To find the domain of the function [tex]\( f(x) = -\sqrt{\frac{1}{x^2 - 4}} \)[/tex], we need to determine the values of [tex]\( x \)[/tex] for which the expression inside the square root is defined and non-negative, and also ensure that the square root itself is defined.
The function inside the square root must be non-negative:
[tex]\[
\frac{1}{x^2 - 4} \ge 0
\][/tex]
First, let's find where the denominator [tex]\( x^2 - 4 \)[/tex] is non-zero because division by zero is undefined. Setting the denominator equal to zero:
[tex]\[
x^2 - 4 = 0
\][/tex]
[tex]\[
x^2 = 4
\][/tex]
[tex]\[
x = \pm 2
\][/tex]
The values [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex] make the denominator zero, therefore, these points are not in the domain.
Next, we investigate the sign of [tex]\( x^2 - 4 \)[/tex] to determine where the fraction [tex]\( \frac{1}{x^2 - 4} \)[/tex] is non-negative. Consider the inequality [tex]\( x^2 - 4 > 0 \)[/tex]:
[tex]\[
x^2 > 4
\][/tex]
Taking the square root of both sides:
[tex]\[
|x| > 2
\][/tex]
This implies:
[tex]\[
x > 2 \quad \text{or} \quad x < -2
\][/tex]
In these regions ([tex]\( x > 2 \)[/tex] and [tex]\( x < -2 \)[/tex]), the expression [tex]\( x^2 - 4 \)[/tex] is positive, and therefore [tex]\( \frac{1}{x^2 - 4} \)[/tex] is positive.
Thus, the domain of the function [tex]\( f(x) = -\sqrt{\frac{1}{x^2 - 4}} \)[/tex] is all [tex]\( x \)[/tex] such that:
[tex]\[
x > 2 \quad \text{or} \quad x < -2
\][/tex]
In interval notation, this is expressed as:
[tex]\[
(-\infty, -2) \cup (2, \infty)
\][/tex]
Therefore, the domain is [tex]\( \boxed{(-\infty, -2) \cup (2, \infty)} \)[/tex].
