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 \( f(x)=59(0.9)^{x} \). The average rate of change over \( 1 \leq x \leq 3 \) is \( \square \). The averagervate of change over \( 4 \leq x \leq 6 \) is \( \square \). The rate of change \( \square \) as \( x \) increases. (Round to three decimal places as needed.)

To find the average rate of change over the specified intervals, we start by calculating the function values at the endpoints of each interval. For \( 1 \leq x \leq 3 \): - \( f(1) = 59(0.9)^{1} = 53.1 \) - \( f(3) = 59(0.9)^{3} = 47.879 \) The average rate of change over \( 1 \leq x \leq 3 \) is calculated as: \[ \frac{f(3) - f(1)}{3 - 1} = \frac{47.879 - 53.1}{2} = \frac{-5.221}{2} = -2.6105 \, (\text{rounded to } -2.611) \] For \( 4 \leq x \leq 6 \): - \( f(4) = 59(0.9)^{4} = 43.0911 \) - \( f(6) = 59(0.9)^{6} = 38.4721 \) The average rate of change over \( 4 \leq x \leq 6 \) is calculated as: \[ \frac{f(6) - f(4)}{6 - 4} = \frac{38.4721 - 43.0911}{2} = \frac{-4.6190}{2} = -2.3095 \, (\text{rounded to } -2.310) \] Now we can fill in the blanks: The average rate of change over \( 1 \leq x \leq 3 \) is approximately \( -2.611 \). The average rate of change over \( 4 \leq x \leq 6 \) is approximately \( -2.310 \). The rate of change decreases as \( x \) increases. So, the final answer will be: The average rate of change over \( 1 \leq x \leq 3 \) is \(-2.611\). The average rate of change over \( 4 \leq x \leq 6 \) is \(-2.310\). The rate of change decreases as \( x \) increases.