given f(x)=x²+1/1-x², x ≠ ±1 and g(x)=x-1/x+1,x ≠ -1 the function (gof)(x)=

Subject : Math
Class : X Senior High School
Chapter : Composition Function and Inverse Function

Answer :
f(x) = (x² + 1)/(1 - x²)
g(x) = (x - 1)/(x + 1)

The Function of Gof(x) is ....
Gof(x) = g(f(x))
Gof(x) = [(x² + 1)/(1 - x²) - 1]/[(x² + 1)/(1 - x²) + 1]
Gof(x) = x²

Solved By : GuruMIPA

gof(x) = {(x²+1 / 1-x²) - 1} / {(x²+1 / 1-x²) + 1}
= {(x²+1 / 1-x²)-(1-x² / 1-x²)} / {(x²+1 / 1-x²)+(1-x² / 1-x²)}
= {(x²+1-1+x²)/(1-x²)} / {(x²+1+1-x²)/(1-x²)}
= {(2x²)/(1-x²)} / {(2)/(1-x²)}
= {2x² / (1-x²)} × {(1-x²)/2}
= 2x²/2
= x²