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consider the concentration, c, (in mg/liter) of a drug in the blood as …

Question

consider the concentration, c, (in mg/liter) of a drug in the blood as a function of the amount of drug given, x, and the time since injection, t. for 0 ≤ x ≤ 5 mg and t ≥ 0 hours, we have c = f(x,t)=30te^{-(5 - x)t}.

f(3,5)=

give a practical interpretation of your answer: f(3,5) is
a. the change in concentration of a 3 mg dose in the blood 5 hours after injection.
b. the amount of a 5 mg dose in the blood 3 hours after injection.
c. the concentration of a 3 mg dose in the blood 5 hours after injection.
d. the change in concentration of a 5 mg dose in the blood 3 hours after injection.
e. the amount of a 3 mg dose in the blood 5 hours after injection.
f. the concentration of a 5 mg dose in the blood 3 hours after injection.

Explanation:

Step1: Substitute values into function

Given $C = f(x,t)=30te^{-(5 - x)t}$, substitute $x = 3$ and $t = 5$ into the function. We get $f(3,5)=30\times5\times e^{-(5 - 3)\times5}$.

Step2: Simplify the exponent part

First, calculate the exponent: $-(5 - 3)\times5=-2\times5=- 10$. So the expression becomes $f(3,5)=150e^{-10}$.

Step3: Calculate the numerical value

We know that $e^{-10}=\frac{1}{e^{10}}\approx\frac{1}{22026.47}$. Then $f(3,5)=150\times\frac{1}{e^{10}}\approx150\times4.54\times10^{-5}=0.00681$.

For the interpretation part, since $x$ represents the amount of drug given (in mg) and $t$ represents the time since injection (in hours), and $C = f(x,t)$ represents the concentration of the drug in the blood. When $x = 3$ and $t = 5$, $f(3,5)$ is the concentration of a 3 - mg dose in the blood 5 hours after injection.

Answer:

$f(3,5)\approx0.00681$
C. the concentration of a 3 mg dose in the blood 5 hours after injection.