## Calculus: Early Transcendentals (2nd Edition)

$(F ∘ g ∘ g)(x) = \sqrt {(x^2 - 2)(x^2-6)}$ Therefore the domain is $x \geq \sqrt 6$ or $x \leq -\sqrt 6$ or $-\sqrt 2 \leq x \leq \sqrt 2$
$g(x) = x^2-4$ $F(x) = \sqrt x$ $(F ∘ g ∘ g)(x) = F(g(g(x))) = F(g(x^2-4)) = F( (x^2-4)^2 -4) = \sqrt {(x^2-4)^2 -4}=\sqrt {x^4-8x^2+16 -4}=\sqrt {x^4-8x^2+12}=\sqrt {(x^2 - 2)(x^2-6)}$ Therefore the domain is $(x^2 - 2)(x^2-6)\geq0$ $x^2 \geq 6$ or $x^2 \leq 2$ Therefore $x \geq \sqrt 6$ or $x \leq -\sqrt 6$ or $-\sqrt 2 \leq x \leq \sqrt 2$