What is the range of the function f(x) = sqrt(x)?

• A. y ≤ 0
• B. y ∈ R
• C. y ≥ 0
• D. y > 0

Answer: (C)

The range is all nonnegative outputs. Since the square root yields the nonnegative root, f(x) = sqrt(x) never produces a negative number. At x = 0 you get 0, and as x increases, sqrt(x) increases without bound, so every nonnegative value is attained (for any y ≥ 0, taking x = y² gives f(x) = y). Therefore, the range is y ≥ 0. The other descriptions don’t fit: negative values never occur, so y ≤ 0 is incorrect; not all real numbers occur since negatives are missing, so y ∈ R is incorrect; and 0 is produced, so y > 0 would miss that value.