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

Show that `f: [-1, 1 ] to R`, given by `f(x) = (x)/((x + 2))` is one-o

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