Solution
= Solution
Because $f''$ exists everywhere, $f'$ is differentiable everywhere and therefore continuous everywhere; <continuity> of $f''$ is not required. For $x<y$, apply the <mean value theorem> to $f'$ on $[x,y]$. It gives $\xi\in(x,y)$ with
$$
f'(y)-f'(x)=(y-x)f''(\xi)\geq0.
$$
For $x=y$ equality is immediate. Hence \b[$f'$ is nondecreasing: $x\leq y$ implies $f'(x)\leq f'(y)$].