Let $f$ be a twice-differentiable function on $\mathbb{R}$ such that $f''$ is continuous. Prove that $f(x) f''(x) < 0$ cannot hold for all $x.$ Let $f$ be a twice-differentiable function on $\mathbb{R}$ such that $f''$ is continuous. Prove that
$f(x) f''(x) < 0$ cannot hold for all $x.$
I have been able to think of specific examples of $f(x)$ in which $f(x)f''(x) <0$ does not hold, but I have not been able to come up with specific values of $x$ for which $f(x)f''(x)<0$ does not hold.
Any help is greatly appreciated!

Assume wlog that $f''(0)>0$. Then $f(0)<0$ and as neither factor can change sign,  $f''(x)>0>f(x)$ for all $x\in\Bbb R$ and $f$ is strictly convex. Pick $a<b$ with $f(a)\ne f(b)$. Then the line through $(a,f(a))$ and $(b,f(b))$ intersects the $x$-axis. By convexity, $f$ is above this line outside $[a,b]$, hence must assume positive values.
