# Proving the existence of point $z$ s.t. $f^{(n)}(z) = 0$

Suppose that $f$ is $n$ times differentiable on an interval $I$ and there are $n + 1$ points $x_0, x_1, \ldots, x_n \in I, x_0 < x_1 < \cdots < x_n$, such that $f(x_0) = f(x_1) = \cdots = f(x_n) = 0$. Prove that there exists a point $z \in I$ such that $f^{(n)}(z) = 0$.

I am trying to solve this, but other than using then using Rolle's Theorem, I am not sure how to proceed.

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Use the Mean Value Theorem $n$ times:

By the Mean Value Theorem, there are $n$ points $y_0\in(x_0,x_1)$, $y_1\in(x_1,x_2)$, $\dots$, $y_{n-1}\in(x_{n-1},x_n)$ so that $f'(y_0)=f'(y_1)=\dots=f'(y_{n-1})=0$.

By the Mean Value Theorem, there are $n-1$ points $z_0\in(y_0,y_1)$, $z_1\in(y_1,y_2)$, $\dots$, $z_{n-2}\in(y_{n-2},y_{n-1})$ so that $f''(z_0)=f''(z_1)=\dots=f''(z_{n-2})=0$.

Repeat until you have $1$ point $w_0$ so that $f^{(n)}(w_0)=0$.

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You know from Rolle’s theorem that it’s true when $n=1$. Try it for $n=2$; Rolle’s theorem gives you points $y_0$ and $y_1$ such that $x_0<y_0<x_1<y_1<x_2$ and $f\;'(y_0)=f\;'(y_1)=0$. Can you now apply Rolle’s theorem to $f\;'$ on some interval to get something useful?
In order to generalize the hint from $2$ to $n$, you’ll need to use mathematical induction.