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] i.e., `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 = f(x) for same x ∈ [–1, 1]

⇒ `y = x/(x + 2)`

⇒ xy + 2y = x

⇒ x(1 – y) = 2y

⇒ `x = (2y)/(1 - y), y ≠ 1`

Now, let us define g: Range f → [–1, 1] as

`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`