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If $f: (0, \infty)\to (0, \infty)$ is increasing, is it true that the function $x\longmapsto f'(x) \cdot x^2 $ is increasing? We can assume that $f$ is twice differentiable.

Can someone provide a counter-example, a function $f$ which is increasing and positive, but $f'(a)\cdot a^2 < f'(b)\cdot b^2$, for some $ a>b $ ?

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$$ f(x) = 1-e^{-x}. $$ We have $x^2 f'(x)>0$ for all $x>0$ and $x^2 f'(x)\to0$ as $x\to\infty$, so $x^2 f'(x)$ cannot continue increasing. (It increases for small enough values of $x$.)

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