# f functions and domain/range

• Mar 20th 2012, 11:16 PM
tankertert
f functions and domain/range
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.

also not sure if posted in the right forum?
• Mar 20th 2012, 11:29 PM
princeps
Re: f functions and domain/range
Quote:

Originally Posted by tankertert
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.

also not sure if posted in the right forum?

$\displaystyle f(f(x))=\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1}=\frac{\frac{x}{x-1}}{\frac{1}{x-1}}=x$

$\displaystyle \text{domain } : x \in (-\infty,+\infty)$

$\displaystyle \text{range } : f(f(x)) \in (-\infty,+\infty)$
• Mar 21st 2012, 07:20 AM
mastermind2007
Re: f functions and domain/range
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.
• Mar 29th 2012, 12:51 PM
HallsofIvy
Re: f functions and domain/range
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.
• Mar 29th 2012, 03:25 PM
skeeter
Re: f functions and domain/range
graph of f[f(x)] ... note the discontinuity at x = 1