Since [latex]f[\/latex] is increasing over the interval [latex](-1,0)[\/latex] and the interval [latex](0,1)[\/latex], [latex]f[\/latex] does not have a local extremum at [latex]x=0[\/latex].<\/p>\r\n<p>Since [latex]f[\/latex] is increasing over the interval [latex](0,1)[\/latex] and decreasing over the interval [latex](1,\\infty ), \\, f[\/latex] has a local maximum at [latex]x=1[\/latex].<\/p>\r\n<p>The analytical results agree with the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"363\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210921\/CNX_Calc_Figure_04_05_009.jpg\" alt=\"The function f(x) = 5x1\/3 \u2013 x5\/3 is graphed. It decreases to its local minimum at x = \u22121, increases to x = 1, and then decreases after that.\" width=\"363\" height=\"272\" \/> Figure 4. The function f has a local minimum at [latex]x=-1[\/latex] and a local maximum at [latex]x=1[\/latex].[\/caption]\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_number=1]288394[\/ohm_question]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_number=1]288395[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Use the first derivative to determine where a function is going up or down, and identify points that might be local highs or lows<\/li>\n<li>Apply the second derivative to find out where a function curves upward or downward and locate points where this curvature changes<\/li>\n<li>Use the second derivative test to determine if a point on a graph is the highest or lowest within specific sections<\/li>\n<\/ul>\n<\/section>\n<h2>The First Derivative Test<\/h2>\n<p id=\"fs-id1165043093996\">Corollary 3 of the Mean Value Theorem showed that if the derivative of a function is positive over an interval [latex]I[\/latex] then the function is increasing over [latex]I[\/latex]. On the other hand, if the derivative of the function is negative over an interval [latex]I[\/latex], then the function is decreasing over [latex]I[\/latex] as shown in the following figure.<\/p>\n<figure id=\"attachment_3274\" aria-describedby=\"caption-attachment-3274\" style=\"width: 487px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3274 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/18170255\/9f1308c124662c7cc7ef9debf1112c9c47e96df6.png\" alt=\"This figure is broken into four figures labeled a, b, c, and d. Figure a shows a function increasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f\u2019 &gt; 0. In other words, f is increasing. Figure b shows a function increasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f\u2019 &gt; 0. In other words, f is increasing. Figure c shows a function decreasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f\u2019 &lt; 0. In other words, f is decreasing. Figure d shows a function decreasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f\u2019 &lt; 0. In other words, f is decreasing.\" width=\"487\" height=\"548\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/18170255\/9f1308c124662c7cc7ef9debf1112c9c47e96df6.png 487w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/18170255\/9f1308c124662c7cc7ef9debf1112c9c47e96df6-267x300.png 267w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/18170255\/9f1308c124662c7cc7ef9debf1112c9c47e96df6-65x73.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/18170255\/9f1308c124662c7cc7ef9debf1112c9c47e96df6-225x253.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/18170255\/9f1308c124662c7cc7ef9debf1112c9c47e96df6-350x394.png 350w\" sizes=\"(max-width: 487px) 100vw, 487px\" \/><figcaption id=\"caption-attachment-3274\" class=\"wp-caption-text\">Figure 1. Both functions are increasing over the interval [latex](a,b)[\/latex]. At each point [latex]x[\/latex], the derivative [latex]f^{\\prime}(x)&gt;0[\/latex]. Both functions are decreasing over the interval [latex](a,b)[\/latex]. At each point [latex]x[\/latex], the derivative [latex]f^{\\prime}(x)&lt;0[\/latex].<\/figcaption><\/figure>\n<p id=\"fs-id1165042319627\">A continuous function [latex]f[\/latex] has a local maximum at point [latex]c[\/latex] if and only if [latex]f[\/latex] switches from increasing to decreasing at point [latex]c[\/latex]. Similarly, [latex]f[\/latex] has a local minimum at [latex]c[\/latex] if and only if [latex]f[\/latex] switches from decreasing to increasing at [latex]c[\/latex].<\/p>\n<p>If [latex]f[\/latex] is a continuous function over an interval [latex]I[\/latex] containing [latex]c[\/latex] and differentiable over [latex]I[\/latex], except possibly at [latex]c[\/latex], the only way [latex]f[\/latex] can switch from increasing to decreasing (or vice versa) at point [latex]c[\/latex] is if [latex]{f}^{\\prime }[\/latex] changes sign as [latex]x[\/latex] increases through [latex]c.[\/latex]<\/p>\n<p>If [latex]f[\/latex] is differentiable at [latex]c,[\/latex] the only way that [latex]{f}^{\\prime }.[\/latex] can change sign as [latex]x[\/latex] increases through [latex]c[\/latex] is if [latex]f^{\\prime}(c)=0[\/latex].<\/p>\n<p>Therefore, for a function [latex]f[\/latex] that is continuous over an interval [latex]I[\/latex] containing [latex]c[\/latex] and differentiable over [latex]I[\/latex], except possibly at [latex]c[\/latex], the only way [latex]f[\/latex] can switch from increasing to decreasing (or vice versa) is if [latex]f^{\\prime}(c)=0[\/latex] or [latex]f^{\\prime}(c)[\/latex] is undefined.<\/p>\n<p>Consequently, to locate local extrema for a function [latex]f[\/latex], we look for points [latex]c[\/latex] in the domain of [latex]f[\/latex] such that [latex]f^{\\prime}(c)=0[\/latex] or [latex]f^{\\prime}(c)[\/latex] is undefined. Recall that such points are called critical points of [latex]f[\/latex]. Note that [latex]f[\/latex] need not have a local extrema at a critical point. The critical points are candidates for local extrema only.<\/p>\n<p>In Figure 2, we show that if a continuous function [latex]f[\/latex] has a local extremum, it must occur at a critical point, but a function may not have a local extremum at a critical point. We show that if [latex]f[\/latex] has a local extremum at a critical point, then the sign of [latex]f^{\\prime}[\/latex] switches as [latex]x[\/latex] increases through that point.<\/p>\n<figure style=\"width: 867px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210914\/CNX_Calc_Figure_04_05_002.jpg\" alt=\"A function f(x) is graphed. It starts in the second quadrant and increases to x = a, which is too sharp and hence f\u2019(a) is undefined. In this section f\u2019 &gt; 0. Then, f decreases from x = a to x = b (so f\u2019 &lt; 0 here), before increasing at x = b. It is noted that f\u2019(b) = 0. While increasing from x = b to x = c, f\u2019 &gt; 0. The function has an inversion point at c, and it is marked f\u2019(c) = 0. The function increases some more to d (so f\u2019 &gt; 0), which is the global maximum. It is marked that f\u2019(d) = 0. Then the function decreases and it is marked that f\u2019 &gt; 0.\" width=\"867\" height=\"429\" \/><figcaption class=\"wp-caption-text\">Figure 2. The function [latex]f[\/latex] has four critical points: [latex]a,b,c[\/latex], and [latex]d[\/latex]. The function [latex]f[\/latex] has local maxima at [latex]a[\/latex] and [latex]d[\/latex], and a local minimum at [latex]b[\/latex]. The function [latex]f[\/latex] does not have a local extremum at [latex]c[\/latex]. The sign of [latex]f^{\\prime}[\/latex] changes at all local extrema.<\/figcaption><\/figure>\n<p id=\"fs-id1165043095850\">Using Figure 2, we summarize the main results regarding local extrema.<\/p>\n<ul id=\"fs-id1165043257290\">\n<li>If a continuous function [latex]f[\/latex] has a local extremum, it must occur at a critical point [latex]c[\/latex].<\/li>\n<li>The function has a local extremum at the critical point [latex]c[\/latex] if and only if the derivative [latex]f^{\\prime}[\/latex] switches sign as [latex]x[\/latex] increases through [latex]c[\/latex].<\/li>\n<li>Therefore, to test whether a function has a local extremum at a critical point [latex]c[\/latex], we must determine the sign of [latex]f^{\\prime}(x)[\/latex] to the left and right of [latex]c[\/latex].<\/li>\n<\/ul>\n<p id=\"fs-id1165043262520\">This result is known as the <strong>first derivative test<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">first derivative test<\/h3>\n<p id=\"fs-id1165043067802\">Suppose that [latex]f[\/latex] is a continuous function over an interval [latex]I[\/latex] containing a critical point [latex]c[\/latex]. If [latex]f[\/latex] is differentiable over [latex]I[\/latex], except possibly at point [latex]c[\/latex], then [latex]f(c)[\/latex] satisfies one of the following descriptions:<\/p>\n<ol id=\"fs-id1165043068012\" style=\"list-style-type: lower-roman;\">\n<li>If [latex]f^{\\prime}[\/latex] changes sign from positive when [latex]x < c[\/latex] to negative when [latex]x > c[\/latex], then [latex]f(c)[\/latex] is a local maximum of [latex]f[\/latex].<\/li>\n<li>If [latex]f^{\\prime}[\/latex] changes sign from negative when [latex]x < c[\/latex] to positive when [latex]x >[\/latex], then [latex]f(c)[\/latex] is a local minimum of [latex]f[\/latex].<\/li>\n<li>If [latex]f^{\\prime}[\/latex] has the same sign for [latex]x < c[\/latex] and [latex]x>c[\/latex], then [latex]f(c)[\/latex] is neither a local maximum nor a local minimum of [latex]f[\/latex].<\/li>\n<\/ol>\n<\/section>\n<p id=\"fs-id1165043354060\">We can summarize the first derivative test as a strategy for locating local extrema.<\/p>\n<section class=\"textbox questionHelp\">\n<p><strong>How to: Use the First Derivative Test<\/strong><\/p>\n<p id=\"fs-id1165042349982\">To determine local extrema of a function [latex]f(x)[\/latex] that is continuous on an interval [latex][a,b][\/latex], follow these steps:<\/p>\n<ol>\n<li style=\"list-style-type: none;\">\n<ol>\n<li><strong>Identify Critical Points:<\/strong> Find all critical points of [latex]f(x)[\/latex] within the interval and note these points along with the endpoints [latex]a[\/latex] and [latex]b[\/latex].<\/li>\n<li><strong>Divide the Interval:<\/strong> Segment the interval [latex][a,b][\/latex] using the critical points as boundaries, creating smaller subintervals.<\/li>\n<li><strong>Analyze the Sign of [latex]f\u2032(x)[\/latex]<\/strong>: Evaluate the derivative [latex]f\u2032(x)[\/latex] within each subinterval. A change in the sign of [latex]f\u2032(x)[\/latex] across a critical point indicates a local extremum at that point:\n<ul>\n<li>If [latex]f\u2032(x)[\/latex] changes from positive to negative, [latex]f(x)[\/latex] has a local maximum at the critical point.<\/li>\n<li>If [latex]f\u2032(x)[\/latex] changes from negative to positive, [latex]f(x)[\/latex] has a local minimum at the critical point.<\/li>\n<li>If [latex]f\u2032(x)[\/latex] does not change sign, [latex]f(x)[\/latex] does not have a local extremum at that critical point.<\/li>\n<\/ul>\n<\/li>\n<li><strong>Compare Values:<\/strong> Evaluate [latex]f(x)[\/latex] at the critical points and endpoints to identify the absolute maximum and minimum over the interval.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox recall\">\n<p>Recall that when talking about local extrema, <em>the value<\/em> of the extremum is the [latex]y[\/latex] value and <em>the location<\/em> of the extremum is the [latex]x[\/latex] value.<\/p>\n<\/section>\n<p>Now let\u2019s look at how to use this strategy to locate all local extrema for particular functions.<\/p>\n<section class=\"textbox example\">\n<p>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 example\">\n<p>Use the first derivative test to find the location of all local extrema for [latex]f(x)=5x^{\\frac{1}{3}}-x^{\\frac{5}{3}}[\/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-id1165043109175\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043109175\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043109175\"><strong>Step 1<\/strong>. The derivative is<\/p>\n<div id=\"fs-id1165043066545\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{5}{3}x^{-2\/3}-\\frac{5}{3}x^{2\/3}=\\frac{5}{3x^{2\/3}}-\\frac{5x^{2\/3}}{3}=\\frac{5-5x^{4\/3}}{3x^{2\/3}}=\\frac{5(1-x^{4\/3})}{3x^{2\/3}}[\/latex].<\/div>\n<p>The derivative [latex]f^{\\prime}(x)=0[\/latex] when [latex]1-x^{4\/3}=0[\/latex]. Therefore, [latex]f^{\\prime}(x)=0[\/latex] at [latex]x=\\pm 1[\/latex].<\/p>\n<p>The derivative [latex]f^{\\prime}(x)[\/latex] is undefined at [latex]x=0[\/latex]. Therefore, we have three critical points: [latex]x=0[\/latex], [latex]x=1[\/latex], and [latex]x=-1[\/latex].<\/p>\n<p>Consequently, divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the smaller intervals [latex](\u2212\\infty ,-1), \\, (-1,0), \\, (0,1)[\/latex], and [latex](1,\\infty )[\/latex].<\/p>\n<p id=\"fs-id1165042354821\"><strong>Step 2<\/strong>: Since [latex]f^{\\prime}[\/latex] is continuous over each subinterval, it suffices to choose a test point [latex]x[\/latex] in each of the intervals from step 1 and determine the sign of [latex]f^{\\prime}[\/latex] at each of these points. The points [latex]x=-2, \\, x=-\\frac{1}{2}, \\, x=\\frac{1}{2}[\/latex], and [latex]x=2[\/latex] are test points for these intervals.<\/p>\n<table id=\"fs-id1165043032734\" class=\"unnumbered\" summary=\"This table has five 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) = 5(1 \u2013 x4\/3)\/(3x2\/3) at Test Point, and Conclusion. Below the header, the first column reads (\u2212\u221e, \u22121), (\u22121, 0), (0, 1), and (1, \u221e). The second column reads x = \u22122, x = \u22121\/2, x = 1\/2, and x = 2. The third column reads (+)(\u2212)\/(+) = \u2212, (+)(+)\/(+) = +, (+)(+)\/(+) = +, and (+)(\u2212)\/(+) = \u2212. The fourth column reads f is decreasing, f is increasing, f is increasing, and f is decreasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{5(1-x^{4\/3})}{3x^{2\/3}}[\/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]\\frac{(+)(\u2212)}{+}=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](-1,0)[\/latex]<\/td>\n<td>[latex]x=-\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{(+)(+)}{+}=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](0,1)[\/latex]<\/td>\n<td>[latex]x=\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{(+)(+)}{+}=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]\\frac{(+)(\u2212)}{+}=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Step 3<\/strong>: Since [latex]f[\/latex] is decreasing over the interval [latex](\u2212\\infty ,-1)[\/latex] and increasing over the interval [latex](-1,0)[\/latex], [latex]f[\/latex] has a local minimum at [latex]x=-1[\/latex].<\/p>\n<p>Since [latex]f[\/latex] is increasing over the interval [latex](-1,0)[\/latex] and the interval [latex](0,1)[\/latex], [latex]f[\/latex] does not have a local extremum at [latex]x=0[\/latex].<\/p>\n<p>Since [latex]f[\/latex] is increasing over the interval [latex](0,1)[\/latex] and decreasing over the interval [latex](1,\\infty ), \\, f[\/latex] has a local maximum at [latex]x=1[\/latex].<\/p>\n<p>The analytical results agree with the following graph.<\/p>\n<figure style=\"width: 363px\" 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\/11210921\/CNX_Calc_Figure_04_05_009.jpg\" alt=\"The function f(x) = 5x1\/3 \u2013 x5\/3 is graphed. It decreases to its local minimum at x = \u22121, increases to x = 1, and then decreases after that.\" width=\"363\" height=\"272\" \/><figcaption class=\"wp-caption-text\">Figure 4. The function f has a local minimum at [latex]x=-1[\/latex] and a local maximum at [latex]x=1[\/latex].<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288394\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288394&theme=lumen&iframe_resize_id=ohm288394&source=tnh&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<section class=\"textbox tryIt\">\n<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><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":25,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.5 Derivatives and the Shape of a Graph\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":652,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"4.5 Derivatives and the Shape of a Graph","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/308"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/308\/revisions"}],"predecessor-version":[{"id":4133,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/308\/revisions\/4133"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/652"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/308\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=308"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=308"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=308"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=308"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}