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Question Number 96586 by  M±th+et+s last updated on 02/Jun/20

$$findtheminimumvalueof$$$$f\left(x\right)=x^x$$$$forx\in {\mathbb{R}}^{+}$$

Answered by 1549442205 last updated on 02/Jun/20

$$\mathrm{y}={\mathrm{x}}^{\mathrm{x}}\Rightarrow \mathrm{lny}=\mathrm{xlnx}\Rightarrow \frac{{\mathrm{y}}^{\prime }}{\mathrm{y}}=\mathrm{lnx}+1$$$$\Rightarrow {\mathrm{y}}^{\prime }={\mathrm{x}}^{\mathrm{x}}(\mathrm{lnx}+1)=0\iff \mathrm{lnx}=-1$$$$\iff \mathrm{x}={\mathrm{e}}^{-1}.{\mathrm{y}}^{″}=\frac{{\mathrm{x}}^{\mathrm{x}}}{\mathrm{x}}+(\mathrm{lnx}+1).{\mathrm{x}}^{\mathrm{x}}(\mathrm{lnx}+1)$$$$={\mathrm{x}}^{\mathrm{x}}(\frac{1}{\mathrm{x}}+{(\mathrm{lnx}+1)}^2).{\mathrm{y}}^{″}\left({\mathrm{e}}^{-1}\right)=$$$${\mathrm{e}}^{-{\mathrm{e}}^{-1}}.\mathrm{e}={\mathrm{e}}^{1-{\mathrm{e}}^{-1}}>0.\mathrm{Hence}$$$$\mathrm{miny}=\mathrm{minf}\left(\mathrm{x}\right)=\mathrm{f}\left({\mathrm{e}}^{-1}\right)={\mathrm{e}}^{-{\mathrm{e}}^{-1}}$$

Commented by  M±th+et+s last updated on 03/Jun/20

$$nicework$$

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