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

Correct answer: \((1,\infty)\)

For \(f(x)=\frac{1}{\sqrt{x-1}}\), the radicand must be non-negative, but the square root is in the denominator, so it cannot be zero. Thus \(x-1>0\), giving \(x>1\). Since \(g(x)=x^2\) is defined for every real number, the domain of \((f+g)\) is the intersection of the two domains: \((1,\infty)\). The endpoint \(x=1\) is excluded because it makes the denominator zero. Exam tip: when a square root occurs in a denominator, the radicand must be strictly positive.