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

(N/A) The given function is $f(x) = \frac{\log x}{x}$.
First,we find the first derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} = \frac{1 - \log x}{x^2}$.
To find the critical points,set $f'(x) = 0$:
$1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
Next,we find the second derivative $f''(x)$:
$f''(x) = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{-3x + 2x \log x}{x^4} = \frac{2 \log x - 3}{x^3}$.
Now,evaluate $f''(x)$ at $x = e$:
$f''(e) = \frac{2 \log e - 3}{e^3} = \frac{2(1) - 3}{e^3} = \frac{-1}{e^3}$.
Since $f''(e) = \frac{-1}{e^3} < 0$,by the second derivative test,the function $f(x)$ has a local maximum at $x = e$.