Extended Rolle's Theorem

Theorem
Let $f: D \to \R$ be differentiable on an interval $I \subset \R$.

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

Let $f \left({x_i}\right) = 0$ for $i = 0, \ldots, n$.

Then:
 * $\exists \xi_i \in \left({x_i \,.\,.\, x_{i+1}}\right): f' \left({\xi_i}\right) = 0$ for all $i = 0, \ldots, n-1$

Proof
Since f is differentiable on $I$, f is differentiable on $\left[{x_i \,.\,.\, x_{i+1}}\right]$ for $i = 0, \ldots, n-1$, so a fortiori f is continuous on the closed interval $\left[{x_i \,.\,.\, x_{i+1}}\right]$ and differentiable on the open interval $\left({x_i \,.\,.\, x_{i+1}}\right)$.

For $i = 0, \ldots, n$ we have $f \left({x_i}\right) = 0$.

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

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