To find the average rate of change of the function $f(x) = x^3 - 2$ over the interval $[-1, 1]$, we use the formula for the average rate of change:

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

where $a = -1$ and $b = 1$.

1. Evaluate $f(x)$ at $a = -1$: $f(-1) = (-1)^3 - 2 = -1 - 2 = -3$

2. Evaluate $f(x)$ at $b = 1$: $f(1) = (1)^3 - 2 = 1 - 2 = -1$

3. Plug the values into the average rate of change formula: $\text{Average Rate of Change} = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{-1 - (-3)}{1 + 1} = \frac{-1 + 3}{2} = \frac{2}{2} = 1$

So, the average rate of change of the function $f(x) = x^3 - 2$ over the interval $[-1, 1]$ is 1.

Let me know if you'd like a deeper explanation or if you have more questions!

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Follow-up Questions:

1. How would the average rate of change change if the interval was $[0, 2]$?
2. Can we find the instantaneous rate of change for this function at any point?
3. How does the concept of average rate of change relate to the derivative of a function?
4. What are some real-world applications of the average rate of change?
5. How would the average rate of change change if the function were quadratic, say $f(x) = x^2 - 2$?

Tip: When dealing with polynomials, remember that the average rate of change between two points can also be viewed as the slope of the secant line between those points on the curve.