Function that gets n times zero value, is it derivative gets n-1 times 0 value?

Question

I faced the following question while practising exams in calculus :

$f(x)$ is a function that is continuous in $[a,b]$ and differentiate in $(a,b)$. If $f(x)$ gets exactly $n$ times zero value (means $f(x) = 0$ occur $n$ times), then it's derivative $f'(x)$ gets $n-1$ times zero value.

On one hand, I couldn't find any function that it's derivative gets $n-2$ times zero value. But on the other hand, I don't know how to prove that.

Thanks!!

Answer 1

Its derivative $f'$ has at least $n-1$ zeroes. Because between any two zeroes of $f$, there is at least one zero of $f'$, by Rolle's Theorem.