`f(x) = x +1`

`g(x) = sqrt(x-2)`

Before proceeding to find `f@g`, we should find the Domain and Range of f(x) and g(x) separately.

f(x) = x+1

This is a simple function which has no complex situations.

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

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

`g(x) = sqrt(x-2)`

In g(x), the value inside the squareroot cannot be negative at any moment, since squareroot od negative values are complex, not real numbers. (Remember we are talking about real functions here, ie.. real input and real answer)

Therefore,
for g(x), `x =gt 2` for its Domain.

Also g(x) will only output positive numbers or zero, since it is `+sqrt(x-2)` not `-sqrt(x-2)` .
 So the result will be positive real numbers including zero.

Domain of g(x) `= {x in R, x =gt 2}`

Range of g(x) `= {x in R_0^+}` or `= {x in R, x =gt 0}`

Now `f@g = g(x) +1`

`f@g = sqrt(x-2)+1`

The domain of f@g is the same as g(x). Therefore,

Domain of `f@g` `= {x in R, x=gt2}`

We know always, `sqrt(x-2) =gt0`  (Range of g(x))

Therefore,

`sqrt(x-2)+1 =gt1` for every x in its domain.

The range of f@g is then,

Range of `f@g = {x in R, x =gt1}`

Domain of `f@g` `= {x in R, x=gt2}`

Range of `f@g` `= {x in R, x =gt1}`