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The problem reads:

"**5. El tamaño de una población está modelada por** \( P(x) = 5000 + 500x - 50x^2 \), **donde x es el número de años después del 2001. Calcule la tasa de crecimiento en** \( x = 2 \)."

### Solution:
To calculate the rate of growth at \( x = 2 \), we need to find the derivative of \( P(x) \) with respect to \( x \), which gives us the rate of change of the population. Then, we will evaluate this derivative at \( x = 2 \).

1. **Given function**:  
   \[
   P(x) = 5000 + 500x - 50x^2
   \]
2. **Differentiate \( P(x) \)** with respect to \( x \):  
   \[
   P'(x) = \frac{d}{dx}[5000 + 500x - 50x^2] = 500 - 100x
   \]
3. **Evaluate the derivative at \( x = 2 \)**:  
   \[
   P'(2) = 500 - 100(2) = 500 - 200 = 300
   \]

### Conclusion:
The rate of growth of the population at \( x = 2 \) (which is the year 2003) is **300 individuals per year**.

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Would you like a more detailed explanation of the steps, or do you have any other questions?

Here are 5 related questions you might find useful:
1. What does the second derivative of this function represent?
2. How would you interpret the population if the growth rate were negative at a given year?
3. How do we find the year in which the population reaches its maximum size?
4. Can you solve for the value of \( x \) when the population growth becomes zero?
5. What is the initial population size in the year 2001?

**Tip**: The first derivative gives the rate of change, while the second derivative indicates the acceleration or deceleration of that change.

El tamaño de una población está modelada por P(x) = 5000 + 500x - 50x^2, donde x es el número de años después del 2001. Calcule la tasa de crecimiento en x = 2.

Learn how to calculate the rate of population growth for the function P(x) = 5000 + 500x - 50x^2 at x = 2. This problem uses basic calculus principles, focusing on taking the derivative to find the rate of change. Suitable for high school and early university students.

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