To determine which of these functions have an average rate of change equal to 0 on the interval $x = -2$ to $x = 2$, let's calculate the average rate of change (AROC) for each function.

The formula for the AROC is: $\text{AROC} = \frac{f(b) - f(a)}{b - a},$ where $a = -2$ and $b = 2$.

Analysis for each function:

1. For $f(x) = 7 - 3x$: $f(-2) = 7 - 3(-2) = 7 + 6 = 13, \quad f(2) = 7 - 3(2) = 7 - 6 = 1.$ $\text{AROC} = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{1 - 13}{4} = \frac{-12}{4} = -3.$ Not zero.

2. For $f(x) = 7x$: $f(-2) = 7(-2) = -14, \quad f(2) = 7(2) = 14.$ $\text{AROC} = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{14 - (-14)}{4} = \frac{28}{4} = 7.$ Not zero.

3. For $f(x) = 7 - 3x^2$: $f(-2) = 7 - 3(-2)^2 = 7 - 3(4) = 7 - 12 = -5, \quad f(2) = 7 - 3(2)^2 = 7 - 12 = -5.$ $\text{AROC} = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{-5 - (-5)}{4} = \frac{0}{4} = 0.$ AROC is zero.

4. For $f(x) = 7$: $f(-2) = 7, \quad f(2) = 7.$ $\text{AROC} = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{7 - 7}{4} = \frac{0}{4} = 0.$ AROC is zero.

Conclusion:

The functions with an average rate of change equal to 0 on the interval are:

• $f(x) = 7 - 3x^2$
• $f(x) = 7$

Let me know if you want a deeper explanation!

Follow-Up Questions:

1. How does the formula for the average rate of change relate to the slope of a line?
2. Why do quadratic and constant functions often have special average rate of change properties?
3. What happens to the AROC for $f(x) = 7x^2$ over the interval $[-2, 2]$?
4. Can you predict the AROC of a cubic function $f(x) = ax^3$ over symmetric intervals?
5. How does the AROC of $f(x)$ change if $f(x)$ is multiplied by a constant?

Tip:

For constant functions $f(x) = c$, the average rate of change is always 0 since $f(x)$ doesn't depend on $x$!