The functions f and g are defined for all x€R \{ -1,0,1} by f(x) = (x+1)/(1-x) and g(x)=1/x . Show that (f○g)^-1 (x)= (f○g)(x). How to do this ?

General approach is to transform both sides of the equality until they start appearing similar.

Consider the right hand side, we have:

##(f@g)(x) = f(g(x)) = (1/x +1)/(1-1/x) = (x+1)/(x-1)##

(defined since ##x=1## is not in the defined interval) (1)

Consider the left hand side, we let: ##(f@g)^-1(x) = y## where ##(f@g)(y) = x##

(according to definition of inverse function) From that, we can have : ##(f@g)(y) = (y+1)/(y-1) = x## ##xy - x = y + 1 ## ##y = (x+1)/(x-1)##= ##(f@g) ^-1 (x) ##

(defined since x = 1 is not included in the defined interval) (2) Thus, from (1) & (2), Q.E.D