Answer
Proof given above.
Work Step by Step
$f'' \gt 0 \Rightarrow f'$ is differentiable on $(a,b) $ (and continuous on $[a,b].$)
Assume that the opposite of the statement is true
(that $f'$ has more than one zero in $[a,b]$. )
Let $c_{1},c_{2}\in[a,b]$ be two such zeros of $f'.$
Then, by Rolle's theorem (which applies for $f'$ on $[c_{1},c_{2}]\subset[a,b]$), since $f'(c_{1})=f'(c_{2})=0$, there must be a $c_{3}\in(c_{1},c_{2})$ such that $f''(c_{3})=0.$
This is a contradiction with ($f''$(x) $ \gt 0$ for all $x\in[a,b]$ ), because $c_{3}\in[a,b].$
Our assumption is wrong, which means that the statement is true: $f^\prime$ has at most one zero on the interval.
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The same logic applies for the case when $f'' \ \lt \ 0$ on $[a,b].$
(Make the same assumption, arrive at the same contradiction.)