hi
let f(x) = x/(x-1). determine f(f(x)) simplify answer, and find domain/range of f(f(x))
is the question i got.
help please!
also not sure if posted in the right forum?
This might not be exactly true.
The domain of $\displaystyle f(x) = \frac{x}{x-1} $ is $\displaystyle \mathbb{R} \setminus \{1\} $
and its range is really $\displaystyle \mathbb{R} $ (infinity shall be included in the real numbers).
But then again, f(f(x)) has the above mentioned domain (since the image of $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R} $ is $\displaystyle \mathbb{R} $),
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 $\displaystyle \mathbb{R} \setminus \{1\} $.
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.