## Calculus with Applications (10th Edition)

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
$f(x) =x^{2}-6x+4$ $f'(x)=2x-6$ $f'(x)=0 \rightarrow 2x-6 =0 \rightarrow x=3$ 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.