If the demand for a product is given by \( D(p)=\frac{1000}{\sqrt{p}}-1 \) what is the average rate of change of demand when \( p \) increases within the following values? (Give exact answers. Do not round.) \( \begin{array}{l}\text { (a) } 4 \text { to } 16 \\ -200 / 9 \\ \text { (b) } 16 . . .16 . t o .100 \\ \text { Your answer cannot be understood or graded. More Information } \\ \text { Entaction, integer, or exact decimal. Do not approximate. }\end{array} \).

Let's compute the **average rate of change** of the demand function \( D(p) = \frac{1000}{\sqrt{p}} - 1 \) over the specified intervals. The **average rate of change** of a function \( D(p) \) from \( p = a \) to \( p = b \) is given by: \[ \text{Average Rate of Change} = \frac{D(b) - D(a)}{b - a} \] ### Part (a): From \( p = 4 \) to \( p = 16 \) 1. **Compute \( D(4) \):** \[ D(4) = \frac{1000}{\sqrt{4}} - 1 = \frac{1000}{2} - 1 = 500 - 1 = 499 \] 2. **Compute \( D(16) \):** \[ D(16) = \frac{1000}{\sqrt{16}} - 1 = \frac{1000}{4} - 1 = 250 - 1 = 249 \] 3. **Calculate the Average Rate of Change:** \[ \text{Average Rate of Change} = \frac{249 - 499}{16 - 4} = \frac{-250}{12} = -\frac{125}{6} \] **Exact Answer:** \( -\frac{125}{6} \) ### Part (b): From \( p = 16 \) to \( p = 100 \) 1. **Compute \( D(16) \):** *(Already computed above)* \[ D(16) = 249 \] 2. **Compute \( D(100) \):** \[ D(100) = \frac{1000}{\sqrt{100}} - 1 = \frac{1000}{10} - 1 = 100 - 1 = 99 \] 3. **Calculate the Average Rate of Change:** \[ \text{Average Rate of Change} = \frac{99 - 249}{100 - 16} = \frac{-150}{84} = -\frac{25}{14} \] **Exact Answer:** \( -\frac{25}{14} \) ### Summary of Answers: - **(a)** From \( p = 4 \) to \( p = 16 \): \( -\frac{125}{6} \) - **(b)** From \( p = 16 \) to \( p = 100 \): \( -\frac{25}{14} \)