Answer
f(x) is decreasing on $(-\infty, 3)$ and increasing on $(3,+\infty)$.
On $(-\infty,+\infty)$ f(x) achieves a minimum value of -5 at x=3
Work Step by Step
$f(x) =x^{2}-6x+4$
$f'(x)=2x-6$
$f'(x)=0 \rightarrow 2x-6 =0 \rightarrow x=3$
Thus, f(x) is decreasing on $(-\infty, 3)$ and increasing on $(3,+\infty)$
On $(-\infty,+\infty)$ f(x) achieves a minimum value of -5 at x=3.