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If f: (-∞, 3] → [7, ∞) with f(x) = x² - 6x + 16, which statement is true?

1. f⁻¹(x) = 3 + sqrt(x - 7)
2. f⁻¹(x) = 3 - sqrt(x - 7)
3. f⁻¹(x) = 1/(x² - 6x + 16)
4. f is many-one

Correct answer: f⁻¹(x) = 3 - sqrt(x - 7)

Solution

f(x) = (x-3)² + 7. On (-∞, 3] the function is one-one (decreasing branch) with range [7, ∞), so it is invertible. Solving y = (x-3)² + 7: x = 3 ± sqrt(y-7). Since x <= 3 requires x-3 <= 0, choose the minus sign: f⁻¹(x) = 3 - sqrt(x - 7).

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