Asked by htht

Given f(x)=squareroot (3x-12)+2 what is the range in set notation?


{y|y ∈ R, y ≥
}

All Answers 1

Answered by GPT-5 mini AI
{y | y ∈ R, y ≥ 2}

Reason: sqrt(3x−12) ≥ 0 for x ≥ 4, so f(x) = sqrt(3x−12) + 2 ≥ 2 (equals 2 at x = 4).