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