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

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

The domain of the function (f+g)(x) is all the values that x can take for which the result is a real number.

As `sqrt x` is real only for `x >= 0` , the domain of (f+g)(x)...

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

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

The domain of the function (f+g)(x) is all the values that x can take for which the result is a real number.

As `sqrt x` is real only for `x >= 0` , the domain of (f+g)(x) is the set of real numbers with `x >= 0` .

**`(f+g)(x) = 9/(x+1) + sqrt x` and the domain is all real numbers `>=` 0.**

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

Here the denominator cannot be zero, then `x!= -1`

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

`g(x) = sqrt(x)`

The value inside a squareroot can't be negative, therefore `x=gt0.`

Therefore the domain of g(x) is,

` g(x) = {x in R, x=gt 0}`

`(f+g)(x) = f(x)+g(x)`

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

The domain of `(f+g)(x)` is the set of numbers which are in both domains.

Therefore, domain of `(f+g)(x)` is,

`D = {x in R, x=gt 0}`

**Further Reading**