Question

Show that f : [−1, 1] → R , given by is one-one. Find the inverse of the function f : [−1, 1] → Range f . (Hint: For y ∈Range f , y = , for some x in [−1, 1], i.e., )

Answer 1

The function f( x )= x x+2 is provided in the range f:[ −1,1 ]→R

To check the function is one-one, consider f( x )=f( y ) .

f( x )=f( y ) x x+2 = y y+2 xy+2x=xy+2y x=y

Therefore, every element on domain has a different image. So, the function f is one-one function.

Since, the input values of the given function lies in the range R. So, the function is onto.

The function is one-one and onto, so the inverse of the function f( x ) exist.

Consider a function g that lies in the range of function f→[ −1,1 ] as the inverse of f and y be an element in the range of f .

The expression to find the inverse of the function is,

y=f( x ) y= x x+2 x( 1−y )=2y x= 2y 1−y

The value of x exists for every value except when y=1 .

The function g: for the range f→[ −1,1 ] is,

g( y )= 2y 1−y →y≠1

The values of ( gof )( x ) to find the range of the function gof is

( gof )( x )=g( f( x ) ) =g( x x+2 ) = 2( x x+2 ) 1− x x+2 = 2x x+2−x

( gof )( x )= 2x 2 =x

The values of ( fog )( x ) to find the range of the function fog is,

( fog )( y )=f( g( y ) ) =f( 2y 1−y ) = 2y 1−y 2y 1−y +2 = 2y 2y+2−2y ( fog )( y )= 2y 2 =y

The range of the function gof= I [ −1,1 ] and fog= I Range:f

The inverse of the function f is g .

f −1 =g

Substitute the value in the above equation.

f −1 ( y )= 2y 1−y →y≠1

Thus, the function is one-one and the inverse of the function is f −1 ( y )= 2y 1−y .