## Calculus: Early Transcendentals (2nd Edition)

$f ∘ g ∘ G =|(\frac{1}{(x-2)})^2 - 4|$ Therefore the domain is $x\ne2$
$f(x) = |x|$ $G(x) = \frac{1}{(x-2)}$ $g(x) = x^2-4$ $f ∘ g ∘ G = f(g(G(x))) = f( g(\frac{1}{(x-2)})) = f((\frac{1}{(x-2)})^2 - 4)=|(\frac{1}{(x-2)})^2 - 4|$ Therefore the domain is $x-2\ne0$ $x\ne2$