Let $f : [a, b] \to \Bbb R$ be $n$ times differentiable and have $n+1$ distinct roots (i.e. solutions of $f(x) = 0$) in $[a,b]$. Show that there is an $x \in [a, b]$ s. t. the $n^{\text{th}}$ derivative of $f$ has a root in $[a,b]$.
I know that you need to use Rolle's Theorem for this problem. Do I prove by induction? So for the base case $(n=1)$ then $f$ has $2$ distinct roots, then by Rolle's Theorem $f'(x)$ must have a root? How would I do the inductive case? Thanks
Between each of the $n$ pairs of neighbouring roots, there is an $x$ such that $f'(x)=0$. Hence there are at least $n$ values of $x$ that $f'(x) = 0$. By inductive assumption, the $(n-1)$th derivative of $f'(x)$ has a root in $[a,b]$.
Suppose $f$ is $n$ times differentiable and has $n+1$ distinct roots in $[a,b]$. Then, by Rolle's theorem, between each consecutive root there must be a point where $f'$ is $0$. Therefore, $f'$ is an $n-1$ times differentiable function with $n$ distinct roots in $[a,b]$.
