Answer
$$
f(x)=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-6x+3; \quad\quad[-4,4]
$$
Critical numbers are: $-2$ and $3$.
The absolute maximum, $\frac{31}{3} \approx 10.3333 $, occurs when $x=-2, $
and the absolute minimum, $-\frac{21}{2} \approx -10.5 $ occurs when $x=3.$
Work Step by Step
$$
f(x)=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-6x+3; \quad\quad[-4,4]
$$
First look for critical numbers in the interval $(-4,4)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=\frac{1}{3}(3)x^{2}-\frac{1}{2}(2)x-6(1)+(0);\\
&=x^{2}-x-6
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =x^{2}-x-6=0\\
&= (x-3)(x+2)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (x-3) =0 \quad \text {or} \quad (x+2) =0 \\
\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x =3 \quad\text {or} \quad x =-2 \\
\end{aligned}
$$
So, critical numbers are: $-2$ and $3$.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\text{when } x=-4: f(-4)=\frac{1}{3}(-4)^{3}-\frac{1}{2}(-4)^{2}-6(-4)+3\\ =-\frac{7}{3} \approx -2.3333 \\
\text{when } x=-2: f(-2)=\frac{1}{3}(-2)^{3}-\frac{1}{2}(-2)^{2}-6(-2)+3\\ =\frac{31}{3} \approx 10.3333 \\
\text{when } x=3: f(3)=\frac{1}{3}(3)^{3}-\frac{1}{2}(3)^{2}-6(3)+3\\ =-\frac{21}{2} \approx -10.5 \\
\text{when } x=4: f(4)=\frac{1}{3}(4)^{3}-\frac{1}{2}(4)^{2}-6(4)+3\\ =-\frac{23}{3} \approx -7.6666 \\
$$
We conclude that, the absolute maximum, $\frac{31}{3} \approx 10.3333 $, occurs when $x=-2 $ and the absolute minimum, $-\frac{21}{2} \approx -10.5 $ occurs when $x=3.$
