{"id":402,"date":"2021-02-04T01:59:54","date_gmt":"2021-02-04T01:59:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=402"},"modified":"2022-03-16T05:45:07","modified_gmt":"2022-03-16T05:45:07","slug":"the-mean-value-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/the-mean-value-theorem\/","title":{"raw":"The Mean Value Theorem","rendered":"The Mean Value Theorem"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Explain the meaning of Rolle\u2019s theorem<\/li>\r\n \t<li>Describe the significance of the Mean Value Theorem<\/li>\r\n \t<li>State three important consequences of the Mean Value Theorem<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Rolle\u2019s Theorem<\/h2>\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[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<div id=\"fs-id1165042955477\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Rolle\u2019s Theorem<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165042609174\">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\r\n<\/div>\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1165043430680\">Let [latex]k=f(a)=f(b)[\/latex]. We consider three cases:<\/p>\r\n\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)&lt;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)&lt;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<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. 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[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<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\r\n<div id=\"fs-id1165043010693\" class=\"textbook exercises\">\r\n<h3>Example: Using Rolle\u2019s Theorem<\/h3>\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\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<div id=\"fs-id1165042329638\" class=\"exercise\">[reveal-answer q=\"fs-id1165043257912\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043257912\"]\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.[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<div class=\"wp-caption-text\"><\/div><\/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.[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]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042986791\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nVerify 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.\r\n<div>[reveal-answer q=\"4366285\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"4366285\"]\r\n<p id=\"fs-id1165043120638\">Find all values [latex]c[\/latex], where [latex]f^{\\prime}(c)=0[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042367594\" class=\"exercise\">[reveal-answer q=\"fs-id1165043098113\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043098113\"]\r\n<p id=\"fs-id1165043098113\">[latex]c=2[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Using Rolle\u2019s Theorem and the above Try It.\r\n\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\"><\/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\/4.4MeanValueTheorem118to339_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.4 Mean Value Theorem\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<h2>The Mean Value Theorem and Its Meaning<\/h2>\r\n<p id=\"fs-id1165043423200\">Rolle\u2019s theorem is a special case of the Mean Value Theorem. In Rolle\u2019s theorem, we consider differentiable functions [latex]f[\/latex]defined on a closed interval\u00a0[latex][a,b][\/latex] with [latex]f(a)=f(b)[\/latex].\u00a0The Mean Value Theorem generalizes Rolle\u2019s theorem by considering functions that do not necessarily have equal value at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle\u2019s theorem\u00a0(Figure 5).\u00a0The Mean Value Theorem states that if [latex]f[\/latex] is continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex], then there exists a point [latex]c \\in (a,b)[\/latex] such that the tangent line to the graph of [latex]f[\/latex] at [latex]c[\/latex] is parallel to the secant line connecting [latex](a,f(a))[\/latex] and [latex](b,f(b))[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"452\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210855\/CNX_Calc_Figure_04_04_010.jpg\" alt=\"A vaguely sinusoidal function y = f(x) is drawn. On the x-axis, a, c1, c2, and b are marked. On the y-axis, f(a) and f(b) are marked. The function f(x) starts at (a, f(a)), decreases to c1, increases to c2, and then decreases to (b, f(b)). A secant line is drawn between (a, f(a)) and (b, f(b)), and it is noted that this line has slope (f(b) \u2013 f(a))\/(b \u2212 a). The tangent lines at c1 and c2 are drawn, and these lines are parallel to the secant line. It is noted that the slopes of these tangent lines are f\u2019(c1) and f\u2019(c2), respectively.\" width=\"452\" height=\"293\" \/> Figure 5. The Mean Value Theorem says that for a function that meets its conditions, at some point the tangent line has the same slope as the secant line between the ends. For this function, there are two values [latex]c_1[\/latex] and [latex]c_2[\/latex] such that the tangent line to [latex]f[\/latex] at [latex]c_1[\/latex] and [latex]c_2[\/latex] has the same slope as the secant line.[\/caption]\r\n<div id=\"fs-id1165043099288\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Mean Value Theorem<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165042357018\">Let [latex]f[\/latex] be continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex]. Then, there exists at least one point [latex]c \\in (a,b)[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1165043066505\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042556204\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1165043096684\">The proof follows from Rolle\u2019s theorem by introducing an appropriate function that satisfies the criteria of Rolle\u2019s theorem. Consider the line connecting [latex](a,f(a))[\/latex] and [latex](b,f(b))[\/latex]. Since the slope of that line is<\/p>\r\n\r\n<div id=\"fs-id1165042514053\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042970671\">and the line passes through the point [latex](a,f(a))[\/latex], the equation of that line can be written as<\/p>\r\n\r\n<div id=\"fs-id1165042326168\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042989371\">Let [latex]g(x)[\/latex] denote the vertical difference between the point [latex](x,f(x))[\/latex] and the point [latex](x,y)[\/latex] on that line. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165043100161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)=f(x)-\\left[\\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)\\right][\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"315\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210858\/CNX_Calc_Figure_04_04_011.jpg\" alt=\"A vaguely sinusoidal function y = f(x) is drawn. On the x-axis, a and b are marked. On the y-axis, f(a) and f(b) are marked. The function f(x) starts at (a, f(a)), decreases, then increases, and then decreases to (b, f(b)). A secant line is drawn between (a, f(a)) and (b, f(b)), and it is noted that this line has equation y = ((f(b) \u2013 f(a))\/(b \u2212 a)) (x \u2212 a) + f(x). A line is drawn between the maximum of f(x) and the secant line and it is marked g(x).\" width=\"315\" height=\"272\" \/> Figure 6. The value [latex]g(x)[\/latex] is the vertical difference between the point [latex](x,f(x))[\/latex] and the point [latex](x,y)[\/latex] on the secant line connecting [latex](a,f(a))[\/latex] and [latex](b,f(b)).[\/latex][\/caption]<\/div>\r\n<p id=\"fs-id1165042892921\">Since the graph of [latex]f[\/latex] intersects the secant line when [latex]x=a[\/latex] and [latex]x=b[\/latex], we see that [latex]g(a)=0=g(b)[\/latex]. Since [latex]f[\/latex] is a differentiable function over [latex](a,b)[\/latex], [latex]g[\/latex] is also a differentiable function over [latex](a,b)[\/latex]. Furthermore, since [latex]f[\/latex] is continuous over [latex][a,b][\/latex], [latex]g[\/latex] is also continuous over [latex][a,b][\/latex]. Therefore, [latex]g[\/latex] satisfies the criteria of Rolle\u2019s theorem. Consequently, there exists a point [latex]c \\in (a,b)[\/latex] such that [latex]g^{\\prime}(c)=0[\/latex]. Since<\/p>\r\n\r\n<div id=\"fs-id1165042979927\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=f^{\\prime}(x)-\\dfrac{f(b)-f(a)}{b-a}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042569046\">we see that<\/p>\r\n\r\n<div id=\"fs-id1165042991202\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(c)=f^{\\prime}(c)-\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042369562\">Since [latex]g^{\\prime}(c)=0[\/latex], we conclude that<\/p>\r\n\r\n<div id=\"fs-id1165042608728\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n<p id=\"fs-id1165042639297\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<p id=\"fs-id1165043131938\">In the next example, we show how the Mean Value Theorem can be applied to the function [latex]f(x)=\\sqrt{x}[\/latex] over the interval [latex][0,9][\/latex]. The method is the same for other functions, although sometimes with more interesting consequences.<\/p>\r\n\r\n<div id=\"fs-id1165042979713\" class=\"textbook exercises\">\r\n<h3>Example: Verifying that the Mean Value Theorem Applies<\/h3>\r\nFor [latex]f(x)=\\sqrt{x}[\/latex] over the interval [latex][0,9][\/latex], show that [latex]f[\/latex] satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value [latex]c \\in (0,9)[\/latex] such that [latex]f^{\\prime}(c)[\/latex] is equal to the slope of the line connecting [latex](0,f(0))[\/latex] and [latex](9,f(9))[\/latex]. Find these values [latex]c[\/latex] guaranteed by the Mean Value Theorem.\r\n<div id=\"fs-id1165042478873\" class=\"exercise\">[reveal-answer q=\"fs-id1165043395556\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043395556\"]\r\n<p id=\"fs-id1165043395556\">We know that [latex]f(x)=\\sqrt{x}[\/latex] is continuous over [latex][0,9][\/latex] and differentiable over [latex](0,9)[\/latex]. Therefore, [latex]f[\/latex] satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value [latex]c \\in (0,9)[\/latex] such that [latex]f^{\\prime}(c)[\/latex] is equal to the slope of the line connecting [latex](0,f(0))[\/latex] and [latex](9,f(9))[\/latex] (Figure 7). To determine which value(s) of [latex]c[\/latex] are guaranteed, first calculate the derivative of [latex]f[\/latex]. The derivative [latex]f^{\\prime}(x)=\\frac{1}{2\\sqrt{x}}[\/latex]. The slope of the line connecting [latex](0,f(0))[\/latex] and [latex](9,f(9))[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1165043251015\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(9)-f(0)}{9-0}=\\dfrac{\\sqrt{9}-\\sqrt{0}}{9-0}=\\dfrac{3}{9}=\\dfrac{1}{3}[\/latex]<\/div>\r\n<p id=\"fs-id1165043096971\">We want to find [latex]c[\/latex] such that [latex]f^{\\prime}(c)=\\frac{1}{3}[\/latex]. That is, we want to find [latex]c[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1165042375801\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{1}{2\\sqrt{c}}=\\dfrac{1}{3}[\/latex]<\/div>\r\nSolving this equation for [latex]c[\/latex], we obtain [latex]c=\\frac{9}{4}[\/latex]. At this point, the slope of the tangent line equals the slope of the line joining the endpoints.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"829\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210901\/CNX_Calc_Figure_04_04_006.jpg\" alt=\"The function f(x) = the square root of x is graphed from (0, 0) to (9, 3). There is a secant line drawn from (0, 0) to (9, 3). At point (9\/4, 3\/2), there is a tangent line that is drawn, and this line is parallel to the secant line.\" width=\"829\" height=\"459\" \/> Figure 7. The slope of the tangent line at [latex]c=\\frac{9}{4}[\/latex] is the same as the slope of the line segment connecting [latex](0,0)[\/latex] and [latex](9,3)[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165042374751\">One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 hr down a straight road with an average velocity of 45 mph. Let [latex]s(t)[\/latex] and [latex]v(t)[\/latex] denote the position and velocity of the car, respectively, for [latex]0 \\le t \\le 1[\/latex] hr. Assuming that the position function [latex]s(t)[\/latex] is differentiable, we can apply the Mean Value Theorem to conclude that, at some time [latex]c \\in (0,1)[\/latex], the speed of the car was exactly<\/p>\r\n\r\n<div id=\"fs-id1165042647085\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(c)=s^{\\prime}(c)=\\dfrac{s(1)-s(0)}{1-0}=45[\/latex] mph<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1165042332061\" class=\"textbook exercises\">\r\n<h3>Example: Mean Value Theorem and Velocity<\/h3>\r\n<p id=\"fs-id1165043312536\">If a rock is dropped from a height of 100 ft, its position [latex]t[\/latex] seconds after it is dropped until it hits the ground is given by the function [latex]s(t)=-16t^2+100[\/latex].<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Determine how long it takes before the rock hits the ground.<\/li>\r\n \t<li>Find the average velocity [latex]v_{\\text{avg}}[\/latex] of the rock for when the rock is released and the rock hits the ground.<\/li>\r\n \t<li>Find the time [latex]t[\/latex] guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is [latex]v_{\\text{avg}}[\/latex].<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165042332064\" class=\"exercise\">[reveal-answer q=\"fs-id1165042373172\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042373172\"]\r\n<ol id=\"fs-id1165042373172\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>When the rock hits the ground, its position is [latex]s(t)=0[\/latex]. Solving the equation [latex]-16t^2+100=0[\/latex] for [latex]t[\/latex], we find that [latex]t=\\pm \\frac{5}{2}[\/latex] sec.\u00a0Since we are only considering [latex]t \\ge 0[\/latex], the ball will hit the ground [latex]\\frac{5}{2}[\/latex] sec after it is dropped.<\/li>\r\n \t<li>The average velocity is given by\r\n<div id=\"fs-id1165043354673\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v_{\\text{avg}}=\\frac{s(5\/2)-s(0)}{5\/2-0}=\\frac{1-100}{5\/2}=-40[\/latex] ft\/sec<\/div><\/li>\r\n \t<li>The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time [latex]t[\/latex] such that [latex]v(t)=s^{\\prime}(t)=v_{\\text{avg}}=-40[\/latex] ft\/sec.\u00a0Since [latex]s(t)[\/latex] is continuous over the interval [latex][0,5\/2][\/latex] and differentiable over the interval [latex](0,5\/2)[\/latex], by the Mean Value Theorem, there is guaranteed to be a point [latex]c \\in (0,5\/2)[\/latex] such that\r\n<div id=\"fs-id1165043390900\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s^{\\prime}(c)=\\frac{s(5\/2)-s(0)}{5\/2-0}=-40[\/latex]<\/div>\r\nTaking the derivative of the position function [latex]s(t)[\/latex], we find that [latex]s^{\\prime}(t)=-32t[\/latex]. Therefore, the equation reduces to [latex]s^{\\prime}(c)=-32c=-40[\/latex]. Solving this equation for [latex]c[\/latex], we have [latex]c=\\frac{5}{4}[\/latex]. Therefore, [latex]\\frac{5}{4}[\/latex] sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: -40 ft\/sec.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"454\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210904\/CNX_Calc_Figure_04_04_007.jpg\" alt=\"The function s(t) = \u221216t2 + 100 is graphed from (0, 100) to (5\/2, 0). There is a secant line drawn from (0, 100) to (5\/2, 0). At the point corresponding to x = 5\/4, there is a tangent line that is drawn, and this line is parallel to the secant line.\" width=\"454\" height=\"272\" \/> Figure 8. At time [latex]t=\\frac{5}{4}[\/latex] sec, the velocity of the rock is equal to its average velocity from the time it is dropped until it hits the ground.[\/caption]\r\n<div class=\"wp-caption-text\"><\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Mean Value Theorem and Velocity.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/meMefcJWbrQ?controls=0&amp;start=931&amp;end=1103&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\/4.4MeanValueTheorem931to1103_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.4 Mean Value Theorem\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose a ball is dropped from a height of 200 ft. Its position at time [latex]t[\/latex] is [latex]s(t)=-16t^2+200[\/latex]. Find the time [latex]t[\/latex] when the instantaneous velocity of the ball equals its average velocity.\r\n\r\n[reveal-answer q=\"9003136\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"9003136\"]\r\n\r\nFirst, determine how long it takes for the ball to hit the ground. Then, find the average velocity of the ball from the time it is dropped until it hits the ground.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042631892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042631892\"]\r\n\r\n[latex]\\dfrac{5}{2\\sqrt{2}}[\/latex] sec\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]6147[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Corollaries of the Mean Value Theorem<\/h2>\r\n<p id=\"fs-id1165042640871\">Let\u2019s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections.<\/p>\r\n<p id=\"fs-id1165042640877\">At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x[\/latex] in some interval [latex]I[\/latex], then [latex]f(x)[\/latex] is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly.<\/p>\r\n\r\n<div id=\"fs-id1165042645708\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Corollary 1: Functions with a Derivative of Zero<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165042645714\">Let [latex]f[\/latex] be differentiable over an interval [latex]I[\/latex]. If [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f(x)[\/latex] is constant for all [latex]x \\in I[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043251996\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1165043252002\">Since [latex]f[\/latex] is differentiable over [latex]I[\/latex], [latex]f[\/latex] must be continuous over [latex]I[\/latex]. Suppose [latex]f(x)[\/latex] is not constant for all [latex]x[\/latex] in [latex]I[\/latex]. Then there exist [latex]a,b \\in I[\/latex], where [latex]a \\ne b[\/latex] and [latex]f(a) \\ne f(b)[\/latex]. Choose the notation so that [latex]a&lt;b[\/latex]. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165042708535\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(b)-f(a)}{b-a} \\ne 0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042708255\">Since [latex]f[\/latex] is a differentiable function, by the Mean Value Theorem, there exists [latex]c \\in (a,b)[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1165042418104\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043259845\">Therefore, there exists [latex]c \\in I[\/latex] such that [latex]f^{\\prime}(c) \\ne 0[\/latex], which contradicts the assumption that [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex].<\/p>\r\n<p id=\"fs-id1165042707831\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<p id=\"fs-id1165042707834\">From the example above, it follows that if two functions have the same derivative, they differ by, at most, a constant.<\/p>\r\n\r\n<div id=\"fs-id1165042707841\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Corollary 2: Constant Difference Theorem<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165042707847\">If [latex]f[\/latex] and [latex]g[\/latex] are differentiable over an interval [latex]I[\/latex] and [latex]f^{\\prime}(x)=g^{\\prime}(x)[\/latex] for all [latex]x \\in I[\/latex], then [latex]f(x)=g(x)+C[\/latex] for some constant [latex]C[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042707738\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1165042707744\">Let [latex]h(x)=f(x)-g(x)[\/latex]. Then, [latex]h^{\\prime}(x)=f^{\\prime}(x)-g^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex]. By Corollary 1, there is a constant [latex]C[\/latex] such that [latex]h(x)=C[\/latex] for all [latex]x \\in I[\/latex]. Therefore, [latex]f(x)=g(x)+C[\/latex] for all [latex]x \\in I[\/latex].<\/p>\r\n<p id=\"fs-id1165043424693\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<p id=\"fs-id1165043424696\">The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function [latex]f[\/latex] is increasing over [latex]I[\/latex] if [latex]f(x_1)&lt;f(x_2)[\/latex] whenever [latex]x_1&lt;x_2[\/latex], whereas [latex]f[\/latex] is decreasing over [latex]I[\/latex] if [latex]f(x_1)&gt;f(x_2)[\/latex] whenever [latex]x_1&lt;x_2[\/latex]. Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure 9). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph.<\/p>\r\nThis fact is important because it means that for a given function [latex]f[\/latex], if there exists a function [latex]F[\/latex] such that [latex]F^{\\prime}(x)=f(x)[\/latex]; then, the only other functions that have a derivative equal to [latex]f[\/latex] are [latex]F(x)+C[\/latex] for some constant [latex]C[\/latex]. We discuss this result in more detail later in the chapter.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img id=\"19\" src=\"https:\/\/openstax.org\/resources\/1ae209b72e21942c9811cc8e33e8be432261ed5f\" alt=\"A vaguely sinusoidal function f(x) is graphed. It increases from somewhere in the second quadrant to (a, f(a)). In this section it is noted that f\u2019 &gt; 0. Then in decreases from (a, f(a)) to (b, f(b)). In this section it is noted that f\u2019 &lt; 0. Finally, it increases to the right of (b, f(b)) and it is noted in this section that f\u2019 &gt; 0\" width=\"731\" height=\"302\" \/> Figure 9. If a function has a positive derivative over some interval [latex]I[\/latex], then the function increases over that interval [latex]I[\/latex]; if the derivative is negative over some interval [latex]I[\/latex], then the function decreases over that interval [latex]I[\/latex].[\/caption]\r\n<div id=\"fs-id1165043217934\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Corollary 3: Increasing and Decreasing Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165043217941\">Let [latex]f[\/latex] be continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1165043217981\">\r\n \t<li>If [latex]f^{\\prime}(x)&gt;0[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f[\/latex] is an increasing function over [latex][a,b][\/latex].<\/li>\r\n \t<li>If [latex]f^{\\prime}(x)&lt;0[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f[\/latex] is a decreasing function over [latex][a,b][\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043327297\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1165043327302\">We will prove 1.; the proof of 2. is similar. Suppose [latex]f[\/latex] is not an increasing function on [latex]I[\/latex]. Then there exist [latex]a[\/latex] and [latex]b[\/latex] in [latex]I[\/latex] such that [latex]a&lt;b[\/latex], but [latex]f(a) \\ge f(b)[\/latex]. Since [latex]f[\/latex] is a differentiable function over [latex]I[\/latex], by the Mean Value Theorem there exists [latex]c \\in (a,b)[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1165043327400\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043327455\">Since [latex]f(a) \\ge f(b)[\/latex], we know that [latex]f(b)-f(a) \\le 0[\/latex]. Also, [latex]a&lt;b[\/latex] tells us that [latex]b-a&gt;0[\/latex]. We conclude that<\/p>\r\n\r\n<div id=\"fs-id1165042651499\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a} \\le 0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042651558\">However, [latex]f^{\\prime}(x)&gt;0[\/latex] for all [latex]x \\in I[\/latex]. This is a contradiction, and therefore [latex]f[\/latex] must be an increasing function over [latex]I[\/latex].<\/p>\r\n<p id=\"fs-id1165042651604\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Explain the meaning of Rolle\u2019s theorem<\/li>\n<li>Describe the significance of the Mean Value Theorem<\/li>\n<li>State three important consequences of the Mean Value Theorem<\/li>\n<\/ul>\n<\/div>\n<h2>Rolle\u2019s Theorem<\/h2>\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<div style=\"width: 897px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<div id=\"fs-id1165042955477\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Rolle\u2019s Theorem<\/h3>\n<hr \/>\n<p id=\"fs-id1165042609174\">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<\/div>\n<h3>Proof<\/h3>\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<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. 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<div style=\"width: 335px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\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<div id=\"fs-id1165043010693\" class=\"textbook exercises\">\n<h3>Example: Using Rolle\u2019s Theorem<\/h3>\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 id=\"fs-id1165042329638\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043257912\">Show Solution<\/span><\/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.\n<div style=\"width: 497px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\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.\n<div style=\"width: 427px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042986791\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\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>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q4366285\">Hint<\/span><\/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>\n<div id=\"fs-id1165042367594\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043098113\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Using Rolle\u2019s Theorem and the above Try It.<\/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\"><\/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\/4.4MeanValueTheorem118to339_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.4 Mean Value Theorem&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<h2>The Mean Value Theorem and Its Meaning<\/h2>\n<p id=\"fs-id1165043423200\">Rolle\u2019s theorem is a special case of the Mean Value Theorem. In Rolle\u2019s theorem, we consider differentiable functions [latex]f[\/latex]defined on a closed interval\u00a0[latex][a,b][\/latex] with [latex]f(a)=f(b)[\/latex].\u00a0The Mean Value Theorem generalizes Rolle\u2019s theorem by considering functions that do not necessarily have equal value at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle\u2019s theorem\u00a0(Figure 5).\u00a0The Mean Value Theorem states that if [latex]f[\/latex] is continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex], then there exists a point [latex]c \\in (a,b)[\/latex] such that the tangent line to the graph of [latex]f[\/latex] at [latex]c[\/latex] is parallel to the secant line connecting [latex](a,f(a))[\/latex] and [latex](b,f(b))[\/latex].<\/p>\n<div style=\"width: 462px\" 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\/11210855\/CNX_Calc_Figure_04_04_010.jpg\" alt=\"A vaguely sinusoidal function y = f(x) is drawn. On the x-axis, a, c1, c2, and b are marked. On the y-axis, f(a) and f(b) are marked. The function f(x) starts at (a, f(a)), decreases to c1, increases to c2, and then decreases to (b, f(b)). A secant line is drawn between (a, f(a)) and (b, f(b)), and it is noted that this line has slope (f(b) \u2013 f(a))\/(b \u2212 a). The tangent lines at c1 and c2 are drawn, and these lines are parallel to the secant line. It is noted that the slopes of these tangent lines are f\u2019(c1) and f\u2019(c2), respectively.\" width=\"452\" height=\"293\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. The Mean Value Theorem says that for a function that meets its conditions, at some point the tangent line has the same slope as the secant line between the ends. For this function, there are two values [latex]c_1[\/latex] and [latex]c_2[\/latex] such that the tangent line to [latex]f[\/latex] at [latex]c_1[\/latex] and [latex]c_2[\/latex] has the same slope as the secant line.<\/p>\n<\/div>\n<div id=\"fs-id1165043099288\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Mean Value Theorem<\/h3>\n<hr \/>\n<p id=\"fs-id1165042357018\">Let [latex]f[\/latex] be continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex]. Then, there exists at least one point [latex]c \\in (a,b)[\/latex] such that<\/p>\n<div id=\"fs-id1165043066505\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165042556204\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1165043096684\">The proof follows from Rolle\u2019s theorem by introducing an appropriate function that satisfies the criteria of Rolle\u2019s theorem. Consider the line connecting [latex](a,f(a))[\/latex] and [latex](b,f(b))[\/latex]. Since the slope of that line is<\/p>\n<div id=\"fs-id1165042514053\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042970671\">and the line passes through the point [latex](a,f(a))[\/latex], the equation of that line can be written as<\/p>\n<div id=\"fs-id1165042326168\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042989371\">Let [latex]g(x)[\/latex] denote the vertical difference between the point [latex](x,f(x))[\/latex] and the point [latex](x,y)[\/latex] on that line. Therefore,<\/p>\n<div id=\"fs-id1165043100161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)=f(x)-\\left[\\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)\\right][\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div>\n<div 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\/11210858\/CNX_Calc_Figure_04_04_011.jpg\" alt=\"A vaguely sinusoidal function y = f(x) is drawn. On the x-axis, a and b are marked. On the y-axis, f(a) and f(b) are marked. The function f(x) starts at (a, f(a)), decreases, then increases, and then decreases to (b, f(b)). A secant line is drawn between (a, f(a)) and (b, f(b)), and it is noted that this line has equation y = ((f(b) \u2013 f(a))\/(b \u2212 a)) (x \u2212 a) + f(x). A line is drawn between the maximum of f(x) and the secant line and it is marked g(x).\" width=\"315\" height=\"272\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. The value [latex]g(x)[\/latex] is the vertical difference between the point [latex](x,f(x))[\/latex] and the point [latex](x,y)[\/latex] on the secant line connecting [latex](a,f(a))[\/latex] and [latex](b,f(b)).[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042892921\">Since the graph of [latex]f[\/latex] intersects the secant line when [latex]x=a[\/latex] and [latex]x=b[\/latex], we see that [latex]g(a)=0=g(b)[\/latex]. Since [latex]f[\/latex] is a differentiable function over [latex](a,b)[\/latex], [latex]g[\/latex] is also a differentiable function over [latex](a,b)[\/latex]. Furthermore, since [latex]f[\/latex] is continuous over [latex][a,b][\/latex], [latex]g[\/latex] is also continuous over [latex][a,b][\/latex]. Therefore, [latex]g[\/latex] satisfies the criteria of Rolle\u2019s theorem. Consequently, there exists a point [latex]c \\in (a,b)[\/latex] such that [latex]g^{\\prime}(c)=0[\/latex]. Since<\/p>\n<div id=\"fs-id1165042979927\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=f^{\\prime}(x)-\\dfrac{f(b)-f(a)}{b-a}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042569046\">we see that<\/p>\n<div id=\"fs-id1165042991202\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(c)=f^{\\prime}(c)-\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042369562\">Since [latex]g^{\\prime}(c)=0[\/latex], we conclude that<\/p>\n<div id=\"fs-id1165042608728\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\n<p id=\"fs-id1165042639297\">[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1165043131938\">In the next example, we show how the Mean Value Theorem can be applied to the function [latex]f(x)=\\sqrt{x}[\/latex] over the interval [latex][0,9][\/latex]. The method is the same for other functions, although sometimes with more interesting consequences.<\/p>\n<div id=\"fs-id1165042979713\" class=\"textbook exercises\">\n<h3>Example: Verifying that the Mean Value Theorem Applies<\/h3>\n<p>For [latex]f(x)=\\sqrt{x}[\/latex] over the interval [latex][0,9][\/latex], show that [latex]f[\/latex] satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value [latex]c \\in (0,9)[\/latex] such that [latex]f^{\\prime}(c)[\/latex] is equal to the slope of the line connecting [latex](0,f(0))[\/latex] and [latex](9,f(9))[\/latex]. Find these values [latex]c[\/latex] guaranteed by the Mean Value Theorem.<\/p>\n<div id=\"fs-id1165042478873\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043395556\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043395556\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043395556\">We know that [latex]f(x)=\\sqrt{x}[\/latex] is continuous over [latex][0,9][\/latex] and differentiable over [latex](0,9)[\/latex]. Therefore, [latex]f[\/latex] satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value [latex]c \\in (0,9)[\/latex] such that [latex]f^{\\prime}(c)[\/latex] is equal to the slope of the line connecting [latex](0,f(0))[\/latex] and [latex](9,f(9))[\/latex] (Figure 7). To determine which value(s) of [latex]c[\/latex] are guaranteed, first calculate the derivative of [latex]f[\/latex]. The derivative [latex]f^{\\prime}(x)=\\frac{1}{2\\sqrt{x}}[\/latex]. The slope of the line connecting [latex](0,f(0))[\/latex] and [latex](9,f(9))[\/latex] is given by<\/p>\n<div id=\"fs-id1165043251015\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(9)-f(0)}{9-0}=\\dfrac{\\sqrt{9}-\\sqrt{0}}{9-0}=\\dfrac{3}{9}=\\dfrac{1}{3}[\/latex]<\/div>\n<p id=\"fs-id1165043096971\">We want to find [latex]c[\/latex] such that [latex]f^{\\prime}(c)=\\frac{1}{3}[\/latex]. That is, we want to find [latex]c[\/latex] such that<\/p>\n<div id=\"fs-id1165042375801\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{1}{2\\sqrt{c}}=\\dfrac{1}{3}[\/latex]<\/div>\n<p>Solving this equation for [latex]c[\/latex], we obtain [latex]c=\\frac{9}{4}[\/latex]. At this point, the slope of the tangent line equals the slope of the line joining the endpoints.<\/p>\n<div style=\"width: 839px\" 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\/11210901\/CNX_Calc_Figure_04_04_006.jpg\" alt=\"The function f(x) = the square root of x is graphed from (0, 0) to (9, 3). There is a secant line drawn from (0, 0) to (9, 3). At point (9\/4, 3\/2), there is a tangent line that is drawn, and this line is parallel to the secant line.\" width=\"829\" height=\"459\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. The slope of the tangent line at [latex]c=\\frac{9}{4}[\/latex] is the same as the slope of the line segment connecting [latex](0,0)[\/latex] and [latex](9,3)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042374751\">One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 hr down a straight road with an average velocity of 45 mph. Let [latex]s(t)[\/latex] and [latex]v(t)[\/latex] denote the position and velocity of the car, respectively, for [latex]0 \\le t \\le 1[\/latex] hr. Assuming that the position function [latex]s(t)[\/latex] is differentiable, we can apply the Mean Value Theorem to conclude that, at some time [latex]c \\in (0,1)[\/latex], the speed of the car was exactly<\/p>\n<div id=\"fs-id1165042647085\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(c)=s^{\\prime}(c)=\\dfrac{s(1)-s(0)}{1-0}=45[\/latex] mph<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1165042332061\" class=\"textbook exercises\">\n<h3>Example: Mean Value Theorem and Velocity<\/h3>\n<p id=\"fs-id1165043312536\">If a rock is dropped from a height of 100 ft, its position [latex]t[\/latex] seconds after it is dropped until it hits the ground is given by the function [latex]s(t)=-16t^2+100[\/latex].<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Determine how long it takes before the rock hits the ground.<\/li>\n<li>Find the average velocity [latex]v_{\\text{avg}}[\/latex] of the rock for when the rock is released and the rock hits the ground.<\/li>\n<li>Find the time [latex]t[\/latex] guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is [latex]v_{\\text{avg}}[\/latex].<\/li>\n<\/ol>\n<div id=\"fs-id1165042332064\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042373172\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042373172\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042373172\" style=\"list-style-type: lower-alpha;\">\n<li>When the rock hits the ground, its position is [latex]s(t)=0[\/latex]. Solving the equation [latex]-16t^2+100=0[\/latex] for [latex]t[\/latex], we find that [latex]t=\\pm \\frac{5}{2}[\/latex] sec.\u00a0Since we are only considering [latex]t \\ge 0[\/latex], the ball will hit the ground [latex]\\frac{5}{2}[\/latex] sec after it is dropped.<\/li>\n<li>The average velocity is given by\n<div id=\"fs-id1165043354673\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v_{\\text{avg}}=\\frac{s(5\/2)-s(0)}{5\/2-0}=\\frac{1-100}{5\/2}=-40[\/latex] ft\/sec<\/div>\n<\/li>\n<li>The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time [latex]t[\/latex] such that [latex]v(t)=s^{\\prime}(t)=v_{\\text{avg}}=-40[\/latex] ft\/sec.\u00a0Since [latex]s(t)[\/latex] is continuous over the interval [latex][0,5\/2][\/latex] and differentiable over the interval [latex](0,5\/2)[\/latex], by the Mean Value Theorem, there is guaranteed to be a point [latex]c \\in (0,5\/2)[\/latex] such that\n<div id=\"fs-id1165043390900\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s^{\\prime}(c)=\\frac{s(5\/2)-s(0)}{5\/2-0}=-40[\/latex]<\/div>\n<p>Taking the derivative of the position function [latex]s(t)[\/latex], we find that [latex]s^{\\prime}(t)=-32t[\/latex]. Therefore, the equation reduces to [latex]s^{\\prime}(c)=-32c=-40[\/latex]. Solving this equation for [latex]c[\/latex], we have [latex]c=\\frac{5}{4}[\/latex]. Therefore, [latex]\\frac{5}{4}[\/latex] sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: -40 ft\/sec.<\/p>\n<div style=\"width: 464px\" 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\/11210904\/CNX_Calc_Figure_04_04_007.jpg\" alt=\"The function s(t) = \u221216t2 + 100 is graphed from (0, 100) to (5\/2, 0). There is a secant line drawn from (0, 100) to (5\/2, 0). At the point corresponding to x = 5\/4, there is a tangent line that is drawn, and this line is parallel to the secant line.\" width=\"454\" height=\"272\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. At time [latex]t=\\frac{5}{4}[\/latex] sec, the velocity of the rock is equal to its average velocity from the time it is dropped until it hits the ground.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Mean Value Theorem and Velocity.<\/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=931&amp;end=1103&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\/4.4MeanValueTheorem931to1103_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.4 Mean Value Theorem&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose a ball is dropped from a height of 200 ft. Its position at time [latex]t[\/latex] is [latex]s(t)=-16t^2+200[\/latex]. Find the time [latex]t[\/latex] when the instantaneous velocity of the ball equals its average velocity.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9003136\">Hint<\/span><\/p>\n<div id=\"q9003136\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, determine how long it takes for the ball to hit the ground. Then, find the average velocity of the ball from the time it is dropped until it hits the ground.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042631892\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042631892\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\dfrac{5}{2\\sqrt{2}}[\/latex] sec<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm6147\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=6147&theme=oea&iframe_resize_id=ohm6147&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Corollaries of the Mean Value Theorem<\/h2>\n<p id=\"fs-id1165042640871\">Let\u2019s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections.<\/p>\n<p id=\"fs-id1165042640877\">At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x[\/latex] in some interval [latex]I[\/latex], then [latex]f(x)[\/latex] is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly.<\/p>\n<div id=\"fs-id1165042645708\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Corollary 1: Functions with a Derivative of Zero<\/h3>\n<hr \/>\n<p id=\"fs-id1165042645714\">Let [latex]f[\/latex] be differentiable over an interval [latex]I[\/latex]. If [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f(x)[\/latex] is constant for all [latex]x \\in I[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165043251996\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1165043252002\">Since [latex]f[\/latex] is differentiable over [latex]I[\/latex], [latex]f[\/latex] must be continuous over [latex]I[\/latex]. Suppose [latex]f(x)[\/latex] is not constant for all [latex]x[\/latex] in [latex]I[\/latex]. Then there exist [latex]a,b \\in I[\/latex], where [latex]a \\ne b[\/latex] and [latex]f(a) \\ne f(b)[\/latex]. Choose the notation so that [latex]a<b[\/latex]. Therefore,<\/p>\n<div id=\"fs-id1165042708535\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(b)-f(a)}{b-a} \\ne 0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042708255\">Since [latex]f[\/latex] is a differentiable function, by the Mean Value Theorem, there exists [latex]c \\in (a,b)[\/latex] such that<\/p>\n<div id=\"fs-id1165042418104\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043259845\">Therefore, there exists [latex]c \\in I[\/latex] such that [latex]f^{\\prime}(c) \\ne 0[\/latex], which contradicts the assumption that [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex].<\/p>\n<p id=\"fs-id1165042707831\">[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1165042707834\">From the example above, it follows that if two functions have the same derivative, they differ by, at most, a constant.<\/p>\n<div id=\"fs-id1165042707841\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Corollary 2: Constant Difference Theorem<\/h3>\n<hr \/>\n<p id=\"fs-id1165042707847\">If [latex]f[\/latex] and [latex]g[\/latex] are differentiable over an interval [latex]I[\/latex] and [latex]f^{\\prime}(x)=g^{\\prime}(x)[\/latex] for all [latex]x \\in I[\/latex], then [latex]f(x)=g(x)+C[\/latex] for some constant [latex]C[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042707738\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1165042707744\">Let [latex]h(x)=f(x)-g(x)[\/latex]. Then, [latex]h^{\\prime}(x)=f^{\\prime}(x)-g^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex]. By Corollary 1, there is a constant [latex]C[\/latex] such that [latex]h(x)=C[\/latex] for all [latex]x \\in I[\/latex]. Therefore, [latex]f(x)=g(x)+C[\/latex] for all [latex]x \\in I[\/latex].<\/p>\n<p id=\"fs-id1165043424693\">[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1165043424696\">The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function [latex]f[\/latex] is increasing over [latex]I[\/latex] if [latex]f(x_1)<f(x_2)[\/latex] whenever [latex]x_1<x_2[\/latex], whereas [latex]f[\/latex] is decreasing over [latex]I[\/latex] if [latex]f(x_1)>f(x_2)[\/latex] whenever [latex]x_1<x_2[\/latex]. Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure 9). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph.<\/p>\n<p>This fact is important because it means that for a given function [latex]f[\/latex], if there exists a function [latex]F[\/latex] such that [latex]F^{\\prime}(x)=f(x)[\/latex]; then, the only other functions that have a derivative equal to [latex]f[\/latex] are [latex]F(x)+C[\/latex] for some constant [latex]C[\/latex]. We discuss this result in more detail later in the chapter.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"19\" src=\"https:\/\/openstax.org\/resources\/1ae209b72e21942c9811cc8e33e8be432261ed5f\" alt=\"A vaguely sinusoidal function f(x) is graphed. It increases from somewhere in the second quadrant to (a, f(a)). In this section it is noted that f\u2019 &gt; 0. Then in decreases from (a, f(a)) to (b, f(b)). In this section it is noted that f\u2019 &lt; 0. Finally, it increases to the right of (b, f(b)) and it is noted in this section that f\u2019 &gt; 0\" width=\"731\" height=\"302\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9. If a function has a positive derivative over some interval [latex]I[\/latex], then the function increases over that interval [latex]I[\/latex]; if the derivative is negative over some interval [latex]I[\/latex], then the function decreases over that interval [latex]I[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165043217934\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Corollary 3: Increasing and Decreasing Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1165043217941\">Let [latex]f[\/latex] be continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex].<\/p>\n<ol id=\"fs-id1165043217981\">\n<li>If [latex]f^{\\prime}(x)>0[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f[\/latex] is an increasing function over [latex][a,b][\/latex].<\/li>\n<li>If [latex]f^{\\prime}(x)<0[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f[\/latex] is a decreasing function over [latex][a,b][\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043327297\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1165043327302\">We will prove 1.; the proof of 2. is similar. Suppose [latex]f[\/latex] is not an increasing function on [latex]I[\/latex]. Then there exist [latex]a[\/latex] and [latex]b[\/latex] in [latex]I[\/latex] such that [latex]a<b[\/latex], but [latex]f(a) \\ge f(b)[\/latex]. Since [latex]f[\/latex] is a differentiable function over [latex]I[\/latex], by the Mean Value Theorem there exists [latex]c \\in (a,b)[\/latex] such that<\/p>\n<div id=\"fs-id1165043327400\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043327455\">Since [latex]f(a) \\ge f(b)[\/latex], we know that [latex]f(b)-f(a) \\le 0[\/latex]. Also, [latex]a<b[\/latex] tells us that [latex]b-a>0[\/latex]. We conclude that<\/p>\n<div id=\"fs-id1165042651499\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a} \\le 0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042651558\">However, [latex]f^{\\prime}(x)>0[\/latex] for all [latex]x \\in I[\/latex]. This is a contradiction, and therefore [latex]f[\/latex] must be an increasing function over [latex]I[\/latex].<\/p>\n<p id=\"fs-id1165042651604\">[latex]_\\blacksquare[\/latex]<\/p>\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-402\">\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>4.4 Mean Value Theorem. <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":13,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-402","chapter","type-chapter","status-publish","hentry"],"part":48,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/402","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":25,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/402\/revisions"}],"predecessor-version":[{"id":4918,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/402\/revisions\/4918"}],"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\/402\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=402"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=402"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=402"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=402"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}