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If $\mathrm{f}\left(\mathrm{x}_{1}\right)-\mathrm{f}\left(\mathrm{x}_{2}\right)=\mathrm{f}\left(\frac{\mathrm{x}_{1}-\mathrm{x}_{2}}{1-\mathrm{x}_{1} \mathrm{x}_{2}}\right)$ for $\mathrm{x}_{1}, \mathrm{x}_{2} \in(-1,1)$, then what
is $\mathrm{f}(\mathrm{x})$ equal to?
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is $\mathrm{f}(\mathrm{x})$ equal to?
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Verified Answer
The correct answer is:
$\ln \left(\frac{1-x}{1+x}\right)$
$f\left(x_{1}\right)-f\left(x_{2}\right)=f\left(\frac{x_{1}-x_{2}}{1-x_{1} x_{2}}\right)$
$x_{1}, x_{2} \in(-1,1)$
then $f(x)=\log \frac{(1-x)}{(1+x)}$
$\begin{aligned} f\left(x_{1}\right)=\log \frac{1-x_{1}}{1+x_{1}} & f\left(x_{2}\right)=\log \frac{1-x_{2}}{1+x_{2}} \\ f\left(x_{1}\right)-f\left(x_{2}\right) &=\log \frac{1-x_{1}}{1+x_{1}}-\log \frac{1-x_{2}}{1+x_{2}} \\=& \log \frac{\left(1-x_{1}\right)}{\left(1+x_{1}\right)} \times \frac{\left(1+x_{2}\right)}{\left(1-x_{2}\right)} \\=& \log \frac{\left(1-x_{1}+x_{2}-x_{1} x_{2}\right)}{\left(1+x_{1}-x_{2}-x_{1} x_{2}\right)} \\=& \log \frac{\left(1-x_{1} x_{2}\right)-\left(x_{1}-x_{2}\right)}{\left(1-x_{1} x_{2}\right)+\left(x_{1}-x_{2}\right)} \\ f\left(x_{1}\right)-f\left(x_{2}\right)=f\left(\frac{x_{1}-x_{2}}{1-x_{1} x_{2}}\right) \end{aligned}$
$x_{1}, x_{2} \in(-1,1)$
then $f(x)=\log \frac{(1-x)}{(1+x)}$
$\begin{aligned} f\left(x_{1}\right)=\log \frac{1-x_{1}}{1+x_{1}} & f\left(x_{2}\right)=\log \frac{1-x_{2}}{1+x_{2}} \\ f\left(x_{1}\right)-f\left(x_{2}\right) &=\log \frac{1-x_{1}}{1+x_{1}}-\log \frac{1-x_{2}}{1+x_{2}} \\=& \log \frac{\left(1-x_{1}\right)}{\left(1+x_{1}\right)} \times \frac{\left(1+x_{2}\right)}{\left(1-x_{2}\right)} \\=& \log \frac{\left(1-x_{1}+x_{2}-x_{1} x_{2}\right)}{\left(1+x_{1}-x_{2}-x_{1} x_{2}\right)} \\=& \log \frac{\left(1-x_{1} x_{2}\right)-\left(x_{1}-x_{2}\right)}{\left(1-x_{1} x_{2}\right)+\left(x_{1}-x_{2}\right)} \\ f\left(x_{1}\right)-f\left(x_{2}\right)=f\left(\frac{x_{1}-x_{2}}{1-x_{1} x_{2}}\right) \end{aligned}$
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