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 [latex][a,b][\/latex] with [latex]f(a)=f(b)[\/latex]. The 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 (Figure 5).\u00a0<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"452\"]![\"A](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210855\/CNX_Calc_Figure_04_04_010.jpg\")
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\r\n\r\n
The 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 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 Proof<\/strong><\/p>\r\n 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 <\/p>\r\n and the line passes through the point [latex](a,f(a))[\/latex], the equation of that line can be written as<\/p>\r\n <\/p>\r\n 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 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 <\/p>\r\n we see that<\/p>\r\n <\/p>\r\n Since [latex]g^{\\prime}(c)=0[\/latex], we conclude that<\/p>\r\n [latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n 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 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>\r\n\r\n\r\n[reveal-answer q=\"fs-id1165043395556\"]Show Solution[\/reveal-answer] 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 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 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>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"829\"] One application that helps illustrate the Mean Value Theorem involves velocity.<\/p>\r\n Suppose we drive a car for [latex]1[\/latex] hr down a straight road with an average velocity of [latex]45[\/latex] mph.<\/p>\r\n 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 If a rock is dropped from a height of [latex]100[\/latex] 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 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}(t)=-32c=-40[\/latex]. Solving this equation for [latex]t[\/latex], we have [latex]t=\\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: [latex] -40[\/latex] ft\/sec.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"454\"] Watch the following video to see the worked solution to this example.<\/p>\r\nMean Value Theorem<\/h3>\r\n
[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n<\/section>\r\n
\r\n[latex]\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n[latex]y=\\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)[\/latex]<\/div>\r\n[latex]g(x)=f(x)-\\left[\\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)\\right][\/latex]<\/div>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"315\"]![\"A](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210858\/CNX_Calc_Figure_04_04_011.jpg\")
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]\r\n\r\n\r\n[latex]g^{\\prime}(x)=f^{\\prime}(x)-\\dfrac{f(b)-f(a)}{b-a}[\/latex],<\/div>\r\n[latex]g^{\\prime}(c)=f^{\\prime}(c)-\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n
\r\n[hidden-answer a=\"fs-id1165043395556\"]\r\n\r\n\r\n[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[latex]\\dfrac{1}{2\\sqrt{c}}=\\dfrac{1}{3}[\/latex]<\/div>\r\n![\"The](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210901\/CNX_Calc_Figure_04_04_006.jpg\")
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]\r\n[\/hidden-answer]<\/section>\r\n[latex]v(c)=s^{\\prime}(c)=\\dfrac{s(1)-s(0)}{1-0}=45[\/latex] mph<\/div>\r\n<\/section>\r\n\r\n\t
\r\n[hidden-answer a=\"fs-id1165042373172\"]\r\n\r\n\r\n\r\n\t
\r\n[latex]v_{\\text{avg}}=\\frac{s(5\/2)-s(0)}{5\/2-0}=\\frac{1-100}{5\/2}=-40[\/latex] ft\/sec<\/div>\r\n<\/li>\r\n\t
\r\n[latex]s^{\\prime}(t)=\\frac{s(5\/2)-s(0)}{5\/2-0}=-40[\/latex]<\/div>\r\n![\"The](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210904\/CNX_Calc_Figure_04_04_007.jpg\")
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<\/li>\r\n<\/ol>\r\n