Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $ ?

share|cite|improve this question

1 Answer 1

$$ f(x) = 1-e^{-x}. $$ 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 $x$.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.