Answer
$$
f(x)=\frac{1}{3}x^{3}+\frac{3}{2}x^{2}-4x+1; \quad\quad[-5,2]
$$
Critical numbers are: $-4$ and $1$.
The absolute maximum, $\frac{59}{3} \approx 19.666 $, occurs when $x=-4 $,
and the absolute minimum, $-\frac{7}{6} \approx -1.1666 $ occurs when $x=1.$
Work Step by Step
$$
f(x)=\frac{1}{3}x^{3}+\frac{3}{2}x^{2}-4x+1; \quad\quad[-5,2]
$$
First look for critical numbers in the interval $(-5,2)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=\frac{1}{3}(3)x^{2}+\frac{3}{2} (2)x-4(1)+(0);\\
&=x^{2}+3x-4
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =x^{2}+3x-4=0\\
&= (x+4)(x-1)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (x+4) =0 \quad \text {or} \quad (x-1) =0 \\
\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x =-4 \quad\text {or} \quad x =1 \\
\end{aligned}
$$
So, critical numbers are: $-4$ and $1$.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\text{when } x=-5: f(-5)=\frac{1}{3}(-5)^{3}+\frac{3}{2}(-5)^{2}-4(-5)+1\\ =\frac{101}{6} \approx 16.83333 \\
\text{when } x=-4: f(-4)=\frac{1}{3}(-4)^{3}+\frac{3}{2}(-4)^{2}-4(-4)+1\\ =\frac{59}{3} \approx 19.666 \\
\text{when } x=1: f(1)=\frac{1}{3}(1)^{3}+\frac{3}{2}(1)^{2}-4(1)+1\\ =-\frac{7}{6} \approx -1.1666 \\
\text{when } x=2: f(2)=\frac{1}{3}(2)^{3}+\frac{3}{2}(2)^{2}-4(2)+1\\ =\frac{5}{3} \approx 1.6666 \\
$$
We conclude that, the absolute maximum, $\frac{59}{3} \approx 19.666 $, occurs when $x=-4 $ and the absolute minimum, $-\frac{7}{6} \approx -1.1666 $ occurs when $x=1.$
