Proof

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

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

Case 1: If $f(x)=0$ for all $x \in (a,b)$, then $f^{\prime}(x)=0$ for all $x \in (a,b)$.

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 $x \in (a,b)$ such that $f(x)>k$, 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 $c \in (a,b)$. Because $f$ has a maximum at an interior point $c$, and $f$ is differentiable at $c$, by Fermat’s theorem, $f^{\prime}(c)=0$.

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

□

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 $f(x)=|x|-1$ is continuous over $[-1,1]$ and $f(-1)=0=f(1)$, but $f^{\prime}(c) \ne 0$ for any $c \in (-1,1)$ as shown in the following figure.

Figure 2. Since $f(x)=|x|-1$ 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 $c \in (-1,1)$ 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 $(a,f(a))$ and $(b,f(b))$. Since the slope of that line is

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

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

Figure 6. 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 $g(a)=0=g(b)$. Since $f$ is a differentiable function over $(a,b)$, $g$ is also a differentiable function over $(a,b)$. 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 $c \in (a,b)$ such that $g^{\prime}(c)=0$. Since

we see that

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

□

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.

Verifying that the Mean Value Theorem Applies

For $f(x)=\sqrt{x}$ 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 $(0,f(0))$ and $(9,f(9))$. Find these values $c$ guaranteed by the Mean Value Theorem.

Answer:

We know that $f(x)=\sqrt{x}$ is continuous over $[0,9]$ and differentiable over $(0,9)$. Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c \in (0,9)$ such that $f^{\prime}(c)$ is equal to the slope of the line connecting $(0,f(0))$ and $(9,f(9))$ ((Figure)). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f$. The derivative $f^{\prime}(x)=\frac{1}{2\sqrt{x}}$. 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 7. The slope of the tangent line at $c=9/4$ 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 hr down a straight road with an average velocity of 45 mph. Let $s(t)$ and $v(t)$ denote the position and velocity of the car, respectively, for $0 \le t \le 1$ hr. Assuming that the position function $s(t)$ is differentiable, we can apply the Mean Value Theorem to conclude that, at some time $c \in (0,1)$, the speed of the car was exactly

Proof

Since $f$ is differentiable over $I$, $f$ must be continuous over $I$. Suppose $f(x)$ is not constant for all $x$ in $I$. Then there exist $a,b \in I$, where $a \ne b$ and $f(a) \ne f(b)$. 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 $f^{\prime}(c) \ne 0$, which contradicts the assumption that $f^{\prime}(x)=0$ for all $x \in I$.

□

From (Figure), it follows that if two functions have the same derivative, they differ by, at most, a constant.

Proof

Let $h(x)=f(x)-g(x)$. Then, $h^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)=0$ 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 $x \in I$.

□

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 $f(x_1)<f(x_2)$ whenever $x_1<x_2$, whereas $f$ is decreasing over $I$ if $f(x_1)>f(x_2)$ 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)). 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 $f$, if there exists a function $F$ such that $F^{\prime}(x)=f(x)$; 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.

Figure 9. 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 1.; the proof of 2. is similar. Suppose $f$ is not an increasing function on $I$. Then there exist $a$ and $b$ in $I$ such that $a<b$, but $f(a) \ge f(b)$. Since $f$ is a differentiable function over $I$, by the Mean Value Theorem there exists $c \in (a,b)$ such that

Since $f(a) \ge f(b)$, we know that $f(b)-f(a) \le 0$. Also, $a<b$ tells us that $b-a>0$. We conclude that

However, $f^{\prime}(x)>0$ for all $x \in I$. This is a contradiction, and therefore $f$ must be an increasing function over $I$.

□