If ${\rm{f}}\left( {{{\rm{x}}_1}} \right) - {\rm{f}}\left( {{{\rm{x}}_2}} \right) = {\rm{f}}\left( {\frac{{{{\rm{x}}_1} - {{\rm{x}}_2}}}{{1 - {{\rm{x}}_1}{{\rm{x}}_2}}}} \right)$ for x1, x2 ∈ (-1, 1), then what is f(x) equal to?

Question

If ${\rm{f}}\left( {{{\rm{x}}_1}} \right) - {\rm{f}}\left( {{{\rm{x}}_2}} \right) = {\rm{f}}\left( {\frac{{{{\rm{x}}_1} - {{\rm{x}}_2}}}{{1 - {{\rm{x}}_1}{{\rm{x}}_2}}}} \right)$ for x1, x2 ∈ (-1, 1), then what is f(x) equal to?

TuteeHUB · Accepted Answer

${\rm{In\;}}\left( {\frac{{2 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right)$

TuteeHUB · Answer

${\rm{In\;}}\left( {\frac{{1 - {\rm{x}}}}{{1 + {\rm{x}}}}} \right)$

TuteeHUB · Answer

${\tan ^{ - 1}}\left( {\frac{{1 - {\rm{x}}}}{{1 + {\rm{x}}}}} \right)$

TuteeHUB · Answer

${\tan ^{ - 1}}\left( {\frac{{1 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right)$

1.	If \({\rm{f}}\left( {{{\rm{x}}_1}} \right) - {\rm{f}}\left( {{{\rm{x}}_2}} \right) = {\rm{f}}\left( {\frac{{{{\rm{x}}_1} - {{\rm{x}}_2}}}{{1 - {{\rm{x}}_1}{{\rm{x}}_2}}}} \right)\) for x1, x2 ∈ (-1, 1), then what is f(x) equal to?
A.	\({\rm{In\;}}\left( {\frac{{1 - {\rm{x}}}}{{1 + {\rm{x}}}}} \right)\)
B.	\({\rm{In\;}}\left( {\frac{{2 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right)\)
C.	\({\tan ^{ - 1}}\left( {\frac{{1 - {\rm{x}}}}{{1 + {\rm{x}}}}} \right)\)
D.	\({\tan ^{ - 1}}\left( {\frac{{1 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right)\)
Answer» B. \({\rm{In\;}}\left( {\frac{{2 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right)\)

If \({\rm{f}}\left( {{{\rm{x}}_1}} \right) - {\rm{f}}\left( {{{\rm{x}}_2}} \right) = {\rm{f}}\left( {\frac{{{{\rm{x}}_1} - {{\rm{x}}_2}}}{{1 - {{\rm{x}}_1}{{\rm{x}}_2}}}} \right)\) for x1, x2 ∈ (-1, 1), then what is f(x) equal to?

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