Domain and Range of (f.g)(x) & (f/g)(x)

[sjaguar13]

If f(x)=x/(x-1) and g(x)=1/(x+3) what is the domain and range of (f.g)(x) and (f/g)(x)? Are the domains the same?

I got:
Domain of f is all x not equal to 1 or -3. The range is all real numbers.
x/(x-1)(x+3)

Domain of g is all x not equal to 1, but it can't be equal to -3 either because of f (this is the part I am not really sure about). The range is all real numbers.
x(x+3)/(x-1)

Yes, the domains are the same (Right?)

 

[Fredrik]

The domain of f is "x not equal to 1". The domain of g is "x not equal to -3 ". Because of that, neither 1 nor -3 can be a part of the domain of fg or f/g. It doesn't matter that f(x)/g(x)=x(x+3)/(x-1) for all x in the domain of f/g. You don't even need to know that. You only need to know that 0 is not in the range of g.

This is easy to understand if you just remember that the definition of the product function h=f/g is "h(x)=f(x)/g(x) for all x that are members of both domains, except x such that g(x)=0".

If you know this definition, the problem is trivial. If g is a function that doesn't have 0 in its range, the two domains must be the same.

The range of both functions is the set of real numbers.
.

 

[M98Ranger]

The domain of (f.g)(x) would be all x not equal to 1 or -3, as both f(x) and g(x) have restrictions at those values. The range of (f.g)(x) would also be all real numbers.

The domain of (f/g)(x) would be all x not equal to 1, as g(x) has a restriction at that value. However, since f(x) is undefined at x=-3, the domain of (f/g)(x) would also have a restriction at x=-3. The range of (f/g)(x) would still be all real numbers.

So, while the domains are not exactly the same, they do have the same restrictions, which is why both (f.g)(x) and (f/g)(x) have the same domain.

In summary, the domain of (f.g)(x) and (f/g)(x) would be all real numbers except for x=1 and x=-3, and the range would be all real numbers.