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 1239\
(round to the nearest whole number as needed.)\
c. the limiting size of the infected population is \\(\square\\)\
(round to the nearest whole number as needed.)

Explanation:

Step1: Recall logistic growth limit

For a logistic growth function \( f(t)=\frac{L}{1 + ae^{-kt}} \), the limiting size (carrying capacity) is \( L \), as \( t\to\infty \), \( e^{-kt}\to0 \), so \( f(t)\to\frac{L}{1 + 0}=L \).

Step2: Identify \( L \) in given function

Given \( f(t)=\frac{117000}{1 + 5100e^{-t}} \), compare with \( \frac{L}{1 + ae^{-kt}} \). Here, \( L = 117000 \).

Answer:

117000