Answer
$$
f(x)=\frac{2}{3} x^{3}-x^{2}-4x+2
$$
(a)
The critical numbers are: $-1 , 2$.
(b)
The function is increasing on interval $(-\infty, -1 ) \text { and } (2, \infty ) $
(c)
The function is decreasing on interval $( -1 , 2 ).$
Work Step by Step
$$
f(x)=\frac{2}{3} x^{3}-x^{2}-4x+2
$$
First, find the points where the derivative $f^{\prime }$ is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=\frac{2}{3} (3) x^{2}-(2)x-4 \\
&=2 x^{2}-2x-4 \\
&= 2(x^{2}-x-2)\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x) =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =2(x^{2}-x-2) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
f^{\prime}(x=2(x-2)(x+1)&=0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x =-1 \quad &\text {and} \quad x =2 \\
\end{aligned}
$$
(a)
The critical numbers are: $-1 , 2$.
Now, we can use the first derivative test.
Check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, -1 ), \quad ( -1,2), \quad \text {and } ( 2, \infty) .
$$
(1)
Test a number in the interval $(-\infty, -1)$ say $-2$:
$$
\begin{aligned}
f^{\prime}(-2) &=2((-2)^{2}-(-2)-2) \\
&=8 \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, -1 )$
(2)
Test a number in the interval $( -1 ,2 )$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=2((0)^{2}-(0)-2) \\
&=-4 \\
& \lt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( -1 , 2),$
(3)
Test a number in the interval $(2, \infty)$ say $3$:
$$
\begin{aligned}
f^{\prime}(3) &=2((3)^{2}-(3)-2) \\
&=8 \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(2, \infty ).$
So,
(b) The function is increasing on interval $(-\infty, -1 ) \text { and } (2, \infty ) $
(c) The function is decreasing on interval $( -1 , 2 ).$
