Question

the concentration c(t) of a drug in a patients bloodstream t hours after administration is given by c(t)=\frac{6t}{8 + t^{2}} where c(t) is in milligrams per liter. during what time interval will the concentration be greater than 1 milligram per liter? answer using interval notation. question 15 score: 13.74/15 answered: 14/15 question help: ebook written example submit question

Explanation:

Step1: Set up the inequality

We want to find when $C(t)=\frac{6t}{t^{2}+8}> 1$. Cross - multiply to get $6t>t^{2}+8$ (since $t^{2}+8>0$ for all real $t$). Rearrange it to the quadratic inequality $t^{2}-6t + 8<0$.

Step2: Factor the quadratic

Factor $t^{2}-6t + 8$ as $(t - 2)(t - 4)<0$.

Step3: Find the roots

Set $(t - 2)(t - 4)=0$. The roots are $t = 2$ and $t = 4$.

Step4: Determine the solution interval

We test the intervals $(-\infty,2)$, $(2,4)$ and $(4,\infty)$. For $t\in(-\infty,2)$, let $t = 1$, then $(1 - 2)(1 - 4)=(-1)\times(-3)=3>0$. For $t\in(2,4)$, let $t = 3$, then $(3 - 2)(3 - 4)=(1)\times(-1)=-1<0$. For $t\in(4,\infty)$, let $t = 5$, then $(5 - 2)(5 - 4)=3\times1 = 3>0$. So the solution of the inequality $(t - 2)(t - 4)<0$ is $2

Answer:

$(2,4)$