Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s 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’s 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’s theorem (Figure 5).
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.
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].
Mean Value Theorem
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
The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting [latex](a,f(a))[/latex] and [latex](b,f(b))[/latex]. Since the slope of that line is
[latex]\dfrac{f(b)-f(a)}{b-a}[/latex]
and the line passes through the point [latex](a,f(a))[/latex], the equation of that line can be written as
[latex]y=\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)[/latex]
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,
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]
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’s theorem. Consequently, there exists a point [latex]c \in (a,b)[/latex] such that [latex]g^{\prime}(c)=0[/latex]. Since
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.
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.
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
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
[latex]\dfrac{1}{2\sqrt{c}}=\dfrac{1}{3}[/latex]
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.
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].
One application that helps illustrate the Mean Value Theorem involves velocity.
Suppose we drive a car for [latex]1[/latex] hr down a straight road with an average velocity of [latex]45[/latex] 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
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].
Determine how long it takes before the rock hits the ground.
Find the average velocity [latex]v_{\text{avg}}[/latex] of the rock for when the rock is released and the rock hits the ground.
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].
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. Since 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.
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. Since [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
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.
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.
Watch the following video to see the worked solution to this example.
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.
Watch the following video to see the worked solution to the two previous examples.
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.
