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

Correct answer: \(x>1\)

In \(f(x)\), \(\sqrt{x-1}\) is in the denominator. Thus, the radicand must satisfy \(x-1\geq 0\), and the denominator must not be zero, so \(x-1\neq 0\). Therefore, \(x>1\). Since \(g(x)=x\) is defined for all real numbers, the domain of \((f+g)\) is the intersection of the two domains, namely \((1,\infty)\). Exam tip: For a square root in the denominator, require the expression under the root to be strictly positive.