Answer
Relative minimum: $(e, e)$
Point of inflection: $(e^{2}, \displaystyle \frac{e^{2}}{2})$
Work Step by Step
$y=\displaystyle \frac{x}{\ln x}$
Logarithmic functions have a restriction on the domain, the argument must be positive. Also, the denominator must not be zero ($x\neq 1)$
Domain:$ (0, 1)\cup (1,\infty)$
Use the quotient rule to find $y^{\prime}$ and $y^{\prime\prime}$
$y^{\prime}=\displaystyle \frac{(\ln x)(1)-(x)(1/x)}{(\ln x)^{2}}=\frac{\ln x-1}{(\ln x)^{2}}$
$[(\ln x)^{2}]^{\prime}$= ... chain rule ...$= 2\displaystyle \ln x\cdot(\ln x)^{\prime}=\frac{2\ln x}{x}$
$y^{\prime\prime}=\displaystyle \frac{(\ln x)^{2}(1/x)-(\ln x-1)(\frac{2\ln x}{x})}{(\ln x)^{4}}$
$=\displaystyle \frac{\frac{\ln x}{x}[\ln x-(\ln x-1)(2)]}{(\ln x)^{4}}$
$=\displaystyle \frac{2-\ln x}{x(\ln x)^{3}}$
$y^{\prime}=0$ when $x=e$ (critical number)
For $x=e,\ y=\displaystyle \frac{e}{\ln e}=e >0$
relative minimum; $(e,e)$
$y^{\prime\prime}=0$ when $x=e^{2}$.
At $x=e^{2},\ y=\displaystyle \frac{e^{2}}{\ln e^{2}}=\frac{e^{2}}{2}.$
Inflection point: $(e^{2}, \displaystyle \frac{e^{2}}{2})$
