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)` .

Explanation

In `f : [-1, 1] to R , f(x) = (x)/( x+ 2)`
Let `x, y in [-1, 1]`
and `f(x) = f(y)`
`rArr (x)/(x + 2) = (y)/(y +2) rArr xy + 2y = xy = 2y`
`rArr 2x = 2y rArr x =y `
`therefore f ` is one- one.
Let `f(x) = y` where `y in R`
`rArr (x)/(x + 2) = y rArr x = xy + 2y `
`rArr x (1 -y) = 2y rArr x = (2y )/(1-y)`
`therefore ` Range of `f= R- {1}`
Let, In `g : ` range of ` f to [-1, 1]` is defined as `g(y) = ( 2y )/(1-y), y ne 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`
and `(fog) (x) = f [g(x) ] = f((2x)/( 1-x))`
`" " ((2x)/( 1-x))/(( 2x)/(1-x)+ 2)= (2x)/( 2x + 2 - 2x) = (2x)/(2) = x `
`therefore gof = fog = I_R`
`rArr f ^(-1) = g`
`rArr f^(-1)(y) = (2y)/( 1-y ) , y ne 1`