Question

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)`

Answer 1

`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`