# Math Help - Define "domain" in terms of algebra of functions.

1. ## What is "domain" in terms of algebra of functions?

Hi, I'm taking notes to help me with my functions homework tonight and I'm not clear about the part about "domains". I was hoping someone here could clarify what it means.

Example: Let f(x) = 1/x-2 and g(x)= square root of x. Solve f+g (x) and find the domain.

What exactly does "domain" mean and how exactly do I find it?

Please assume that I am a math dummy and explain clearly and step by step if you can. Thank you in advance.

2. ## Re: Define "domain" in terms of algebra of functions.

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, $0\le x\le 5$, and $y= x^2$. 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 the understanding that the domain is the set of all numbers for which the formula can be 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 number except x= 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 $\{x| x\ne 2\}$ For $g(x)= \sqrt{x}$, the number whose square is x, because no square of a real number is ever negative, we must have $x\ge 0$ and that is the domain of g: $\{x| x\ge 0\}$.

"f+ g" is the function defined by "for each x, calculate f(x) and g(x) and add them". We cannot do 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 we can calculate 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, $\{x| x\ge 0 and x\ne 2\}$.

(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.)

3. ## Re: Define "domain" in terms of algebra of functions.

Thanks for the informative response. You were right in your first assumption, it was actually f(x)=1/(x-2). Sorry about that. This does actually help clear things up.