Let's start by analyzing both functions and finding their domains.

1. For $f(x) = \sqrt{3 - x}$:

• The square root function $\sqrt{x}$ is only defined when the argument is non-negative.
• Therefore, we need $3 - x \geq 0$, which simplifies to: $x \leq 3$
• Hence, the domain of $f(x)$ is all $x$ such that $x \leq 3$, which in interval notation is: $\text{Domain of } f(x): (-\infty, 3]$

2. For $g(x) = \sqrt{x^2 - 4}$:

• Again, the square root function is only defined when the argument is non-negative.
• We need $x^2 - 4 \geq 0$, which simplifies to: $x^2 \geq 4$ This inequality holds when $x \leq -2$ or $x \geq 2$.
• So, the domain of $g(x)$ is $x \in (-\infty, -2] \cup [2, \infty)$.

3. For $(f + g)(x)$:

• The function $(f + g)(x)$ will be defined wherever both $f(x)$ and $g(x)$ are defined.
• Therefore, the domain of $f + g$ is the intersection of the domains of $f(x)$ and $g(x)$.

Intersection of Domains:

• Domain of $f(x)$: $(-\infty, 3]$
• Domain of $g(x)$: $(-\infty, -2] \cup [2, \infty)$

The intersection of these two domains is:

• For $(-\infty, 3]$ and $(-\infty, -2]$, the overlap is $(-\infty, -2]$.
• For $(-\infty, 3]$ and $[2, \infty)$, the overlap is $[2, 3]$.

Thus, the domain of $f + g$ is: $(-\infty, -2] \cup [2, 3]$

4. Conclusion:

• $(f + g)(x) = \sqrt{3 - x} + \sqrt{x^2 - 4}$
• Domain of $f + g$: $(-\infty, -2] \cup [2, 3]$

Would you like further details on this? Here are some related questions you might find useful:

1. What is the graph of $f(x) = \sqrt{3 - x}$?
2. How do you solve inequalities involving square roots?
3. Can the domain of $f(x)$ and $g(x)$ overlap at specific points?
4. What happens when the square root is complex (i.e., negative inside)?
5. How do you compute the sum of two functions in general?

Tip: Always check for restrictions such as non-negativity for functions like square roots and logarithms.