# Show that f : [–1

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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