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

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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.
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