Prove/Disprove: Inverse Function g(x)=x-\frac{1}{x},x>0

[haha1234]

Homework Statement

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

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

Homework Equations

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

 

[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

 

[Mark44]

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

 

[SammyS]

Homework Statement

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

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

Homework Equations

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

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