The task is to find the domain of the given functions:

1. $f(x) = \frac{x + 3}{x - 4}$
2. $f(x) = \sqrt{x - 1} + \sqrt{x - 3}$

Let's determine the domain for each function:

1. Function: $f(x) = \frac{x + 3}{x - 4}$

The domain of a rational function excludes values that make the denominator zero. For this function, we need to find when the denominator is zero:

$x - 4 = 0 \implies x = 4$

Thus, the domain is all real numbers except $x = 4$.

Domain: $x \in \mathbb{R}, x \neq 4$.

2. Function: $f(x) = \sqrt{x - 1} + \sqrt{x - 3}$

The domain of this function depends on the values inside the square roots being non-negative. We set up the conditions:

1. For $\sqrt{x - 1}$: $x - 1 \geq 0 \implies x \geq 1$

2. For (