If \(f(x)=\sqrt{x+2}+\sqrt{8-x}\), what is the domain of \(f\)?

Correct answer: \([-2,8]\)

For both square roots to be real, each radicand must be non-negative. From \(x+2\ge 0\), we get \(x\ge -2\), and from \(8-x\ge 0\), we get \(x\le 8\). Hence, the intersection is \([-2,8]\). At \(x=-2\) and \(x=8\), one square root becomes zero, so both endpoints are included; therefore, option B is incorrect. Exam tip: For radical functions, require every radicand to be \(\ge 0\) and intersect all resulting conditions.