What is the domain of f(x) = √(x − 3) + 1/(x − 5)?

Correct answer: [3, ∞) \ {5}

The domain of a real-valued expression must satisfy every restriction at the same time. For √(x − 3), the radicand must be non-negative, so x − 3 ≥ 0 and therefore x ≥ 3. The denominator x − 5 cannot be zero, so x ≠ 5. Taking the intersection of these conditions gives [3, ∞) with 5 removed, written [3, ∞) \ {5}. Thus option A is correct. Notice that x = 3 is allowed because √0 is defined, which rules out option B. Values less than 3 are excluded by the square root, so option D is too broad. Option C unnecessarily excludes all values below 5. The governing concept is intersecting the domains of the component expressions rather than treating either restriction in isolation.