What is the range of (f(x)=\sqrt{x^2-2x+10})?

Correct answer: ([3,\infty))

Direct answer: A, [3,∞). Rewrite the expression inside the square root: x^2-2x+10=(x-1)^2+9. Since a square is at least 0, the inside is at least 9. The square root is therefore at least √9=3. Equality occurs at x=1, so 3 is included; as |x-1| grows, the expression grows without bound, so every value above 3 is possible. Option A is correct. Option B [10,∞) forgets to take the square root of the minimum 9. Option C [0,∞) is too broad and ignores the positive lower bound. Option D all real numbers is impossible because a principal square root is never negative. Memory cue: complete the square first, then apply the square-root restriction.