Answer
$$
f(x)=x^{4}-32x^{2}-7; \quad\quad[-5,6]
$$
C ritical numbers are: $0, -4$ and $4$.
The absolute maximum, $137 $ occurs when $x=-4, x=4 $ and the absolute minimum, $-263 $ occurs when $x=6.$
Work Step by Step
$$
f(x)=x^{4}-32x^{2}-7; \quad\quad[-5,6]
$$
First look for critical numbers in the interval $(-5,6)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=(4)x^{3}-32(2)x-(0);\\
&=4x^{3}-64x
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =4x^{3}-64x=0\\
& 4x(x^{2}-16)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 4x (x+4)(x-4)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0 \quad \text {or} \quad (x+4) =0 \quad \text {or} \quad (x-4) =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0 \quad \text {or} \quad x =-4 \quad \text {or} \quad x =4
\end{aligned}
$$
So, critical numbers are: $0, -4$ and $4$.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\text{when } x=-5: f(-5)=(-5)^{4}-32(-5)^{2}-7 =-182 \\
\text{when } x=-4: f(-4)=(-4)^{4}-32(-4)^{2}-7 =-263 \\
\text{when } x=0: f(0)=(0)^{4}-32(0)^{2}-7 =-7 \\
\text{when } x=4: f(4)=(4)^{4}-32(4)^{2}-7 =-263 \\
\text{when } x=6: f(6)=(6)^{4}-32(6)^{2}-7 =137 \\
$$
We conclude that, the absolute maximum, $137 $ occurs when $x=-4, x=4 $ and the absolute minimum, $-263 $ occurs when $x=6.$
