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Prove that if $p > 0$, then there exists a constant $C_p$ such that for $x > 0$,
$\hspace{5cm}$ $x^p$ $\leq$ $C_pe^x$

The book gave a hint: examine the derivative of $\hspace{.1cm}$ $x^pe^{-x}$

This problem is in the Mean Value Theorem chapter in our book. I really have no idea where to start or what part of the MVT to use.

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  • $\begingroup$ Perhaps they mean for you to note that $x \mapsto x^p e^{-x}$ has a maximum in $[0, \infty)$? $\endgroup$ – copper.hat Nov 12 '16 at 18:10
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Let $f_p(x)=x^pe^{-x}$ for $x>0\;,\; p>0$.

$$f'_p(x)=x^{p-1}(p-x)e^{-x}$$

the maximum is reached at $x=p$, and is

$$\sup_{x>0}f_p(x)=p^pe^{-p}=C_p$$

$\implies$

$$\forall x>0\;\; x^pe^{-x}\leq C_p$$

$\implies$

$$\forall x>0\;\; x^p\leq C_p e^x.$$

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