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.

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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:

$$
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
$$

where $$ a = -2 $$ and $$ b = 2 $$.

First, calculate $$ f(2) $$ and $$ f(-2) $$:

$$
f(2) = (2)^2 + 2 \cos(2^2) = 4 + 2 \cos(4)
$$
$$
f(-2) = (-2)^2 + 2 \cos((-2)^2) = 4 + 2 \cos(4)
$$

Notice that $$ f(2) = f(-2) $$, so:

$$
f(2) - f(-2) = (4 + 2 \cos(4)) - (4 + 2 \cos(4)) = 0
$$

Therefore, the average rate of change is:

$$
\frac{f(2) - f(-2)}{2 - (-2)} = \frac{0}{4} = 0
$$

**Answer:** 0
The average rate of change of $$ f(x) $$ on $$[-2, 2]$$ is 0.

The function is defined by
. What is the average rate of
change of on the interval ? You may use a
calculator and round your answer to the nearest
thousandth.

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The function is defined by . What is the average rate of change of on the interval ? You may use a calculator and round your answer to the nearest thousandth.