#### Question

Write the maximum value of f(x) = \[\frac{\log x}{x}\], if it exists .

#### Solution

\[\text { Given }: \hspace{0.167em} f\left( x \right) = \frac{\log x}{x}\]

\[ \Rightarrow f'\left( x \right) = \frac{1 - \log x}{x^2}\]

\[\text { For a local maxima or a local minima, we must have } \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow \frac{1 - \log x}{x^2} = 0\]

\[ \Rightarrow 1 - \log x = 0\]

\[ \Rightarrow \log x = 1\]

\[ \Rightarrow \log x = \log e\]

\[ \Rightarrow x = e\]

\[\text { Now,} \]

\[f''\left( x \right) = \frac{- x - 2x\left( 1 - \log x \right)}{x^4} = \frac{- 3x - 2x \log x}{x^4}\]

\[\text { At }x = e: \]

\[f''\left( e \right) = \frac{- 3e - 2e \log e}{e^4} = \frac{- 5}{e^3} < 0\]

\[\text { So, x = e is a point of local maximum }. \]

\[\text { Thus, the local maximum value is given by}\]

\[f\left( e \right) = \frac{\log e}{e} = \frac{1}{e}\]