Answer
f(x) achieves a minimum value of -85 at $x=\pm 3$ and a maximum value of -4 at $x=0$.
Work Step by Step
$f(x) =x^{4}-18x^{2}-4$
$f'(x)=4x^{3}-36x$
$f'(x)=0 \rightarrow 4x^{3}-36x=0 \rightarrow x=\pm 3, x=0$
Thus, f(x) is decreasing on $(-\infty, -3) \cup (0,3)$ and increasing on $(-3,0) \cup (3,+\infty)$
f(0) = -4
f(-3) = -85
f(3) = -85
f(x) achieves a minimum value of -85 at $x=\pm 3$ and a maximum value of -4 at $x=0$.
