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Question Number 15234 by mrW1 last updated on 08/Jun/17

$$\mathrm{A}\mathrm{question}\mathrm{related}\mathrm{to}\mathrm{Q}.15184$$$$\mathrm{Find}\mathrm{the}\mathrm{maximum}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)={\left(\ln \mathrm{x}\right)}^{\frac{1}{\mathrm{x}}}$$

Commented by mrW1 last updated on 09/Jun/17

$$\mathrm{the}\mathrm{domain}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)={\left(\ln \mathrm{x}\right)}^{\frac{1}{\mathrm{x}}}\mathrm{is}[1,+\mathrm{\infty })$$$$\mathrm{let}\mathrm{t}=\frac{1}{\mathrm{x}},\mathrm{t}\in (0,1]$$$$\mathrm{y}={\left(\ln \frac{1}{\mathrm{t}}\right)}^{\mathrm{t}}={(-\ln \mathrm{t})}^{\mathrm{t}}$$$$\ln \mathrm{y}=\mathrm{tln}(-\ln \mathrm{t})$$$$\frac{1}{\mathrm{y}}\times \frac{\mathrm{dy}}{\mathrm{dt}}=\ln (-\ln \mathrm{t})+\mathrm{t}\times \frac{1}{-\ln \mathrm{t}}\times \frac{-1}{\mathrm{t}}=\ln (-\ln \mathrm{t})+\frac{1}{\ln \mathrm{t}}$$$$\frac{\mathrm{dy}}{\mathrm{dt}}={(-\ln \mathrm{t})}^{\mathrm{t}}[\ln (-\ln \mathrm{t})+\frac{1}{\ln \mathrm{t}}]=0$$$$\Rightarrow \ln (-\ln \mathrm{t})+\frac{1}{\ln \mathrm{t}}=0$$$$\Rightarrow \mathrm{t}=0.1714907907\left(\mathrm{through}\mathrm{graph}\right)$$$$\Rightarrow {\mathrm{y}}_{\max }=1.102147$$

Commented by mrW1 last updated on 09/Jun/17

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