Identify the domain of $f(x)=\frac{\sqrt{x^2-4}}{x^2-4x+3}$

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We need to determine the domain of f(x) = sqrt(x^2 - 4) / (x^2 - 4x + 3). First, ensure the expression under the square root is nonnegative, since we cannot take the square root of a negative number in the real numbers. The condition is x^2 - 4 >= 0, which factors to (x - 2)(x + 2) >= 0. This inequality holds when x <= -2 or x >= 2. Next, ensure the denominator is not zero, because division by zero is undefined. The denominator x^2 - 4x + 3 factors to (x - 1)(x - 3). Thus we must have x ≠ 1 and x ≠ 3. Now take the intersection of the two conditions: x <= -2 or x >= 2, with x ≠ 1 and x ≠ 3. Since 1 is not in the regions x <= -2 or x >= 2 anyway, we only need to exclude x = 3 from the x >= 2 branch. This yields: (-∞, -2] ∪ [2, 3) ∪ (3, ∞). To check edge cases: at x = -2, the numerator is sqrt((-2)^2 - 4) = sqrt(0 + 0) = 0, so the value is 0 divided by a nonzero number, which is 0, so x = -2 is allowed. At x = 2, the numerator is sqrt(4 - 4) = 0, again giving 0, and the denominator is not zero, so x = 2 is allowed. At x = 3, the denominator is zero, so x = 3 is excluded. Therefore the domain is exactly (-∞, -2] ∪ [2, 3) ∪ (3, ∞). If the provided option is d. Dom(f) = (-∞, -2] ∪ [2, 3) ∪ (3, ∞), then this option accurately describes the domain described above.

ENG1090 - MUM S1 2025 Mock Final Exam

Identify the domain of \(f(x)=\frac{\sqrt{x^2-4}}{x^2-4x+3}\)

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