Answer
Proof of statement is given below.
A generalization of the statement would be:
If $f^{(n)}$ is continuous on $[a,b]$, and $f$ has $n+1$ zeros in $[a,b]$, then $f^{(n)}$ has at least one zero in $[a,b]$.
Work Step by Step
Let $z_{1}, z_{2},z_{3}$ be zeros of $f(x)$ in $[a,b]$, ordered from least to greatest.
f is differentiable $\Rightarrow$ f is continuous,
so we can apply Rolle's Theorem on $f(x)$ over $[z_{1},z_{2}]$
( because $f$ is continuous on $[z_1,z_2]$, differentiable on $(z_1,z_2)$ and $f(z_{1})=f(z_{2})=0$)
There exists a value $c_{1}\in(z_{1},z_{2})$ such that $f'(c_{1})=0$
Applying Rolle's Theorem on $f(x)$ over $[z_{2},z_{3}]$
There exists a value $c_{2}\in(z_{2},z_{3})$ such that $f'(c_{2})=0$
Now we apply Rolle's Theorem on $f'(x)$ over $[c_{1},c_{2}]$:
there exists a $c\in(c_{1},c_{2})$ such that $f''(c)=0$
which proves the statement.
A generalization of the statement would be:
If $f^{(n)}$ is continuous on $[a,b]$, and $f$ has $n+1$ zeros in $[a,b]$, then $f^{(n)}$ has at least one zero in $[a,b]$.
