Proof

Let [latex]k=f\left(a\right)=f\left(b\right).[/latex] We consider three cases:

1. [latex]f\left(x\right)=k[/latex] for all [latex]x\in \left(a,b\right).[/latex]
2. There exists [latex]x\in \left(a,b\right)[/latex] such that [latex]f\left(x\right)>k.[/latex]
3. There exists [latex]x\in \left(a,b\right)[/latex] such that [latex]f\left(x\right)<k.[/latex]

Case 1: If [latex]f\left(x\right)=0[/latex] for all [latex]x\in \left(a,b\right),[/latex] then [latex]f\prime \left(x\right)=0[/latex] for all [latex]x\in \left(a,b\right).[/latex]

Case 2: Since [latex]f[/latex] is a continuous function over the closed, bounded interval [latex]\left[a,b\right],[/latex] by the extreme value theorem, it has an absolute maximum. Also, since there is a point [latex]x\in \left(a,b\right)[/latex] such that [latex]f\left(x\right)>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 \left(a,b\right).[/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’s theorem, [latex]f\prime \left(c\right)=0.[/latex]

Case 3: The case when there exists a point [latex]x\in \left(a,b\right)[/latex] such that [latex]f\left(x\right)<k[/latex] is analogous to case 2, with maximum replaced by minimum.

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An important point about Rolle’s 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\left(x\right)=|x|-1[/latex] is continuous over [latex]\left[-1,1\right][/latex] and [latex]f\left(-1\right)=0=f\left(1\right),[/latex] but [latex]f\prime \left(c\right)\ne 0[/latex] for any [latex]c\in \left(-1,1\right)[/latex] as shown in the following figure.

Since [latex]f\left(x\right)=|x|-1[/latex] is not differentiable at [latex]x=0,[/latex] the conditions of Rolle’s theorem are not satisfied. In fact, the conclusion does not hold here; there is no [latex]c\in \left(-1,1\right)[/latex] such that [latex]f\prime \left(c\right)=0.[/latex]

Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points [latex]c[/latex] where [latex]f\prime \left(c\right)=0.[/latex]

Proof

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]\left(a,f\left(a\right)\right)[/latex] and [latex]\left(b,f\left(b\right)\right).[/latex] Since the slope of that line is

and the line passes through the point [latex]\left(a,f\left(a\right)\right),[/latex] the equation of that line can be written as

Let [latex]g\left(x\right)[/latex] denote the vertical difference between the point [latex]\left(x,f\left(x\right)\right)[/latex] and the point [latex]\left(x,y\right)[/latex] on that line. Therefore,

The value [latex]g\left(x\right)[/latex] is the vertical difference between the point [latex]\left(x,f\left(x\right)\right)[/latex] and the point [latex]\left(x,y\right)[/latex] on the secant line connecting [latex]\left(a,f\left(a\right)\right)[/latex] and [latex]\left(b,f\left(b\right)\right).[/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\left(a\right)=0=g\left(b\right).[/latex] Since [latex]f[/latex] is a differentiable function over [latex]\left(a,b\right),[/latex] [latex]g[/latex] is also a differentiable function over [latex]\left(a,b\right).[/latex] Furthermore, since [latex]f[/latex] is continuous over [latex]\left[a,b\right],[/latex] [latex]g[/latex] is also continuous over [latex]\left[a,b\right].[/latex] Therefore, [latex]g[/latex] satisfies the criteria of Rolle’s theorem. Consequently, there exists a point [latex]c\in \left(a,b\right)[/latex] such that [latex]g\prime \left(c\right)=0.[/latex] Since

we see that

Since [latex]g\prime \left(c\right)=0,[/latex] we conclude that

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In the next example, we show how the Mean Value Theorem can be applied to the function [latex]f\left(x\right)=\sqrt{x}[/latex] over the interval [latex]\left[0,9\right].[/latex] The method is the same for other functions, although sometimes with more interesting consequences.

Verifying that the Mean Value Theorem Applies

For [latex]f\left(x\right)=\sqrt{x}[/latex] over the interval [latex]\left[0,9\right],[/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 \left(0,9\right)[/latex] such that [latex]{f}^{\prime }\left(c\right)[/latex] is equal to the slope of the line connecting [latex]\left(0,f\left(0\right)\right)[/latex] and [latex]\left(9,f\left(9\right)\right).[/latex] Find these values [latex]c[/latex] guaranteed by the Mean Value Theorem.

We know that [latex]f\left(x\right)=\sqrt{x}[/latex] is continuous over [latex]\left[0,9\right][/latex] and differentiable over [latex]\left(0,9\right).[/latex] Therefore, [latex]f[/latex] satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value [latex]c\in \left(0,9\right)[/latex] such that [latex]{f}^{\prime }\left(c\right)[/latex] is equal to the slope of the line connecting [latex]\left(0,f\left(0\right)\right)[/latex] and [latex]\left(9,f\left(9\right)\right)[/latex] ([link]). To determine which value(s) of [latex]c[/latex] are guaranteed, first calculate the derivative of [latex]f.[/latex] The derivative [latex]{f}^{\prime }\left(x\right)=\frac{1}{\left(2\sqrt{x}\right)}.[/latex] The slope of the line connecting [latex]\left(0,f\left(0\right)\right)[/latex] and [latex]\left(9,f\left(9\right)\right)[/latex] is given by

[latex]\frac{f\left(9\right)-f\left(0\right)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9-0}=\frac{3}{9}=\frac{1}{3}.[/latex]

We want to find [latex]c[/latex] such that [latex]{f}^{\prime }\left(c\right)=\frac{1}{3}.[/latex] That is, we want to find [latex]c[/latex] such that

[latex]\frac{1}{2\sqrt{c}}=\frac{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.

The slope of the tangent line at [latex]c=9\text{/}4[/latex] is the same as the slope of the line segment connecting [latex]\left(0,0\right)[/latex] and [latex]\left(9,3\right).[/latex]

One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 h down a straight road with an average velocity of 45 mph. Let [latex]s\left(t\right)[/latex] and [latex]v\left(t\right)[/latex] denote the position and velocity of the car, respectively, for [latex]0\le t\le 1[/latex] h. Assuming that the position function [latex]s\left(t\right)[/latex] is differentiable, we can apply the Mean Value Theorem to conclude that, at some time [latex]c\in \left(0,1\right),[/latex] the speed of the car was exactly

Proof

Since [latex]f[/latex] is differentiable over [latex]I,[/latex] [latex]f[/latex] must be continuous over [latex]I.[/latex] Suppose [latex]f\left(x\right)[/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\left(a\right)\ne f\left(b\right).[/latex] Choose the notation so that [latex]a<b.[/latex] Therefore,

Since [latex]f[/latex] is a differentiable function, by the Mean Value Theorem, there exists [latex]c\in \left(a,b\right)[/latex] such that

Therefore, there exists [latex]c\in I[/latex] such that [latex]{f}^{\prime }\left(c\right)\ne 0,[/latex] which contradicts the assumption that [latex]{f}^{\prime }\left(x\right)=0[/latex] for all [latex]x\in I.[/latex]

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From [link], it follows that if two functions have the same derivative, they differ by, at most, a constant.

Proof

Let [latex]h\left(x\right)=f\left(x\right)-g\left(x\right).[/latex] Then, [latex]{h}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)-{g}^{\prime }\left(x\right)=0[/latex] for all [latex]x\in I.[/latex] By Corollary 1, there is a constant [latex]C[/latex] such that [latex]h\left(x\right)=C[/latex] for all [latex]x\in I.[/latex] Therefore, [latex]f\left(x\right)=g\left(x\right)+C[/latex] for all [latex]x\in I.[/latex]

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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\left({x}_{1}\right)<f\left({x}_{2}\right)[/latex] whenever [latex]{x}_{1}<{x}_{2},[/latex] whereas [latex]f[/latex] is decreasing over [latex]I[/latex] if [latex]f{\left(x\right)}_{1}>f\left({x}_{2}\right)[/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 ([link]). 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.

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 }\left(x\right)=f\left(x\right);[/latex] then, the only other functions that have a derivative equal to [latex]f[/latex] are [latex]F\left(x\right)+C[/latex] for some constant [latex]C.[/latex] We discuss this result in more detail later in the chapter.

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]

Proof

We will prove i.; the proof of ii. 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\left(a\right)\ge f\left(b\right).[/latex] Since [latex]f[/latex] is a differentiable function over [latex]I,[/latex] by the Mean Value Theorem there exists [latex]c\in \left(a,b\right)[/latex] such that

Since [latex]f\left(a\right)\ge f\left(b\right),[/latex] we know that [latex]f\left(b\right)-f\left(a\right)\le 0.[/latex] Also, [latex]a<b[/latex] tells us that [latex]b-a>0.[/latex] We conclude that

However, [latex]{f}^{\prime }\left(x\right)>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]

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