The population of an unknown bacteria in an experimental culture is estimated by the population function, N(x)=123001+10e−0.14xN(x)=123001+10e−0.14x, where x is the number of hours after the experiment begins.

How fast is the population changing at the end of 33 hours? Round your answer to the nearest integer.

To determine how fast the population is changing at the end of 33 hours, we need to find the derivative of the population function \( N(x) = \frac{12300}{1 + 10e^{-0.14x}} \) and then evaluate it at \( x = 33 \).

1. **Find the derivative \( N'(x) \):**

Given:
\[ N(x) = \frac{12300}{1 + 10e^{-0.14x}} \]

Let's denote:
\[ u = 1 + 10e^{-0.14x} \]
So:
\[ N(x) = \frac{12300}{u} \]

To find \( N'(x) \), we use the chain rule and the quotient rule:
\[ N'(x) = 12300 \cdot \left( \frac{-u'}{u^2} \right) \]

First, find \( u' \):
\[ u = 1 + 10e^{-0.14x} \]
\[ u' = 10 \cdot (-0.14e^{-0.14x}) \]
\[ u' = -1.4e^{-0.14x} \]

Now substitute \( u \) and \( u' \) into the expression for \( N'(x) \):
\[ N'(x) = 12300 \cdot \left( \frac{-(-1.4e^{-0.14x})}{(1 + 10e^{-0.14x})^2} \right) \]
\[ N'(x) = 12300 \cdot \left( \frac{1.4e^{-0.14x}}{(1 + 10e^{-0.14x})^2} \right) \]

2. **Evaluate \( N'(x) \) at \( x = 33 \):**

Substitute \( x = 33 \) into \( N'(x) \):
\[ N'(33) = 12300 \cdot \left( \frac{1.4e^{-0.14 \cdot 33}}{(1 + 10e^{-0.14 \cdot 33})^2} \right) \]

First, calculate the exponent:
\[ -0.14 \cdot 33 = -4.62 \]
\[ e^{-4.62} \approx 0.0099 \]

Now calculate the terms:
\[ 1 + 10 \cdot 0.0099 \approx 1.099 \]
\[ (1 + 10 \cdot 0.0099)^2 \approx 1.208 \]

Then,
\[ N'(33) = 12300 \cdot \left( \frac{1.4 \cdot 0.0099}{1.208} \right) \]
\[ N'(33) = 12300 \cdot \left( \frac{0.01386}{1.208} \right) \]
\[ N'(33) = 12300 \cdot 0.01147 \]
\[ N'(33) \approx 141.08 \]

Rounding to the nearest integer:
\[ N'(33) \approx 141 \]

So, the population is changing at a rate of approximately 141 individuals per hour at the end of 33 hours.

Would you like any further details or have any questions on this solution?

Here are 8 related questions that might interest you:

1. How do you apply the chain rule in differentiation?
2. What is the significance of the chain rule and quotient rule in calculus?
3. How can you find the second derivative of the population function \( N(x) \)?
4. What are the critical points of the function \( N(x) = \frac{12300}{1 + 10e^{-0.14x}} \)?
5. How would you determine the inflection points of the population function?
6. How does the exponential decay \( e^{-0.14x} \) affect the population growth?
7. What is the initial population growth rate at \( x = 0 \)?
8. How can you find the maximum population in the long term for the given function?

**Tip:** When dealing with exponential functions, remember to keep track of significant digits to ensure the accuracy of your calculations.

Learn how to calculate the rate of change in population using the exponential function N(x) = 12300 / (1 + 10e^(-0.14x)). This problem involves differentiation using the chain rule and quotient rule, suitable for college-level mathematics.

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