Q. Show that the function given by f(x)=\frac{log x}{x} has maximum at x = e.

Ans:
f(x)=\frac{log x}{x}
f'(x)=\frac{\frac{1}{x}.x - 1.log x}{x^2}
or f'(x)=\frac{1 - log x}{x^2}
now f'(x) = 0,
\frac{1 - log x}{x^2} = 0
1 - log x = 0
log x = 1
\implies x = e
thus, f'(x) = 0 at x = e
Now f''(x)=\frac{\frac{-1}{x}.x^2 - 2x(1-log x)}{x^4}
or f''(x)=\frac{-3 + 2log x}{x^3}
or f''(e)=\frac{-3 + 2log e}{e^3}
or f''(e)=\frac{-1}{e^3}
\implies f''(e) < 0
Therefore, by second derivative test x = e is a point of local maxima.
\therefore f(x) has maximum value at x = e.
Hence proved.