Answer
$x=\displaystyle \frac{1}{2},$
relative minimum at $(\displaystyle \frac{1}{2},\frac{5}{2})$
Work Step by Step
Critical points of $f$ are points that are either stationary or singular
(for which either $f'(c)=0$ or $f'(c)$ is undefined, but $f(c)$ is)
$f(x)=2x^{2}-2x+3$
$f'(x)=4x-2\qquad $defined for all x.
$f'(x)=0$
$4x-2=0$
$4x=2$
$ x=\displaystyle \frac{1}{2}\qquad$ ... critical point: $(\displaystyle \frac{1}{2},\frac{5}{2})$
Second-derivative test:
$f''(x)=4$
$ f''(\displaystyle \frac{1}{2})=4 \gt 0$, so f has a relative minimum at $x=1/2.$
