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


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


Indeed,


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



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.

 A: This answer contains: (i) first, another method; (ii) a comment on your proof technique; (iii) an alternative approach, related to your proof (at least more than (i)) but that I find less confusing and risky.

A possible approach.


*

*For any $x >0$, we have $f(x) = \frac{1+\lvert x\rvert}{x} = \frac{1+x}{x} = 1+\frac{1}{x}$. It is immediate that $f$ is continuous and decreasing on $(0,\infty)$ (since $x>0\mapsto \frac{1}{x}$ is), that $f(x) > 1$ for all $x>0$, and since
$$
f(x) \xrightarrow[x\to0^+]{} +\infty,\qquad f(x) \xrightarrow[x\to+\infty]{} 1
$$
we get by the intermediate value theorem that $f(\mathbb{R}_+^\ast) = (1,\infty)$.

*For any $x <0$, we have $f(x) = \frac{1+\lvert x\rvert}{x} = \frac{1-x}{x} = -1+\frac{1}{x}$. It is again immediate that $f$ is continuous and decreasing on $(-\infty,0)$ (since $x<0\mapsto \frac{1}{x}$ is), that $f(x) < -1$ for all $x<0$, and since
$$
f(x) \xrightarrow[x\to0^-]{} -\infty,\qquad f(x) \xrightarrow[x\to-\infty]{} -1
$$
we get by the intermediate value theorem that $f(\mathbb{R}_-^\ast) = (-\infty,-1)$.

*By combining the two, $f(\mathbb{R}^\ast) = (-\infty,-1)\cup (1,+\infty)=\mathbb{R}\setminus[-1,1]$.
Edit: as @Laurent Duval remarked, $f$ is odd. in particular, we get that $f(\mathbb{R}_-^\ast)=-f(\mathbb{R}_+^\ast)$, i.e. $f(\mathbb{R}_-^\ast)=(-\infty,-1)$, immediately (removing the need for the second "bullet point"). 

A comment on your method, and a suggestion (another method).
While it is a valid way to proceed, the path you chose is rather cumbersome (you have to carry equivalences all the way) and risky (you have to make sure that your equivalences do remain... well, equivalences. Losing an equivalence for an implication by dropping a quantifier or  condition happens very fast.)
I'd suggest first, if you choose this way (i.e., to fix $y$ in the image of $f$ and check conditions on $y$), to avoid equivalences altogether, but instead proceed first by implication (what $y$ must satisfy), then by sufficiency (if $y$ satisfies all the conditions derived in the first step, then indeed here is $x$ such that $f(x)=y$.
For instance:


*

*Necessity: start by observing that $x$ and $f(x)$ always have the same sign, so that any $y\in f(\mathbb{R}_+^\ast)$ must satisfy $y > 0$. [Etc.] Then, at the very end, you'll have that any such $y$ must necessarily be such that $y > 1$.

*Sufficiency: then check it is sufficient: take any $y>1$, and show that for $x\stackrel{\rm def}{=}\frac{1}{y-1} \in \mathbb{R}_+^\ast$ we indeed have $f(x)=y$.

*Conclusion: Every $y\in f(\mathbb{R}_+^\ast)$ satisfies $y>1$, and converly every $y>1$ is in $f(\mathbb{R}_+^\ast)$. Therefore $f(\mathbb{R}_+^\ast) = (1,\infty)$.
A: This answer is meant to visualize how the image of $f$ changes with the $1+$ part. If you study it on $\mathbb{R}^*_+$, $f(x) = 1+\frac{1}{x}$. Noting $g(x) = \frac{1}{x}$, you easily get $g(\mathbb{R}^*_+) = \mathbb{R}^*_+$. Hence with a lousy notation $f(\mathbb{R}^*_+) = \mathbb{R}^*_+ +1 = (0+1,+\infty+1)= (1,+\infty)$. 
Function $f$ is odd. Hence $f(\mathbb{R}^*_-) = -\left(\mathbb{R}^*_+ +1\right) = (-\infty,-1)$. Finally, $f(\mathbb{R}^*)= (-\infty,-1)\cup (1,+\infty)$, as shown by @Clement C.
A: Just for $f(\mathbb{\mathbb{R}^{*}_{+}})$ you have $f(x )=\frac{1+x}{x}$, a continuous function for which $$\lim_{x\to 0}=+\infty$$ and $$\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\frac{\frac 1x+1}{1}=1$$ Besides for all $x\gt 0$ $$\frac{1+x}{x}\gt 1$$ Hence $$f(\mathbb{\mathbb{R}^{*}_{+}})=\mathbb{R}^{*}_{+}\setminus(0,1)$$
