What is the domain of the function f(x) = 1/√(x² − 9)?

Correct answer: (−∞, −3) ∪ (3, ∞)

For a real-valued function, the expression under a square root must be non-negative. Here the square root is also in the denominator, so its value cannot be zero. Consequently, the stricter condition is x² − 9 > 0, not merely x² − 9 ≥ 0. Factoring gives (x − 3)(x + 3) > 0. A sign analysis shows this inequality holds when x < −3 or x > 3. Therefore the domain is (−∞, −3) ∪ (3, ∞), which is option A. The endpoints −3 and 3 must be excluded because they make the denominator zero. Option B incorrectly includes those endpoints, option C makes the radicand non-positive, and option D excludes only the endpoints while admitting values between them.