Answer
f(x) is decreasing on $(-\infty, 2)$ and increasing on $(2,+\infty)$.
On $(-\infty,+\infty)$ f(x) achieves a minimum value of -7 at x=2
Work Step by Step
$f(x) =2x^{2}-8x+1$
$f'(x)=4x-8$
$f'(x)=0 \rightarrow 4x-8 =0 \rightarrow x=2$
Thus, f(x) is decreasing on $(-\infty, 2)$ and increasing on $(2,+\infty)$
On $(-\infty,+\infty)$ f(x) achieves a minimum value of -7 at x=2
