The Mean Value Theorem

Main idea

If your average speed on a trip was 60 mph, then at some moment you were going exactly 60. For any smooth function, somewhere between $a$ and $b$ the instantaneous rate of change equals the average rate of change.

f'({\htmlData{sym=1}{c}}) = {\htmlData{sym=0}{\frac{f(b) - f(a)}{b - a}}} \quad\text{for some } {\htmlData{sym=1}{c}} \in (a, b)

What the symbols mean

$\frac{f(b) - f(a)}{b - a}$
Average rate
The secant slope across the whole interval.
$c$
Somewhere inside
The theorem guarantees it exists but doesn't say where.

Explanation

The Mean Value Theorem simply says that at some moment, how fast you're going right now matches your average for the whole trip. ("Mean" is just another word for average.) The formula is three small pieces: $\frac{f(b) - f(a)}{b - a}$ (total change divided by the length of the interval, which is your average rate), $f'(c)$ (your rate at one instant $c$ ), and "for some $c$ " (the theorem promises such an instant exists, without telling you when). On a graph, it says some tangent line runs parallel to the secant line joining the two ends. It's one of those results that sounds obvious once you say it out loud.

Think about why it has to be true. If you averaged 60 mph, you couldn't have been under 60 the entire time, or you'd have covered less ground. You couldn't have been over 60 the entire time either. So unless you held exactly 60 throughout, you were sometimes below and sometimes above, and a speed that changes smoothly can't jump from one side to the other without hitting 60 on the way.

Hypotheses: $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ . Drop either and it can fail ( $|x|$ on $[-1, 1]$ has no point with slope 0, because the corner at 0 lets the slope jump straight from $-1$ to $1$ ).

Rolle's theorem is the special case $f(a) = f(b)$ : then $f'(c) = 0$ somewhere in between. If you end where you started, at some point you were standing still (or turning around).

The speeding ticket. If you drive 100 miles in 1.25 hours, your average speed is 80 mph, so at some instant the speedometer read exactly 80.

Why it matters. It connects derivatives to function values, which proves the facts we use constantly:

If $f'(x) = 0$ everywhere on an interval, $f$ is constant there.
If $f'(x) = g'(x)$ everywhere, then $f$ and $g$ differ by a constant (which is why antiderivatives have a $+C$ ).
If $f' > 0$ , then $f$ is increasing.

See it

Two speed cameras 3 miles apart clock a car taking 3 minutes, an average of 1 mile per minute (60 mph). The car speeds up and slows down along the way, but as you can see, at some moments its tangent line runs parallel to the dashed average line, which means it was going exactly 60 mph. Here that happens twice, at t = 0.75 and t = 2.25 minutes.

Worked examples

Example 1

Find $c$ for $f(x) = x^3 + 2x^2 - x$ on $[-1, 2]$ .

1
The average rate is $\dfrac{f(2) - f(-1)}{3} = \dfrac{14 - 2}{3} = 4$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $c$ for $f(x) = x^2$ on $[1, 3]$ .

Problem 2+5 XP

Find $c$ for $f(x) = \sqrt x$ on $[0, 4]$ .

Problem 3+5 XP

Rolle's theorem applies to $f(x) = x^2 - 4x + 3$ on $[1, 3]$ . Find $c$ .

Problem 4+5 XP

A car goes 150 miles in 2 hours. What speed must it have hit exactly at some moment?

Problem 5+5 XP

If $f'(x) = 0$ for all $x$ and $f(1) = 5$ , what is $f(4)$ ?

Problem 6+5 XP

A tank holds $300$ L at noon and can drain at most $12$ L per minute. What's the least it could hold at $12{:}20$ ?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

How many real solutions does $x^5 + 3x - 7 = 0$ have?

Where it's used

Average-speed cameras that time cars between two points rely on the Mean Value Theorem: an average over the limit proves the car exceeded it at some instant.

Average-Speed CamerasCalc I · Physics

Concavity and the Second Derivative Test Optimization