# Inverse function

1. Feb 15, 2014

### haha1234

1. The problem statement, all variables and given/known data

Prove/Disprove following function being one-to-one.If yes,find its inverse.

g(x)=x-$\frac{1}{x}$,x>0

2. Relevant equations

3. The attempt at a solution
My tutor said that it is one-to-one,but I found that the are two solutions for g-1(x).
Are there any mistakes?
g(x)=x-$\frac{1}{x}$,x>0
x=g-1(x)-$\frac{1}{g^{-1}(x)}$
[g-1(x)]2-xg-1(x)-1=0
g-1(x)=$(x\pm\sqrt{x^2-4(1)(-1)})/2$
THANKS

Last edited by a moderator: Feb 15, 2014
2. Feb 15, 2014

### lurflurf

x>0
so
$$\frac{x\pm\sqrt{x^2+4}}{2}$$
reduces to
$$\frac{x+\sqrt{x^2+4}}{2}$$
keep in mind f is increasing that is
$$\mathrm{f}(x+h)-\mathrm{f}(x)=h \left( 1+\frac{1}{x(x+h)}\right)>0$$
so
f(x+h)=f(x)
implies x=x+h

3. Feb 15, 2014

### Staff: Mentor

I fixed the broken LaTeX near the bottom of post #1. It didn't render correctly because tags were mixed in some itex script.

4. Feb 15, 2014

### SammyS

Staff Emeritus
How are the domain of a function and the range of the inverse of that function related ?