Question

Consider f : R + → [−5, ∞ ) given by f ( x ) = 9 x 2 + 6 x − 5. Show that f is invertible with .

Answer 1

The function provided is f( x )=9 x 2 +6x−5 for the range f: R + →[−5,∞) .

For the function to be invertible, the function should be one-one and onto.

Check whether the function is one-one.

f( a )=9 a 2 +6a−5

f( b )=9 b 2 +6b−5

Now,

f( a )=f( b ) 9 a 2 +6a−5=9 b 2 +6a−5 9 a 2 −9 b 2 +6a−6b=0 9( a 2 − b 2 )+6( a−b )=0

Further simplify.

3( a−b )( a+b )+2( a−b )=0 ( a−b )( 3a+3b+2 )=0

Now, ( a−b )=0 a=b

And ( 3a+3b+2 )=0 3a=−3b−2

f:R→[ −5,∞ ), So x∈R

Therefore, x is always positive. So, function is one-one.

Now, check whether the function is onto.

Consider a function y∈[−5,∞) , such that,

y=9 x 2 +6x−5 y= ( 3x+1 ) 2 −6 3x+1= y+6 ≥0 x= y+6 −1 3

As the particular value of the function exists in the range provided, so the function is onto.

Hence, the f function is one-one and onto, so f −1 exists.

( gof )( x )=g( f( x ) ) =g( 9 x 2 +6x−5 ) =g( ( 3x+1 ) 2 −6 ) = ( 3x+1 ) 2 −6+6 −1 3

( gof )( x )= 3x+1−1 3 =x

( fog )( y )=f( g( y ) ) =f( y+6 −1 3 ) = ( 3( y+6 −1 3 )+1 ) 2 −6 = ( y+6 ) 2 −6

( fog )( y )=y+6−6 =y

Therefore,

gof= I R fog= I [−5,∞)

So the function f is invertible.

y=9 x 2 +6x−5 y= ( 3x+1 ) 2 −6 3x+1= y+6 ≥0 x= y+6 −1 3

Further solve the above equation.

g( y )= y+6 −1 3 f −1 ( y )= y+6 −1 3

Thus, the inverse of the function is f −1 ( y )= y+6 −1 3 .