Simple Functions, Range

How do I show that the range of $f(x)=\frac{x}{(x+1)}$ is $\mathbb{R}$ - {1}?

Informally, I tried saying basically that if f(x) were to produce a value of 1, then $\frac{x}{x+1}=1$ would need to hold true. Solve it and it obviously does not hold. But if you let $y \in \mathbb{R}$ - {1}, then when you solve for $\frac{x}{(x+1)}=y$, you find it holds. Answer being $x=\frac{y}{(y-1)}$.

Is this logic strong enough for a formal proof, or is there a better way?

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try this basic solution : try to solve $f(x)=y$ for $x$, and $x\neq -1$ and $y$ is a real number.

$\displaystyle \frac{x}{x+1}=y \Longleftrightarrow (1-y)x-y=0$

assuming $y\neq 0$, the above equation has the solution : $\displaystyle x=\frac{y}{1-y}\in \mathbb{R}$

but if $y=1$ then $-y=0$ which is absurd.

so $f(x)=y$ has real solution for $x$ only if $y\in \mathbb{R}-\{1\}$.

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First of all, show that the function $f(x)=\frac{x}{x+1}:\mathbb R-\{-1\}\to \mathbb R$ is an one-one function. Secondly, $$f(x)=\frac{x}{x+1}:\mathbb R-\{-1\}\to \text{Im}(\mathbb R-\{-1\})\subset\mathbb R$$ is an onto function, so it is a bijection between $\mathbb R-\{-1\}$ and its image. I mean, there is an inverse for it, say $f^{-1}:\text{Im}(\mathbb R-\{-1\})\to \mathbb R-\{-1\}$. Thirdly, find the rule of $f^{-1}$ as you did befor and use this fact that $$\text{dom}(f^{-1})=\text{Im}(f)$$

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clean and neat and thorough +1 – amWhy Feb 4 '13 at 2:23