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

Correct answer: (\mathbb{R}-{-2,1})

The direct answer is option A: \(\mathbb{R}\setminus\{-2,1\}\). A rational function is defined only when its denominator is nonzero. For \(f(x)=1/(x-1)\), we need \(x-1\ne0\), so \(x\ne1\). For \(g(x)=1/(x+2)\), we need \(x+2\ne0\), so \(x\ne-2\). The sum \((f+g)(x)\) exists only where both functions exist, so we take the common domain: all real numbers except -2 and 1. Option A states exactly this. Option B removes only 1 and wrongly allows -2, where \(g\) has division by zero. Option C removes only -2 and wrongly allows 1, where \(f\) is undefined. Option D lists only the excluded numbers, but a domain contains the allowed inputs, not merely the forbidden ones. Even if the fractions are combined algebraically, the original functions are not defined at these points. Memory cue: for addition, intersect the domains; for denominators, never allow zero.