For f(x) = √(x − 3) + 1, which statement about the domain is true?

For f(x) = √(x − 3) + 1, which statement about the domain is true?

• A. x ≥ 3
• B. x ≤ 3
• C. All real numbers
• D. x > 3

Answer: (A)

The domain of a square root function is determined by where the expression inside the root is nonnegative. For f(x) = √(x − 3) + 1, the radicand is x − 3, so we require x − 3 ≥ 0. This simplifies to x ≥ 3. The +1 outside the root doesn’t change which x-values are allowed; it only shifts the graph upward by 1. So at x = 3, f(3) = √0 + 1 = 1, which is defined. The other options would either allow negative values inside the root, exclude x = 3, or ignore the nonnegativity restriction entirely. Therefore, x is at least 3.

The domain of a square root function is determined by where the expression inside the root is nonnegative. For f(x) = √(x − 3) + 1, the radicand is x − 3, so we require x − 3 ≥ 0. This simplifies to x ≥ 3. The +1 outside the root doesn’t change which x-values are allowed; it only shifts the graph upward by 1. So at x = 3, f(3) = √0 + 1 = 1, which is defined. The other options would either allow negative values inside the root, exclude x = 3, or ignore the nonnegativity restriction entirely. Therefore, x is at least 3.