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Question
Write the maximum value of f(x) = \[\frac{\log x}{x}\], if it exists .
Solution
\[\text { Given }: \hspace{0.167em} f\left( x \right) = \frac{\log x}{x}\]
\[ \Rightarrow f'\left( x \right) = \frac{1 - \log x}{x^2}\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \frac{1 - \log x}{x^2} = 0\]
\[ \Rightarrow 1 - \log x = 0\]
\[ \Rightarrow \log x = 1\]
\[ \Rightarrow \log x = \log e\]
\[ \Rightarrow x = e\]
\[\text { Now,} \]
\[f''\left( x \right) = \frac{- x - 2x\left( 1 - \log x \right)}{x^4} = \frac{- 3x - 2x \log x}{x^4}\]
\[\text { At }x = e: \]
\[f''\left( e \right) = \frac{- 3e - 2e \log e}{e^4} = \frac{- 5}{e^3} < 0\]
\[\text { So, x = e is a point of local maximum }. \]
\[\text { Thus, the local maximum value is given by}\]
\[f\left( e \right) = \frac{\log e}{e} = \frac{1}{e}\]
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