# $f$ has zero of multiplicity $m$ at $\alpha$ and $k$ at $\beta$, where $m+k-1=n$. Prove that $f^{(n)}$ has at least one zero in $(\alpha, \beta)$.

Let $f\in C^n[a,b]$. Suppose that $f$ has a zero of multiplicity $m$ at $\alpha$ and a root of multiplicity $k$ at $\beta$, where $m\geq1$, $k\geq1$, and $m+k-1=n$. Prove that $f^{(n)}$ has at least one zero in $(\alpha, \beta)$.

I know we can use Rolle's Theorem. I'm not sure how. Any solutions/hints are greatly appreciated.

The key fact is this: if $f$ is differentiable and has $n$ zeroes on $[a,b]$ counting multiplicities, then $f'$ has at least $n-1$ zeroes on $[a,b]$, counting multiplicities. The statement you want to prove then follows by induction.
To prove the above fact, let $k$ be the number of distinct zeroes. Taking the derivative, we lose $k$ zeros (counting multiplicities), and gain at least $k-1$ by Rolle's theorem. So the total goes down at most by $1$.