Let f: [3, ∞) → R be defined by f(x) = x² – 6x +

Solution

We are given the function: f(x) = x² – 6x + 11, with domain x ∈ [3, ∞) We are to find the range of f(x). Given: f(x) = x² – 6x + 11 = (x – 3)² + 2 This is a perfect square shifted: It's a parabola opening upward, with minimum value at x = 3. At x = 3: f(3) = (3 – 3)² + 2 = 0 + 2 = 2 As x increases, (x – 3)² increases, so f(x) increases. Hence: Minimum value = 2 Maximum = ∞ (since x → ∞ ⇒ f(x) → ∞) So: Range = [2, ∞)