AP Calculus AB and BC

When Rolle's Theorem Fails: Three Counterexamples

Rolle's Theorem needs three things: continuity on the closed interval, differentiability on the open one, and equal values at the two endpoints. Drop any single one and there is a function with no horizontal tangent anywhere inside. All three failures are shown below.

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. 1

   f is continuous on the closed interval [a, b]

   $$f \in C[a, b]$$

   This is the hypothesis that binds the endpoints to the rest of the graph. Differentiability on the open interval already gives continuity inside, so what continuity adds here is the promise that the function does not jump as it arrives at $a$ or $b$.

   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

   The endpoint values now agree, $f(0) = f(1) = 0$, so the third hypothesis holds. The function is differentiable throughout $(0,1)$, so the second holds. Only continuity at $x = 1$ fails, and that alone is enough: $f'(x) = 1$ everywhere inside, never 0.

2. 2

   f is differentiable on the open interval (a, b)

   $$f'(x) \text{ exists for all } x \in (a, b)$$

   The conclusion names $f'(c)$, so a function with no derivative at the one place a horizontal tangent could occur has nothing to offer. A single corner is enough, and the corner tends to sit exactly where the turning point is.

   Drop it and the theorem fails

   $$f(x) = |x|$$

   f(x) = |x| on [-1, 1]

   Continuity holds and the endpoints agree, $f(-1) = f(1) = 1$. The turning point is at the origin, which is precisely where the derivative fails to exist. Everywhere else the slope is $\pm 1$. The graph turns around without ever being flat.

3. 3

   The endpoint values are equal, f(a) = f(b)

   $$f(a) = f(b)$$

   This is what forces the graph to come back. Without it there is nothing making the function turn around, and a function that only ever climbs has no flat spot to offer. The Mean Value Theorem is what you get when you drop this one and ask for the secant slope instead of zero.

   Drop it and the theorem fails

   $$f(x) = x$$

   f(x) = x on [0, 1]

   As bland as a function gets: continuous everywhere, differentiable everywhere, and utterly without a horizontal tangent. The only failing hypothesis is $f(0) = 0 \ne 1 = f(1)$. The Mean Value Theorem still applies here and correctly promises a point where $f'(c) = 1$, which is every point.

All 3 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

1. The Extreme Value Theorem applies, since $f$ is continuous on a closed bounded interval, so $f$ attains a maximum and a minimum on $[a,b]$.
2. If both the maximum and the minimum occur at the endpoints, then since $f(a) = f(b)$ they are equal, so $f$ is constant and $f'(c) = 0$ at every interior point.
3. Otherwise at least one of them occurs at an interior point $c$.
4. At an interior extremum where the derivative exists, the derivative must be 0, by the interior extremum theorem. Differentiability on $(a,b)$ supplies exactly that, and $c$ is the point.

What it does not say