Because the domain needs g (x) >; =0, so 0 =

So the range of y=√g(x) is:

2) the domain is1-x2 >; =0, that is-1 =

f(x)=x+√( 1-x^2)>； = x & gt=- 1

f(x)^2= 1+2x√( 1-x^2)

Pass the inequality:-(A 2+B 2)/2 =

Get:- 1/2 =

Therefore: 0 =

So there is: -√ 2 =

Therefore, the range is: [- 1, √2]