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

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

Here, $a = -3$ and $b = 3$.

Step 1: Find $f(3)$

$f(3) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$

Step 2: Find $f(-3)$

$f(-3) = 2(-3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$

Step 3: Apply the Formula

$\text{Average Rate of Change} = \frac{17 - 17}{3 - (-3)} = \frac{0}{6} = 0$

Final Answer:

The average rate of change is 0.

Would you like me to explain why the average rate of change is zero or go over related concepts?

Here are 5 related questions to expand your understanding:

1. What does it mean graphically when the average rate of change is zero over an interval?
2. How is the average rate of change different from the instantaneous rate of change?
3. How would the average rate of change differ for the function $f(x) = 2x^2 + 3$ over the same interval?
4. What is the derivative of $f(x) = 2x^2 - 1$, and how does it relate to the rate of change?
5. Can a function have a zero average rate of change over an interval but still be increasing or decreasing at points within that interval?

Tip:
For quadratic functions, if the interval is symmetric around the vertex, the average rate of change could be zero due to symmetry!