Answer
relative minimum: $(1, \displaystyle \frac{1}{2})$
Work Step by Step
$y=\displaystyle \frac{x^{2}}{2}-\ln x$
Logarithmic functions have a restriction on the domain, the argument must be positive.
Domain: $x>0 $
$y^{\prime}=x-\displaystyle \frac{1}{x}=\frac{x^{2}-1}{x}=\frac{(x+1)(x-1)}{x}$
Critical numbers: 0, $-1$, 1
Since 0 and $-1$ are not in the domain,
we have one critical number, $x=1$.
Second derivative test:
$y^{\prime\prime}=(x-\displaystyle \frac{1}{x})^{\prime}=1-(-x^{-2})=1+\frac{1}{x^{2}},$
which is always positive.
So, at x=1
$y=\displaystyle \frac{1}{2}-\ln 1=\frac{1}{2}-0=\frac{1}{2}$,
we have a
relative minimum: $(1, \displaystyle \frac{1}{2})$
