Answer
relative maximum: $( e, \displaystyle \frac{1}{e})$
point of inflection: ( $e^{3/2}$ , $\displaystyle \frac{3}{2e^{3/2}}$ )
Work Step by Step
$y=\displaystyle \frac{\ln x}{x}$
Logarithmic functions have a restriction on the domain, the argument must be positive.
Domain: $x>0 $
Critical numbers$ (y^{\prime}=0$ or undefined):
$y^{\prime}=$ ... quotient rule ... =$ \displaystyle \frac{\frac{1}{x}\cdot x-1\cdot\ln x}{x^{2}}=\frac{1-\ln x}{x^{2}}$
$1-\ln x=0$
$\ln x=1$
$x=e$
0 is not in the domain, so we have one critical number, $x=e$ .
Second derivative test:
$y^{\prime\prime}=(\displaystyle \frac{1-\ln x}{x^{2}})^{\prime}=\frac{(-\frac{1}{x})\cdot x^{2}-2x(1-\ln x)}{x^{4}}$
$=\displaystyle \frac{x(-1-2+2\ln x}{x^{4}}$
$=\displaystyle \frac{2\ln x-3}{x^{3}}$
which, for x=$e$, is $\displaystyle \frac{2-3}{e^{3}}$, negative
So, at $x=e$
$y=\displaystyle \frac{\ln e}{e}=\frac{1}{e},$
we have a
relative maximum: $( e, \displaystyle \frac{1}{e})$
Points of inflection: ($y^{\prime\prime}=0)$
$y^{\prime\prime}=\displaystyle \frac{2\ln x-3}{x^{3}}=0$
$2\ln x-3=0$
$\ln x=3/2$
$x=e^{3/2}$
at $x=e^{3/2}$, y=$\displaystyle \frac{\ln e^{3/2}}{e^{3/2}}=\frac{3}{2e^{3/2}}$,
so at ( $e^{3/2}$ , $\displaystyle \frac{3}{2e^{3/2}}$ ) there is a point of inflection
