{"id":405,"date":"2021-02-04T02:00:23","date_gmt":"2021-02-04T02:00:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=405"},"modified":"2022-03-16T05:45:56","modified_gmt":"2022-03-16T05:45:56","slug":"the-first-derivative-test-and-concavity","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/the-first-derivative-test-and-concavity\/","title":{"raw":"The First Derivative Test and Concavity","rendered":"The First Derivative Test and Concavity"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Explain how the sign of the first derivative affects the shape of a function\u2019s graph<\/li>\r\n \t<li>State the first derivative test for critical points<\/li>\r\n \t<li>Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function\u2019s graph<\/li>\r\n \t<li>Explain the concavity test for a function over an open interval<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165043199295\" class=\"bc-section 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=\"\" align=\"aligncenter\" width=\"487\"]<img id=\"2\" src=\"https:\/\/openstax.org\/resources\/9f1308c124662c7cc7ef9debf1112c9c47e96df6\" 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<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]. 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] 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]. 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. 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].<\/p>\r\nNote that [latex]f[\/latex] need not have a local extrema at a critical point. The critical points are candidates for local extrema only. 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.\r\n\r\n[caption id=\"\" align=\"aligncenter\" 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<p id=\"fs-id1165043095850\">Using Figure 2, we summarize the main results regarding local extrema.<\/p>\r\n\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\r\n<div id=\"fs-id1165042327970\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">First Derivative Test<\/h3>\r\n\r\n<hr \/>\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\r\n<ol id=\"fs-id1165043068012\">\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;c[\/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<\/div>\r\n<p id=\"fs-id1165043354060\">We can summarize the first derivative test as a strategy for locating local extrema.<\/p>\r\n\r\n<div id=\"fs-id1165043183681\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy: Using the First Derivative Test<\/h3>\r\n<p id=\"fs-id1165042349982\">Consider a function [latex]f[\/latex] that is continuous over an interval [latex]I[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1165042905754\">\r\n \t<li>Find all critical points of [latex]f[\/latex] and divide the interval [latex]I[\/latex] into smaller intervals using the critical points as endpoints.<\/li>\r\n \t<li>Analyze the sign of [latex]f^{\\prime}[\/latex] in each of the subintervals. If [latex]f^{\\prime}[\/latex] is continuous over a given subinterval (which is typically the case), then the sign of [latex]f^{\\prime}[\/latex] in that subinterval does not change and, therefore, can be determined by choosing an arbitrary test point [latex]x[\/latex] in that subinterval and by evaluating the sign of [latex]f^{\\prime}[\/latex] at that test point. Use the sign analysis to determine whether [latex]f[\/latex] is increasing or decreasing over that interval.<\/li>\r\n \t<li>Use the first derivative test and the results of step 2 to determine whether [latex]f[\/latex] has a local maximum, a local minimum, or neither at each of the critical points.<\/li>\r\n<\/ol>\r\n<\/div>\r\nRecall from Chapter 4.3 that when talking about local extrema, <em>the value<\/em> of the extremum is the y value and <em>the location<\/em> of the extremum is the x value.\r\n\r\nNow let\u2019s look at how to use this strategy to locate all local extrema for particular functions.\r\n<div id=\"fs-id1165043423022\" class=\"textbook exercises\">\r\n<h3>Example: Using the First Derivative Test to Find Local Extrema<\/h3>\r\nUse 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.\r\n<div id=\"fs-id1165042973802\" class=\"exercise\">[reveal-answer q=\"fs-id1165043253635\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043253635\"]\r\n<p id=\"fs-id1165043253635\">Step 1. 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\">Step 2. 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. For example, let\u2019s choose [latex]x=-2, \\, x=0[\/latex], and [latex]x=4[\/latex] as test points.<\/p>\r\n\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\">Step 3. 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]. 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]. 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][\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Using the First Derivative Test to Find Local Extrema.\r\n\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\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]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>.[\/hidden-answer]\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the first derivative test to locate all local extrema for [latex]f(x)=\u2212x^3+\\frac{3}{2}x^2+18x[\/latex].\r\n\r\n[reveal-answer q=\"6192003\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"6192003\"]\r\n\r\nFind all critical points of [latex]f[\/latex] and determine the signs of [latex]f^{\\prime}(x)[\/latex] over particular intervals determined by the critical points.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043281485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043281485\"]\r\n\r\n[latex]f[\/latex] has a local minimum at -2 and a local maximum at 3.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043312782\" class=\"textbook exercises\">\r\n<h3>Example: Using the First Derivative Test<\/h3>\r\nUse 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.\r\n<div id=\"fs-id1165042640287\" class=\"exercise\">[reveal-answer q=\"fs-id1165043109175\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043109175\"]\r\n<p id=\"fs-id1165043109175\">Step 1. The derivative is<\/p>\r\n\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\nThe 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]. 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]. 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].\r\n<p id=\"fs-id1165042354821\">Step 2: 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\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\nStep 3: 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]. 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]. 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]. The analytical results agree with the following graph.\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][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the first derivative test to find all local extrema for [latex]f(x)=\\sqrt[3]{x-1}[\/latex].\r\n\r\n[reveal-answer q=\"25579900\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"25579900\"]\r\n\r\nThe only critical point of [latex]f[\/latex] is [latex]x=1[\/latex].\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042982077\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042982077\"]\r\n\r\n[latex]f[\/latex] has no local extrema because [latex]f^{\\prime}[\/latex] does not change sign at [latex]x=1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20222[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]16201[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Concavity and Points of Inflection<\/h2>\r\n<p id=\"fs-id1165042517819\">We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the <strong>concavity<\/strong> of the function.<\/p>\r\n<p id=\"fs-id1165043256949\">Figure 5(a) shows a function [latex]f[\/latex] with a graph that curves upward. As [latex]x[\/latex] increases, the slope of the tangent line increases. Thus, since the derivative increases as [latex]x[\/latex] increases, [latex]f^{\\prime}[\/latex] is an increasing function. We say this function [latex]f[\/latex] is <strong>concave up<\/strong>. Figure 5(b) shows a function [latex]f[\/latex] that curves downward. As [latex]x[\/latex] increases, the slope of the tangent line decreases. Since the derivative decreases as [latex]x[\/latex] increases, [latex]f^{\\prime}[\/latex] is a decreasing function. We say this function [latex]f[\/latex] is<strong> concave down<\/strong>.<\/p>\r\n\r\n<div id=\"fs-id1165042559132\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042559135\">Let [latex]f[\/latex] be a function that is differentiable over an open interval [latex]I[\/latex]. If [latex]f^{\\prime}[\/latex] is increasing over [latex]I[\/latex], we say [latex]f[\/latex] is concave up over [latex]I[\/latex]. If [latex]f^{\\prime}[\/latex] is decreasing over [latex]I[\/latex], we say [latex]f[\/latex] is concave down over [latex]I[\/latex].<\/p>\r\n\r\n<\/div>\r\n[caption id=\"\" align=\"aligncenter\" width=\"633\"]<img id=\"13\" src=\"https:\/\/openstax.org\/resources\/5c0c6292e60d9ab5f29417baa39af97e6ff3c053\" 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 both are increasing, but the one taken further to the right is increasing more. It is noted that f\u2019 is increasing and f is concave up. Figure b shows a function increasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are increasing, but the one taken further to the right is increasing less. It is noted that f\u2019 is decreasing and f is concave down. Figure c shows a function decreasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are decreasing, but the one taken further to the right is decreasing less. It is noted that f\u2019 is increasing and f is concave up. Figure d shows a function decreasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are decreasing, but the one taken further to the right is decreasing more. It is noted that f\u2019 is decreasing and f is concave down.\" width=\"633\" height=\"673\" data-media-type=\"image\/jpeg\" \/> Figure 5. (a), (c) Since [latex]f^{\\prime}[\/latex] is increasing over the interval [latex](a,b)[\/latex], we say [latex]f[\/latex] is concave up over [latex](a,b)[\/latex]. (b), (d) Since [latex]f^{\\prime}[\/latex] is decreasing over the interval [latex](a,b)[\/latex], we say [latex]f[\/latex] is concave down over [latex](a,b)[\/latex].[\/caption]In general, without having the graph of a function [latex]f[\/latex], how can we determine its concavity? By definition, a function [latex]f[\/latex] is concave up if [latex]f^{\\prime}[\/latex] is increasing. From Corollary 3, we know that if [latex]f^{\\prime}[\/latex] is a differentiable function, then [latex]f^{\\prime}[\/latex] is increasing if its derivative [latex]f^{\\prime \\prime}(x)&gt;0[\/latex]. Therefore, a function [latex]f[\/latex] that is twice differentiable is concave up when [latex]f^{\\prime \\prime}(x)&gt;0[\/latex]. Similarly, a function [latex]f[\/latex] is concave down if [latex]f^{\\prime}[\/latex] is decreasing. We know that a differentiable function [latex]f^{\\prime}[\/latex] is decreasing if its derivative [latex]f^{\\prime \\prime}(x)&lt;0[\/latex]. Therefore, a twice-differentiable function [latex]f[\/latex] is concave down when [latex]f^{\\prime \\prime}(x)&lt;0[\/latex]. Applying this logic is known as the<strong> concavity test.<\/strong>\r\n<div id=\"fs-id1165043082168\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Test for Concavity<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165043088519\">Let [latex]f[\/latex] be a function that is twice differentiable over an interval [latex]I[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1165042925616\">\r\n \t<li>If [latex]f^{\\prime \\prime}(x)&gt;0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f[\/latex] is concave up over [latex]I[\/latex].<\/li>\r\n \t<li>If [latex]f^{\\prime \\prime}(x)&lt;0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f[\/latex] is concave down over [latex]I[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"fs-id1165043195362\">We conclude that we can determine the concavity of a function [latex]f[\/latex] by looking at the second derivative of [latex]f[\/latex]. In addition, we observe that a function [latex]f[\/latex] can switch concavity (Figure 6). However, a continuous function can switch concavity only at a point [latex]x[\/latex] if [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. Consequently, to determine the intervals where a function [latex]f[\/latex] is concave up and concave down, we look for those values of [latex]x[\/latex] where [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. When we have determined these points, we divide the domain of [latex]f[\/latex] into smaller intervals and determine the sign of [latex]f^{\\prime \\prime}[\/latex] over each of these smaller intervals. If [latex]f^{\\prime \\prime}[\/latex] changes sign as we pass through a point [latex]x[\/latex], then [latex]f[\/latex] changes concavity. It is important to remember that a function [latex]f[\/latex] may not change concavity at a point [latex]x[\/latex] even if [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. If, however, [latex]f[\/latex] does change concavity at a point [latex]a[\/latex] and [latex]f[\/latex] is continuous at [latex]a[\/latex], we say the point [latex](a,f(a))[\/latex] is an <strong>inflection point<\/strong> of [latex]f[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165043120988\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043120992\">If [latex]f[\/latex] is continuous at [latex]a[\/latex] and [latex]f[\/latex] changes concavity at [latex]a[\/latex], the point [latex](a,f(a))[\/latex] is an inflection point of [latex]f[\/latex].<\/p>\r\n\r\n<\/div>\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img id=\"16\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/7abb392a249b7e8a60b3285996e777d73855d665\" alt=\"A sinusoidal function is shown that has been shifted into the first quadrant. The function starts decreasing, so f\u2019 &lt; 0 and f\u2019\u2019 &gt; 0. The function reaches the local minimum and starts increasing, so f\u2019 &gt; 0 and f\u2019\u2019 &gt; 0. It is noted that the slope is increasing for these two intervals. The function then reaches an inflection point (a, f(a)) and from here the slop is decreasing even though the function continues to increase, so f\u2019 &gt; 0 and f\u2019\u2019 &lt; 0. The function reaches the maximum and then starts decreasing, so f\u2019 &lt; 0 and f\u2019\u2019 &lt; 0.\" width=\"731\" height=\"401\" data-media-type=\"image\/jpeg\" \/> Figure 6. Since [latex]f^{\\prime \\prime}(x)&gt;0[\/latex] for [latex]x&lt;a[\/latex], the function [latex]f[\/latex] is concave up over the interval [latex](\u2212\\infty,a)[\/latex]. Since [latex]f^{\\prime \\prime}(x)&lt;0[\/latex] for [latex]x&gt;a[\/latex], the function [latex]f[\/latex] is concave down over the interval [latex](a,\\infty)[\/latex]. The point [latex](a,f(a))[\/latex] is an inflection point of [latex]f[\/latex].[\/caption]\r\n<div id=\"fs-id1165042558993\" class=\"textbook exercises\">\r\n<h3>Example: Testing for Concavity<\/h3>\r\nFor the function [latex]f(x)=x^3-6x^2+9x+30[\/latex], determine all intervals where [latex]f[\/latex] is concave up and all intervals where [latex]f[\/latex] is concave down. List all inflection points for [latex]f[\/latex]. Use a graphing utility to confirm your results.\r\n<div id=\"fs-id1165042558995\" class=\"exercise\">[reveal-answer q=\"fs-id1165043306699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043306699\"]\r\n<p id=\"fs-id1165043306699\">To determine concavity, we need to find the second derivative [latex]f^{\\prime \\prime}(x)[\/latex]. The first derivative is [latex]f^{\\prime}(x)=3x^2-12x+9[\/latex], so the second derivative is [latex]f^{\\prime \\prime}(x)=6x-12[\/latex]. If the function changes concavity, it occurs either when [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. Since [latex]f^{\\prime \\prime}[\/latex] is defined for all real numbers [latex]x[\/latex], we need only find where [latex]f^{\\prime \\prime}(x)=0[\/latex]. Solving the equation [latex]6x-12=0[\/latex], we see that [latex]x=2[\/latex] is the only place where [latex]f[\/latex] could change concavity. We now test points over the intervals [latex](\u2212\\infty ,2)[\/latex] and [latex](2,\\infty)[\/latex] to determine the concavity of [latex]f[\/latex]. The points [latex]x=0[\/latex] and [latex]x=3[\/latex] are test points for these intervals.<\/p>\r\n\r\n<table id=\"fs-id1165042960171\" class=\"unnumbered\" summary=\"This table has three 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\u2019(x) = 6x \u2013 12 at Test Point, and Conclusion. Below the header, the first column reads (\u2212\u221e, 2) and (2, \u221e). The second column reads x = 0 and x = 3. The third column reads \u2212 and +. The fourth column reads f is concave down and f is concave up.\">\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 \\prime}(x)=6x-12[\/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 ,2)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex]-[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](2,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=3[\/latex]<\/td>\r\n<td>[latex]+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave up.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042329993\">We conclude that [latex]f[\/latex] is concave down over the interval [latex](\u2212\\infty ,2)[\/latex] and concave up over the interval [latex](2,\\infty)[\/latex]. Since [latex]f[\/latex] changes concavity at [latex]x=2[\/latex], the point [latex](2,f(2))=(2,32)[\/latex] is an inflection point. Figure 7\u00a0confirms the analytical results.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"379\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210931\/CNX_Calc_Figure_04_05_011.jpg\" alt=\"The function f(x) = x3 \u2013 6x2 + 9x + 30 is graphed. The inflection point (2, 32) is marked, and it is roughly equidistant from the two local extrema.\" width=\"379\" height=\"385\" \/> Figure 7. The given function has a point of inflection at [latex](2,32)[\/latex] where the graph changes concavity.[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Testing for Concavity.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/PBeo4ZJ-FGY?controls=0&amp;start=600&amp;end=713&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]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\/4.5DerivativesAndTheShapeOfAGraph600to713_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>.[\/hidden-answer]\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFor [latex]f(x)=\u2212x^3+\\frac{3}{2}x^2+18x[\/latex], find all intervals where [latex]f[\/latex] is concave up and all intervals where [latex]f[\/latex] is concave down.\r\n\r\n[reveal-answer q=\"33881102\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"33881102\"]\r\n\r\nFind where [latex]f^{\\prime \\prime}(x)=0[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042369709\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042369709\"]\r\n\r\n[latex]f[\/latex] is concave up over the interval [latex](\u2212\\infty ,\\frac{1}{2})[\/latex] and concave down over the interval [latex](\\frac{1}{2},\\infty )[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043163933\">We now summarize, in the box below, the information that the first and second derivatives of a function [latex]f[\/latex] provide about the graph of [latex]f[\/latex], and illustrate this information in Figure 8.<\/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.\"><caption>What Derivatives Tell Us about Graphs<\/caption>\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<div><\/div>\r\n<div id=\"fs-id1165042355740\" class=\"bc-section section\">[caption id=\"\" align=\"aligncenter\" width=\"901\"]<img id=\"20\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/b4fd75d3ef022de52e9f3b782e4d8908f12a7e23\" alt=\"A function is graphed in the first quadrant. It is broken up into four sections, with the breaks coming at the local minimum, inflection point, and local maximum, respectively. The first section is decreasing and concave up; here, f\u2019 &lt; 0 and f\u2019\u2019 &gt; 0. The second section is increasing and concave up; here, f\u2019 &gt; 0 and f\u2019\u2019 &gt; 0. The third section is increasing and concave down; here, f\u2019 &gt; 0 and f\u2019\u2019 &lt; 0. The fourth section is increasing and concave down; here, f\u2019 &lt; 0 and f\u2019\u2019 &lt; 0.\" width=\"901\" height=\"457\" data-media-type=\"image\/jpeg\" \/> Figure 8. Consider a twice-differentiable function [latex]f[\/latex] over an open interval [latex]I[\/latex]. If [latex]f^{\\prime}(x)&gt;0[\/latex] for all [latex]x \\in I[\/latex], the function is increasing over [latex]I[\/latex]. If [latex]f^{\\prime}(x)&lt;0[\/latex] for all [latex]x \\in I[\/latex], the function is decreasing over [latex]I[\/latex]. If [latex]f^{\\prime \\prime}(x)&gt;0[\/latex] for all [latex]x \\in I[\/latex], the function is concave up. If [latex]f^{\\prime \\prime}(x)&lt;0[\/latex] for all [latex]x \\in I[\/latex], the function is concave down on [latex]I[\/latex].[\/caption]<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Explain how the sign of the first derivative affects the shape of a function\u2019s graph<\/li>\n<li>State the first derivative test for critical points<\/li>\n<li>Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function\u2019s graph<\/li>\n<li>Explain the concavity test for a function over an open interval<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165043199295\" class=\"bc-section 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<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"2\" src=\"https:\/\/openstax.org\/resources\/9f1308c124662c7cc7ef9debf1112c9c47e96df6\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\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]. 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] 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]. 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. 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].<\/p>\n<p>Note that [latex]f[\/latex] need not have a local extrema at a critical point. The critical points are candidates for local extrema only. 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<div style=\"width: 877px\" 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\/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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\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<div id=\"fs-id1165042327970\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">First Derivative Test<\/h3>\n<hr \/>\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\">\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>c[\/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<\/div>\n<p id=\"fs-id1165043354060\">We can summarize the first derivative test as a strategy for locating local extrema.<\/p>\n<div id=\"fs-id1165043183681\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy: Using the First Derivative Test<\/h3>\n<p id=\"fs-id1165042349982\">Consider a function [latex]f[\/latex] that is continuous over an interval [latex]I[\/latex].<\/p>\n<ol id=\"fs-id1165042905754\">\n<li>Find all critical points of [latex]f[\/latex] and divide the interval [latex]I[\/latex] into smaller intervals using the critical points as endpoints.<\/li>\n<li>Analyze the sign of [latex]f^{\\prime}[\/latex] in each of the subintervals. If [latex]f^{\\prime}[\/latex] is continuous over a given subinterval (which is typically the case), then the sign of [latex]f^{\\prime}[\/latex] in that subinterval does not change and, therefore, can be determined by choosing an arbitrary test point [latex]x[\/latex] in that subinterval and by evaluating the sign of [latex]f^{\\prime}[\/latex] at that test point. Use the sign analysis to determine whether [latex]f[\/latex] is increasing or decreasing over that interval.<\/li>\n<li>Use the first derivative test and the results of step 2 to determine whether [latex]f[\/latex] has a local maximum, a local minimum, or neither at each of the critical points.<\/li>\n<\/ol>\n<\/div>\n<p>Recall from Chapter 4.3 that when talking about local extrema, <em>the value<\/em> of the extremum is the y value and <em>the location<\/em> of the extremum is the x value.<\/p>\n<p>Now let\u2019s look at how to use this strategy to locate all local extrema for particular functions.<\/p>\n<div id=\"fs-id1165043423022\" class=\"textbook exercises\">\n<h3>Example: Using the First Derivative Test to Find Local Extrema<\/h3>\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 id=\"fs-id1165042973802\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043253635\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043253635\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043253635\">Step 1. 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\">Step 2. 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. 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\">Step 3. 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]. 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]. These analytical results agree with the following graph.<\/p>\n<div style=\"width: 465px\" 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\" \/><\/p>\n<p 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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Using the First Derivative Test to Find Local Extrema.<\/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\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">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>.<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the first derivative test to locate all local extrema for [latex]f(x)=\u2212x^3+\\frac{3}{2}x^2+18x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6192003\">Hint<\/span><\/p>\n<div id=\"q6192003\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find all critical points of [latex]f[\/latex] and determine the signs of [latex]f^{\\prime}(x)[\/latex] over particular intervals determined by the critical points.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043281485\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043281485\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f[\/latex] has a local minimum at -2 and a local maximum at 3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043312782\" class=\"textbook exercises\">\n<h3>Example: Using the First Derivative Test<\/h3>\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 id=\"fs-id1165042640287\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043109175\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043109175\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043109175\">Step 1. 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]. 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]. 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\">Step 2: 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>Step 3: 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]. 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]. 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]. The analytical results agree with the following graph.<\/p>\n<div style=\"width: 373px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the first derivative test to find all local extrema for [latex]f(x)=\\sqrt[3]{x-1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q25579900\">Hint<\/span><\/p>\n<div id=\"q25579900\" class=\"hidden-answer\" style=\"display: none\">\n<p>The only critical point of [latex]f[\/latex] is [latex]x=1[\/latex].\n<\/p><\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042982077\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042982077\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f[\/latex] has no local extrema because [latex]f^{\\prime}[\/latex] does not change sign at [latex]x=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20222\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20222&theme=oea&iframe_resize_id=ohm20222&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm16201\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16201&theme=oea&iframe_resize_id=ohm16201&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Concavity and Points of Inflection<\/h2>\n<p id=\"fs-id1165042517819\">We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the <strong>concavity<\/strong> of the function.<\/p>\n<p id=\"fs-id1165043256949\">Figure 5(a) shows a function [latex]f[\/latex] with a graph that curves upward. As [latex]x[\/latex] increases, the slope of the tangent line increases. Thus, since the derivative increases as [latex]x[\/latex] increases, [latex]f^{\\prime}[\/latex] is an increasing function. We say this function [latex]f[\/latex] is <strong>concave up<\/strong>. Figure 5(b) shows a function [latex]f[\/latex] that curves downward. As [latex]x[\/latex] increases, the slope of the tangent line decreases. Since the derivative decreases as [latex]x[\/latex] increases, [latex]f^{\\prime}[\/latex] is a decreasing function. We say this function [latex]f[\/latex] is<strong> concave down<\/strong>.<\/p>\n<div id=\"fs-id1165042559132\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1165042559135\">Let [latex]f[\/latex] be a function that is differentiable over an open interval [latex]I[\/latex]. If [latex]f^{\\prime}[\/latex] is increasing over [latex]I[\/latex], we say [latex]f[\/latex] is concave up over [latex]I[\/latex]. If [latex]f^{\\prime}[\/latex] is decreasing over [latex]I[\/latex], we say [latex]f[\/latex] is concave down over [latex]I[\/latex].<\/p>\n<\/div>\n<div style=\"width: 643px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"13\" src=\"https:\/\/openstax.org\/resources\/5c0c6292e60d9ab5f29417baa39af97e6ff3c053\" 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 both are increasing, but the one taken further to the right is increasing more. It is noted that f\u2019 is increasing and f is concave up. Figure b shows a function increasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are increasing, but the one taken further to the right is increasing less. It is noted that f\u2019 is decreasing and f is concave down. Figure c shows a function decreasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are decreasing, but the one taken further to the right is decreasing less. It is noted that f\u2019 is increasing and f is concave up. Figure d shows a function decreasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are decreasing, but the one taken further to the right is decreasing more. It is noted that f\u2019 is decreasing and f is concave down.\" width=\"633\" height=\"673\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. (a), (c) Since [latex]f^{\\prime}[\/latex] is increasing over the interval [latex](a,b)[\/latex], we say [latex]f[\/latex] is concave up over [latex](a,b)[\/latex]. (b), (d) Since [latex]f^{\\prime}[\/latex] is decreasing over the interval [latex](a,b)[\/latex], we say [latex]f[\/latex] is concave down over [latex](a,b)[\/latex].<\/p>\n<\/div>\n<p>In general, without having the graph of a function [latex]f[\/latex], how can we determine its concavity? By definition, a function [latex]f[\/latex] is concave up if [latex]f^{\\prime}[\/latex] is increasing. From Corollary 3, we know that if [latex]f^{\\prime}[\/latex] is a differentiable function, then [latex]f^{\\prime}[\/latex] is increasing if its derivative [latex]f^{\\prime \\prime}(x)>0[\/latex]. Therefore, a function [latex]f[\/latex] that is twice differentiable is concave up when [latex]f^{\\prime \\prime}(x)>0[\/latex]. Similarly, a function [latex]f[\/latex] is concave down if [latex]f^{\\prime}[\/latex] is decreasing. We know that a differentiable function [latex]f^{\\prime}[\/latex] is decreasing if its derivative [latex]f^{\\prime \\prime}(x)<0[\/latex]. Therefore, a twice-differentiable function [latex]f[\/latex] is concave down when [latex]f^{\\prime \\prime}(x)<0[\/latex]. Applying this logic is known as the<strong> concavity test.<\/strong><\/p>\n<div id=\"fs-id1165043082168\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Test for Concavity<\/h3>\n<hr \/>\n<p id=\"fs-id1165043088519\">Let [latex]f[\/latex] be a function that is twice differentiable over an interval [latex]I[\/latex].<\/p>\n<ol id=\"fs-id1165042925616\">\n<li>If [latex]f^{\\prime \\prime}(x)>0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f[\/latex] is concave up over [latex]I[\/latex].<\/li>\n<li>If [latex]f^{\\prime \\prime}(x)<0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f[\/latex] is concave down over [latex]I[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1165043195362\">We conclude that we can determine the concavity of a function [latex]f[\/latex] by looking at the second derivative of [latex]f[\/latex]. In addition, we observe that a function [latex]f[\/latex] can switch concavity (Figure 6). However, a continuous function can switch concavity only at a point [latex]x[\/latex] if [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. Consequently, to determine the intervals where a function [latex]f[\/latex] is concave up and concave down, we look for those values of [latex]x[\/latex] where [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. When we have determined these points, we divide the domain of [latex]f[\/latex] into smaller intervals and determine the sign of [latex]f^{\\prime \\prime}[\/latex] over each of these smaller intervals. If [latex]f^{\\prime \\prime}[\/latex] changes sign as we pass through a point [latex]x[\/latex], then [latex]f[\/latex] changes concavity. It is important to remember that a function [latex]f[\/latex] may not change concavity at a point [latex]x[\/latex] even if [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. If, however, [latex]f[\/latex] does change concavity at a point [latex]a[\/latex] and [latex]f[\/latex] is continuous at [latex]a[\/latex], we say the point [latex](a,f(a))[\/latex] is an <strong>inflection point<\/strong> of [latex]f[\/latex].<\/p>\n<div id=\"fs-id1165043120988\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1165043120992\">If [latex]f[\/latex] is continuous at [latex]a[\/latex] and [latex]f[\/latex] changes concavity at [latex]a[\/latex], the point [latex](a,f(a))[\/latex] is an inflection point of [latex]f[\/latex].<\/p>\n<\/div>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"16\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/7abb392a249b7e8a60b3285996e777d73855d665\" alt=\"A sinusoidal function is shown that has been shifted into the first quadrant. The function starts decreasing, so f\u2019 &lt; 0 and f\u2019\u2019 &gt; 0. The function reaches the local minimum and starts increasing, so f\u2019 &gt; 0 and f\u2019\u2019 &gt; 0. It is noted that the slope is increasing for these two intervals. The function then reaches an inflection point (a, f(a)) and from here the slop is decreasing even though the function continues to increase, so f\u2019 &gt; 0 and f\u2019\u2019 &lt; 0. The function reaches the maximum and then starts decreasing, so f\u2019 &lt; 0 and f\u2019\u2019 &lt; 0.\" width=\"731\" height=\"401\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Since [latex]f^{\\prime \\prime}(x)&gt;0[\/latex] for [latex]x&lt;a[\/latex], the function [latex]f[\/latex] is concave up over the interval [latex](\u2212\\infty,a)[\/latex]. Since [latex]f^{\\prime \\prime}(x)&lt;0[\/latex] for [latex]x&gt;a[\/latex], the function [latex]f[\/latex] is concave down over the interval [latex](a,\\infty)[\/latex]. The point [latex](a,f(a))[\/latex] is an inflection point of [latex]f[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165042558993\" class=\"textbook exercises\">\n<h3>Example: Testing for Concavity<\/h3>\n<p>For the function [latex]f(x)=x^3-6x^2+9x+30[\/latex], determine all intervals where [latex]f[\/latex] is concave up and all intervals where [latex]f[\/latex] is concave down. List all inflection points for [latex]f[\/latex]. Use a graphing utility to confirm your results.<\/p>\n<div id=\"fs-id1165042558995\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043306699\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043306699\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043306699\">To determine concavity, we need to find the second derivative [latex]f^{\\prime \\prime}(x)[\/latex]. The first derivative is [latex]f^{\\prime}(x)=3x^2-12x+9[\/latex], so the second derivative is [latex]f^{\\prime \\prime}(x)=6x-12[\/latex]. If the function changes concavity, it occurs either when [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. Since [latex]f^{\\prime \\prime}[\/latex] is defined for all real numbers [latex]x[\/latex], we need only find where [latex]f^{\\prime \\prime}(x)=0[\/latex]. Solving the equation [latex]6x-12=0[\/latex], we see that [latex]x=2[\/latex] is the only place where [latex]f[\/latex] could change concavity. We now test points over the intervals [latex](\u2212\\infty ,2)[\/latex] and [latex](2,\\infty)[\/latex] to determine the concavity of [latex]f[\/latex]. The points [latex]x=0[\/latex] and [latex]x=3[\/latex] are test points for these intervals.<\/p>\n<table id=\"fs-id1165042960171\" class=\"unnumbered\" summary=\"This table has three 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\u2019(x) = 6x \u2013 12 at Test Point, and Conclusion. Below the header, the first column reads (\u2212\u221e, 2) and (2, \u221e). The second column reads x = 0 and x = 3. The third column reads \u2212 and +. The fourth column reads f is concave down and f is concave up.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}(x)=6x-12[\/latex] at Test Point<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,2)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex]-[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](2,\\infty )[\/latex]<\/td>\n<td>[latex]x=3[\/latex]<\/td>\n<td>[latex]+[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave up.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042329993\">We conclude that [latex]f[\/latex] is concave down over the interval [latex](\u2212\\infty ,2)[\/latex] and concave up over the interval [latex](2,\\infty)[\/latex]. Since [latex]f[\/latex] changes concavity at [latex]x=2[\/latex], the point [latex](2,f(2))=(2,32)[\/latex] is an inflection point. Figure 7\u00a0confirms the analytical results.<\/p>\n<div style=\"width: 389px\" 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\/11210931\/CNX_Calc_Figure_04_05_011.jpg\" alt=\"The function f(x) = x3 \u2013 6x2 + 9x + 30 is graphed. The inflection point (2, 32) is marked, and it is roughly equidistant from the two local extrema.\" width=\"379\" height=\"385\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. The given function has a point of inflection at [latex](2,32)[\/latex] where the graph changes concavity.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Testing for Concavity.<\/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=600&amp;end=713&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">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\/4.5DerivativesAndTheShapeOfAGraph600to713_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>.<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>For [latex]f(x)=\u2212x^3+\\frac{3}{2}x^2+18x[\/latex], find all intervals where [latex]f[\/latex] is concave up and all intervals where [latex]f[\/latex] is concave down.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q33881102\">Hint<\/span><\/p>\n<div id=\"q33881102\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find where [latex]f^{\\prime \\prime}(x)=0[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042369709\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042369709\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f[\/latex] is concave up over the interval [latex](\u2212\\infty ,\\frac{1}{2})[\/latex] and concave down over the interval [latex](\\frac{1}{2},\\infty )[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165043163933\">We now summarize, in the box below, the information that the first and second derivatives of a function [latex]f[\/latex] provide about the graph of [latex]f[\/latex], and illustrate this information in Figure 8.<\/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<caption>What Derivatives Tell Us about Graphs<\/caption>\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<div><\/div>\n<div id=\"fs-id1165042355740\" class=\"bc-section section\">\n<div style=\"width: 911px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"20\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/b4fd75d3ef022de52e9f3b782e4d8908f12a7e23\" alt=\"A function is graphed in the first quadrant. It is broken up into four sections, with the breaks coming at the local minimum, inflection point, and local maximum, respectively. The first section is decreasing and concave up; here, f\u2019 &lt; 0 and f\u2019\u2019 &gt; 0. The second section is increasing and concave up; here, f\u2019 &gt; 0 and f\u2019\u2019 &gt; 0. The third section is increasing and concave down; here, f\u2019 &gt; 0 and f\u2019\u2019 &lt; 0. The fourth section is increasing and concave down; here, f\u2019 &lt; 0 and f\u2019\u2019 &lt; 0.\" width=\"901\" height=\"457\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. Consider a twice-differentiable function [latex]f[\/latex] over an open interval [latex]I[\/latex]. If [latex]f^{\\prime}(x)&gt;0[\/latex] for all [latex]x \\in I[\/latex], the function is increasing over [latex]I[\/latex]. If [latex]f^{\\prime}(x)&lt;0[\/latex] for all [latex]x \\in I[\/latex], the function is decreasing over [latex]I[\/latex]. If [latex]f^{\\prime \\prime}(x)&gt;0[\/latex] for all [latex]x \\in I[\/latex], the function is concave up. If [latex]f^{\\prime \\prime}(x)&lt;0[\/latex] for all [latex]x \\in I[\/latex], the function is concave down on [latex]I[\/latex].<\/p>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-405\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>4.5 Derivatives and the Shape of a Graph. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":16,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-405","chapter","type-chapter","status-publish","hentry"],"part":48,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/405","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":23,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/405\/revisions"}],"predecessor-version":[{"id":3836,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/405\/revisions\/3836"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/48"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/405\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=405"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=405"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=405"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=405"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}