Answer
See proof below ( $f'$ can not have more than one zero.)
Also, $f'$ can not have more than one zero in the case when $f''\lt 0.$
Work Step by Step
$f''$ is defined on $[a,b] \Rightarrow f'$ is continuous on $[a,b]$ .
Let $f''$ be positive (nonzero) on $[a,b].$
$\text{Assumption:}$ $f'$ has more than one zero in $[a,b]$.
Take two of the zeros $z_{1}$ and $z_{2},$
and apply Rolle's theorem on $f'(x)$ over $[z_{1},z_{2}].$
By the theorem, there exists a $c\in(z_{1},z_{2})$ such that $f''(c)=0$.
This is a contradiction with $\qquad f''(x)\gt 0 $ for all x in $[a,b]$.
So, the assumption was wrong. $f'$ can not have more than one zero.
In the case where $f''$ is negative (nonzero) on $[a,b]$,
the same assumption as above would also lead to a contradiction, proving that $f'$ can not have more than one zero in this case as well.
