Answer :

It is given that f : [–1, 1] R, given by

Now, Let f(x) = f(y)



xy + 2x = xy +2y


2x = 2y


x = y


f is a one- one function.


Now, Let y = , xy = x + 2y so x =


So, for every y in the range there exists x in the domain such that f(x) = y


f is onto function.


f: [-1,1] Range f is one-one and onto


the inverse of the function : f: [-1, 1] Range f exists.


Let g: Range f [-1, 1]be the inverse of range f.


Let y be an arbitrary element of range f.


Since, f :[-1, 1] Range f is onto, we get:


y = f(x) for same x ϵ [-1, 1]



xy +2y = x


x(1 - y) = 2y


x = , y ≠ 1


Now, Let us define g: Range f [-1, 1]


g(y) = , y ≠ 1


Now, (gof)(x) = g(f(x)) =


(fog)(y) = f(g(y)) =


Thus, gof = I[-1,1] and fog = IRange f


f-1 = g


Therefore, f-1(y) = , y ≠ 1


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