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प्रश्न
Show that \[\frac{\log x}{x}\] has a maximum value at x = e ?
उत्तर
\[\text { Here }, \]
\[f\left( x \right) = \frac{\log x}{x}\]
\[ \Rightarrow f'\left( x \right) = \frac{1 - \log x}{x^2}\]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow \frac{1 - \log x}{x^2} = 0\]
\[ \Rightarrow 1 = \log x\]
\[ \Rightarrow \log e = \log x\]
\[ \Rightarrow x = e\]
\[\text { Now,} \]
\[f''\left( x \right) = \frac{x^2 \left( \frac{- 1}{x} \right) - 2x\left( 1 - \log x \right)}{x^4} = \frac{- 3 + 2 \log x}{x^3}\]
\[ \Rightarrow f''\left( e \right) = \frac{- 3 + 2 \log e}{e^3} = \frac{- 1}{e^3} < 0\]
\[\text { So, x = e is the point of local maximum }.\]
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