The problem asks to find the average rate of change of the function:

$f(x) = \frac{4}{-3x - 3}$

on the interval $x = -4$ to $x = 4$.

The average rate of change of a function $f(x)$ on an interval $[a, b]$ is given by:

$\frac{f(b) - f(a)}{b - a}$

In this case, $a = -4$ and $b = 4$.

Step 1: Calculate $f(a)$, where $a = -4$:

$f(-4) = \frac{4}{-3(-4) - 3} = \frac{4}{12 - 3} = \frac{4}{9}$

Step 2: Calculate $f(b)$, where $b = 4$:

$f(4) = \frac{4}{-3(4) - 3} = \frac{4}{-12 - 3} = \frac{4}{-15} = -\frac{4}{15}$

Step 3: Apply the formula for average rate of change:

$\text{Average rate of change} = \frac{f(4) - f(-4)}{4 - (-4)} = \frac{-\frac{4}{15} - \frac{4}{9}}{4 + 4}$

Simplifying the numerator:

$-\frac{4}{15} - \frac{4}{9} = -\left(\frac{4}{15} + \frac{4}{9}\right)$

Find a common denominator:

$\frac{4}{15} + \frac{4}{9} = \frac{4 \times 3}{45} + \frac{4 \times 5}{45} = \frac{12}{45} + \frac{20}{45} = \frac{32}{45}$

Thus:

$-\frac{4}{15} - \frac{4}{9} = -\frac{32}{45}$

Now divide by 8 (since $4 - (-4) = 8$):

$\text{Average rate of change} = \frac{-\frac{32}{45}}{8} = -\frac{32}{45} \times \frac{1}{8} = -\frac{32}{360} = -\frac{8}{90} = -\frac{4}{45}$

So, the average rate of change of the function on the interval $[-4, 4]$ is:

$-\frac{4}{45}$

Let me know if you would like further details or clarification!

Here are 5 related questions:

1. What is the average rate of change of a linear function?
2. How do you interpret the result of the average rate of change in real-world problems?
3. How is the average rate of change different from the instantaneous rate of change?
4. Can the average rate of change be zero for non-constant functions?
5. How does the sign of the average rate of change inform us about the behavior of the function?

Tip: To find the average rate of change of a function, think of it as finding the slope of the secant line between two points on the curve!