4.4 The Mean Value Theorem

The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem.

Rolle’s Theorem

Informally, Rolle’s theorem states that if the outputs of a differentiable function $f$ are equal at the endpoints of an interval, then there must be an interior point $c$ where $f^{'} (c) = 0.$ Figure 4.21 illustrates this theorem.

The figure is divided into three parts labeled a, b, and c. Figure a shows the first quadrant with values a, c, and b marked on the x-axis. A downward-facing parabola is drawn such that its values at a and b are the same. The point c is the global maximum, and it is noted that f’(c) = 0. Figure b shows the first quadrant with values a, c, and b marked on the x-axis. An upward-facing parabola is drawn such that its values at a and b are the same. The point c is the global minimum, and it is noted that f’(c) = 0. Figure c shows the first quadrant with points a, c1, c2, and b marked on the x-axis. One period of a sine wave is drawn such that its values at a and b are equal. The point c1 is the global maximum, and it is noted that f’(c1) = 0. The point c2 is the global minimum, and it is noted that f’(c2) = 0. — Figure 4.21If a differentiable function f satisfies $f (a) = f (b),$ then its derivative must be zero at some point(s) between $a$ and $b .$

Proof

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

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

Case 1: If $f (x) = k$ for all $x \in (a, b),$ then $f^{'} (x) = 0$ 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 has a maximum at an interior point and is differentiable at by Fermat’s theorem,

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.

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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 $f (x) = ∣ x ∣ - 1$ is continuous over $[- 1, 1]$ and $f (- 1) = 0 = f (1),$ but for any as shown in the following figure.

The function f(x) = |x| − 1 is graphed. It is shown that f(1) = f(−1), but it is noted that there is no c such that f’(c) = 0. — Figure 4.22Since $f (x) = ∣ x ∣ - 1$ 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 $f^{'} (c) = 0.$

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

The Mean Value Theorem and Its Meaning

Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions $f$ defined on a closed interval $[a, b]$ with $f (a) = f (b)$ . 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 4.25). The Mean Value Theorem states that if $f$ is continuous over the closed interval $[a, b]$ and differentiable over the open interval $(a, b),$ then there exists a point such that the tangent line to the graph of at is parallel to the secant line connecting and

A vaguely sinusoidal function y = f(x) is drawn. On the x-axis, a, c1, c2, and b are marked. On the y-axis, f(a) and f(b) are marked. The function f(x) starts at (a, f(a)), decreases to c1, increases to c2, and then decreases to (b, f(b)). A secant line is drawn between (a, f(a)) and (b, f(b)), and it is noted that this line has slope (f(b) – f(a))/(b − a). The tangent lines at c1 and c2 are drawn, and these lines are parallel to the secant line. It is noted that the slopes of these tangent lines are f’(c1) and f’(c2), respectively. — Figure 4.25The 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 $c_{1}$ and $c_{2}$ such that the tangent line to $f$ at $c_{1}$ and has the same slope as the secant line.

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

\frac{f ( b ) - f ( a )}{b - a}

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

y = \frac{f ( b ) - f ( a )}{b - a} (x - a) + f (a) .

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

g (x) = f (x) - [\frac{f ( b ) - f ( a )}{b - a} (x - a) + f (a)] .

A vaguely sinusoidal function y = f(x) is drawn. On the x-axis, a and b are marked. On the y-axis, f(a) and f(b) are marked. The function f(x) starts at (a, f(a)), decreases, then increases, and then decreases to (b, f(b)). A secant line is drawn between (a, f(a)) and (b, f(b)), and it is noted that this line has equation y = ((f(b) – f(a))/(b − a)) (x − a) + f(a). A line is drawn between the maximum of f(x) and the secant line and it is marked g(x). — Figure 4.26The 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 $(a, b),$ is also a differentiable function over Furthermore, since is continuous over is also continuous over Therefore, satisfies the criteria of Rolle’s theorem. Consequently, there exists a point such that Since

g^{'} (x) = f^{'} (x) - \frac{f ( b ) - f ( a )}{b - a},

we see that

g^{'} (c) = f^{'} (c) - \frac{f ( b ) - f ( a )}{b - a} .

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

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .

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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 $v (t)$ denote the position and velocity of the car, respectively, for $0 \leq t \leq 1$ h. 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

v (c) = s^{'} (c) = \frac{s ( 1 ) - s ( 0 )}{1 - 0} = 45 mph .

Corollaries of the Mean Value Theorem

Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections.

At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if $f^{'} (x) = 0$ for all $x$ in some interval $I,$ then $f (x)$ is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly.

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 $a \neq = b$ and Choose the notation so that Therefore,

\frac{f ( b ) - f ( a )}{b - a} \neq = 0.

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

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .

Therefore, there exists $c \in I$ such that $f^{'} (c) \neq = 0,$ which contradicts the assumption that $f^{'} (x) = 0$ for all

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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 $h (x) = f (x) - g (x) .$ Then, $h^{'} (x) = f^{'} (x) - g^{'} (x) = 0$ for all By Corollary 1, there is a constant such that for all Therefore, for all

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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 $F^{'} (x) = f (x);$ 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.

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 is decreasing over if whenever 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.

A vaguely sinusoidal function f(x) is graphed. It increases from somewhere in the second quadrant to (a, f(a)). In this section it is noted that f’ > 0. Then in decreases from (a, f(a)) to (b, f(b)). In this section it is noted that f’ < 0. Finally, it increases to the right of (b, f(b)) and it is noted in this section that f’ > 0. — Figure 4.29If 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 $f^{'} (x) > 0$ for all $x \in I$ . By way of contradiction, suppose that $f$ is not an increasing function on $I$ . Then there exist and in such that , but . Since is a differentiable function over , the Mean Value Theorem guarantees that there is some value such that

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .

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

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} < 0.

However, $f^{'} (x) > 0$ 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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Key Concepts

If $f$ is continuous over $[a, b]$ and differentiable over $(a, b)$ and $f (a) = f (b),$ then there exists a point $c \in (a, b)$ such that $f^{'} (c) = 0.$ This is Rolle’s theorem.

Rolle’s Theorem

Proof

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

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

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

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.

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

The Mean Value Theorem and Its Meaning

Proof

\frac{f ( b ) - f ( a )}{b - a}

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

y = \frac{f ( b ) - f ( a )}{b - a} (x - a) + f (a) .

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

g (x) = f (x) - [\frac{f ( b ) - f ( a )}{b - a} (x - a) + f (a)] .

g^{'} (x) = f^{'} (x) - \frac{f ( b ) - f ( a )}{b - a},

we see that

g^{'} (c) = f^{'} (c) - \frac{f ( b ) - f ( a )}{b - a} .

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

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .

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v (c) = s^{'} (c) = \frac{s ( 1 ) - s ( 0 )}{1 - 0} = 45 mph .

Corollaries of the Mean Value Theorem

Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections.

Proof

\frac{f ( b ) - f ( a )}{b - a} \neq = 0.

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

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .

Therefore, there exists $c \in I$ such that $f^{'} (c) \neq = 0,$ which contradicts the assumption that $f^{'} (x) = 0$ for all

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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 $h (x) = f (x) - g (x) .$ Then, $h^{'} (x) = f^{'} (x) - g^{'} (x) = 0$ for all By Corollary 1, there is a constant such that for all Therefore, for all

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Proof

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .

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

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} < 0.

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Key Concepts

If $f$ is continuous over $[a, b]$ and differentiable over $(a, b)$ and $f (a) = f (b),$ then there exists a point $c \in (a, b)$ such that $f^{'} (c) = 0.$ This is Rolle’s theorem.