Answer
f is one-to-one
Work Step by Step
$f(x)=\frac{x+1}{x-1}\\
A\,function\,\,f: \mathbb{R}-\left \{ 1 \right \} \rightarrow \mathbb{R}-\left \{ 1 \right \} \,is\,\,one-to-one\,\Leftrightarrow \\
\forall \,\,x_{1}\,and\,x_{2}\,\,in\,\,\mathbb{R}-\left \{ 1 \right \} \,\,if\,
f(x_{1}) = f(x_{2})\,\,then\,x_{1} = x_{2}.\\
let\,f(x_{1})=f(x_{2})\\
\Rightarrow \frac{x_{1}+1}{x_{1}-1}=\frac{x_{2}+1}{x_{2}-1}\\
\Rightarrow (x_{1}+1)(x_{2}-1)=(x_{2}+1)(x_{1}-1)\\
\Rightarrow x_{2}x_{1}-x_{1}+x_{2}-1=x_{2}x_{1}-x_{2}+x_{1}-1\\
\Rightarrow x_{2}-x_{1}=x_{1}-x_{2}\\
\Rightarrow 2x_{1}=2x_{2}\\
\Rightarrow x_{1}=x_{2}\\
therefore\,f\,is\,one-to-one
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