To find the average rate of change of a function over an interval $[a, b]$, the formula is:

$\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}$

Where $f(x)$ is the function, and $a$ and $b$ are the endpoints of the interval.

The given function is:

$f(x) = x^2 + x - 12$

Let's solve each part:

(a) Over the interval $[0, 4]$

1. Find $f(0)$:

$f(0) = 0^2 + 0 - 12 = -12$

2. Find $f(4)$:

$f(4) = 4^2 + 4 - 12 = 16 + 4 - 12 = 8$

3. Calculate the average rate of change:

$\text{Average rate of change} = \frac{f(4) - f(0)}{4 - 0} = \frac{8 - (-12)}{4 - 0} = \frac{8 + 12}{4} = \frac{20}{4} = 5$

Thus, the average rate of change over $[0, 4]$ is 5.

(b) Over the interval $[-2, 10]$

1. Find $f(-2)$:

$f(-2) = (-2)^2 + (-2) - 12 = 4 - 2 - 12 = -10$

2. Find $f(10)$:

$f(10) = 10^2 + 10 - 12 = 100 + 10 - 12 = 98$

3. Calculate the average rate of change:

$\text{Average rate of change} = \frac{f(10) - f(-2)}{10 - (-2)} = \frac{98 - (-10)}{10 + 2} = \frac{98 + 10}{12} = \frac{108}{12} = 9$

Thus, the average rate of change over $[-2, 10]$ is 9.

Summary:

• (a) The average rate of change over $[0, 4]$ is 5.
• (b) The average rate of change over $[-2, 10]$ is 9.

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Would you like further clarification on these steps or have any questions?

Here are some related questions for further exploration:

1. What is the instantaneous rate of change of $f(x)$ at $x = 0$?
2. How does the average rate of change relate to the slope of a secant line between two points?
3. Can the average rate of change be negative? How would that affect the graph of the function?
4. How do you calculate the average rate of change of a function over a non-constant interval?
5. What does the average rate of change represent in real-life scenarios (e.g., velocity, cost)?

Tip: The average rate of change is the slope of the secant line between two points on the graph of the function!