Answer
$$
f(x)=4 x^{3}-9^{2}-30x+6
$$
(a)
The critical numbers are: $-1 , \frac{5}{2}$.
(b)
The function is increasing on intervals $(-\infty, -1 ) \text { and } (\frac{5}{2}, \infty). $
(c)
The function is decreasing on interval $( -1 , \frac{5}{2} ).$
Work Step by Step
$$
f(x)=4 x^{3}-9^{2}-30x+6
$$
First, find the points where the derivative $f^{\prime }$ is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=4 (3) x^{2}-9 (2)x-30 \\
&=12 x^{2}-18x-30\\
&=6(2 x^{2}-3 x-5) \\
\end{aligned}
$$
Solve the equation $f^{\prime}(x) =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =6(2 x^{2}-3 x-5)&=0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
f^{\prime}(x=6(2x-5)(x+1)&=0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x =-1 \quad &\text {and} \quad x =\frac{5}{2} \\
\end{aligned}
$$
(a)
The critical numbers are: $-1 , \frac{5}{2}$.
Now, we can use the first derivative test.
Check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, -1 ), \quad ( -1, \frac{5}{2}), \quad \text {and } ( \frac{5}{2}, \infty) .
$$
(1)
Test a number in the interval $(-\infty, -1)$ say $-2$:
$$
\begin{aligned}
f^{\prime}(-2) &=6(2 (-2)^{2}-3 (-2)-5) \\
&=54 \\
&\gt 0
\end{aligned}
$$
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 $( -1, \frac{5}{2} )$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=6(2 (0)^{2}-3 (0)-5) \\
&=-30 \\
&\lt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( -1 , \frac{5}{2}),$
(3)
Test a number in the interval $(\frac{5}{2}, \infty)$ say $3$:
$$
\begin{aligned}
f^{\prime}(3) &=6(2 (3)^{2}-3 (3)-5) \\
&=24 \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(\frac{5}{2}, \infty) .$
So,
(b) The function is increasing on intervals $(-\infty, -1 ) \text { and } (\frac{5}{2}, \infty). $
(c) The function is decreasing on interval $( -1 , \frac{5}{2} ).$
