解析:∵y=√(x+1)-1,(x≥0)∴x=(y+1)^2-1=y^2+2y,即f-1(x)=x^2+2x,由x≥0,∴√x+1≥1,得√(x+1)-1≥0
有原函数值域y≥0,即反函数定义域为x≥0