If (f(x)=\sqrt{x^2-1}) and (g(x)=\sqrt{4-x^2}), what is the domain of ((f+g)(x))?

Correct answer: ([-2,-1]\cup[1,2])

The direct answer is option A: \([-2,-1]\cup[1,2]\). For a square root to be real, its expression must be non-negative. For \(f(x)=\sqrt{x^2-1}\), require \(x^2-1\ge0\), so \(x^2\ge1\), giving \(x\le-1\) or \(x\ge1\). Thus the domain of \(f\) is \((-infty,-1]\cup[1,\infty)\). For \(g(x)=\sqrt{4-x^2}\), require \(4-x^2\ge0\), so \(x^2\le4\), giving \(-2\le x\le2\). The domain of \(f+g\) is the common part, or intersection, of both domains: \([-2,-1]\cup[1,2]\). Option A works because every square root is defined there. Option B, \([-1,1]\), contains points where \(x^2-1<0\), so \(f\) is not real. Option C, \([-2,2]\), includes numbers between \(-1\) and \(1\), again invalid for \(f\). Option D is the domain of \(f\) alone and ignores the restriction from \(g\). Memory cue: for a sum, take the intersection of the two domains, not their union.