Answer
$x=2,$
relative minimum at $(2,-3)$
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)
$f(x)=x^{2}-4x+1$
$f'(x)=2x-4\qquad $defined for all x.
$f'(x)=0$
$2x-4=0$
$2x=4$
$ x=2\qquad$ ... critical point: $(2,-3)$
Second-derivative test:
$f''(x)=2$
$ f''(2)=2 \gt 0$, so f has a relative minimum at $x=2.$
