If the demand for a product is given by \( D(p)=\frac{1000}{\sqrt{\rho}}-1 \) what is the average rate of change of demand when \( p \) increases within the following values? (Glve exact answers, Do not round,) (a) 1 to 4 -250 (b) 4 to 16 -15.525

We begin by writing the demand function as \[ D(p)=\frac{1000}{p}-1. \] The average rate of change of \( D(p) \) from \( p=a \) to \( p=b \) is given by \[ \frac{D(b)-D(a)}{b-a}. \] **(a) When \( p \) increases from 1 to 4** 1. Compute \( D(1) \): \[ D(1)=\frac{1000}{1}-1=1000-1=999. \] 2. Compute \( D(4) \): \[ D(4)=\frac{1000}{4}-1=250-1=249. \] 3. The average rate of change is \[ \frac{D(4)-D(1)}{4-1}=\frac{249-999}{3}=\frac{-750}{3}=-250. \] Thus, the exact answer for (a) is \[ -250. \] **(b) When \( p \) increases from 4 to 16** 1. We already have \[ D(4)=249. \] 2. Compute \( D(16) \): \[ D(16)=\frac{1000}{16}-1. \] Notice that \[ \frac{1000}{16}=\frac{125}{2}, \] so \[ D(16)=\frac{125}{2}-1=\frac{125-2}{2}=\frac{123}{2}. \] 3. The average rate of change is \[ \frac{D(16)-D(4)}{16-4}=\frac{\frac{123}{2}-249}{12}. \] Write \( 249 \) as a fraction over 2: \[ 249=\frac{498}{2}, \] so \[ \frac{D(16)-D(4)}{16-4}=\frac{\frac{123-498}{2}}{12}=\frac{-375}{2\cdot 12}=\frac{-375}{24}. \] 4. Simplify the fraction: \[ \frac{-375}{24}=-\frac{125}{8}. \] Thus, the exact answer for (b) is \[ -\frac{125}{8}. \]