Math Core

If you drive 150 miles in 2 hours, your average speed is 75 mph. Common sense says that at some instant your speedometer must have read exactly 75. The mean value theorem turns that common sense into a precise statement about functions, and it is the tool that connects what a function does over an interval to what its derivative does at a single point.

Average rate versus instantaneous rate

You already know two kinds of rate of change for a function $f$ on an interval $[a, b]$ :

The average rate of change is the slope of the secant line through $(a, f(a))$ and $(b, f(b))$ : $\frac{f(b) - f(a)}{b - a}.$
The instantaneous rate of change at $x = c$ is $f'(c)$ , the slope of the tangent line at $c$ .

The mean value theorem says that, for a well-behaved function, some tangent line inside the interval is parallel to the secant line. In other words, at some point the instantaneous rate equals the average rate.

The mean value theorem (MVT)

If $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , then there is at least one number $c$ in $(a, b)$ with

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

Look at the picture below for $f(x) = x^3 - x$ on $[0, 2]$ . The secant line through $(0, 0)$ and $(2, 6)$ has slope $3$ . Slide a line with slope $3$ across the graph and it touches the curve at one point strictly between $0$ and $2$ . That point is the $c$ the theorem promises.

The secant line y = 3x (slope 3) and the parallel tangent line at c = 2/√3 ≈ 1.155 (dashed).Open in grapher →

Why both hypotheses matter

The theorem only works if both conditions hold. On the AP exam you are expected to check them out loud before you use the theorem.

Continuous on $[a, b]$ . A jump or hole lets the function "teleport" without ever having the needed slope. For example, a function could be $0$ on $[0, 1)$ and jump to $5$ at $x = 1$ . The average rate on $[0, 1]$ is $5$ , but the derivative is $0$ everywhere inside.
Differentiable on $(a, b)$ . A corner lets the slope switch suddenly from one value to another and skip the average. The function $f(x) = |x|$ on $[-1, 1]$ has average rate of change $\dfrac{1 - 1}{2} = 0$ , yet $f'(x)$ is only ever $-1$ or $1$ . The corner at $x = 0$ breaks the theorem.

Notice that differentiability is only required on the open interval. The endpoints need continuity, but not a derivative. That is why $f(x) = \sqrt{x}$ on $[0, 4]$ still qualifies even though $f'(0)$ doesn't exist.

Tip

Differentiable implies continuous. So if a function is differentiable on all of $[a, b]$ (for example, any polynomial), both hypotheses hold automatically. Polynomials, $\sin x$ , $\cos x$ and $e^x$ satisfy the MVT on every closed interval. Rational functions and $\ln x$ are fine on intervals that stay inside their domains.

Rolle's theorem: the flat case

When $f(a) = f(b)$ , the secant line is horizontal, so the average rate of change is $0$ . The MVT then promises a horizontal tangent.

Definition

Rolle's theorem

If $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $f(a) = f(b)$ , then there is at least one $c$ in $(a, b)$ with $f'(c) = 0$ .

Rolle's theorem is just the MVT with a level secant line. If a ball is thrown up and lands back at the height it started from, at some moment its vertical velocity was exactly $0$ .

Finding the value of $c$

To find the value (or values) of $c$ guaranteed by the theorem:

Check that $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ .
Compute the average rate of change $\dfrac{f(b) - f(a)}{b - a}$ .
Set $f'(c)$ equal to it and solve.
Keep only the solutions that lie strictly between $a$ and $b$ .

Worked example: Finding c for a polynomial

Find all values of $c$ that satisfy the conclusion of the MVT for $f(x) = x^3 - x$ on $[0, 2]$ .

Solution. $f$ is a polynomial, so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$ . The average rate of change is

\frac{f(2) - f(0)}{2 - 0} = \frac{6 - 0}{2} = 3.

Since $f'(x) = 3x^2 - 1$ , solve $3c^2 - 1 = 3$ , so $c^2 = \dfrac{4}{3}$ and $c = \pm \dfrac{2}{\sqrt{3}}$ . Only the positive value lies in $(0, 2)$ , so $c = \dfrac{2}{\sqrt{3}} \approx 1.155$ .

Worked example: When the theorem does not apply

Let $f(x) = \dfrac{1}{x}$ on $[-1, 1]$ . Is there a $c$ in $(-1, 1)$ with $f'(c)$ equal to the average rate of change?

Solution. The average rate of change is $\dfrac{f(1) - f(-1)}{1 - (-1)} = \dfrac{1 - (-1)}{2} = 1$ . But $f'(x) = -\dfrac{1}{x^2}$ is negative for every $x \ne 0$ , so $f'(c) = 1$ has no solution. This doesn't contradict the MVT: $f$ is not continuous on $[-1, 1]$ because it is undefined at $x = 0$ . The hypotheses fail, so the theorem makes no promise.

The MVT with tables

AP free-response questions often give a table of values instead of a formula. You can't solve for $c$ , but you can still conclude that $c$ exists.

Worked example: Justifying with a table

A differentiable function $h$ gives the temperature, in degrees Celsius, of a cup of coffee $t$ minutes after it is poured.

$t$ (minutes)	$0$	$4$	$10$	$16$
$h(t)$ (°C)	$88$	$76$	$64$	$55$

Must there be a time $t$ with $0 < t < 16$ at which $h'(t) = -2$ ? Justify your answer.

Solution. Look for a pair of table values whose average rate of change is $-2$ . On $[4, 10]$ :

\frac{h(10) - h(4)}{10 - 4} = \frac{64 - 76}{6} = -2.

Because $h$ is differentiable, it is also continuous, so $h$ is continuous on $[4, 10]$ and differentiable on $(4, 10)$ . By the mean value theorem, there is a time $t$ in $(4, 10)$ with $h'(t) = -2$ . So yes, there must be such a time.

Common mistake

Don't skip the hypotheses. An AP justification needs three parts: (1) the function is continuous on the closed interval and differentiable on the open interval (say why, for example "because $h$ is differentiable"), (2) the computed average rate of change, and (3) the conclusion that some $c$ in the open interval has $f'(c)$ equal to that value. Also remember the MVT only guarantees existence. It doesn't tell you where $c$ is, and there may be more than one.

Practice

Practice 1

Find the value of $c$ guaranteed by the mean value theorem for $f(x) = x^2 - 4x + 1$ on $[1, 5]$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find the value of $c$ guaranteed by the mean value theorem for $f(x) = \sqrt{x}$ on $[0, 9]$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Which function satisfies the hypotheses of the mean value theorem on $[-1, 1]$ ?

Practice 4

Find all values of $c$ in $(-3, 3)$ that satisfy the conclusion of the mean value theorem for $f(x) = x^3$ on $[-3, 3]$ .

Separate answers with commas, e.g. 2, -5

Practice 5

The function $f(x) = \cos(2x)$ satisfies the hypotheses of Rolle's theorem on $[0, \pi]$ . Find the value of $c$ in $(0, \pi)$ with $f'(c) = 0$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

A car's position $s(t)$ , in miles, is differentiable. Selected values are shown.

$t$ (hours)	$0$	$2$	$5$	$8$
$s(t)$ (miles)	$0$	$110$	$260$	$380$

Which statement must be true?

Practice 7

Find the value of $c$ guaranteed by the mean value theorem for $f(x) = \ln x$ on $[1, e]$ . Give the exact value or a decimal rounded to three places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Average rate versus instantaneous rate

Why both hypotheses matter

Rolle's theorem: the flat case

Finding the value of ccc

The MVT with tables

Practice

Finding the value of $c$