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Question Number 12015 by Nayon last updated on 09/Apr/17
prove that for all x ∈R,  e^x ≥x^e
$${prove}\:{that}\:{for}\:{all}\:{x}\:\in{R}, \\ $$$${e}^{{x}} \geqslant{x}^{{e}} \\ $$
Answered by mrW1 last updated on 18/Apr/17
at x=e, e^x =x^e   f(x)=e^x   g(x)=x^e =e^(eln x)   f′(x)=e^x   f′′(x)=e^x   g′(x)=e^(eln x) (e/x)  g′′(x)=((e/x))^2 e^(eln x) −e^(eln x) (e/x^2 )=(e^(x+1) /x^2 )(e−1)  f′′(e)=e^e   g′′(e)=(e^(e+1) /e^2 )(e−1)=e^(e−1) (e−1)<e^(e−1) e=e^e   ⇒f′′(e)>g′′(e)  ⇒f(x)≥g(x)
$${at}\:{x}={e},\:{e}^{{x}} ={x}^{{e}} \\ $$$${f}\left({x}\right)={e}^{{x}} \\ $$$${g}\left({x}\right)={x}^{{e}} ={e}^{{e}\mathrm{ln}\:{x}} \\ $$$${f}'\left({x}\right)={e}^{{x}} \\ $$$${f}''\left({x}\right)={e}^{{x}} \\ $$$${g}'\left({x}\right)={e}^{{e}\mathrm{ln}\:{x}} \frac{{e}}{{x}} \\ $$$${g}''\left({x}\right)=\left(\frac{{e}}{{x}}\right)^{\mathrm{2}} {e}^{{e}\mathrm{ln}\:{x}} −{e}^{{e}\mathrm{ln}\:{x}} \frac{{e}}{{x}^{\mathrm{2}} }=\frac{{e}^{{x}+\mathrm{1}} }{{x}^{\mathrm{2}} }\left({e}−\mathrm{1}\right) \\ $$$${f}''\left({e}\right)={e}^{{e}} \\ $$$${g}''\left({e}\right)=\frac{{e}^{{e}+\mathrm{1}} }{{e}^{\mathrm{2}} }\left({e}−\mathrm{1}\right)={e}^{{e}−\mathrm{1}} \left({e}−\mathrm{1}\right)<{e}^{{e}−\mathrm{1}} {e}={e}^{{e}} \\ $$$$\Rightarrow{f}''\left({e}\right)>{g}''\left({e}\right) \\ $$$$\Rightarrow{f}\left({x}\right)\geqslant{g}\left({x}\right) \\ $$

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