Describe fog and gof wherever is possible for the following functions
(i) `f(x)=sqrt(x+3),g(x)=1+x^(2)` (ii) `f(x)=sqrt(x),g(x)=x^(2)-1`

Explanation

(i) Domain of `f` is `[3,oo)`, range of `f` is `[0,oo)`
Domain is `g` is `R`, range of `g` is `[1,oo)`
For go`f(x)`
Since range of `f` is a subset of domain of `g`
`:.` domain of gof is `[-3,oo)` {equal to the domain of `f`}
`gof(x)=g{f(x)}=gsqrt(x+e)=1+(x+3)=x+r`. Range of gof is `[1,oo)`
For `fog(x)`
Since range of `g` is a subset fo domain of `f`
`:.` domain of fog is `R` {equal to the domain of `g`}
`fog(x)=f{g(x)}=f(1+x^(2))=sqrt(x^(2)+r` Range of fog is `[2,oo)`
(ii) `f(x)=sqrt(x),g(x)=x^(2)-1`
Domain of `f` is `[0,oo)` range of `f` is `[0,oo)`
Domain of `g` is `R` range of `g` is`[-1,oo)`
For `gof(x)`
Since range of `f` is a subset of the domain of `g`
`:.` domain of gof is `[0,oo)` and `g{f(x)}=g(sqrt(x))=x-1`. Range of gof is `[-1,oo)`
For `fog(x)`
Since range of `g` is not a subset of the domain of `f`
i.e. `[-1,oo)cancelsub[0,oo)`
`:.` fog is not defined on whole of the domain of `g`
Domain of fog is `{xepsilonR` the domain of `g:g(x)epsilon[0,oo)`, the domain of `f}`.
Thus the domain of fog is `D={xepsilonR:0leg(x)ltoo}`
i.e. `D={xepsilonR:lex^(2)-1}={xepsilonR:xle-1` or `xge1}=(-oo,-1]uu[1,oo)`
`fog(x)=f{g(x)}=f(x^(2)-1)=srt(x^(2)-1)` Its range is `[0,oo)`