If \(f(x)=\frac{x^2-1}{x+1}\) and \(g(x)=x-1\), for which real values of \(x\) do \(f\) and \(g\) have the same value?

Correct answer: For every real \(x\) such that \(x\ne -1\)

Since \(x^2-1=(x-1)(x+1)\), for \(x\ne -1\) we get \(f(x)=\frac{(x-1)(x+1)}{x+1}=x-1=g(x)\). However, the original function \(f\) is undefined at \(x=-1\) because its denominator becomes zero. Therefore, the functions have equal values for every real \(x\ne -1\). Exam tip: after simplifying a rational expression, always retain the domain restriction from the original expression.