Domain:
f(x) has a denominator, which can't be zero. So, its domain is given by
[tex]x^2-1\neq 0 \iff x^2\neq 1 \iff x\neq \pm 1[/tex]
g(x) has a denominator as well. Moreover, it has a root. So, the content of the root can't be negative:
[tex]x-1\geq 0 \iff x \geq 1[/tex]
And the denominator can't be zero:
[tex]\sqrt{x-1}\neq 0 \iff x-1 \neq 0 \iff x \neq 1[/tex]
So, the domain is [tex]x>1[/tex]
Composition:
We have
[tex]f(g(x))=\dfrac{g^2(x)}{g^2(x)-1} = \dfrac{\dfrac{1}{x-1}}{\dfrac{1}{x-1}-1} = \dfrac{\dfrac{1}{x-1}}{\dfrac{1-x+1}{x-1}} = \dfrac{\dfrac{1}{x-1}}{\dfrac{2-x}{x-1}}=\dfrac{1}{2-x}[/tex]
The domain of this function is
[tex]2-x\neq 0 \iff x\neq 2[/tex]
But we also have to remember about the domain of g(x): if g(x) is undefined, we can't compute f(g(x))!
So, the domain of f(g(x)) is
[tex]x>1\ \land x\neq 2[/tex]
