Let $f: \mathbb{R}^{\*}\to \mathbb{R}$ function definied by $f(x )=\dfrac{1+|x|}{x}$

* Find $f(\mathbb{\mathbb{R}^{\*}\_{+}})$ ,$f(\mathbb{\mathbb{R}^{\*}\_{-}})$ ,$f(\mathbb{\mathbb{R}^{\*}}) $

Indeed,

* $f(\mathbb{\mathbb{R}^{\*}\_{+}})$

> 
> note that :
>  Let A,F,F three subsets and Let f : E → F be an arbitrary function with domain E and codomain F.  
> 
>  $$f\in F^{E},\ A\subset E, \quad f(A)=\{f(x)\mid x\in A \}=\{y\in F \mid \exists x \in E \mbox{ such that } f(x)=y \}$$
> 
> 
> 

then $y\in f(\mathbb{\mathbb{R}^{\*}\_{+}})\iff \exists x\in \mathbb{R}^{\*}\_{+} \mbox{ such that } f(x)=y$ 

since $f(x)=y \iff \dfrac{1+|x|}{x}=y \iff \left( \begin{cases} \dfrac{1+x}{x}=y & \\ x>0 \end{cases} \mbox{ or } \begin{cases} \dfrac{1-x}{x}=y & \\ x<0 \end{cases} \right)$
then 

$\begin{align}
y\in f(\mathbb{\mathbb{R}^{\*}\_{+}}) &\iff \dfrac{1+|x|}{x}=y \\
&\iff \left( \begin{cases} \dfrac{1+x}{x}=y & \\ x>0 \end{cases} \mbox{ or } \begin{cases} \dfrac{1-x}{x}=y & \\ x<0 \end{cases} \right)\\
&\iff y> 1 \mbox{ or } ?? \end{align}$

* Is my proof correct? I'm also interested in others methods.