If \({\rm{f}}\left( {\rm{x}} \right) = {\log _{\rm{e}}}\left( {\frac{{1 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right),{\rm{\;g}}\left( {\rm{x}} \right) = \frac{{3{\rm{x}} + {{\rm{x}}^3}}}{{1 + 3{{\rm{x}}^2}}}\) and g ∘ f(t) = g (f (t)), then what is g ∘ \({\rm{f}}\left( {\frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}} \right)\) equal to?

If \({\rm{f}}\left( {\rm{x}} \right) = {\log _{\rm{e}}}\left( {\frac{{

1.	If \({\rm{f}}\left( {\rm{x}} \right) = {\log _{\rm{e}}}\left( {\frac{{1 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right),{\rm{\;g}}\left( {\rm{x}} \right) = \frac{{3{\rm{x}} + {{\rm{x}}^3}}}{{1 + 3{{\rm{x}}^2}}}\) and g ∘ f(t) = g (f (t)), then what is g ∘ \({\rm{f}}\left( {\frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}} \right)\) equal to?
A.	2
B.	1
C.	0
D.	1/2
Answer» C. 0

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