Write an exponential decay function to model the situation. Compare the average rates of change over the given intervals. initial value: 49 decay factor: 0.8 \( 1 \leq x \leq 3 \) and \( 4 \leq x \leq 8 \) The exponential decay function to model the situation is \( f(x)=49=0.8^{x} \). The average rate of change over \( 1 \leq x \leq 3 \) is \( \square \). The average rate of change over \( 4 \leq x \leq 8 \) is \( \square \). The rate of change \( \begin{array}{l}\text { (Round to three decimal places as needed.) }\end{array} \) \( \begin{array}{l}\text { as increases. }\end{array} \)

Let's clarify that the exponential decay function should actually be written as \( f(x) = 49 \cdot 0.8^x \), not \( 49 = 0.8^x \). Now, to calculate the average rates of change for the specified intervals: For the interval \( 1 \leq x \leq 3 \): 1. Calculate \( f(1) \) and \( f(3) \): - \( f(1) = 49 \cdot 0.8^1 = 39.2 \) - \( f(3) = 49 \cdot 0.8^3 = 25.088 \) Now, use the average rate of change formula: \[ \text{Average Rate of Change} = \frac{f(3) - f(1)}{3 - 1} = \frac{25.088 - 39.2}{2} = \frac{-14.112}{2} = -7.056 \] For the interval \( 4 \leq x \leq 8 \): 1. Calculate \( f(4) \) and \( f(8) \): - \( f(4) = 49 \cdot 0.8^4 = 16.384 \) - \( f(8) = 49 \cdot 0.8^8 = 3.2768 \) Apply the average rate of change formula: \[ \text{Average Rate of Change} = \frac{f(8) - f(4)}{8 - 4} = \frac{3.2768 - 16.384}{4} = \frac{-13.1072}{4} = -3.2768 \] So the average rate of change over the interval \( 1 \leq x \leq 3 \) is approximately \(-7.056\) and over \( 4 \leq x \leq 8 \) is approximately \(-3.277\). Now, as we compare these average rates of change, we observe that the rate of change becomes less negative, indicating that the function is decaying at a slower rate as \( x \) increases, which is a characteristic behavior of exponential decay.