If (f(x)=x^2-9) and the codomain is (\mathbb{R}), what is the range on real domain (\mathbb{R})?

Correct answer: ([-9,\infty))

Direct answer: A, [-9,∞). For every real x, x^2 is at least 0. Thus x^2-9 is at least -9. The value -9 is attained at x=0, and every value greater than -9 is possible: if y≥-9, choose x=√(y+9) or its negative. Option A is therefore the actual range. Option B, all real numbers, is the codomain but not the actual output set; negative values below -9 cannot occur. Option C [0,∞) describes x^2 before subtracting 9. Option D (-∞,-9] incorrectly reverses the inequality and gives only values at or below -9. Memory cue: complete the square or use x^2≥0; distinguish codomain from range.