Extended Rolle's Theorem

Theorem
Let $f: D \to \R$ be differentiable on a closed interval $I \subseteq \R$.

Let $x_0 < x_1 < \dots < x_n \in I$.

Let $\map f {x_i} = 0$ for $i = 0, \ldots, n$.

Then for all $i = 0, \ldots, n-1$:
 * $\exists \xi_i \in \openint {x_i} {x_{i + 1} }: \map {f'} {\xi_i} = 0$

Proof
Since $f$ is differentiable on $I$, f is differentiable on $\closedint {x_i} {x_{i + 1} }$ for $i = 0, \ldots, n - 1$.

Thus a fortiori, $f$ is also continuous on the closed interval $\closedint {x_i} {x_{i + 1} }$ and differentiable on the open interval $\openint {x_i} {x_{i + 1} }$.

For $i = 0, \ldots, n$ we have $\map f {x_i} = 0$.

Hence the conditions of Rolle's Theorem are fulfilled on each interval which yields the result.

Comment
If $f$ is differentiable multiple times, this theorem can be applied multiple times recursively until one runs out of zeroes, each interval pair $\openint {x_i} {x_{i + 1} }$, $\openint {x_{i + 1} } {x_{i + 2} }$ becoming $\openint {\xi_i} {\xi_{i + 1} }$ with $\map f {\xi_i} = 0$, $\map f {\xi_{i + 1} } = 0$ and all $\xi \in I$.

Also see

 * Rolle's Theorem