91Ó°ÊÓ

We start by finding \(g \circ f\). This means plugging the function \(f(x)\) inside \(g(x)\): $$ (g \circ f)(x) = g(f(x)) = g(|x|). $$ Now we plug in \(|x|\) into \(g(x) = x^2 - 4\): $$ (g \circ f)(x) = (|x|)^2 - 4 = x^2 - 4. $$ So, \(g \circ f(x) = x^2 - 4\). Since \(g\) and \(f\) both have domain \((-\infty, \infty)\), the domain of \(g \circ f\) is also \((-\infty, \infty)\).

Next, we need to find \(G \circ (g \circ f)\). This means plugging \((g \circ f)(x)\) into the function \(G(x)\): $$ (G \circ (g \circ f))(x) = G((g \circ f)(x)) = G(x^2 - 4). $$ Now we plug in \(x^2 - 4\) into \(G(x) = \frac{1}{x-2}\): $$ (G \circ (g \circ f))(x) = \frac{1}{(x^2 - 4) - 2} = \frac{1}{x^2 - 6}. $$ So, \(G \circ g \circ f(x) = \frac{1}{x^2 - 6}\).

Finally, we need to find the domain of \(G \circ g \circ f(x) = \frac{1}{x^2 - 6}\). The domain of \(G \circ g \circ f\) is the intersection of the domain of \(g \circ f\), which is \((-\infty, \infty)\), and the domain where \(x^2 - 6 \neq 0\). Solving for \(x^2 - 6 = 0\): $$ x^2 - 6 = 0 \Rightarrow x^2 = 6 \Rightarrow x = \pm \sqrt{6}. $$ The values \(x = \pm \sqrt{6}\) are the only ones for which the denominator becomes zero. Therefore, the domain of \(G \circ g \circ f\) is \((-\infty, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, \infty)\). In summary, the composite function \(G \circ g \circ f(x) = \frac{1}{x^2 - 6}\) and its domain is \((-\infty, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, \infty)\).