Answer
$$
C(x)=x^{3}-2x^{2}+8x+50
$$
(a)
The function $C(x)$ is decreasing nowhere .
(b)
The function $C(x)$ is increasing on interval $(-\infty, \infty)$.
Work Step by Step
$$
C(x)=x^{3}-2x^{2}+8x+50
$$
To find the critical numbers, we first find any values of $x$ that make $C^{\prime}(x) =0 $.
Now, solve the equation $C^{\prime}(x)=0 $ to get
$$
\begin{aligned}
C^{\prime} (x)&=3x^{2}-2(2)x+8(1)+(0)=0\\
&=3x^{2}-4x+8=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x_{1,\:2} &=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:3\cdot \:8}}{2\cdot \:3}\\
\Rightarrow\quad\quad\quad\quad\quad\\\
x_{1,\:2} &=\frac{4\pm \sqrt{-80}}{6} \quad\quad\quad \text {not real values}
\end{aligned}
$$
Therefore, there are no critical points. The function either always increases
or always decreases.
Now, we can use the first derivative test. Check the sign of $C^{\prime}(x)$.
Test a number say $0$:
$$
\begin{aligned}
C^{\prime}(0) &=3(0)^{2}-4(0)+8\\
&=8\\
& \gt 0
\end{aligned}
$$
we see that $C^{\prime}(x)$ is positive for all real number $x$, so $C(x)$ is increasing on $(-\infty, \infty)$.
So,
(a)
The function $C(x)$ is decreasing nowhere .
(b)
The function $C(x)$ is increasing on interval $(-\infty, \infty)$.
