Rolle's Theorem

Theorem 7.2 (Rolle's Theorem). Let $f(x)$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, then there is at least one point $c\in(a,b)$ where $f'(c)=0$.

Geometric meaning. If the tangent is imagined moving along the curve from $x=a$ to $x=b$, and the curve starts and ends at the same height ($f(a)=f(b)$), then somewhere in between there must be a point $c\in(a,b)$ at which the tangent is parallel to the $x$-axis.

Working procedure to find the guaranteed $c$: verify $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, check $f(a)=f(b)$, solve $f'(x)=0$, and select the root(s) that actually lie in the open interval $(a,b)$ (a formula may produce extra roots outside the interval, which must be discarded).

Using Rolle's theorem indirectly — counting real roots. Rolle's theorem can also bound the number of real roots of an equation without solving it: for a polynomial $P(x)$, if $\alpha<\beta$ are two of its real zeros, then Rolle's theorem (applied to $P$ on $[\alpha,\beta]$) forces some $\gamma\in(\alpha,\beta)$ with $P'(\gamma)=0$ — i.e. between any two distinct real zeros of $P(x)$ there is a zero of $P'(x)$. Conversely, if $P'(x)$ can be shown to have no zero in some interval, $P(x)$ can have at most one zero there; combined with the Intermediate Value Theorem (a sign change of $P$ across the interval), this pins down that $P$ has exactly one real root there.

Functions for which Rolle's theorem fails to apply (the hypotheses are violated, even if $f(a)=f(b)$ happens to hold):

1. $f(x)=|x|,\ x\in[-1,1]$: $f(-1)=f(1)=1$, but $f$ is not differentiable at $x=0\in(-1,1)$.
2. A function defined piecewise with a jump is not continuous at the jump point, even if the two endpoint values agree. …