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.

$p$ is a polynomial such that $\deg(p)\geq 1$ and $p(x)>0$ whenever $x\geq 0.$ How can one show that there is no function $f$ satisfying the following two properties:

(1) $f(\frac{\pi}{2})=\frac{\pi}{2}$ and $f^'(x)=\frac{1}{x^r+p(f(x))}$ for $x\geq \frac{\pi}{2}$ and $r>1$

(2)$\lim_{x\rightarrow \infty}f(x)=\infty$

share|cite|improve this question

1 Answer 1

up vote 3 down vote accepted

It can be broken in several steps:

  1. $f'(\pi/2)>0$, so that $f$ is increasing in an interval $[\pi/2,\pi/2+\delta)$, $\delta>0$.
  2. $f$ is increasing and $p(f(x))>0$ in $[\pi/2,\infty)$.
  3. $0\le f'(x)\le x^{-r}$ in $[\pi/2,\infty)$.
  4. Integrate the last inequality.
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.