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 ?

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General way is to transfigure twain faces of the resemblingity until they set-on-foot ostensible resembling.

Consider the just laborer face, we entertain:

##([email protected])(x) = f(g(x)) = (1/x +1)/(1-1/x) = (x+1)/(x-1)##

(defined past ##x=1## is not in the defined interim) (1)

Consider the left laborer face, we let: ##([email protected])^-1(x) = y## where ##([email protected])(y) = x##

(according to limitation of inverse discharge) From that, we can entertain : ##([email protected])(y) = (y+1)/(y-1) = x## ##xy - x = y + 1 ## ##y = (x+1)/(x-1)##= ##([email protected]) ^-1 (x) ##

(defined past x = 1 is not interjacent in the defined interim) (2) Thus, from (1) & (2), Q.E.D

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