Consider the function f(x) = sqrt(x^2 - 4x + 3) / (x^2 - 9). What is the domain of f?

Correct answer

A. (-∞, 1] ∪ [3, ∞) excluding x = -3 and x = 3

Principle or equation

The domain of a function with a square root and a denominator is the set of x such that the radicand is nonnegative and the denominator is not zero.

Why this answer is correct

Radicand: x^2 - 4x + 3 = (x - 1)(x - 3) ≥ 0, so x ≤ 1 or x ≥ 3. Denominator: x^2 - 9 = (x - 3)(x + 3) ≠ 0, so x ≠ 3 and x ≠ -3. Since x = 3 is already excluded by the radicand condition? Actually x = 3 makes radicand zero, but denominator zero, so exclude. x = -3 is not in the radicand domain because -3 ≤ 1, but we must exclude. Thus domain: (-∞, 1] ∪ [3, ∞) with x = 3 and x = -3 removed. Since x = 3 is in [3, ∞), remove it; x = -3 is in (-∞, 1], remove it. So (-∞, -3) ∪ (-3, 1] ∪ (3, ∞). But among options, the first one says (-∞, 1] ∪ [3, ∞) excluding x = -3 and x = 3, which is equivalent.

Example

For f(x) = sqrt(x^2 - 1)/(x - 2), domain is (-∞, -1] ∪ [1, 2) ∪ (2, ∞).