Given the function `f(x)=9/(x+1)` and `g(x)=sqrt x` determine (f+g)(x) and its domain.
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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.
thilina-g | College Teacher | (Level 1) Educator
`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}`