Problem

Begin with \((x-3)^{2}\ge~0\) and derive a bound for \(f(x)=(x-3)^{2}+2\). State where equality occurs, classify the included extremum, and express the range in inequality and interval notation.

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What this problem is really about

A real square’s nonnegative bound becomes a bound on the function after the vertical shift is applied. We’ll locate the input where equality occurs, confirm that the boundary output is attained, and write the range using output-variable inequality notation and an interval with the correct closed endpoint.

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The square term is at most 0, so the function has maximum 2. A square is nonnegative, not nonpositive.

(x − 3)² ≥ 0, so f(x) ≥ 2. The minimum 2 is included at x = 3; the range is y ≥ 2, or [2, ∞).

The range is x ≥ 2, or [2, ∞). A range describes output y-values, not input x-values.

The square is strictly positive, so y > 2. At x = 3 the square equals 0, so 2 is included.

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Begin with \((x-3)^{2}\ge~0\) and derive a bound for \(f(x)=(x-3)^{2}+2\). State where equality occurs, classify the included extremum, and express the range in inequality and interval notation.

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