Answer
$$
A(x) =0.003631 x^{3}-0.03746 x^{2} +0.1012 x+0.009 , \quad(0\leq x \leq 5)
$$
(a)
The function is increasing on $ (0, 1.85 ).$
(b)
The function is decreasing on $(1.85 , 5).$
Work Step by Step
$$
A(x) =0.003631 x^{3}-0.03746 x^{2} +0.1012 x+0.009 , \quad(0\leq x \leq 5)
$$
To find any intervals where this function is increasing, set $A^{\prime}(x)=0$
$$
\begin{aligned}
A^{\prime}(x) &= 0.003631 (3) x^{2}-0.03746 (2)x +0.1012 \\
&=0.010893 x^{2}-0.07492 x+0.1012 \\
\:x_{1,\:2} &=\frac{-\left(-0.07492\right)\pm \sqrt{\left(-0.07492\right)^2-4\cdot \:0.010893\cdot \:0.1012}}{2\cdot \:0.010893}\\
\Rightarrow\quad\quad\quad\quad\quad\\\
x_{1} &=\frac{0.07492+\sqrt{0.00120352}}{0.021786} \approx 1.85\\
\text { or }\\
x_{2}&=\frac{0.07492-\sqrt{0.00120352}}{0.021786} \approx 5.03\\
\end{aligned}
$$
Now, we can use the first derivative test. Check the sign of $A^{\prime}(x)$ in the intervals:
$$
(0, 1.85 ), \quad ( 1.85 , 5),
$$
(1)
Test a number in the interval $(0, 1.85 ) $ say $1$:
$$
\begin{aligned}
A^{\prime}(1) &=0.010893 (1)^{2}-0.07492 (1)+0.1012 \\
&=0.037173\\
& \gt 0
\end{aligned}
$$
we see that $A^{\prime}(x)$ is positive in that interval, so $A(x)$ is increasing on $(0, 1.85)$.
(2)
Test a number in the interval $(1.85, 5 ) $ say $2$:
$$
\begin{aligned}
A^{\prime}(2) &=0.010893 (2)^{2}-0.07492 (2)+0.1012 \\
&=-0.005068\\
&\lt 0
\end{aligned}
$$
we see that $A^{\prime}(x)$ is negative in that interval, so $A(x)$ is decreasing on $(1.85, 5 )$.
So,
(a)
The function is increasing on $ (0, 1.85 ).$
(b)
The function is decreasing on $(1.85 , 5).$
