If (f:\mathbb{R}\to\mathbb{R}) is given by (f(x)=\sqrt{x^2+4x+8}), what is the range?

Correct answer: ([2,\infty))

To find the range, first study the expression inside the square root. Completing the square gives x²+4x+8=(x+2)²+4. Since a square is always nonnegative, the smallest possible value of the expression is 4, reached at x=−2. As x varies over all real numbers, the expression can take every value from 4 upward. The square root therefore can take every value from √4=2 upward.

Hence the range is [2,∞), so option A is correct. The output cannot be below 2 because the quantity under the root is never below 4. Also, every number y≥2 is possible: choose x so that (x+2)²=y²−4. Thus there is no upper bound. The supplied answer correctly identifies both endpoints and the unbounded direction.