Answer
relative minimum: $( \displaystyle \frac{1}{e}, -\frac{1}{e})$
Work Step by Step
$y=x\ln x$
Logarithmic functions have a restriction on the domain, the argument must be positive.
Domain: $x>0 $
Critical numbers$ (y^{\prime}=0$ or undefined):
$y^{\prime}=$ ... product rule ... =$ x\displaystyle \cdot\frac{1}{x}+1\cdot\ln x=1+\ln x$
$1+\ln x=0$
$\ln x=-1$
$x=e^{-1}=\displaystyle \frac{1}{e}$
we have one critical number, $x=\displaystyle \frac{1}{e}$ .
Second derivative test:
$y^{\prime\prime}=(1+\displaystyle \ln x)^{\prime}=\frac{1}{x},$
which, for x=$\displaystyle \frac{1}{e}$, is positive.
So, at x=$\displaystyle \frac{1}{e}$
$y=\displaystyle \frac{1}{e}\ln\frac{1}{e}=-\frac{1}{e},$
we have a
relative minimum: $( \displaystyle \frac{1}{e}, -\frac{1}{e})$
