We are given two functions:

$f(x) = \frac{4}{x + 2}, \quad g(x) = \frac{x}{x + 2}$

We are asked to find the following operations and state the domains:

(a) $f + g$

$f(x) + g(x) = \frac{4}{x + 2} + \frac{x}{x + 2} = \frac{4 + x}{x + 2}$ The domain is all real numbers except where the denominator is 0. For both $f(x)$ and $g(x)$, the denominator $x + 2 \neq 0$, so: $x \neq -2$ Thus, the domain is $(-\infty, -2) \cup (-2, \infty)$.

(b) $f - g$

$f(x) - g(x) = \frac{4}{x + 2} - \frac{x}{x + 2} = \frac{4 - x}{x + 2}$ The domain is the same as in part (a), $x \neq -2$, so: $(-\infty, -2) \cup (-2, \infty)$

(c) $f \cdot g$

$f(x) \cdot g(x) = \frac{4}{x + 2} \cdot \frac{x}{x + 2} = \frac{4x}{(x + 2)^2}$ The domain is still $x \neq -2$, so: $(-\infty, -2) \cup (-2, \infty)$

(d) $\frac{f}{g}$

$\frac{f(x)}{g(x)} = \frac{\frac{4}{x + 2}}{\frac{x}{x + 2}} = \frac{4}{x}, \quad x \neq 0$ So, the domain must exclude $x = -2$ (from the original denominator) and $x = 0$ (from this expression): $(-\infty, -2) \cup (-2, 0) \cup (0, \infty)$

(e) $\frac{g}{f}$

$\frac{g(x)}{f(x)} = \frac{\frac{x}{x + 2}}{\frac{4}{x + 2}} = \frac{x}{4}$ The domain excludes only $x = -2$: $(-\infty, -2) \cup (-2, \infty)$

Would you like further details on any specific part or have any questions?

Here are 5 related questions for practice:

1. What happens to the domain if we add or subtract two rational functions?
2. How do you handle finding the domain when multiplying two functions?
3. Why do we need to find where the denominator is zero when determining the domain?
4. How does dividing two functions affect their domain?
5. What is the general strategy for simplifying complex fractions?

Tip: Always check for restrictions in the domain before performing operations on functions!