Answer
The percent of concentration of a drug in the bloodstream $x$ hours after the drug is administered is given by:
$$
K(x) =\frac{4 x}{3 x^{2}+27}
$$
(a)
The function $K(x) $ is increasing on $ (0, 3 ).$
(b)
The function $K(x) $ is decreasing on $(3 , \infty).$
Work Step by Step
The percent of concentration of a drug in the bloodstream $x$ hours after the drug is administered is given by:
$$
K(x) =\frac{4 x}{3 x^{2}+27}
$$
To determine any critical numbers, first use the quotient rule to find the derivative, $K^{\prime}(x)$
$$
\begin{aligned}
K^{\prime}(x) &=\frac{\left(3 x^{2}+27\right) 4-4 x(6 x)}{\left(3 x^{2}+27\right)^{2}} \\
&=\frac{108-12 x^{2}}{\left(3 x^{2}+27\right)^{2}}
\end{aligned}
$$
To find any intervals where this function is increasing or decreasing, set $K^{\prime}(x)=0$
$$
\begin{aligned}
K^{\prime}(x) &= \frac{108-12 x^{2}}{\left(3 x^{2}+27\right)^{2}}=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\\
& 108-12 x^{2}=0\\
& 121(9- x^{2})=0\\
\Rightarrow\quad\quad\quad\quad\quad\\\
& (3- x)(3+x)=0\\
& x=3 \\
\text { or }\\
& x=-3\\
& [ \text {refuse, since $x$ is represents time, must be at least 0}]
\end{aligned}
$$
so $3$ is the only critical number.
Now, we can use the first derivative test. Check the sign of $K^{\prime}(x)$ in the intervals:
$$
(0, 3 ), \quad (3, \infty),
$$
(1)
Test a number in the interval $(0, 3 ) $ say $1$:
$$
\begin{aligned}
K^{\prime}(1) &=\frac{108-12 (1)^{2}}{\left(3 (1)^{2}+27\right)^{2}}\\
&=\frac{8}{75} \\
&\approx 0.1066\\
& \gt 0
\end{aligned}
$$
we see that $K^{\prime}(x)$ is positive in that interval, so $K(x)$ is increasing on $(0, 3)$.
(2)
Test a number in the interval $(3, \infty ) $ say $4$:
$$
\begin{aligned}
K^{\prime}(4) &=\frac{108-12 (4)^{2}}{\left(3 (4)^{2}+27\right)^{2}}\\
&=-\frac{28}{1875} \\
&\approx -0.01493\\
& \lt 0
\end{aligned}
$$
we see that $K^{\prime}(x)$ is negative in that interval, so $K(x)$ is decreasing on $(3, \infty )$.
So,
(a)
The function $K(x) $ is increasing on $ (0, 3 ).$
(b)
The function $K(x) $ is decreasing on $(3 , \infty).$