If (f(x)=\sqrt{x+4}) and (g(x)=\frac{1}{x-1}) then what will be the domain of ((fg)(x))?

Correct answer: \([-4,\infty)-{1})

The direct answer is option A: \([-4,\infty)\setminus\{1\}\). First, \(f(x)=\sqrt{x+4}\) is real only when the radicand is non-negative: \(x+4\ge0\), so \(x\ge-4\). Next, \(g(x)=1/(x-1)\) is defined only when its denominator is not zero: \(x-1\ne0\), so \(x\ne1\). For the product \((fg)(x)=f(x)g(x)\), both functions must be defined, so we take the intersection of these conditions. Thus all numbers from \(-4\) onward are allowed except \(1\). Option A is correct. Option B incorrectly removes \(-4\), although \(\sqrt{0}\) is defined. Option C removes every number below \(1\), which is unnecessary. Option D wrongly allows values below \(-4\) and excludes \(-4\), even though the square root is valid at \(-4\). Memory cue: for a square root use radicand \(\ge0\), and for a denominator use \(\ne0\).