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

Correct answer: (\mathbb{R}-{0})

The direct answer is A: the domain is \(\mathbb{R}\setminus\{0\}\). The domain means all input values for which a function is defined. Here, \(f(x)=1/x\) is undefined when \(x=0\), because division by zero is not allowed. The function \(g(x)=x+1\) is defined for every real number. For a sum \((f+g)(x)=f(x)+g(x)=1/x+x+1\), both parts must be defined at the same time. Therefore every real number except zero is allowed. Option A is correct because it gives \(\mathbb{R}\setminus\{0\}\). Option B, all real numbers, wrongly includes zero. Option C removes \(-1\), but \(x=-1\) is allowed: only \(1/x\) causes a restriction. Option D contains only zero, which is precisely the forbidden input and also leaves out all valid nonzero real numbers. Memory cue: for a sum, take the common domain, and always exclude values that make a denominator zero.