### 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.

Ankit singhBadiya