Question

Show that \[\frac{\log x}{x}\] has a maximum value at x = e ?

Answer 1

\[\text { Here }, \]

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

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

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

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

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

\[ \Rightarrow 1 = \log x\]

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

\[ \Rightarrow x = e\]

\[\text { Now,} \]

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

\[ \Rightarrow f''\left( e \right) = \frac{- 3 + 2 \log e}{e^3} = \frac{- 1}{e^3} < 0\]

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