4.4 The Mean Value Theorem

Proof

Let $k=f(a)=f(b).$ We consider three cases:

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

Case 1: If $f\left(x\right)=k$ for all $x\in \left(a,b\right),$ then $f\prime \left(x\right)=0$ for all $x\in \left(a,b\right).$

Case 2: Since $f$ is a continuous function over the closed, bounded interval $[a,b],$ by the extreme value theorem, it has an absolute maximum. Also, since there is a point $x\in \left(a,b\right)$ such that $f\left(x\right)>k,$ the absolute maximum is greater than $k.$ Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point $c\in \left(a,b\right).$ Because $f$ has a maximum at an interior point $c,$ and $f$ is differentiable at $c,$ by Fermat’s theorem, $f\prime \left(c\right)=0.$

Case 3: The case when there exists a point $x\in \left(a,b\right)$ such that $f\left(x\right)<k$ 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 $f$ is critical. If $f$ is not differentiable, even at a single point, the result may not hold. For example, the function $f\left(x\right)=\left|x\right|-1$ is continuous over $[\mathrm{-1},1]$ and $f\left(\mathrm{-1}\right)=0=f\left(1\right),$ but $f\prime \left(c\right)\ne 0$ for any $c\in (\mathrm{-1},1)$ as shown in the following figure.

Figure 4.22 Since $f(x)=\left|x\right|-1$ is not differentiable at $x=0,$ the conditions of Rolle’s theorem are not satisfied. In fact, the conclusion does not hold here; there is no $c\in (\mathrm{-1},1)$ such that $f\prime (c)=0.$

Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points $c$ where $f\prime (c)=0.$

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 $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$ Since the slope of that line is

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

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

Figure 4.26 The value $g(x)$ is the vertical difference between the point $(x,f(x))$ and the point $(x,y)$ on the secant line connecting $(a,f(a))$ and $(b,f(b)).$

Since the graph of $f$ intersects the secant line when $x=a$ and $x=b,$ we see that $g\left(a\right)=0=g\left(b\right).$ Since $f$ is a differentiable function over $\left(a,b\right),$ $g$ is also a differentiable function over $\left(a,b\right).$ Furthermore, since $f$ is continuous over $[a,b],$ $g$ is also continuous over $[a,b].$ Therefore, $g$ satisfies the criteria of Rolle’s theorem. Consequently, there exists a point $c\in \left(a,b\right)$ such that $g\prime \left(c\right)=0.$ Since

we see that

Since $g\prime (c)=0,$ we conclude that

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

Example 4.15

Verifying that the Mean Value Theorem Applies

For $f\left(x\right)=\sqrt{x}$ over the interval $[0,9],$ show that $f$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $c\in (0,9)$ such that $f^{\prime }(c)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right).$ Find these values $c$ guaranteed by the Mean Value Theorem.

Solution

We know that $f(x)=\sqrt{x}$ is continuous over $[0,9]$ and differentiable over $\left(0,9\right).$ Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c\in \left(0,9\right)$ such that $f^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ (Figure 4.27). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f.$ The derivative $f^{\prime }\left(x\right)=\frac{1}{(2\sqrt{x})}.$ The slope of the line connecting $(0,f(0))$ and $(9,f(9))$ is given by

$$\frac{f(9)-f(0)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9-0}=\frac{3}{9}=\frac{1}{3}.$$

We want to find $c$ such that $f^{\prime }(c)=\frac{1}{3}.$ That is, we want to find $c$ such that

$$\frac{1}{2\sqrt{c}}=\frac{1}{3}.$$

Solving this equation for $c,$ we obtain $c=\frac{9}{4}.$ At this point, the slope of the tangent line equals the slope of the line joining the endpoints.

Figure 4.27 The slope of the tangent line at $c=9\text{/}4$ is the same as the slope of the line segment connecting $(0,0)$ and $(9,3).$

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 $s(t)$ and $v\left(t\right)$ denote the position and velocity of the car, respectively, for $0\le t\le 1$ h. Assuming that the position function $s\left(t\right)$ is differentiable, we can apply the Mean Value Theorem to conclude that, at some time $c\in \left(0,1\right),$ the speed of the car was exactly

Proof

Since $f$ is differentiable over $I,$ $f$ must be continuous over $I.$ Suppose $f\left(x\right)$ is not constant for all $x$ in $I.$ Then there exist $a,b\in I,$ where $a\ne b$ and $f\left(a\right)\ne f\left(b\right).$ Choose the notation so that $a<b.$ Therefore,

Since $f$ is a differentiable function, by the Mean Value Theorem, there exists $c\in (a,b)$ such that

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

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From Corollary 1: Functions with a Derivative of Zero, it follows that if two functions have the same derivative, they differ by, at most, a constant.

Proof

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

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This fact is important because it means that for a given function $f,$ if there exists a function $F$ such that $F^{\prime }\left(x\right)=f\left(x\right);$ then, the only other functions that have a derivative equal to $f$ are $F(x)+C$ for some constant $C.$ We discuss this result in more detail later in the chapter.

The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function $f$ is increasing over $I$ if $f\left(x_1\right)<f\left(x_2\right)$ whenever $x_1<x_2,$ whereas $f$ is decreasing over $I$ if $f{\left(x\right)}_1>f\left(x_2\right)$ whenever $x_1<x_2.$ 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 4.29). 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.

Figure 4.29 If a function has a positive derivative over some interval $I,$ then the function increases over that interval $I;$ if the derivative is negative over some interval $I,$ then the function decreases over that interval $I.$

Proof

We will prove i.; the proof of ii. is similar. Suppose $f$ is continuous and differentiable over an interval $I$ and $f\text{'}\left(x\right)>0$ for all $x\in I$. By way of contradiction, suppose that $f$ is not an increasing function on $I$. Then there exist $a$ and $b$ in $I$ such that $a<b$, but $f\left(a\right)>f\left(b\right)$. Since $f$ is a differentiable function over $I$, the Mean Value Theorem guarantees that there is some value $c\in \left(a,b\right)$ such that

Now $a<b$, so $b-a>0$. Because $f\left(a\right)>f\left(b\right)$, it follows that $f\left(b\right)-f\left(a\right)<0$. It follows that the quotient of $f\left(b\right)-f\left(a\right)$ and $b-a$ is negative. So, for c,

However, $f^{\prime }\left(x\right)>0$ for all $x\in I$ including $c$. This is a contradiction. The assumption that $f$ is not an increasing function on $I$ is false. Therefore, $f$ must be increasing throughout $I$.

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