Solutions

We prove this by induction. Suppose that $f$ is continuous on $[a,b]$. For the base case, suppose $f$ is one-time differentiable on $(a,b)$ and that $f(x_0)=f(x_1)=0$ for some $x_0,x_1\in[a,b]$. By Rolle's theorem, there exists $x\in(a,b)$ such that $f'(x)=0$.

Now, for the induction step suppose the statement holds true for $n$. Let $f$ be $(n+1)$-times differentiable on $(a,b)$ and that $f(x)=0$ for $(n+1)+1$ different $x$ in $[a,b]$. We can then apply Rolle's theorem to the $n+1$ intervals which means that $f'(x)=0$ for $n+1$ different $x$ in $(a,b)$. Let $g(x)=f'(x)$. Since $g$ is $n$-times differentiable on $(a,b)$ and $g(x)=0$ for $n+1$ different $x$ in $(a,b]$, it follows that $g^{(n)}(x)=f^{(n+1)}(x)=0$ for some $x$ in $(a,b)$.