What is the domain of the function \(f(x)=\frac{1}{x^2+4}\)?

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

For every real \(x\), \(x^2\ge 0\), so \(x^2+4\ge 4>0\). Thus the denominator is never zero, and no real value of \(x\) must be excluded; hence the domain is \(\mathbb{R}\). Options B and C are incorrect because at \(x=2\) and \(x=-2\), the denominator equals 8, not 0. Exam tip: for a fractional function, exclude only those input values that make the denominator zero.