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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.

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  • $\begingroup$ You don't need $f'$ to be smooth. Differentiable is enough. $\endgroup$ – Chris Eagle Jan 5 '11 at 19:22
  • $\begingroup$ @Chris, thanks, I fixed it. I did not remember exactly, so I added the condition, just to be sure. $\endgroup$ – mpiktas Jan 5 '11 at 19:35
  • $\begingroup$ thanks for the explanation. $\endgroup$ – Ma.H Jan 5 '11 at 19:45
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Yes. Set $g=f'$. Then $g' = f'' > 0$ and monotonicity of $g=f'$ follows. (By the way: $f$ is convex.)

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