The Mean Value Theorem: Learn It 2

Lumen Learning

The Mean Value Theorem: Learn It 2

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 5).

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 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 $c_1$ and $c_2$ such that the tangent line to $f$ at $c_1$ and $c_2$ has the same slope as the secant line.

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 $c \in (a,b)$ such that the tangent line to the graph of $f$ at $c$ is parallel to the secant line connecting $(a,f(a))$ and $(b,f(b))$ .

Mean Value Theorem

Let $f$ be continuous over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$ . Then, there exists at least one point $c \in (a,b)$ such that

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

$f^{\prime}(c)=\dfrac{f(b)-f(a)}{b-a}$

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}

$\dfrac{f(b)-f(a)}{b-a}$

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)

$y=\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)$

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)]

$g(x)=f(x)-\left[\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)\right]$

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(x). A line is drawn between the maximum of f(x) and the secant line and it is marked g(x). — 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

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

$g^{\prime}(x)=f^{\prime}(x)-\dfrac{f(b)-f(a)}{b-a}$ ,

we see that

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

$g^{\prime}(c)=f^{\prime}(c)-\dfrac{f(b)-f(a)}{b-a}$

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

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

$f^{\prime}(c)=\dfrac{f(b)-f(a)}{b-a}$

$_\blacksquare$

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.

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.

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 7). 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}

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

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}

$\dfrac{1}{2\sqrt{c}}=\dfrac{1}{3}$

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.

The function f(x) = the square root of x is graphed from (0, 0) to (9, 3). There is a secant line drawn from (0, 0) to (9, 3). At point (9/4, 3/2), there is a tangent line that is drawn, and this line is parallel to the secant line. — Figure 7. The slope of the tangent line at $c=\frac{9}{4}$ 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.

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

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

$v(c)=s^{\prime}(c)=\dfrac{s(1)-s(0)}{1-0}=45$ mph

If a rock is dropped from a height of $100$ ft, its position $t$ seconds after it is dropped until it hits the ground is given by the function $s(t)=-16t^2+100$ .

Determine how long it takes before the rock hits the ground.
Find the average velocity $v_{\text{avg}}$ of the rock for when the rock is released and the rock hits the ground.
Find the time $t$ guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is $v_{\text{avg}}$ .

When the rock hits the ground, its position is $s(t)=0$ . Solving the equation $-16t^2+100=0$ for $t$ , we find that $t=\pm \frac{5}{2}$ sec. Since we are only considering $t \ge 0$ , the ball will hit the ground $\frac{5}{2}$ sec after it is dropped.
The average velocity is given by:
$v_{\text{avg}}=\frac{s(5/2)-s(0)}{5/2-0}=\frac{1-100}{5/2}=-40$ ft/sec
The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time $t$ such that $v(t)=s^{\prime}(t)=v_{\text{avg}}=-40$ ft/sec. Since $s(t)$ is continuous over the interval $[0,5/2]$ and differentiable over the interval $(0,5/2)$ , by the Mean Value Theorem, there is guaranteed to be a point $c \in (0,5/2)$ such that
$s^{\prime}(t)=\frac{s(5/2)-s(0)}{5/2-0}=-40$

Taking the derivative of the position function $s(t)$ , we find that $s^{\prime}(t)=-32t$ . Therefore, the equation reduces to $s^{\prime}(t)=-32c=-40$ . Solving this equation for $t$ , we have $t=\frac{5}{4}$ . Therefore, $\frac{5}{4}$ sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: $-40$ ft/sec.

Figure 8. At time $t=\frac{5}{4}$ 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.

You can view the transcript for this segmented clip of “4.4 Mean Value Theorem” here (opens in new window).

Watch the following video to see the worked solution to the two previous examples.