# Thread: prove g is 1-1

1. ## prove g is 1-1

g(x) = x/(1-|x|)

I began by assuming not

so y/(1-|y|) = x/(1-|x|)

but came unstuck when dealing with the modulus signs - do I need to deal with the case where x<0, y>0. If so, how do I do this?

thanks

2. You must have copied it incorrectly.
As written it is not one-to-one.
$\displaystyle g\left( 3 \right) = g\left( {\frac{{ - 3}} {5}} \right) = \frac{{ - 3}} {2}$

3. ahh sorry - its defined from -1 to 1

4. In that case they cannot have different signs.
Note $\displaystyle \left| x \right| < 1\; \Rightarrow \;1 - \left| x \right| > 0$.
So the numerators must have the same sign.

5. of course - I missed the fact it was between -1 and 1

I am also asked to find g((-1,1))

what does this mean? is it g(-1) and g(1)?

I also need inverse of g.. do I need to do this with separate cases for x>0, x<0?

many thanks

6. Originally Posted by James0502
I am also asked to find g((-1,1))
what does this mean? is it g(-1) and g(1)?
First, did you graph the function?
$\displaystyle g[(-1,1)]$ is simply the range of the function.

7. yes, I did.. so the range is infinity, since the graph goes from -inf to + inf?