Find the domain of the function $$f(x) = \frac{1}{x+8} + \frac{1}{\sqrt{x - 8}} + \frac{1}{\sqrt{8 - x}}.$$ Express your answer as a union of intervals.
For the function to be defined, the denominators of all the fractions must be nonzero. The first fraction has a nonzero denominator for all $x \neq -8,$ the second fraction has a nonzero denominator for all $x \neq 8,$ and the third fraction has a nonzero denominator for all $x \neq 8.$ Therefore, the function is defined for all $x$ such that
\[x \neq -8 \quad \text{and} \quad x \neq 8.\]The solution is thus $\boxed{(-\infty,-8) \cup (-8,8) \cup (8,\infty)}.$
To find the domain of the function, we need to identify any values of $x$ that would result in division by zero or the square root of a negative number, as these are not defined.
Let's look at each term separately.
The first term, $\frac{1}{x+8}$, is defined for all values of $x$ except when the denominator, $x+8$, equals zero. So, $x+8 \neq 0 \implies x \neq -8$.
The second term, $\frac{1}{\sqrt{x - 8}}$, is defined for all values of $x$ greater than or equal to 8, as the square root of a negative number is not defined for real numbers.
The third term, $\frac{1}{\sqrt{8 - x}}$, is defined for all values of $x$ less than or equal to 8, as the square root of a negative number is not defined for real numbers.
Combining these conditions, we find that the domain of the function is the set of all values of $x$ that satisfy the following conditions:
$x \neq -8$
$x \geq 8$
$x \leq 8$
This can be expressed as a union of intervals. Breaking down the intervals:
$x \neq -8$: This can be represented by $(-\infty, -8) \cup (-8, \infty)$.
$x \geq 8$: This can be represented by $[8, \infty)$.
$x \leq 8$: This can be represented by $(-\infty, 8]$.
Therefore, the domain of the function is $(-\infty, -8) \cup (-8, 8] \cup [8, \infty)$.
To find the domain of the function, we need to determine the values of $x$ for which the function is defined. In this case, we need to consider three separate conditions:
1. The expression $\frac{1}{x+8}$ is defined for all real numbers except when the denominator is equal to zero. So, $x+8 \neq 0$. Solving this inequality, we get $x \neq -8$. Therefore, the first condition is $x \neq -8$.
2. The expression $\frac{1}{\sqrt{x - 8}}$ is defined when the radicand (the expression inside the square root) is greater than or equal to zero. So, $x - 8 \geq 0$. Solving this inequality, we get $x \geq 8$. Therefore, the second condition is $x \geq 8$.
3. The expression $\frac{1}{\sqrt{8 - x}}$ is defined when the radicand is greater than or equal to zero. So, $8 - x \geq 0$. Solving this inequality, we get $x \leq 8$. Therefore, the third condition is $x \leq 8$.
Now, we consider the intersection of all these conditions. From the first condition $x \neq -8$, we can conclude that the values of $x$ which satisfy this condition are all real numbers except $x = -8$. From the second condition $x \geq 8$, we can conclude that the values of $x$ which satisfy this condition are all real numbers greater than or equal to $8$. From the third condition $x \leq 8$, we can conclude that the values of $x$ which satisfy this condition are all real numbers less than or equal to $8$.
Combining all these conditions, we find that the domain of the function is the union of the intervals $(\infty, -8) \cup (-8, 8) \cup (8, \infty)$.