What is the range of f(x) = √(25 − (x − 1)²)?

Correct answer: [0, 5]

The governing concept is the range of a square-root function. For f(x) = √(25 − (x − 1)²), the expression under the radical must be non-negative, so (x − 1)² ≤ 25. The quantity (x − 1)² can vary from 0 to 25, making the radicand vary from 25 down to 0. Since a principal square root is always non-negative, f(x) varies from √0 = 0 to √25 = 5, and both endpoints are attained: x = 6 or −4 gives 0, while x = 1 gives 5. Therefore the range is [0, 5]. Option B incorrectly includes negative outputs, option C confuses the input shift with output values, and option D uses the radicand rather than its square root.