{"id":305,"date":"2023-09-20T22:48:57","date_gmt":"2023-09-20T22:48:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-mean-value-theorem\/"},"modified":"2025-04-16T15:17:59","modified_gmt":"2025-04-16T15:17:59","slug":"the-mean-value-theorem-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-mean-value-theorem-learn-it-1\/","title":{"raw":"The Mean Value Theorem: Learn It 1","rendered":"The Mean Value Theorem: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Use Rolle\u2019s theorem and the Mean Value Theorem to show how functions behave between two points<\/li>\r\n\t<li>Discuss three key implications of the Mean Value Theorem for understanding function behavior<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The Mean Value Theorem<\/h2>\r\n<p>The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let\u2019s start with a special case of the Mean Value Theorem, called Rolle\u2019s theorem.<\/p>\r\n<h3>Rolle\u2019s Theorem<\/h3>\r\n<p id=\"fs-id1165042565539\">Informally, <strong>Rolle\u2019s theorem<\/strong> states that if the outputs of a differentiable function [latex]f[\/latex] are equal at the endpoints of an interval, then there must be an interior point [latex]c[\/latex] where [latex]f^{\\prime}(c)=0[\/latex]. Figure 1 illustrates this theorem.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"887\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210845\/CNX_Calc_Figure_04_04_009.jpg\" alt=\"The figure is divided into three parts labeled a, b, and c. Figure a shows the first quadrant with values a, c, and b marked on the x-axis. A downward-facing parabola is drawn such that its values at a and b are the same. The point c is the global maximum, and it is noted that f\u2019(c) = 0. Figure b shows the first quadrant with values a, c, and b marked on the x-axis. An upward-facing parabola is drawn such that its values at a and b are the same. The point c is the global minimum, and it is noted that f\u2019(c) = 0. Figure c shows the first quadrant with points a, c1, c2, and b marked on the x-axis. One period of a sine wave is drawn such that its values at a and b are equal. The point c1 is the global maximum, and it is noted that f\u2019(c1) = 0. The point c2 is the global minimum, and it is noted that f\u2019(c2) = 0.\" width=\"887\" height=\"311\" \/> Figure 1. If a differentiable function f satisfies [latex]f(a)=f(b)[\/latex], then its derivative must be zero at some point(s) between [latex]a[\/latex] and [latex]b[\/latex].[\/caption]\r\n\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">Rolle\u2019s theorem<\/h3>\r\n<p>Let [latex]f[\/latex] be a continuous function over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex] such that [latex]f(a)=f(b)[\/latex]. There then exists at least one [latex]c \\in (a,b)[\/latex] such that [latex]f^{\\prime}(c)=0[\/latex].<\/p>\r\n<\/section>\r\n<section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n<hr \/>\r\n<p id=\"fs-id1165043430680\">Let [latex]k=f(a)=f(b)[\/latex]. We consider three cases:<\/p>\r\n<ol id=\"fs-id1165043086290\">\r\n\t<li>[latex]f(x)=k[\/latex] for all [latex]x \\in (a,b)[\/latex].<\/li>\r\n\t<li>There exists [latex]x \\in (a,b)[\/latex] such that [latex]f(x)&gt;k[\/latex].<\/li>\r\n\t<li>There exists [latex]x \\in (a,b)[\/latex] such that [latex]f(x) < k[\/latex].<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165042640192\">Case 1: If [latex]f(x)=k[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in (a,b)[\/latex].<\/p>\r\n<p id=\"fs-id1165042974297\">Case 2: Since [latex]f[\/latex] is a continuous function over the closed, bounded interval [latex][a,b][\/latex], by the extreme value theorem, it has an absolute maximum. Also, since there is a point [latex]x \\in (a,b)[\/latex] such that [latex]f(x)&gt;k[\/latex], the absolute maximum is greater than [latex]k[\/latex]. Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point [latex]c \\in (a,b)[\/latex]. Because [latex]f[\/latex] has a maximum at an interior point [latex]c[\/latex], and [latex]f[\/latex] is differentiable at [latex]c[\/latex], by Fermat\u2019s theorem, [latex]f^{\\prime}(c)=0[\/latex].<\/p>\r\n<p id=\"fs-id1165043354252\">Case 3: The case when there exists a point [latex]x \\in (a,b)[\/latex] such that [latex]f(x) < k[\/latex] is analogous to case 2, with maximum replaced by minimum.<\/p>\r\n<p id=\"fs-id1165042333238\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n<p id=\"fs-id1165043111636\">An important point about Rolle\u2019s theorem is that the differentiability of the function [latex]f[\/latex] is critical. If [latex]f[\/latex] is not differentiable, even at a single point, the result may not hold.<\/p>\r\n<p>For example, the function [latex]f(x)=|x|-1[\/latex] is continuous over [latex][-1,1][\/latex] and [latex]f(-1)=0=f(1)[\/latex], but [latex]f^{\\prime}(c) \\ne 0[\/latex] for any [latex]c \\in (-1,1)[\/latex] as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210847\/CNX_Calc_Figure_04_04_002.jpg\" alt=\"The function f(x) = |x| \u2212 1 is graphed. It is shown that f(1) = f(\u22121), but it is noted that there is no c such that f\u2019(c) = 0.\" width=\"325\" height=\"265\" \/> Figure 2. Since [latex]f(x)=|x|-1[\/latex] is not differentiable at [latex]x=0[\/latex], the conditions of Rolle\u2019s theorem are not satisfied. In fact, the conclusion does not hold here; there is no [latex]c \\in (-1,1)[\/latex] such that [latex]f^{\\prime}(c)=0[\/latex].[\/caption]\r\n\r\n\r\n<p id=\"fs-id1165043013898\">Let\u2019s now consider functions that satisfy the conditions of Rolle\u2019s theorem and calculate explicitly the points [latex]c[\/latex] where [latex]f^{\\prime}(c)=0[\/latex].<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165043118608\">For each of the following functions, verify that the function satisfies the criteria stated in Rolle\u2019s theorem and find all values [latex]c[\/latex] in the given interval where [latex]f^{\\prime}(c)=0[\/latex].<\/p>\r\n<ol id=\"fs-id1165043005323\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>[latex]f(x)=x^2+2x[\/latex] over [latex][-2,0][\/latex]<\/li>\r\n\t<li>[latex]f(x)=x^3-4x[\/latex] over [latex][-2,2][\/latex]<\/li>\r\n<\/ol>\r\n\r\n\r\n[reveal-answer q=\"fs-id1165043257912\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043257912\"]\r\n\r\n\r\n<ol id=\"fs-id1165043257912\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>Since [latex]f[\/latex] is a polynomial, it is continuous and differentiable everywhere. In addition, [latex]f(-2)=0=f(0)[\/latex]. Therefore, [latex]f[\/latex] satisfies the criteria of Rolle\u2019s theorem. We conclude that there exists at least one value [latex]c \\in (-2,0)[\/latex] such that [latex]f^{\\prime}(c)=0[\/latex]. Since [latex]f^{\\prime}(x)=2x+2=2(x+1)[\/latex], we see that [latex]f^{\\prime}(c)=2(c+1)=0[\/latex] implies [latex]c=-1[\/latex] as shown in the following graph.\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210849\/CNX_Calc_Figure_04_04_003.jpg\" alt=\"The function f(x) = x2 +2x is graphed. It is shown that f(0) = f(\u22122), and a dashed horizontal line is drawn at the absolute minimum at (\u22121, \u22121).\" width=\"487\" height=\"312\" \/> Figure 3. This function is continuous and differentiable over [latex][-2,0][\/latex], [latex]f^{\\prime}(c)=0[\/latex] when [latex]c=-1[\/latex].[\/caption]\r\n<\/li>\r\n\t<li>As in part a., [latex]f[\/latex] is a polynomial and therefore is continuous and differentiable everywhere. Also, [latex]f(-2)=0=f(2)[\/latex]. That said, [latex]f[\/latex] satisfies the criteria of Rolle\u2019s theorem. Differentiating, we find that [latex]f^{\\prime}(x)=3x^2-4[\/latex]. Therefore, [latex]f^{\\prime}(c)=0[\/latex] when [latex]x=\\pm \\frac{2}{\\sqrt{3}}[\/latex]. Both points are in the interval [latex][-2,2][\/latex], and, therefore, both points satisfy the conclusion of Rolle\u2019s theorem as shown in the following graph.\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210853\/CNX_Calc_Figure_04_04_004.jpg\" alt=\"The function f(x) = x3 \u2013 4x is graphed. It is obvious that f(2) = f(\u22122) = f(0). Dashed horizontal lines are drawn at x = \u00b12\/square root of 3, which are the local maximum and minimum.\" width=\"417\" height=\"572\" \/> Figure 4. For this polynomial over [latex][-2,2][\/latex], [latex]f^{\\prime}(c)=0[\/latex] at [latex]x=\\pm 2\/\\sqrt{3}[\/latex].[\/caption]\r\n<\/li>\r\n<\/ol>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Verify that the function [latex]f(x)=2x^2-8x+6[\/latex] defined over the interval [latex][1,3][\/latex] satisfies the conditions of Rolle\u2019s theorem. Find all points [latex]c[\/latex] guaranteed by Rolle\u2019s theorem.<\/p>\r\n\r\n\r\n[reveal-answer q=\"4366285\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"4366285\"]\r\n\r\n\r\n<p id=\"fs-id1165043120638\">Find all values [latex]c[\/latex], where [latex]f^{\\prime}(c)=0[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n\r\n\r\n[reveal-answer q=\"fs-id1165043098113\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043098113\"]\r\n\r\n\r\n<p id=\"fs-id1165043098113\">[latex]c=2[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox watchIt\">\r\n<p>Watch the following video to see the worked solution to the two previous examples.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/meMefcJWbrQ?controls=0&amp;start=118&amp;end=339&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]<br \/>\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.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.4MeanValueTheorem118to339_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.4 Mean Value Theorem\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Use Rolle\u2019s theorem and the Mean Value Theorem to show how functions behave between two points<\/li>\n<li>Discuss three key implications of the Mean Value Theorem for understanding function behavior<\/li>\n<\/ul>\n<\/section>\n<h2>The Mean Value Theorem<\/h2>\n<p>The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let\u2019s start with a special case of the Mean Value Theorem, called Rolle\u2019s theorem.<\/p>\n<h3>Rolle\u2019s Theorem<\/h3>\n<p id=\"fs-id1165042565539\">Informally, <strong>Rolle\u2019s theorem<\/strong> states that if the outputs of a differentiable function [latex]f[\/latex] are equal at the endpoints of an interval, then there must be an interior point [latex]c[\/latex] where [latex]f^{\\prime}(c)=0[\/latex]. Figure 1 illustrates this theorem.<\/p>\n<figure style=\"width: 887px\" 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\/11210845\/CNX_Calc_Figure_04_04_009.jpg\" alt=\"The figure is divided into three parts labeled a, b, and c. Figure a shows the first quadrant with values a, c, and b marked on the x-axis. A downward-facing parabola is drawn such that its values at a and b are the same. The point c is the global maximum, and it is noted that f\u2019(c) = 0. Figure b shows the first quadrant with values a, c, and b marked on the x-axis. An upward-facing parabola is drawn such that its values at a and b are the same. The point c is the global minimum, and it is noted that f\u2019(c) = 0. Figure c shows the first quadrant with points a, c1, c2, and b marked on the x-axis. One period of a sine wave is drawn such that its values at a and b are equal. The point c1 is the global maximum, and it is noted that f\u2019(c1) = 0. The point c2 is the global minimum, and it is noted that f\u2019(c2) = 0.\" width=\"887\" height=\"311\" \/><figcaption class=\"wp-caption-text\">Figure 1. If a differentiable function f satisfies [latex]f(a)=f(b)[\/latex], then its derivative must be zero at some point(s) between [latex]a[\/latex] and [latex]b[\/latex].<\/figcaption><\/figure>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">Rolle\u2019s theorem<\/h3>\n<p>Let [latex]f[\/latex] be a continuous function over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex] such that [latex]f(a)=f(b)[\/latex]. There then exists at least one [latex]c \\in (a,b)[\/latex] such that [latex]f^{\\prime}(c)=0[\/latex].<\/p>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1165043430680\">Let [latex]k=f(a)=f(b)[\/latex]. We consider three cases:<\/p>\n<ol id=\"fs-id1165043086290\">\n<li>[latex]f(x)=k[\/latex] for all [latex]x \\in (a,b)[\/latex].<\/li>\n<li>There exists [latex]x \\in (a,b)[\/latex] such that [latex]f(x)>k[\/latex].<\/li>\n<li>There exists [latex]x \\in (a,b)[\/latex] such that [latex]f(x) < k[\/latex].<\/li>\n<\/ol>\n<p id=\"fs-id1165042640192\">Case 1: If [latex]f(x)=k[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in (a,b)[\/latex].<\/p>\n<p id=\"fs-id1165042974297\">Case 2: Since [latex]f[\/latex] is a continuous function over the closed, bounded interval [latex][a,b][\/latex], by the extreme value theorem, it has an absolute maximum. Also, since there is a point [latex]x \\in (a,b)[\/latex] such that [latex]f(x)>k[\/latex], the absolute maximum is greater than [latex]k[\/latex]. Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point [latex]c \\in (a,b)[\/latex]. Because [latex]f[\/latex] has a maximum at an interior point [latex]c[\/latex], and [latex]f[\/latex] is differentiable at [latex]c[\/latex], by Fermat\u2019s theorem, [latex]f^{\\prime}(c)=0[\/latex].<\/p>\n<p id=\"fs-id1165043354252\">Case 3: The case when there exists a point [latex]x \\in (a,b)[\/latex] such that [latex]f(x) < k[\/latex] is analogous to case 2, with maximum replaced by minimum.<\/p>\n<p id=\"fs-id1165042333238\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<p id=\"fs-id1165043111636\">An important point about Rolle\u2019s theorem is that the differentiability of the function [latex]f[\/latex] is critical. If [latex]f[\/latex] is not differentiable, even at a single point, the result may not hold.<\/p>\n<p>For example, the function [latex]f(x)=|x|-1[\/latex] is continuous over [latex][-1,1][\/latex] and [latex]f(-1)=0=f(1)[\/latex], but [latex]f^{\\prime}(c) \\ne 0[\/latex] for any [latex]c \\in (-1,1)[\/latex] as shown in the following figure.<\/p>\n<figure style=\"width: 325px\" 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\/11210847\/CNX_Calc_Figure_04_04_002.jpg\" alt=\"The function f(x) = |x| \u2212 1 is graphed. It is shown that f(1) = f(\u22121), but it is noted that there is no c such that f\u2019(c) = 0.\" width=\"325\" height=\"265\" \/><figcaption class=\"wp-caption-text\">Figure 2. Since [latex]f(x)=|x|-1[\/latex] is not differentiable at [latex]x=0[\/latex], the conditions of Rolle\u2019s theorem are not satisfied. In fact, the conclusion does not hold here; there is no [latex]c \\in (-1,1)[\/latex] such that [latex]f^{\\prime}(c)=0[\/latex].<\/figcaption><\/figure>\n<p id=\"fs-id1165043013898\">Let\u2019s now consider functions that satisfy the conditions of Rolle\u2019s theorem and calculate explicitly the points [latex]c[\/latex] where [latex]f^{\\prime}(c)=0[\/latex].<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1165043118608\">For each of the following functions, verify that the function satisfies the criteria stated in Rolle\u2019s theorem and find all values [latex]c[\/latex] in the given interval where [latex]f^{\\prime}(c)=0[\/latex].<\/p>\n<ol id=\"fs-id1165043005323\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]f(x)=x^2+2x[\/latex] over [latex][-2,0][\/latex]<\/li>\n<li>[latex]f(x)=x^3-4x[\/latex] over [latex][-2,2][\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043257912\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043257912\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165043257912\" style=\"list-style-type: lower-alpha;\">\n<li>Since [latex]f[\/latex] is a polynomial, it is continuous and differentiable everywhere. In addition, [latex]f(-2)=0=f(0)[\/latex]. Therefore, [latex]f[\/latex] satisfies the criteria of Rolle\u2019s theorem. We conclude that there exists at least one value [latex]c \\in (-2,0)[\/latex] such that [latex]f^{\\prime}(c)=0[\/latex]. Since [latex]f^{\\prime}(x)=2x+2=2(x+1)[\/latex], we see that [latex]f^{\\prime}(c)=2(c+1)=0[\/latex] implies [latex]c=-1[\/latex] as shown in the following graph.<br \/>\n<figure style=\"width: 487px\" 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\/11210849\/CNX_Calc_Figure_04_04_003.jpg\" alt=\"The function f(x) = x2 +2x is graphed. It is shown that f(0) = f(\u22122), and a dashed horizontal line is drawn at the absolute minimum at (\u22121, \u22121).\" width=\"487\" height=\"312\" \/><figcaption class=\"wp-caption-text\">Figure 3. This function is continuous and differentiable over [latex][-2,0][\/latex], [latex]f^{\\prime}(c)=0[\/latex] when [latex]c=-1[\/latex].<\/figcaption><\/figure>\n<\/li>\n<li>As in part a., [latex]f[\/latex] is a polynomial and therefore is continuous and differentiable everywhere. Also, [latex]f(-2)=0=f(2)[\/latex]. That said, [latex]f[\/latex] satisfies the criteria of Rolle\u2019s theorem. Differentiating, we find that [latex]f^{\\prime}(x)=3x^2-4[\/latex]. Therefore, [latex]f^{\\prime}(c)=0[\/latex] when [latex]x=\\pm \\frac{2}{\\sqrt{3}}[\/latex]. Both points are in the interval [latex][-2,2][\/latex], and, therefore, both points satisfy the conclusion of Rolle\u2019s theorem as shown in the following graph.<br \/>\n<figure style=\"width: 417px\" 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\/11210853\/CNX_Calc_Figure_04_04_004.jpg\" alt=\"The function f(x) = x3 \u2013 4x is graphed. It is obvious that f(2) = f(\u22122) = f(0). Dashed horizontal lines are drawn at x = \u00b12\/square root of 3, which are the local maximum and minimum.\" width=\"417\" height=\"572\" \/><figcaption class=\"wp-caption-text\">Figure 4. For this polynomial over [latex][-2,2][\/latex], [latex]f^{\\prime}(c)=0[\/latex] at [latex]x=\\pm 2\/\\sqrt{3}[\/latex].<\/figcaption><\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Verify that the function [latex]f(x)=2x^2-8x+6[\/latex] defined over the interval [latex][1,3][\/latex] satisfies the conditions of Rolle\u2019s theorem. Find all points [latex]c[\/latex] guaranteed by Rolle\u2019s theorem.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q4366285\">Hint<\/button><\/p>\n<div id=\"q4366285\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043120638\">Find all values [latex]c[\/latex], where [latex]f^{\\prime}(c)=0[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043098113\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043098113\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043098113\">[latex]c=2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\">\n<p>Watch the following video to see the worked solution to the two previous examples.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/meMefcJWbrQ?controls=0&amp;start=118&amp;end=339&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/button><\/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\/4.4MeanValueTheorem118to339_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.4 Mean Value Theorem\" here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":20,"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\":\"4.4 Mean Value Theorem\",\"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":"4.4 Mean Value Theorem","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\/305"}],"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":10,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/305\/revisions"}],"predecessor-version":[{"id":4698,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/305\/revisions\/4698"}],"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\/305\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=305"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=305"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=305"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=305"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}