To find the domain of the function $g(x) = \frac{1}{x + 5}$, we need to determine the values of $x$ for which the function is defined.

The function $g(x)$ is a rational function, and rational functions are undefined where the denominator is zero. In this case, the denominator is $x + 5$. So, we need to find the values of $x$ that make the denominator zero:

$x + 5 = 0$

Solving for $x$, we get:

$x = -5$

This means that the function $g(x) = \frac{1}{x + 5}$ is undefined at $x = -5$. Therefore, the domain of $g(x)$ is all real numbers except $x = -5$.

Domain of $g(x)$:

$\boxed{(-\infty, -5) \cup (-5, \infty)}$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the domain of $f(x) = \frac{2x + 1}{x^2 - 4}$?
2. How do you find the domain of a square root function, such as $h(x) = \sqrt{x - 3}$?
3. What are the steps to determine the range of a rational function like $g(x) = \frac{1}{x+5}$?
4. How does the presence of an absolute value in the denominator, such as $k(x) = \frac{1}{|x| + 2}$, affect the domain?
5. What happens to the domain if we modify $g(x) = \frac{1}{x + 5}$ to include an additional term, such as $g(x) = \frac{1}{x + 5} + 2$?

Tip: For rational functions, always check where the denominator is zero to determine the domain.