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f(x)=sqrt(x-1) and g(x)=2-x^2
Therefore f(g(x))=sqrt(2-x^2-1)
=sqrt(1-x^2) ....... (What makes you think only one works?)
You know that square root of a negative number is imaginary.
So 1-x^2 > 0
ie x^2<1
ie -1<=x<=1
Thats the domain
Range is 0<=y<=1

I guess I was trying to satisfy the domain of f(x) too. I see now that the range of g(x) has to be the domain of f(x). For g(x) >= 1, x^2 <= 1. So, the domain is [-1,1]. Well, I got half the domain right. I hope it was only worth 1 point on the test.