What is the real domain of the function \(f(x)=\frac{1}{x^2+1}\)?

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

For every real \(x\), \(x^2\ge 0\), so \(x^2+1\ge 1>0\). Thus the denominator is never zero, and the function is defined for every real number. Therefore, its domain is \(\mathbb{R}\). Option B is incorrect because at \(x=0\), the denominator is \(1\), not zero. Exam tip: For a rational function, exclude only those values that make the denominator zero.