If (f(x)=\sqrt{x^2-4}) and (g(x)=\frac{1}{x-5}), what is the domain of (f+g)?

Correct answer: ( (-\infty,-2]\cup[2,\infty)-{5} )

The direct answer is option A: \(((-\infty,-2]\cup[2,\infty))\setminus\{5\}\). For \(f(x)=\sqrt{x^2-4}\), require \(x^2-4\ge0\). Factoring gives \((x-2)(x+2)\ge0\), which holds when \(x\le-2\) or \(x\ge2\). For \(g(x)=1/(x-5)\), require \(x-5\ne0\), so \(x\ne5\). The domain of a sum is the intersection of the two domains, so remove 5 from the two outer intervals. Option A states this exactly. Option B uses the interval between -2 and 2, but there the square-root radicand is negative except at endpoints. Option C correctly uses the outer regions but wrongly excludes the endpoints, where the radicand is zero and the square root is defined. Option D excludes only 5 and wrongly includes numbers where \(f\) is not real. Memory cue: solve the square-root inequality first, then remove denominator zeros.