This might not be exactly true.
The domain of is
and its range is really (infinity shall be included in the real numbers).
But then again, f(f(x)) has the above mentioned domain (since the image of is ),
and then you map the real numbers again, in the same way, to itself. And again, the value is not defined at x = 1 (you can really not define it there, because
the left and right limits are different).
So I'd say it's a little tricky here and would nevertheless say that the domain is .
Of course you have as effective function the identity mapping, however since you cannot divide by zero, the simplification holds ONLY IF x is NOT 1.
That's the point.
I may be corrected if I'm wrong, however I think that this is the correct answer.
Yes, mastermind is correct. f(f(x))= (x/(x-1))/(1/(x-1)) for all x for which it is defined- all x except 1. That reduces to f(x)= x for those values of x- x not equal to 1.
A simpler example of that is (x- 1)/(x^2- 1)= (x- 1)/((x- 1)(x+ 1))= 1/(x+ 1) for all x except 1. The left side is not defined for x= 1 but the right side is so they are not equal for x= 1.