If (f(x)=\frac{1}{x^2-4}) and (g(x)=x+2), what is the domain of ((fg)(x))?

Correct answer: (\mathbb{R}-{-2,2})

For a product \((fg)(x)=f(x)g(x)\), the domain is the intersection of the domains of the two original functions. The function \(f(x)=\frac{1}{x^2-4}\) is undefined when its denominator is zero. Factoring gives \(x^2-4=(x-2)(x+2)\), so \(x=2\) and \(x=-2\) must be excluded. The function \(g(x)=x+2\) is defined for every real number, so the product domain is \(\mathbb R\setminus\{-2,2\}\). Option A is correct.

Although multiplication simplifies formally to \(\frac{x+2}{(x-2)(x+2)}=\frac{1}{x-2}\) for allowed values, the original function was not defined at \(x=-2\). Cancellation does not restore a missing input. Thus both excluded values remain excluded from the domain.