The most basic definition of "function" is that it is a set of ordered pair with the property that there are NOT two ordered pairs in the set with the same first member. It would be possible to present a function by listing all the pairs. For example {(1, 3), (0, 3), (2, -1), (4, 1)} is such a funtion. The "domain" is the set of all numbers that are first members of a pair in that set. Here, the domain is {0, 1, 2, 4}.

However, most "interesting" functions contain an infinite number of pairs so, rather than listing them, we tell what values can be first members of the pairs and some "rule" or "formula" for finding the second member. For example, I can define a function by saying that it is the set of pairs {(x, y)} where x can be any real number, [itex]0\le x\le 5[/itex], and [itex]y= x^2[/itex]. In that case, the domain is the set of all real numbers, greater than or equal to 0 and less than or equal to 5, just as stated. Most often we give just the formula with theunderstandingthat the domain is the set of all numbers for which the formulacanbe calculated. In the case of your function f, which I will assume was f(x)= 1/(x- 2) (what you wrote "f(x)= 1/x- 2 is really f(x)= (1/x)- 2), I can calculate the corresponding value of the function for every real numberexceptx= 2. I cannot calculate a function for x= 2 because then x- 2= 0 and we cannot divide by 0. The domain of x is For , the number whose square is x, because no square of a real number is ever negative, we must have and that is the domain of g: .

"f+ g" is the function defined by "for each x, calculate f(x) and g(x) and add them". Wecannotdo that if x= 2 because we cannot calculate f(x). We also cannot do that for x< 0 because we cannot calculate g(x). As long as wecancalculate f(x) and g(x) we certainly can add them so those are the only problems. The domain of f+ g is "all non-negative real numbers except 2" or, in set notation, .

(To clarify "we can certainly add them", suppose the function were f(x)/(g(x)- 3) rather than just f+ g. We cannot evaluate f(2), as above, and we cannot evaluate g at any negative number. But, while we can evaluate both f(9) and g(9), g(9)- 3= 3- 3= 0 and we cannot divide by 0. The domain of "f(x)/(g(x)- 3)" with g(x) and f(x) as above is "all non-negative real numbers except 2 and 9".)

(If you really did mean f(x)= (1/x)- 2, the domain of f is "all real numbers except 0" and the domain of f+ g is "all positive real numbers".

(I keep saying "real numbers" because I am assuming that, with a problem of this level, you are not dealing with complex numbers.)