Answer
Relative minimum: $(4e^{-1/2}, -8e^{-1})$
Point of inflection: $(4e^{-3/2}, -24e^{-3})$
Work Step by Step
$y=x^{2}\displaystyle \ln\frac{x}{4}$.
Domain $x>0$
[$\displaystyle \ln\frac{x}{4}]^{\prime}=$ ... chain rule ... $= \displaystyle \frac{1}{\frac{x}{4}}\cdot\frac{1}{4}=\frac{1}{x}$
$y^{\prime}=x^{2}(\displaystyle \frac{1}{x})+2x\ln\frac{x}{4}=x(1+2\ln\frac{x}{4})$
$y^{\prime\prime}=1+2\displaystyle \ln\frac{x}{4}+2x(\frac{1}{x})=3+2\ln\frac{x}{4}$
$y^{\prime}=0$ when x=0 (not in the domain) or
$1+2\displaystyle \ln\frac{x}{4}=0$
$-1=2\displaystyle \ln\frac{x}{4}$
$\displaystyle \ln\frac{x}{4}=-\frac{1}{2}$
$x=4e^{-1/2}$
For $x=4e^{-1/2}$,
$y=(4e^{-1/2})^{2}\displaystyle \ln\frac{4e^{-1/2}}{4}=16e^{-1}(-\frac{1}{2})=-8e^{-1}$
$y^{\prime\prime}=3+2\displaystyle \ln\frac{4e^{-1/2}}{4}=3+2(-\frac{1}{2})>0, $
Relative minimum: $(4e^{-1/2}, -8e^{-1})$
$y^{\prime\prime}=0$ when
$3+2\displaystyle \ln\frac{x}{4}=0$
$2\displaystyle \ln\frac{x}{4}=-3$
$\displaystyle \ln\frac{x}{4}=-\frac{3}{2}$
$\displaystyle \frac{x}{4}=e^{-3/2}$
$x=4e^{-3/2}$
For $x=4e^{-3/2}$
$y=(4e^{-3/2})^{2}\displaystyle \ln\frac{4e^{-3/2}}{4}=16e^{-3}(-\frac{3}{2})=-24e^{-3}$
Point of inflection: $(4e^{-3/2}, -24e^{-3})$
