Show that *f*: [−1, 1] → **R**, given by f(x) = `x/(x + 2)` is one-one. Find the inverse of the function *f*: [−1, 1] → Range *f*.

(Hint: For y in Range f, y = `f(x) = x/(x +2)` for some x in [-1, 1] ie x = `2y/(1-y)`

#### Solution

*`f*: [−1, 1] → R is given as `f(x) = x/(x + 2)`

Let *f*(*x*) = *f*(*y*).

`=> x/(x + 2) = y/(y +2)`

=> xy + 2x = xy + 2y

=> 2x = 2y

=> x = y

∴ *f* is a one-one function.

It is clear that *f*: [−1, 1] → Range *f* is onto.

∴ *f*: [−1, 1] → Range *f* is one-one and onto and therefore, the inverse of the function:

*f*: [−1, 1] → Range *f *exists.

Let *g*: Range *f* → [−1, 1] be the inverse of *f*.

Let* y* be an arbitrary element of range *f*.

Since *f*: [−1, 1] → Range *f* is onto, we have:

`=> y = x/(x + 2)`

=> xy + 2y = x

=> x(1-y)= 2y

`=> x = (2y)/(1-y), y !=1`

`g(y) = (2y)/(1-y), y != 1`

Now `(gof) (x) = g(f(x)) = g(x/(x+2))= (2(x/(x+2)))/(1-x/(x+2)) = (2x)/(x + 2 - x) = 2x/2 = x`

`(fog)(y) = f(g(y)) = f("2y"/(1-y)) = ((2y)/((1-y)))/(((2y)/(1-y)) +2) = (2y)/(2y + 2 -2y) = (2y)/2 = y`

`:. gof = I_(-1,1) and fog - I_"Range f"`

`:. f^(-1) = g`

`=> f^(-1) (y) = "2y"/(1 - y), y != 1`