The function \( f \) is defined by \( f(x)=x^{2}+2 \cos \left(x^{2}\right) \). What is the average rate of change of \( f \) on the interval \( [-2,2] \) ? You may use a calculator and round your answer to the nearest thousandth.

To find the average rate of change of the function \( f(x) = x^2 + 2 \cos(x^2) \) on the interval \([-2, 2]\), we use the formula for average rate of change: \[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \] Here, \( a = -2 \) and \( b = 2 \). First, we need to calculate \( f(-2) \) and \( f(2) \): 1. **Calculating \( f(-2) \)**: \[ f(-2) = (-2)^2 + 2 \cos((-2)^2) = 4 + 2 \cos(4) \] Using a calculator, \( \cos(4) \approx -0.6536 \): \[ f(-2) \approx 4 + 2(-0.6536) \approx 4 - 1.3072 \approx 2.6928 \] 2. **Calculating \( f(2) \)**: \[ f(2) = (2)^2 + 2 \cos((2)^2) = 4 + 2 \cos(4) \] Thus, \[ f(2) \approx 4 + 2(-0.6536) \approx 4 - 1.3072 \approx 2.6928 \] Now that we have both values, we can calculate the average rate of change: \[ \text{Average Rate of Change} = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{2.6928 - 2.6928}{2 + 2} = \frac{0}{4} = 0 \] Therefore, the average rate of change of \( f \) on the interval \( [-2,2] \) is approximately \( 0.000 \) when rounded to the nearest thousandth.