Which of the following correctly describes the domain and range of the real-valued function \(f(x)=\frac{x^2+1}{x^2-1}\)?

Correct answer: Domain \(\mathbb{R}\setminus\{-1,1\}\); Range \(( -\infty,-1]\cup(1,\infty)\)

The denominator must be non-zero: \(x^2-1\ne0\), so \(x\ne\pm1\). Hence, the domain is \(\mathbb{R}\setminus\{-1,1\}\). For the range, let \(y=\frac{x^2+1}{x^2-1}\). Then \(x^2=\frac{y+1}{y-1}\). Since \(x^2\ge0\) and \(y\ne1\), we must have \(y\le-1\) or \(y>1\). The value \(y=-1\) occurs at \(x=0\), whereas \(y=1\) is never attained. Option B incorrectly treats excluded domain values as excluded range values. Exam tip: for rational functions, find denominator restrictions first, then determine range separately.