Prove that $f^{(n)}(x) = 0$ for some $x$. 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 ?
 A: 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.
A: The induction hypothesis is that if f is (n-1) times differentiable and f has n distinct zeros then $f^(n-1)(x)=0 for some x.
Now we show for when f is n times differentiable and f has n+1 distinct zeros. By Rolle's Theorem applied n times (between the n+1 distinct zeros of f) we have that f' has n zeros and is n-1 times differentiable. Applying the induction hypothesis completes the induction.
A: Hint: Between any two roots of a differentiable function there exists a root of the derived function.
A: Rolle's Theorem applied $n$ times give us $n$ zeros of the new function $f'$. Apply the same argument to $f'$ then $f''$ has $n-1$ zeros, and $f'''$ must has $n-2$ zeros by aplying Rolle to $f''$, etc. This is the inductive approach. Can you see how is the inductive process? Which is the induction hypotesis?, and which is the inductive step?
"Suppose that $f^{n−1}(x)=0$ is true" is not enough, suppose that $f^{n−1}(x)=0$ for a quantity of values of $x$, which must be this quantity?
A: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be an $n$-times differentiable function with $n+1$ roots, $x_0 < x_1 < \dots < x_{n}$.
Applying Rolle's theorem, there exists a $c_i$ in the open interval $(x_i,x_{i+1})$ such that $Df(c_i) = 0$ for $i=0,1,\dots,n-1$.
As the intervals $(x_i,x_{i+1})$ are disjoint, $Df$ is $(n-1)$ times differentiable with $n$ distinct roots. 
Thus, by the inductive hypothesis, there is an $x \in \mathbb{R}$ such that $D_{n-1}Df(x) = D_nf(x)=0$.
