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? 6,000 b. what was the maximum amount of bacteria? 12,000

Explanation:

Part A

Step 1: Identify the start time

At the start, \( x = 0 \) (since \( x \) represents time, and the start is \( t = 0 \)).

Step 2: 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 \( e^{0}=1 \), we have:

$$ f(0)=\frac{12000}{1 + 1}=\frac{12000}{2}=6000 $$

Part B

Step 1: Analyze the limit as \( x \to \infty \)

To find the maximum number of bacteria, we consider the limit of \( f(x) \) as \( x \) approaches infinity.

Step 2: Evaluate the limit

As \( x \to \infty \), \( e^{-x}=\frac{1}{e^{x}}\to 0 \). So:

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

Answer:

s:
A. The number of bacteria at the start is \(\boldsymbol{6000}\).
B. The maximum amount of bacteria is \(\boldsymbol{12000}\).