If f(x)=(x²−1)/(x+1) and g(x)=x−1, on which domain are f and g equal?

Correct answer: x≠−1

The governing concept is equality of functions together with preservation of the original domain. Factor the numerator of f: x²−1=(x−1)(x+1). Whenever x≠−1, cancellation is valid and f(x)=[(x−1)(x+1)]/(x+1)=x−1, which equals g(x). At x=−1, the original formula for f has denominator zero, so f is undefined. The cancelled expression x−1 may have a numerical value there, but cancellation does not repair a missing point in the original domain. Thus the functions agree on all real x except −1, represented by x≠−1 or ℝ\{−1}. Option A is correct. x=1 is permitted, so option B is wrong. All real numbers incorrectly includes −1, and option D is far too restrictive because equality holds at infinitely many other inputs.