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What $=\frac{f(x)+1}{f(x)-1}+x$ is equal to ?
Consider the function $f(x)=\frac{x-1}{x+1}$.
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Consider the function $f(x)=\frac{x-1}{x+1}$.
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Given $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-1}{\mathrm{x}+1}$
Applying componendo and dividendo, we get
$\frac{\mathrm{f}(\mathrm{x})+1}{\mathrm{f}(\mathrm{x})-1}=\frac{\mathrm{x}-1+\mathrm{x}+1}{\mathrm{x}-1-\mathrm{x}-1}$
$\Rightarrow \frac{\mathrm{f}(\mathrm{x})+1}{\mathrm{f}(\mathrm{x})-1}=-\mathrm{x}$
Now, $\frac{\mathrm{f}(\mathrm{x})+1}{\mathrm{f}(\mathrm{x})-1}+\mathrm{x}=-\mathrm{x}+\mathrm{x}=0$
Applying componendo and dividendo, we get
$\frac{\mathrm{f}(\mathrm{x})+1}{\mathrm{f}(\mathrm{x})-1}=\frac{\mathrm{x}-1+\mathrm{x}+1}{\mathrm{x}-1-\mathrm{x}-1}$
$\Rightarrow \frac{\mathrm{f}(\mathrm{x})+1}{\mathrm{f}(\mathrm{x})-1}=-\mathrm{x}$
Now, $\frac{\mathrm{f}(\mathrm{x})+1}{\mathrm{f}(\mathrm{x})-1}+\mathrm{x}=-\mathrm{x}+\mathrm{x}=0$
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