The function `f(x)=9/(x+1)` and `g(x)=sqrt x` .

g(f(x)) = `sqrt (9/(x+1))`

The domain of g(f(x)) is all values of x for which the resulting expression is real.

`(x+1) >= 0`

=> `x >= -1`

**The function `g(f(x)) = sqrt (9/(x+1))` and the domain is all real values of x such that**` x >= -1`

`g(x) = sqrt(x)`

Domain of g(x) is, `{x in R, x=gt0}`

The range of g(x) = `{x inR, x=gt0}`

`f(x) = 9/(x+1)`

Domain of f(x) = `{x in R, x!=1}`

Range of f(x) = `{x in R}`

f`(g(x)) = 9/(g(x)+1)`

`f(g(x)) = 9/(sqrt(x)+1)`

I would find g(f(x)) now.

`g(f(x)) = sqrt(f(x))`

`g(f(x)) = sqrt(9/(x+1))`

In this function, the denominator of the expression inside the squareroot cannot be either zero or negative.

Therefore `x+1 gt 0`

`x gt-1,`

Therefore domain of` g(f(x)) = {x in R, xgt-1}`

**Further Reading**