AP Calculus AB and BC

When the Mean Value Theorem Fails: Counterexamples

Q: Why does the Mean Value Theorem need differentiability only on the open interval?

Because the conclusion only ever asks about a point strictly inside. Requiring differentiability at the endpoints too would rule out perfectly good cases like f(x) = \sqrt{1-x^2} on [-1, 1], whose derivative blows up at both ends while the theorem still applies.

Q: Is |x| a counterexample to the Mean Value Theorem?

It is a counterexample to the theorem with the differentiability hypothesis removed, which is the useful way to read it. With the hypothesis in place, |x| simply does not qualify on any interval containing 0, so the theorem says nothing about it and nothing is contradicted.

The Mean Value Theorem needs f continuous on the closed interval and differentiable on the open one. Drop continuity at a single endpoint and the promised point disappears. Drop differentiability at one interior corner and it disappears too. Neither condition is decoration.

Mean Value Theorem

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

If f is continuous on the closed interval and differentiable on the open interval inside it, then somewhere strictly inside, the instantaneous rate of change equals the average rate of change across the whole interval.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1
f is continuous on the closed interval [a, b]
$f \in C[a, b]$
Differentiability on the open interval already forces continuity strictly inside, so this hypothesis is really about the two endpoints. It is what ties the secant line to the function: without it, $f(a)$ and $f(b)$ can be any numbers at all, and the average rate of change stops describing the function.
Drop it and the theorem fails
$f(x) = \begin{cases} x & 0 \le x < 1 \\ 0 & x = 1 \end{cases}$
f(x) = x on [0, 1), with f(1) redefined as 0
This function is differentiable at every point of the open interval, with $f'(x) = 1$ throughout. But moving the single endpoint value to $f(1) = 0$ makes the average rate of change $(0 - 0)/(1 - 0) = 0$ , and the derivative is never 0. One point, changed once, and the guarantee is gone.
2
f is differentiable on the open interval (a, b)
$f'(x) \text{ exists for all } x \in (a, b)$
The conclusion is a statement about $f'$ , so the derivative has to exist somewhere for the theorem to point at. One bad point is enough to break it, because the proof needs the derivative at whichever interior point the argument lands on, and you do not get to choose that point in advance.
Drop it and the theorem fails
$f(x) = |x|$
f(x) = |x| on [-1, 1]
Absolute value is continuous everywhere, so the first hypothesis holds. The endpoints give $f(-1) = f(1) = 1$ , so the average rate of change is 0. But $f'(x)$ is $-1$ to the left of the origin and $+1$ to the right, and at the origin itself the derivative does not exist. There is no point where the derivative is 0, because there is no point where the derivative is anything other than plus or minus one.

All 2 counterexamples above are checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

Subtract off the secant line. Define $g(x) = f(x) - L(x)$ , where $L$ is the straight line through $(a, f(a))$ and $(b, f(b))$ .
By construction $g(a) = g(b) = 0$ , and $g$ inherits continuity on $[a,b]$ and differentiability on $(a,b)$ from $f$ , because $L$ is a polynomial and has both everywhere.
Apply Rolle's Theorem to $g$ : there is a $c$ in $(a,b)$ with $g'(c) = 0$ .
Since $g'(x) = f'(x) - L'(x)$ and $L'$ is the constant slope $(f(b)-f(a))/(b-a)$ , that $c$ is exactly the point the Mean Value Theorem promises.

What it does not say

It tells you where c is.

It does not. The theorem is pure existence: it says at least one such point is in there, and gives you no way to find it. On a free-response question, an answer that names a specific $c$ without solving for it is asserting something the theorem never provides.

There is exactly one such c.

There can be many. For $f(x) = \sin x$ on $[0, 2\pi]$ the average rate of change is 0, and the derivative is 0 at both $\pi/2$ and $3\pi/2$ . The theorem guarantees at least one, never exactly one.

If the conclusion holds, the hypotheses must have held.

The converse is false. A function can happen to hit its average rate of change while failing a hypothesis somewhere else on the interval. The hypotheses are sufficient, not necessary.

The conclusion on its own

With every hypothesis in place, this is what you are handed, and nothing more: $\exists\, c \in (a,b)$

Frequently asked questions

Why does the Mean Value Theorem need differentiability only on the open interval?

Because the conclusion only ever asks about a point strictly inside. Requiring differentiability at the endpoints too would rule out perfectly good cases like $f(x) = \sqrt{1-x^2}$ on $[-1, 1]$ , whose derivative blows up at both ends while the theorem still applies.

Is |x| a counterexample to the Mean Value Theorem?

It is a counterexample to the theorem with the differentiability hypothesis removed, which is the useful way to read it. With the hypothesis in place, $|x|$ simply does not qualify on any interval containing 0, so the theorem says nothing about it and nothing is contradicted.