The Mean Value Theorem and the Extreme Value Theorem

Quick answer

The Mean Value Theorem says that if f is continuous on [a, b] and differentiable on (a, b), then at some point c in between, the instantaneous rate f′(c) equals the average rate (f(b) − f(a))/(b − a): somewhere the tangent line is parallel to the secant. Rolle's theorem is the case f(a) = f(b). The Extreme Value Theorem says a continuous function on a closed interval reaches a maximum and a minimum. Together they explain why f′ = 0 forces f to be constant and f′ > 0 forces f to increase.

What you'll learn

State and apply Rolle's theorem and the Mean Value Theorem
Check the hypotheses before applying either theorem
State the Extreme Value Theorem and recognize when it fails
Use the Mean Value Theorem to justify facts about functions

Rolle’s theorem

If a smooth curve starts and ends at the same height, somewhere in between it must level off.

Rolle's theorem

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

By the Extreme Value Theorem below, $f$ reaches a maximum and a minimum on $[a, b]$ . If both are at the endpoints, $f$ is constant and $f' = 0$ everywhere. Otherwise one of them is inside, and at an interior peak or valley of a differentiable function the tangent is horizontal.

The Mean Value Theorem

Tilt Rolle’s picture, and the horizontal tangent becomes a tangent parallel to the secant line.

Mean Value Theorem

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then for at least one $c$ in $(a, b)$ ,

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

y = x³ − x
secant, slope 3
tangent at c, slope 3

A tangent parallel to the secant

Why the theorem is true

Subtract the secant line from $f$ . Let $s(x)$ be the secant through $\big(a, f(a)\big)$ and $\big(b, f(b)\big)$ , and set $g(x) = f(x) - s(x)$ .

At both endpoints $f$ and $s$ agree, so $g(a) = g(b) = 0$ . And $g$ is continuous and differentiable wherever $f$ is. Rolle’s theorem gives a $c$ with $g'(c) = 0$ , which means $f'(c) = s'(c)$ : the slope of the curve equals the slope of the secant. The Mean Value Theorem is Rolle’s theorem viewed from a tilted angle.

In terms of motion: if a car covers $90$ miles in $1.5$ hours, its average speed is $60$ mph, and at some instant its speedometer read exactly $60$ .

What the theorem proves

The Mean Value Theorem turns information about $f'$ into information about $f$ :

If $f'(x) = 0$ on an interval, then $f$ is constant there: any two values differ by $f'(c)(b - a) = 0$ .
If $f'(x) = g'(x)$ on an interval, then $f$ and $g$ differ by a constant.
If $f'(x) > 0$ on an interval, then $f$ is increasing: for $a < b$ , $f(b) - f(a) = f'(c)(b - a) > 0$ .

The Extreme Value Theorem

Extreme Value Theorem

If $f$ is continuous on the closed interval $[a, b]$ , then $f$ attains an absolute maximum value and an absolute minimum value on $[a, b]$ .

Both hypotheses are needed. On the open interval $(0, 1)$ , $f(x) = x$ has no maximum: it gets arbitrarily close to $1$ but never reaches it. And a function with a jump or an asymptote can skip over the value it seems to approach.

Worked examples

Common mistakes

Practice problems

Find the $c$ guaranteed by the Mean Value Theorem for $f(x) = x^2$ on $[1, 3]$ .

Answer

$c = 2$

Full solution

Secant slope: $\tfrac{9 - 1}{3 - 1} = 4$ . $f'(c) = 2c = 4$ gives $c = 2$ , which is in $(1, 3)$ .
Verify Rolle’s theorem for $f(x) = x^2 - 4x$ on $[0, 4]$ .

Answer

$c = 2$

Full solution

$f$ is a polynomial, and $f(0) = f(4) = 0$ . $f'(c) = 2c - 4 = 0$ gives $c = 2$ , in $(0, 4)$ .
Find the $c$ for $f(x) = \sqrt{x}$ on $[0, 4]$ .

Answer

$c = 1$

Full solution

$f$ is continuous on $[0, 4]$ and differentiable on $(0, 4)$ . Secant slope: $\tfrac{2 - 0}{4} = \tfrac{1}{2}$ . $\tfrac{1}{2\sqrt{c}} = \tfrac{1}{2}$ gives $c = 1$ .
Does the Mean Value Theorem apply to $f(x) = \tfrac{1}{x}$ on $[-1, 1]$ ? Explain.

Answer

No; $f$ is not continuous on $[-1, 1]$ .

Full solution

$f$ is undefined at $0$ . In fact the secant slope is $\tfrac{1 - (-1)}{2} = 1$ , but $f'(x) = -\tfrac{1}{x^2}$ is always negative — no $c$ works.
A runner covers $10$ km in $50$ minutes. Explain why at some moment the runner’s speed was exactly $0.2$ km per minute.

Answer

The average speed is $0.2$ km per minute, and the Mean Value Theorem guarantees an instant at that speed.

Full solution

Position is continuous and differentiable in time. The average rate is $\tfrac{10}{50} = 0.2$ km/min, so $s'(c) = 0.2$ at some instant $c$ .
$f(1) = 3$ and $f'(x) \le 2$ for all $x$ . What is the largest possible value of $f(4)$ ?

Answer

$9$

Full solution

$f(4) - f(1) = f'(c)(4 - 1) \le 2 \cdot 3 = 6$ , so $f(4) \le 9$ .
Does $f(x) = x^3 + x + 1$ have more than one real root? Use Rolle’s theorem.

Answer

No, exactly one

Full solution

It has at least one root, since it goes from negative to positive values. If it had two, Rolle’s theorem would give a $c$ with $f'(c) = 0$ . But $f'(x) = 3x^2 + 1 > 0$ everywhere.
Does $f(x) = \tfrac{1}{x}$ attain a maximum on $(0, 1]$ ? On $[1, 2]$ ?

Answer

Not on $(0, 1]$ ; yes on $[1, 2]$ , at $x = 1$ .

Full solution

On $(0, 1]$ the values grow without bound as $x \to 0^+$ . On $[1, 2]$ , $f$ is continuous on a closed interval, so the Extreme Value Theorem applies; the maximum is $f(1) = 1$ .
$f'(x) = g'(x)$ for all $x$ , $f(0) = 5$ and $g(0) = 2$ . Find $f(x) - g(x)$ .

Answer

$3$

Full solution

Equal derivatives mean $f - g$ is constant. At $0$ it is $5 - 2 = 3$ .
A student applies Rolle’s theorem to $f(x) = x^{2/3}$ on $[-1, 1]$ and concludes $f'(c) = 0$ for some $c$ . What went wrong?

Hint

Is $f$ differentiable at $0$ ?

Answer

$f$ has a cusp at $0$ , so it is not differentiable on $(-1, 1)$ and the theorem does not apply.

Full solution

$f(-1) = f(1) = 1$ , but $f'(x) = \tfrac{2}{3}x^{-1/3}$ is undefined at $0$ and never $0$ elsewhere.

The conclusion fails exactly because a hypothesis does.

Frequently asked questions

What does the Mean Value Theorem say?

If f is continuous on [a, b] and differentiable on (a, b), there is a c in (a, b) with f′(c) = (f(b) − f(a))/(b − a): the instantaneous rate equals the average rate somewhere.

What is Rolle's theorem?

The special case with f(a) = f(b): the average rate is 0, so f′(c) = 0 for some c in (a, b), a horizontal tangent.

What is the Extreme Value Theorem?

A function continuous on a closed interval [a, b] attains an absolute maximum and an absolute minimum on that interval.

Why do the hypotheses matter?

Without them the conclusions can fail: |x| on [−1, 1] has equal endpoint values but no horizontal tangent, because it is not differentiable at 0.

Does the Mean Value Theorem say where c is?

No. It guarantees that such a c exists in (a, b); finding it is a separate calculation, and there may be several.

What to learn next

Increasing, Decreasing and the First Derivative TestUse the sign of f′ to find where a function rises and falls, locate critical points, and classify local maxima and minima with the first derivative test.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Frequently asked questions

What does the Mean Value Theorem say?

What is Rolle's theorem?

What is the Extreme Value Theorem?

Why do the hypotheses matter?

Does the Mean Value Theorem say where c is?

What to learn next

More Calculus lessons