 Solution

Apply Rolle's theorem on each $[x_i,x_{i+1}]$. This gives $n$ ordered zeros of the derivative $f^{\prime }$ strictly between the original $n+1$ zeros of $f$. Apply Rolle's theorem to $f^{\prime }$ between each consecutive pair of its zeros to obtain $n-1$ ordered zeros of $f^{\prime \prime }$. Continue in this way.

To justify the iteration, for $j<n$, the derivative $f^{(j)}$ is differentiable throughout $(a,b)$ and hence continuous there by differentiability implies continuity. Every new closed interval lies strictly inside $(a,b)$, so Rolle's theorem applies at each step. After $j$ steps there are $n+1-j$ zeros of $f^{(j)}$. At $j=n$ this givesThis is the distinct-zero case of the Rolle root count with multiplicities.