# Is $f'(x)$ monotonically increasing if $f''(x) >0$?

Given $f:(0,\infty) \rightarrow \mathbb{R}$ and $f''(x)>0\forall x \in (0,\infty)$. Is it correct to say that $f'(x)$ is a monotonically increasing function? Can I correctly assume that for any $a,b \in (0,\infty)$ $f'(a) > f('b)$ if $a > b$?

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Mean value theorem says yes. For each $a>0$, $b>0$ there exists $c\in(a,b)$ with $f'(b)=f'(a)+f''(c)(b-a)$, since $f''(x)$ is positive for all $x>0$ we get that $f'(b)\ge f'(a)$ which gives us the required statement.
You don't need $f'$ to be smooth. Differentiable is enough. – Chris Eagle Jan 5 '11 at 19:22
Yes. Set $g=f'$. Then $g' = f'' > 0$ and monotonicity of $g=f'$ follows. (By the way: $f$ is convex.)