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

Correct answer: \([8,\infty)\)

For a real-valued function, the expression inside each square root must be non-negative. Thus, \(x-2\ge 0\) gives \(x\ge 2\), and \(x-8\ge 0\) gives \(x\ge 8\). Their intersection is \(x\ge 8\), so the domain is \([8,\infty)\). Option D is incorrect because the function is defined at \(x=8\), so 8 must be included. Exam tip: For a function containing multiple radicals, find the domain by taking the intersection of all the resulting conditions.