What is the domain of the function \(f(x)=\sqrt{x+2}+\sqrt{6-x}\)?

Correct answer: \([-2,6]\)

For real values, the expression inside each square root must be non-negative. From \(x+2\ge 0\), we get \(x\ge -2\), and from \(6-x\ge 0\), we get \(x\le 6\). Hence, the intersection of the two conditions is \([-2,6]\), so option A is correct. Option C wrongly excludes \(-2\) and \(6\); at these endpoints, one square root is zero, which is valid. Exam tip: for a function involving square roots, require every radicand to be \(\ge 0\) and intersect the resulting conditions.