Rolle's theorem

Statement of Rolle's theorem

Rolle's theorem. Let $f$ be a function continuous on $[a \,;\, b]$, differentiable on $\left]a \,;\, b\right[$, with $a < b$ and $f(a) = f(b)$. Then there exists at least one real number $c$ in $\left]a \,;\, b\right[$ such that $f'(c) = 0$.

Graphically: if the curve comes back to the same height, it has a horizontal tangent somewhere. Michel Rolle stated this result in 1691, for polynomials; the general version is an achievement of the 19th century.

Notice the asymmetry of the hypotheses: continuity on the closed interval, differentiability only on the open one. This is what allows the theorem to be applied to $x \mapsto \sqrt{1 - x^2}$ on $[-1 \,;\, 1]$, whose derivative blows up at the endpoints.

Proof

The function $f$ is continuous on the closed bounded interval $[a \,;\, b]$: by the extreme value theorem, it has a maximum $M$ and a minimum $m$ there, both attained.

First case: $M = m$. The function is constant, its derivative is zero everywhere, and any $c$ in $\left]a \,;\, b\right[$ works.

Second case: $M > m$. Since $f(a) = f(b)$, at least one of the two values $M$ and $m$ differs from $f(a)$. Suppose for example $M \neq f(a) = f(b)$. The maximum is attained at a point $c$ which is therefore neither $a$ nor $b$: it is interior to the interval. The function $f$ is differentiable there, and Fermat's lemma gives $f'(c) = 0$.

The reasoning is identical if it is the minimum that differs from $f(a)$.

Every hypothesis matters

Hypothesis removed	Counterexample on $[0 \,;\, 1]$	Conclusion
continuity on $[a \,;\, b]$	$f(x) = x$ for $x < 1$ and $f(1) = 0$	$f' = 1$ wherever it exists
differentiability on $\left]a \,;\, b\right[$	$f(x) = \left\lvert x - \tfrac{1}{2} \right\rvert$	$f'$ never vanishes
$f(a) = f(b)$	$f(x) = x$	$f' = 1$ everywhere

An immediate application: counting roots

Between two consecutive roots of $f$, the derivative $f'$ vanishes at least once — just apply Rolle with $f(a) = f(b) = 0$.

Thus, if a polynomial of degree $n$ has $n$ distinct real roots, its derivative has at least $n - 1$, hence exactly $n - 1$ since it has degree $n - 1$. This is the derivative version of the theorem, very useful for locating roots.

Classic pitfall

Concluding that the point $c$ is unique. The theorem only provides one, and there may be many: for $f(x) = \sin x$ on $[0 \,;\, 4\pi]$, the derivative vanishes four times. Writing "the real number $c$ such that $f'(c) = 0$" is a wording mistake; you must say "a real number $c$".