To find the domain of a function, we need to consider values of $x$ that make the function undefined.
For the first term, $\frac{\sqrt{x - 2}}{\sqrt{5}}$, the function is undefined when $x - 2 < 0$, which means $x < 2$. However, there are no restrictions on the numerator because we are taking the square root of $x - 2$, which is always nonnegative. So the domain of the first term is $x \ge 2$.
For the second term, $\sqrt{(x - 1)(x + 2)(x + 4)}$, the function is undefined when the expression inside the square root becomes negative. This occurs when the product $(x - 1)(x + 2)(x + 4)$ is negative. Since we are taking the square root of this product, the expression must be nonnegative. This means we need to find when the product is zero.
The term $x - 1$ is equal to 0 when $x = 1$, the term $x + 2$ is equal to 0 when $x = -2$, and the term $x + 4$ is equal to 0 when $x = -4$. So the product $(x - 1)(x + 2)(x + 4)$ is equal to 0 when $x = -4$, $x = -2$, or $x = 1$.
Therefore, the domain of the second term is $x \neq -4, -2, 1$. In other words, $x$ can be any real number except $x = -4, -2,$ or $1$.
Since the domain of $f(x)$ is the intersection of the domains of each term, we have $x \ge 2$ and $x \neq -4, -2, 1$. Therefore, the domain of $f(x)$ is $x \in \boxed{[2, \infty)} \setminus \{-4, -2, 1\}$.
Find the domain of the function $f(x) = \frac{\sqrt{x-2}}{\sqrt{5}} + \sqrt{(x - 1)(x + 2)(x + 4)}.
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