Question

the logistic growth function $f(t) = \frac{117,000}{1 + 5100e^{-t}}$ describes the number of people, $f(t)$, who have become ill with influenza $t$ weeks after its initial outbreak in a particular community.\
\
a. how many people became ill with the flu when the epidemic began?\
b. how many people were ill by the end of the fourth week?\
c. what is the limiting size of the population that becomes ill?\
\
\
a. the number of people initially infected is 23.\
(round to the nearest whole number as needed.)\
\
b. the number of people infected after 4 weeks is \\(\square\\).\
(round to the nearest whole number as needed.)

Explanation:

Step1: Substitute t = 4 into the function

We have the logistic growth function \( f(t)=\frac{117000}{1 + 5100e^{-t}} \). For part b, we need to find \( f(4) \), so substitute \( t = 4 \) into the function: \( f(4)=\frac{117000}{1 + 5100e^{-4}} \).

Step2: Calculate the exponent part

First, calculate \( e^{-4} \). We know that \( e^{-4}=\frac{1}{e^{4}}\approx\frac{1}{54.5982}\approx0.0183156 \).

Step3: Calculate the denominator

Next, calculate the denominator \( 1 + 5100e^{-4} \). Substitute the value of \( e^{-4} \) we just found: \( 1+5100\times0.0183156 = 1 + 93.40956=94.40956 \).

Step4: Calculate the final value

Now, divide 117000 by the denominator: \( f(4)=\frac{117000}{94.40956}\approx1239.27 \). Rounding to the nearest whole number, we get 1239.

Answer:

1239