Let's analyze each function step by step.

1. $f(x) = 3 - x$

• Domain of $f(x)$:

  • Since $f(x) = 3 - x$ is a linear function, there are no restrictions on the value of $x$ (i.e., no denominators, square roots, or logarithms that could create undefined points).
  • The domain is all real numbers $x \in (-\infty, \infty)$.

• Domain of $f(x)$ in interval notation:

  • $(-\infty, \infty)$

2. $g(x) = \frac{1}{x^2 - 4}$

• Domain of $g(x)$:

  • This is a rational function. For rational functions, the denominator must not be zero.
  • So, find when the denominator is zero: $x^2 - 4 = 0$ Solving this: $x^2 = 4 \implies x = \pm 2$ Therefore, $g(x)$ is undefined at $x = 2$ and $x = -2$.

• Domain of $g(x)$ in interval notation:

  • The domain is all real numbers except $x = 2$ and $x = -2$, so the domain in interval notation is: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

Final Answer:

• Domain of $f(x) = 3 - x$: $(-\infty, \infty)$
• Domain of $g(x) = \frac{1}{x^2 - 4}$: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

Would you like further details or clarifications on these functions?

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Here are five related questions:

1. What is the range of $f(x) = 3 - x$?
2. How would the domain change if $f(x)$ had a square root?
3. Can you explain how vertical asymptotes relate to the domain of $g(x)$?
4. What happens to the domain of $g(x)$ if we change the denominator to $x^2 - 9$?
5. How does adding a constant to $g(x)$ affect its domain?

Tip: Always check the denominator of a rational function for zero values to avoid undefined points!