The average rate of change of a function $f(x)$ over an interval $[a, b]$ is given by the formula:

$\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}$

In this case, the demand function is $D(p) = \frac{1000}{\sqrt{p}} - 1$, and we need to calculate the average rate of change for two different intervals of $p$.

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(a) From $p = 1$ to $p = 4$

1. Evaluate $D(1)$: $D(1) = \frac{1000}{\sqrt{1}} - 1 = 1000 - 1 = 999$

2. Evaluate $D(4)$: $D(4) = \frac{1000}{\sqrt{4}} - 1 = \frac{1000}{2} - 1 = 500 - 1 = 499$

3. Average rate of change from $p = 1$ to $p = 4$: $\text{Average Rate of Change} = \frac{D(4) - D(1)}{4 - 1} = \frac{499 - 999}{4 - 1} = \frac{-500}{3} = -\frac{500}{3}$

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(b) From $p = 4$ to $p = 16$

1. Evaluate $D(4)$: $D(4) = 499 \quad \text{(as calculated above)}$

2. Evaluate $D(16)$: $D(16) = \frac{1000}{\sqrt{16}} - 1 = \frac{1000}{4} - 1 = 250 - 1 = 249$

3. Average rate of change from $p = 4$ to $p = 16$: $\text{Average Rate of Change} = \frac{D(16) - D(4)}{16 - 4} = \frac{249 - 499}{16 - 4} = \frac{-250}{12} = -\frac{125}{6}$

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Final Answers:

• (a) The average rate of change from $p = 1$ to $p = 4$ is $-\frac{500}{3}$.
• (b) The average rate of change from $p = 4$ to $p = 16$ is $-\frac{125}{6}$.

Would you like further clarification or details on the process?

Here are 5 related questions to expand your understanding:

1. How would the average rate of change change for smaller intervals of $p$?
2. What happens to the demand if $p$ approaches zero?
3. How would you interpret the negative rate of change in the context of demand and price?
4. What is the instantaneous rate of change of the demand at $p = 4$?
5. Can the demand function $D(p)$ ever equal zero for any $p$?

Tip: The average rate of change gives insight into the overall trend between two points, while the instantaneous rate of change (given by derivatives) tells us how the function behaves at a specific point.