Answer
f(x) achieves a maximum value of -7 at $x= \pm 2$.
Work Step by Step
$f(x) =x^{4}-8x^{2}+9$
$f'(x)=4x^{3}-16x$
$f'(x)=0 \rightarrow 4x^{3}-16x=0 \rightarrow x=\pm 2$
Thus, f(x) is increasing on $(-\infty, -2) \cup (2,+\infty)$ and decreasing on $(-2,2)$
f(-2) = -7
f(2) = -7
f(x) achieves a maximum value of -7 at $x= \pm 2$.
