Answer
$$
f(x)=\frac{2}{3} x^{3}-x^{2}-24x-4
$$
(a)
The critical numbers are: $-3 , 4$.
(b)
The function is increasing on intervals $(-\infty, -3 ) \text { and } (4, \infty ). $
(c)
The function is decreasing on interval $( -3 , 4 ).$
Work Step by Step
$$
f(x)=\frac{2}{3} x^{3}-x^{2}-24x-4
$$
First, find the points where the derivative $f^{\prime }$ is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=\frac{2}{3} (3) x^{2}-(2)x-24 \\
&=2 x^{2}-2x-24 \\
&= 2(x^{2}-x-12)\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x) =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =2(x^{2}-x-12) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
f^{\prime}(x=2(x+3)(x-4)&=0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x =-3 \quad &\text {and} \quad x =4 \\
\end{aligned}
$$
(a)
The critical numbers are: $-3 , 4$.
Now, we can use the first derivative test.
Check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, -3 ), \quad ( -3,4), \quad \text {and } ( 4, \infty) .
$$
(1)
Test a number in the interval $(-\infty, -3)$ say $-4$:
$$
\begin{aligned}
f^{\prime}(-4) &=2((-4)^{2}-(-4)-12) \\
&=16 \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, -3 )$
(2)
Test a number in the interval $( -3 ,4 )$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=2((0)^{2}-(0)-12 )\\
&=-24 \\
&\lt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( -3 , 4),$
(3)
Test a number in the interval $(4, \infty)$ say $5$:
$$
\begin{aligned}
f^{\prime}(5) &=2((5)^{2}-(5)-12) \\
&=16 \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(4, \infty ).$
So,
(b) The function is increasing on intervals $(-\infty, -3 ) \text { and } (4, \infty ) $
(c) The function is decreasing on interval $( -3 , 4 ).$
