Question

The concentration C of a certain drug in a patient's bloodstream t hours after injection is given by

$C\left(t\right)=\frac{t}{2t^2+1}$

Part (a): Find the horizontal asymptote of $C\left(t\right)$. What happens to the concentration of the drug as t increases?

Part (b): Using your graphing utility, graph $C=C\left(t\right)$.

Part (c): Determine the time at which the concentration is highest.

Short answer

Part (a): The horizontal asymptote of $C\left(t\right)$ is t-axis.

When t increases, the concentration of the drug is given$C\left(t\right)\to 0$.

Part (b): On plotting the graph, we get,

Part (c): Concentration is highest at $0.71$hours after injection.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given function,

$C\left(t\right)=\frac{t}{2t^2+1}$

The horizontal of this function can be given by,

localid="1646069009980" $=\underset{x\to \infty }{\lim }C\left(t\right)$

Using the long division method,

localid="1646069014524" $=\underset{x\to \infty }{\lim }\frac{1}{2t+{\displaystyle \frac{1}{t}}}$

We have the horizontal asymptote of the given function as $y=0$as $x\to \infty$.

Part (b) Step 1. Plot the graph.

Plot the function $C=C\left(t\right)$,

Part (c) Step 1. Determine the time.

Consider the given function,

$C\left(t\right)=\frac{t}{2t^2+1}$

From the graph, it can be said that the concentration of the drug is highest at the time $0.71$hours after injection is$0.35$.