The problem asks to find the average rate of change of the function $g(x) = \frac{1}{x + 3}$ over the interval $[5, 5 + h]$, where $h \neq 0$.

Step-by-Step Solution:

The average rate of change of a function $g(x)$ over an interval $[a, b]$ is given by the formula: $\text{Average Rate of Change} = \frac{g(b) - g(a)}{b - a}$ Here, $a = 5$ and $b = 5 + h$.

1. Evaluate $g(a) = g(5)$: $g(5) = \frac{1}{5 + 3} = \frac{1}{8}$

2. Evaluate $g(b) = g(5 + h)$: $g(5 + h) = \frac{1}{(5 + h) + 3} = \frac{1}{h + 8}$

3. Substitute into the average rate of change formula: $\text{Average Rate of Change} = \frac{\frac{1}{h + 8} - \frac{1}{8}}{(5 + h) - 5}$ Simplify the denominator: $(5 + h) - 5 = h$ So, the expression becomes: $\text{Average Rate of Change} = \frac{\frac{1}{h + 8} - \frac{1}{8}}{h}$

4. Simplify the numerator: The difference of fractions in the numerator is: $\frac{1}{h + 8} - \frac{1}{8} = \frac{8 - (h + 8)}{8(h + 8)} = \frac{8 - h - 8}{8(h + 8)} = \frac{-h}{8(h + 8)}$

5. Substitute back into the expression: $\text{Average Rate of Change} = \frac{\frac{-h}{8(h + 8)}}{h}$ Simplify this by dividing by $h$: $\text{Average Rate of Change} = \frac{-h}{8(h + 8)} \times \frac{1}{h} = \frac{-1}{8(h + 8)}$

Final Answer:

The average rate of change is: $\boxed{\frac{-1}{8(h + 8)}}$

Would you like further clarification or a breakdown of any specific part?

Here are 5 related questions:

1. How do you interpret the average rate of change geometrically?
2. What is the derivative of $g(x) = \frac{1}{x+3}$?
3. How does the average rate of change compare to the instantaneous rate of change?
4. What happens to the average rate of change as $h$ approaches 0?
5. How would the formula change if the function had a different denominator, like $g(x) = \frac{1}{x + 5}$?

Tip: The average rate of change formula is the same as the slope of the secant line connecting two points on the graph.