# A bit of clarification on the domain of a composite function

Recently when I reviewed something about the composite function for my calculus exam, I remembered I had been thinking a question for quite a long time (maybe I was going into a dead end) since I was in high school.

I was thinking whether f(x)=1/(1/x) and g(x)=x are the same function or not.
It seems like f(x) is essentially g(x) but according to some kind of definition:

" The composition of two functions f and g is the function h = f ◦ g defined by h(x) = f(g(x)), for all x in the domain of g such that g(x) is in the domain of f. "

Then if I take p(x)=q(x)=1/x (just to avoid confusion), then the composite function is f(x)=p(q(x)), for all x in the domain of q such that q(x) is in the domain of p(x). While in this case, q(x) or the range of q(x), namely (-∞,0)∪(0,∞), is always in the domain of p(x), and all x in the domain of q is also (-∞,0)∪(0,∞), then should the domain of f(x) be (-∞,0)∪(0,∞)? If so, f(x) and g(x) have different domains (the domain of g(x) is R obviously). Does it mean f(x) and g(x) are different functions?
(Of course I could have used just p(x)=1/x and said f(x)=p(p(x)), but it's just for avoiding some confusion.
Hope you will get that idea.)

fresh_42
Mentor
You are right, ##f## and ##g## are different functions, simply because ##f(0)## isn't defined, whereas ##g(0)## is.
They both coincide on ##\mathbb{R}-\{0\}##, and in case we would only consider functions on the multiplicative part of the reals, that is ##\mathbb{R}^*=\mathbb{R}-\{0\}## they are indeed equal. But this is a bit artificial in this situation. On the entire real number line, we have two straights through the origin, but ##f(x)## has a gap at ##x=0##. It is called a removable singularity, because we can insert just one point and get ##g(x)##.

mfb
Mentor
To fully define a function you have to define its domain and codomain. If you say the domain of g is the whole set of real numbers then f and g will be different. But that is not the only choice. You can define both f and g only for positive real x for example, and make them identical.

Okay. Thanks a lot. That really helps.

But why can I still plot the point at origin intersecting with f(x) in Desmos (a graphing calculator)?

mfb
Mentor
It might simplify the expression before determining its possible domain.