Question:

Key Concept: Since $f''(x) > 0$, $f$ is convex on $[a, b]$. A convex function with $f(a) = f(b) = 0$ satisfies $f(x) \leq$ the chord from $(a, 0)$ to $(b, 0)$, which is identically 0. Combined with convexity: $f(x) \leq 0$ for $x \in (a, b)$... but $f''(x) > 0$ gives $f$ is concave upward, so values lie below the chord $\equiv 0$. Hence $f(x) \leq 0$; strictly $f(x) < 0$ (cannot be zero throughout unless $f \equiv 0$, contradicting $f''(x) > 0$).