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

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

The direct answer is option A: \(\mathbb{R}\setminus\{-1,1\}\). For \(f(x)=2x/(x^2-1)\), the denominator must be nonzero: \(x^2-1=(x-1)(x+1)\ne0\), so \(x\ne1,-1\). In \(g(x)=1/(x-1)+1/(x+1)\), both denominators also require \(x\ne1\) and \(x\ne-1\). The difference \(f-g\) is defined only on the common domain, so the allowed set is all real numbers except -1 and 1. Option A is correct. Option B incorrectly allows both denominator zeros. Option C removes 0, but 0 is allowed: the denominators are then -1 and 1. Option D gives only the forbidden points rather than the allowed domain. Simplifying the expression might create cancellation, but cancellation cannot restore values where the original functions were undefined. Memory cue: for sums or differences, intersect the original domains before simplifying.