What is the domain of the function \(f(x)=\sqrt{x-1}+\sqrt{x+2}\)?

Correct answer: \([1,\infty)\)

For a real-valued function, the expression inside each square root must be non-negative. Thus, \(x-1\ge 0\) gives \(x\ge 1\), while \(x+2\ge 0\) gives \(x\ge -2\). The intersection of these conditions is \(x\ge 1\), so the domain is \([1,\infty)\). Option B considers only the condition for \(\sqrt{x+2}\) and ignores the stricter condition from \(\sqrt{x-1}\). Exam tip: For a function with multiple square roots, impose every radicand condition and take their intersection.