Let $x=c$ be the $x$ coordinate of absolute max of $f(x)$ on $[a,b]$. (This point exists by the extreme value theorem). I will show that $f(c) = 0$. Since $f(a) = 0$ and $c$ is the absolute max, $f(c)\geq 0$. By Fermat's theorem, we know f'(c) = 0. Hence, we learn that f(c) = f''(c)\geq 0.
Now, assume for a contradiction that $f(c) > 0$, so f''(c) > 0 and $c\neq a$ and $c\neq b$. I claim that for $x$ close enough to $c$, but bigger than it, that $f(x) > f(c)$, contradicting maximality of $f(c)$.
Since f'' is continuous, for $x$ close enough to $c$ say, within $\delta$, we have f''(x) > 0. On the interval where $c, f'(x) \geq 0 with equality only at $x=c$. This follows from the Mean value theorem applied to f', because if f'(x)\leq 0 for a point $x\in(c,c+\delta)$, then by the MVT, f''(d) \leq 0 for some $d\in(c,c+\delta)$, giving a contradiction.
From this, it follows that $f(x)>f(c)$ for $x\in(c,c+\delta)$, because, again by the MVT, we have \frac{f(x)-f(c)}{x-c} = f'(d) > 0 for some $d\in(c,c+\delta)$, so, $f(x) - f(c) > 0$.
Thus, we contradict maximality of $f(c)$. From this contradiction, we deduce $f(c) = 0$ is the maximum of the function. Now, repeat a similar argument to $-f$ (changing the interval $(c,c+\delta)$ to $(c-\delta, c)$) to deduce the minimum value of $f$ is $0$. From this it follows that $f$ is identically $0$.