Answer
$$
f(x)=x^{3}-3x^{2}-24x+5; \quad\quad[-3,6]
$$
Critical numbers are: $-2$ and $4$.
We conclude that, the absolute maximum, $33$, occurs when $x=-2 $ and the absolute minimum, $-75 $ occurs when $x=4.$
Work Step by Step
$$
f(x)=x^{3}-3x^{2}-24x+5; \quad\quad[-3,6]
$$
First look for critical numbers in the interval $(-3,6)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=(3)x^{2}-3(2)x-24(1)+(0)\\
&=3x^{2}-6x-24
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =3x^{2}-6x-24=0\\
&= (3x+6)(x-4)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (3x+6) =0 \quad \text {or} \quad (x-4) =0 \\
\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 3x =-6 \quad\text {or} \quad x =4 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x =-2 \quad \text {or} \quad x =4.\\
\end{aligned}
$$
So, critical numbers are: $-2$ and $4$.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\text{when } x=0: f(-3)=(-3)^{3}-3(-3)^{2}-24(-3)+5=23 \\
\text{when } x=1: f(-2)=(-2)^{3}-3(-2)^{2}-24(-2)+5=33 \\
\text{when } x=3: f(4)=(4)^{3}-3(4)^{2}-24(4)+5=-75\\
\text{when } x=5: f(6)=(6)^{3}-3(6)^{2}-24(6)+5=-31 \\
$$
We conclude that, the absolute maximum, $33$, occurs when $x=-2 $ and the absolute minimum, $-75 $ occurs when $x=4.$
