Answer
$$
f(x)=x^{4}-18x^{2}+1; \quad\quad[-4,4]
$$
Critical numbers are: $0$, $-3$ and $3$.
The absolute maximum, $1 $, occurs when $x=0 $ and
the absolute minimum, $-80$ occurs when $x=-3$ and $ 3.$
Work Step by Step
$$
f(x)=x^{4}-18x^{2}+1; \quad\quad[-4,4]
$$
First look for critical numbers in the interval $(-4,4)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=(4)x^{3}-18(2)x+(0);\\
&=4x^{3}-36x
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =4x^{3}-36x=0\\
&= 4x(x^{2}-9)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
&= 4x(x-3)(x+3)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x =0 \quad \text {or} \quad (x-3) =0 \quad \text {or} \quad (x+3) =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x =0 \quad \text {or} \quad x =3 \quad \text {or} \quad x=-3
\end{aligned}
$$
So, critical numbers are: $0$, $-3$ and $3$.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\text{when } x=-4: f(-4)=(-4)^{4}-18(-4)^{2}+1 =-31 \\
\text{when } x=-3: f(-3)=(-3)^{4}-18(-3)^{2}+1 =-80 \\
\text{when } x=0: f(0)=(0)^{4}-18(0)^{2}+1 =1 \\
\text{when } x=3: f(3)=(3)^{4}-18(3)^{2}+1 =-80 \\
\text{when } x=4: f(4)=(4)^{4}-18(4)^{2}+1 =-31 \\
$$
We conclude that, the absolute maximum, $1 $, occurs when $x=0 $ and the absolute minimum, $-80$ occurs when $x=-3, 3.$
