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If $\mathrm{f}(\mathrm{x})=\frac{x-2}{x+2}, x \neq-2$, then what is $\mathrm{f}^{-1}(\mathrm{x})$ equal to?
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Verified Answer
The correct answer is:
$\frac{2(1+x)}{1-x}$
$f(x) \mathrm{y}=4 \frac{x-2}{x+2}, x \neq-2$
$\frac{y}{1}=\frac{x-2}{x+2} \Rightarrow \frac{y+1}{y-1}=\frac{x-2+x+2}{x-2-x-2}$
$\Rightarrow \frac{y+1}{y-1}=\frac{2 x}{-4}$
$\Rightarrow \frac{y+1}{y-1}=\frac{-x}{2} \Rightarrow x=-2\left(\frac{y+1}{y-1}\right)$
Now, $\mathrm{y}=\frac{-2(x+1)}{x-1}=\frac{2(x+1)}{1-x}$
$\frac{y}{1}=\frac{x-2}{x+2} \Rightarrow \frac{y+1}{y-1}=\frac{x-2+x+2}{x-2-x-2}$
$\Rightarrow \frac{y+1}{y-1}=\frac{2 x}{-4}$
$\Rightarrow \frac{y+1}{y-1}=\frac{-x}{2} \Rightarrow x=-2\left(\frac{y+1}{y-1}\right)$
Now, $\mathrm{y}=\frac{-2(x+1)}{x-1}=\frac{2(x+1)}{1-x}$
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