Write an exponential decay function to model the situation. Compare the average rates of change over the given intervals. initial value: 59 decay factor: 0.9 \( 1 \leq x \leq 3 \) and \( 4 \leq x \leq 6 \) The exponential decay function to model the situation is \( f(x)=\square \).

To model the given situation using an exponential decay function and compare the average rates of change over the specified intervals, follow the steps below: ### **1. Exponential Decay Function** The general form of an exponential decay function is: \[ f(x) = A \cdot (D)^x \] Where: - \( A \) is the initial value. - \( D \) is the decay factor. - \( x \) is the independent variable (e.g., time). Given: - **Initial Value (\( A \))**: 59 - **Decay Factor (\( D \))**: 0.9 Substituting the given values into the general form: \[ f(x) = 59 \cdot (0.9)^x \] **Thus, the exponential decay function is:** \[ f(x) = 59(0.9)^x \] ### **2. Average Rate of Change** The **average rate of change** of a function \( f(x) \) over an interval \([a, b]\) is calculated using the formula: \[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \] We'll calculate this for the two intervals: \( 1 \leq x \leq 3 \) and \( 4 \leq x \leq 6 \). #### **a. Interval \( 1 \leq x \leq 3 \)** 1. **Calculate \( f(1) \):** \[ f(1) = 59 \cdot (0.9)^1 = 59 \cdot 0.9 = 53.1 \] 2. **Calculate \( f(3) \):** \[ f(3) = 59 \cdot (0.9)^3 = 59 \cdot 0.729 = 43.011 \] 3. **Average Rate of Change:** \[ \frac{f(3) - f(1)}{3 - 1} = \frac{43.011 - 53.1}{2} = \frac{-10.089}{2} = -5.0445 \] **Rounded to two decimal places:** \[ \approx -5.04 \] #### **b. Interval \( 4 \leq x \leq 6 \)** 1. **Calculate \( f(4) \):** \[ f(4) = 59 \cdot (0.9)^4 = 59 \cdot 0.6561 = 38.7099 \] 2. **Calculate \( f(6) \):** \[ f(6) = 59 \cdot (0.9)^6 = 59 \cdot 0.531441 = 31.353 \] 3. **Average Rate of Change:** \[ \frac{f(6) - f(4)}{6 - 4} = \frac{31.353 - 38.7099}{2} = \frac{-7.3569}{2} = -3.67845 \] **Rounded to two decimal places:** \[ \approx -3.68 \] ### **3. Summary** - **Exponential Decay Function:** \[ f(x) = 59(0.9)^x \] - **Average Rates of Change:** - **For \( 1 \leq x \leq 3 \):** \[ \approx -5.04 \] - **For \( 4 \leq x \leq 6 \):** \[ \approx -3.68 \] This comparison shows that the average rate of decay is decreasing (i.e., becoming less negative) as \( x \) increases, which is consistent with the nature of exponential decay.