Answer
The percent of concentration of the drug in the bloodstream $t$ hours later given by:
$$
K(t) =\frac{5 t}{t^{2}+1}
$$
(a)
The function $K(t) $ is increasing on $ (0, 1 ).$
(b)
The function $K(t) $ is decreasing on $(1 , \infty).$
Work Step by Step
The percent of concentration of the drug in the bloodstream $t$ hours later given by:
$$
K(t) =\frac{5 t}{t^{2}+1}
$$
To determine any critical numbers, first use the quotient rule to find the derivative, $K^{\prime}(t)$
$$
\begin{aligned}
K^{\prime}(t) &=\frac{5\left(t^{2}+1\right)-2 t(5 t)}{\left(t^{2}+1\right)^{2}} \\
&=\frac{5 t^{2}+5-10 t^{2}}{\left(t^{2}+1\right)^{2}} \\
&=\frac{5-5 t^{2}}{\left(t^{2}+1\right)^{2}}
\end{aligned}
$$
To find any intervals where this function is increasing or decreasing, set $K^{\prime}(t)=0$
$$
\begin{aligned}
K^{\prime}(t) &=\frac{5-5 t^{2}}{\left(t^{2}+1\right)^{2}}=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\\
& 5-5 t^{2}=0\\
& 5(1- t^{2})=0\\
\Rightarrow\quad\quad\quad\quad\quad\\\
& (1- t)(1+t)=0\\
& t=1 \\
\text { or }\\
& t=-1 \\
& [ \text {refuse, since $t$ is represents time, must be at least 0}]
\end{aligned}
$$
so $1$ is the only critical number.
Now, we can use the first derivative test. Check the sign of $K^{\prime}(t)$ in the intervals:
$$
(0, 1 ), \quad (1, \infty),
$$
(1)
Test a number in the interval $(0, 1 ) $ say $0.5$:
$$
\begin{aligned}
K^{\prime}(0.5) &=\frac{5-5 (0.5)^{2}}{\left((0.5)^{2}+1\right)^{2}} \\
&=2.4 \\
& \gt 0
\end{aligned}
$$
we see that $K^{\prime}(t)$ is positive in that interval, so $K(t)$ is increasing on $(0, 1)$.
(2)
Test a number in the interval $(1, \infty ) $ say $2$:
$$
\begin{aligned}
K^{\prime}(2) &=\frac{5-5 (2)^{2}}{\left((2)^{2}+1\right)^{2}} \\
&=-\frac{3}{5} \\
& \approx -0.6\\
& \lt 0
\end{aligned}
$$
we see that $K^{\prime}(t)$ is negative in that interval, so $K(t)$ is decreasing on $(1, \infty )$.
So,
(a)
The function $K(t) $ is increasing on $ (0, 1 ).$
(b)
The function $K(t) $ is decreasing on $(1 , \infty).$
