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We are given a demand function \( x = 200 - p - \frac{4p}{p + 3} \) and the price \( p = 5 \). The goal is to find the rate of change of the demand \( x \) with respect to the price \( p \), which means finding \( \frac{dx}{dp} \), and then evaluate it at \( p = 5 \).

### Steps:

1. Differentiate the function \( x(p) \) with respect to \( p \).
2. Plug in the value \( p = 5 \).
3. Round the answer to two decimal places.

Let me calculate this.The rate of change of the demand \( x \) with respect to the price \( p \) at \( p = 5 \) is approximately \(-1.19\) units per dollar.

Would you like any further details or have any questions?

Here are 5 related questions you might find interesting:
1. How does the rate of change vary for different prices?
2. What is the significance of a negative rate of change in demand?
3. Can the demand function have a positive rate of change for some values of \( p \)?
4. How would the demand change if \( p = 10 \)?
5. What happens to the rate of change as \( p \to \infty \)?

**Tip:** When differentiating a function, remember that terms like \( \frac{4p}{p + 3} \) require the use of the quotient rule.

Use the demand function to find the rate of change in the demand x for the given price p. (Round your answer to two decimal places.)
x = 200 - p - (4p / (p + 3)), p = 5

Explore how to find the rate of change in demand using the function x = 200 - p - (4p / (p + 3)) at p = 5. Learn about key calculus concepts including differentiation and the quotient rule. Suitable for high school students in Grades 11-12.

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