If \(f(x)=x^2-4\) and \(g(x)=\sqrt{x+2}\), what is the domain of \(\left(\frac{f}{g}\right)(x)\)?

Correct answer: \(x>-2\)

The direct answer is option A: \(x>-2\). For \(f(x)=x^2-4\), every real number is allowed. For \(g(x)=\sqrt{x+2}\), the radicand must satisfy \(x+2\ge0\), so \(x\ge-2\). However, in \((f/g)(x)\), this square root is in the denominator, so it must not equal zero. Thus \(\sqrt{x+2}>0\), which requires \(x+2>0\), or \(x>-2\). Option A is correct. Option B includes \(x=-2\), but at that point the denominator is zero. Option C excludes 2 for no valid reason; the numerator may be zero there and the quotient is still defined. Option D includes values below -2, where the square root is not real, and also includes -2. The key difference is that a square root in a numerator may be zero, but a square root in a denominator may not. Memory cue: denominator square root means strictly positive radicand.