Answer
$$
f(x)=4 x^{3}-15x^{2}-72x+5
$$
(a)
The critical numbers are: $-\frac{3}{2} , 4$.
(b)
The function is increasing on intervals $(-\infty, -\frac{3}{2} ) \text { and } (4, \infty ). $
(c)
The function is decreasing on interval $( -\frac{3}{2} , 4 ).$
Work Step by Step
$$
f(x)=4 x^{3}-15x^{2}-72x+5
$$
First, find the points where the derivative $f^{\prime }$ is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=4 (3) x^{2}-15 (2)x-72 \\
&=12 x^{2}-30 x-72 \\
&=3(4 x^{2}-10 x-24) \\
\end{aligned}
$$
Solve the equation $f^{\prime}(x) =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =3(4 x^{2}-10 x-24) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
f^{\prime}(x=3(2x-8)(2x+3)&=0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x =-\frac{3}{2} \quad &\text {and} \quad x =4 \\
\end{aligned}
$$
(a)
The critical numbers are: $-\frac{3}{2} , 4$.
Now, we can use the first derivative test.
Check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, -\frac{3}{2} ), \quad ( -\frac{3}{2}, 4), \quad \text {and } ( 4, \infty) .
$$
(1)
Test a number in the interval $(-\infty, -\frac{3}{2})$ say $-2$:
$$
\begin{aligned}
f^{\prime}(-2) &=3(4 (-2)^{2}-10 (-2)-24) \\
&=36 \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, -\frac{3}{2} )$
(2)
Test a number in the interval $( -\frac{3}{2} , 4 )$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=3(4 (0)^{2}-10 (0)-24) \\
&=-72 \\
& \lt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( -\frac{3}{2} , 4),$
(3)
Test a number in the interval $(4, \infty)$ say $5$:
$$
\begin{aligned}
f^{\prime}(5) &=3(4 (5)^{2}-10 (5)-24) \\
&=78 \\
&\gt 0
\end{aligned}
$$
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, -\frac{3}{2} ) \text { and } (4, \infty ) $
(c) The function is decreasing on interval $( -\frac{3}{2} , 4 ).$
