What is the Mean Value Theorem in simple terms?

Somewhere between the start and the end, the instant rate of change equals the average rate of change. On a graph, some tangent line is parallel to the line joining the end points.

What is the difference between Rolle's theorem and the Mean Value Theorem?

Rolle's theorem is the special case with equal heights at the ends, f(a)=f(b), so the promised tangent line is horizontal: f'(c)=0. The Mean Value Theorem allows any heights.

Why does the Mean Value Theorem need continuity and differentiability?

Without continuity the graph could jump and skip the slope; without differentiability it could have a corner, like |x|, where no tangent line exists.

Can there be more than one value of c?

Yes. The theorem promises at least one. For a wavy graph, several tangent lines can be parallel to the secant line.

Does the Mean Value Theorem tell you where c is?

No, only that it exists. To find it, solve f'(c)=(f(b)-f(a))/(b-a); for some functions that equation can only be solved approximately.

The Mean Value Theorem Explained, With Examples

What does the Mean Value Theorem say?

Suppose you drive $120$ miles in $2$ hours. Your average speed is $60$ miles per hour. You probably went faster at some moments and slower at others, but it seems clear that at some instant your speedometer read exactly $60$ . The Mean Value Theorem is the general rule behind that intuition. (Mean here is an old word for average.)

The Mean Value Theorem

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

with

a < c < b

such that

f^{'} (c) =

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

Why it works

Draw the secant line through the two end points $(a, f (a))$ and $(b, f (b))$ . Its slope is the average rate of change.
Slide a copy of that line up or down, keeping it parallel. The last point where it still touches the graph is a point where the curve is furthest from the secant line.
At that furthest point the graph is momentarily running parallel to the secant line, so its tangent line has the same slope: $f^{'} (c)$ equals the average rate.
Continuity guarantees the furthest point exists (a continuous function on a closed interval has a largest value); differentiability guarantees there's a tangent line there.

f (x) = \sqrt{x}

[1, 9]

: the secant line joining the ends has slope

\frac{1}{4}

, and so does the tangent line at

c = 4

What do the two sides of the equation mean?

Read the equation from right to left. The right side, $\frac{f (b) - f (a)}{b - a}$ , is the change in output divided by the change in input over the whole interval: the average rate of change, or the slope of the secant line. The left side, $f^{'} (c)$ , is the instantaneous rate of change at a single input, the slope of the tangent line there (see what is a derivative?). The theorem promises they are equal at least once.

What is Rolle's theorem?

Rolle's theorem is the special case where the function has the same height at both ends, $f (a) = f (b)$ . Then the average rate of change is $0$ , so the theorem promises a point where $f^{'} (c) = 0$ : a horizontal tangent, at a hilltop or valley floor. If you throw a ball up and catch it at the same height, at some moment its vertical speed was exactly $0$ .

Worked example: Rolle's theorem

For $f (x) = x^{2} - 4 x + 1$ on $[0, 4]$ , check that Rolle's theorem applies and find $c$ .

$f$ is a polynomial, so it's continuous and differentiable everywhere.
$f (0) = 1$ and $f (4) = 16 - 16 + 1 = 1$ : equal heights.
$f^{'} (x) = 2 x - 4$ . Set it to $0$ : $2 c - 4 = 0$ , so $c = 2$ , which is between $0$ and $4$ .

Answer: $c = 2$

How do you find the value of c?

Check the conditions: $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ .
Compute the average rate of change $\frac{f (b) - f (a)}{b - a}$ .
Find $f^{'} (x)$ and solve $f^{'} (c) =$ the average rate.
Keep only solutions strictly between $a$ and $b$ .

Worked example: A polynomial

Find the number $c$ promised by the Mean Value Theorem for $f (x) = x^{2} + 2 x$ on $[0, 3]$ .

Polynomials are continuous and differentiable everywhere, so the conditions hold.
Average rate: $f (0) = 0$ and $f (3) = 9 + 6 = 15$ , so $\frac{15 - 0}{3 - 0} = 5$ .
$f^{'} (x) = 2 x + 2$ . Solve $2 c + 2 = 5$ : $2 c = 3$ , so $c = \frac{3}{2}$ .
$\frac{3}{2}$ is between $0$ and $3$ . ✓

Answer: $c = \frac{3}{2}$

Worked example: A square root

Find $c$ for $f (x) = \sqrt{x}$ on $[1, 9]$ .

$\sqrt{x}$ is continuous on $[1, 9]$ and differentiable for $x > 0$ , so the conditions hold.
Average rate: $\frac{\sqrt{9} - \sqrt{1}}{9 - 1} =$ $\frac{3 - 1}{8} =$ $\frac{1}{4}$ .
$f^{'} (x) =$ $\frac{1}{2 \sqrt{x}}$ . Solve $\frac{1}{2 \sqrt{c}} =$ $\frac{1}{4}$ : cross-multiplying gives $2 \sqrt{c} = 4$ , so $\sqrt{c} = 2$ and $c = 4$ .
$4$ is between $1$ and $9$ . ✓

Answer: $c = 4$

What happens if the conditions aren't met?

Then the conclusion can fail. $f (x) = | x |$ on $[- 1, 1]$ is continuous, and $f (- 1) = f (1) = 1$ , so the average rate of change is $0$ . But the slope is $- 1$ on the left and $1$ on the right; it's never $0$ . The theorem doesn't apply because $| x |$ isn't differentiable at $x = 0$ , where the graph has a corner.

y = | x |

[- 1, 1]

: the secant line is horizontal, but no tangent line is, because of the corner at

0

Common mistake

Giving a value of $c$ outside the interval, or at an endpoint. The theorem promises a $c$ strictly between $a$ and $b$ ; discard any other solutions of $f^{'} (c) =$ average rate.

What is the Mean Value Theorem used for?

Finding $c$ is mostly practice. The theorem's real value is what it lets you conclude about a whole function from facts about its derivative:

A zero derivative means a constant function. If $f^{'} (x) = 0$ for every $x$ in an interval, then for any two points the average rate of change equals some $f^{'} (c) = 0$ , so $f$ has the same value everywhere. This is why two antiderivatives of the same function differ only by a constant, the " $+ C$ " in integration.
Bounding change. If $f (0) = 2$ and $f^{'} (x) \leq 3$ for all $x$ , then $\frac{f (4) - f (0)}{4 - 0} =$ $f^{'} (c) \leq 3$ , so $f (4) - 2 \leq 12$ and $f (4) \leq 14$ .
Increasing and decreasing. A positive derivative on an interval means the function increases there; this is the foundation of curve sketching with derivatives.

Worked example: A speeding ticket

A car passes two toll booths $50$ miles apart, $40$ minutes apart. The speed limit is $65$ mph. Show the driver must have been speeding.

$40$ minutes is $\frac{40}{60} =$ $\frac{2}{3}$ of an hour.
Average speed: $\frac{50}{2 / 3} =$ $50 \cdot \frac{3}{2} =$ $75$ miles per hour.
Position changes smoothly (continuously and differentiably), so by the Mean Value Theorem the instantaneous speed equaled $75$ mph at some moment.
$75 > 65$ .

Answer: At some moment the speedometer read exactly $75$ mph.

The Mean Value Theorem in plain language