TITLE: Prove that $f^{(n)}(x) = 0$ for some $x$.
QUESTION [6 upvotes]: Suppose that $f : \mathbb{R} \rightarrow \mathbb{R}$ is $n$-times differentiable and $f$ has $n+1$ distinct zeros. Prove that $f^{(n)}(x) = 0$ for some $x$.
My attempt: Prove by induction
For $n=1$, we have $f$ is differentiable and has $2$ distinct zeros. Let the two zeros be $a$ and $b$. Hence, we have $f(a)=f(b)$. By Rolle's Theorem, there exists $x \in (a,b)$ such that $f^{\prime}(x)=0$
Suppose that $f^{(n-1)}(x) = 0$ is true. Then I stuck here. Can anyone guide me ?

REPLY [3 votes]: Suppose that any function $g:\mathbb{R}\to\mathbb{R}$, that is $n-1$ differentiable and has (at least) $n$ distinct zeroes satisfies that $g^{(n-1)}(x)=0$ for some $x$. Now our function $f$ is $n$ times differentiable and has $n+1$ distinct zeros, we need to show that $f^{(n)}(x)=0$ for some $x$. 
Consider $g(x)=f'(x)$, which is $n-1$ times differentiable. If we accept that $g$ has $n$ distinct zeroes then we will have that $g^{(n-1)}(x)=0$ for some $x$ but since $g^{(n-1)}(x)=f^{(n)}(x)$, we will be done.
To prove that $g(x)=f'(x)$ has $n$ distinct zeroes, it is simple: Since $f$ is $n$ times differentiable, in particular is differentiable, and since it has $n+1$ distinct zeroes, $x_0<x_1<\ldots<x_{n_1}<x_n$, between any two consecutive zeroes, $x_i<x_{i+1}$ apply Rolle's Theorem to conclude that $f'$ will have (at least) one zero, that is $g$ will have (at least) $n$ distinct zeroes.