The exercise asks to calculate the average rate of change of given functions over specified intervals. The average rate of change of a function $f(x)$ on an interval $[a, b]$ is calculated using the formula:

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

Let's go through the first few exercises to illustrate the solution process:

1. $f(x) = x^3 + 1$

(a) Interval $[2, 3]$

$f(3) = 3^3 + 1 = 27 + 1 = 28, \quad f(2) = 2^3 + 1 = 8 + 1 = 9$ $\text{Average Rate of Change} = \frac{f(3) - f(2)}{3 - 2} = \frac{28 - 9}{1} = 19$

(b) Interval $[-1, 1]$

$f(1) = 1^3 + 1 = 2, \quad f(-1) = (-1)^3 + 1 = -1 + 1 = 0$ $\text{Average Rate of Change} = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{2 - 0}{2} = 1$

2. $g(x) = x^2 - 2x$

(a) Interval $[1, 3]$

$g(3) = 3^2 - 2(3) = 9 - 6 = 3, \quad g(1) = 1^2 - 2(1) = 1 - 2 = -1$ $\text{Average Rate of Change} = \frac{g(3) - g(1)}{3 - 1} = \frac{3 - (-1)}{2} = \frac{4}{2} = 2$

(b) Interval $[-2, 4]$

$g(4) = 4^2 - 2(4) = 16 - 8 = 8, \quad g(-2) = (-2)^2 - 2(-2) = 4 + 4 = 8$ $\text{Average Rate of Change} = \frac{g(4) - g(-2)}{4 - (-2)} = \frac{8 - 8}{6} = 0$

Would you like me to continue solving the remaining problems or explain a specific part of these calculations in more detail? 😊

Here are 5 related questions to deepen your understanding:

1. How is the formula for the average rate of change connected to the concept of a derivative?
2. Why does the average rate of change become zero in Exercise 2(b)?
3. Can you think of a real-world scenario that uses the concept of average rate of change?
4. How can you graphically interpret the average rate of change on a function's graph?
5. What happens to the average rate of change as the interval shrinks to a single point?

Tip: Always double-check your endpoints $a$ and $b$ to ensure correct substitution in the formula!