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