Question

the model for the growth of bacteria is $f(x)=\frac{12,000}{1 + e^{-x}}$.
a. how many bacteria were there at the start?
b. what was the maximum amount of bacteria? 12,000

Explanation:

Part A

Step1: Identify the start time

At the start, \( x = 0 \) (initial time).

Step2: Substitute \( x = 0 \) into the function

The function is \( f(x)=\frac{12000}{1 + e^{-x}} \). Substitute \( x = 0 \):
\( f(0)=\frac{12000}{1 + e^{-0}} \)
Since \( e^{0}=1 \), \( e^{-0}=1 \), so:
\( f(0)=\frac{12000}{1 + 1}=\frac{12000}{2}=6000 \)

Step1: Analyze the limit as \( x \to \infty \)

For the function \( f(x)=\frac{12000}{1 + e^{-x}} \), as \( x \to \infty \), \( e^{-x}=\frac{1}{e^{x}}\to 0 \).

Step2: Calculate the limit

\( \lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{12000}{1 + e^{-x}}=\frac{12000}{1 + 0}=12000 \)

Answer:

6000

Part B (already provided, but for completeness)