Question

Let f : N→N be a function defined as f(x)=`9x^2`+6x−5. Show that f : N→S, where S is the range of f, is invertible. Find the inverse of f and hence find `f^-1`(43) and` f^−1`(163).

Answer 1

Given: f : N→ N is a function defined as           f(x)=9x²+6x−5

Let y=f(x) = 9x²+6x−5

⇒ y = `9x^2`+6x-5

⇒ y =`9x^2`+6x+1-1-5

⇒ y =(`9x^2`+6x+1) - 6

⇒ y =`(3x+1)^2` - 6

⇒ y+6=`(3x+1)^2`

 

⇒`sqrt( y+6)=3x+1`                  (∵ y ∈ N)

⇒`sqrt(y+6)-1=3x`

⇒x=`(sqrt(y+6)-1)/3`   

⇒g (y) = `(sqrt(y+6)-1)/3`             [Let x = g (y)]

Now,

fog (y) = f [g (y) ]

=f `((sqrt( y+6)-1)/3)`

=9`((sqrt (y+6)-1)/3)^2` +6`((sqrt (y+6)-1)/3)`-5

=9`((y+6-2sqrt(y+6)+1)/9)`+2`(sqrt(y+6)-1)`-5

=y+6-2`sqrt (y+6)+1+2sqrt(y+6)-2-5`

=y

=`I_Y`, Identify funtion

 

gof (x) = g [f(x)]

=g `(9x^2`+6x-5)

=`(sqrt((9x^2+6x-5)+6)-1)/3`

=`(sqrt((9x^2+6x+1))-1)/3`

=`(sqrt((3x+1)^2)-1)/3`

=`((3x+1)-1)/3`

=`(3x)/3`

=x

=`I_x`, Identify function

 

Since, fog(y) and gof(x) are identify function .

Thus, f is invertible.

So,`f^-1`(x)=g (x)=`(sqrt (x+6)-1)/3`.

Now,

`f^-1`(43)=`(sqrt(43+6)-1)/3`=`(sqrt(49)-1)/3` =`(7-1)/3`=`6/3`=2 And `f^1`(163)=`(sqrt(163+6)-1)/3`=`(sqrt(169)-1)/3`= `(13-1)/3`=`12/3`=4.

 

 

 

Answer 2

We have,

f : N→ is a function defined as f (x) = 9x² + 6x-5.

Let y = f (x) = 9x² + 6x - 5

⇒ y = 9x²+ 6x − 5

⇒ y = 9x²+ 6x + 1−1−5

⇒ y = (9x²+ 6x +1) − 6

⇒ y = ( 3x+1)² − 6

⇒ y + 6 = (3x+1)²

⇒ `sqrt(y+6) = 3x +1` (∵ y ∈ N)

⇒ `sqrt (y+6) -1 = 3x`

⇒ `x =(sqrt(y + 6) - 1)/3`

⇒ `g (y) = (sqrt(y+6)-1)/3` [Let x = g (y) ]

Now,

fog (y) = f [ g (y) ]

= `f ((sqrt(y+6) -1)/3)`

= `9((sqrt(y+6) -1)/3)^2 + 6 ((sqrt(y+6) -1)/3) -5`

= `9 ((y +6 - 2 sqrt(y +6) +1 )/9) +2 (sqrt(y+6)-1)-5`

= ` y+6 -2 sqrt(y+6)+1+2 sqrt(y+6)-2 -5`

= y

= IY, Identity function

gof (x) = g [f(x)]

= `g (9x^2 + 6x -5)`

= `(sqrt((9x^2 + 6x -5)+6)-1)/3`

= `(sqrt((9x^2 + 6x +1))-1)/3`

= `(sqrt((3x +1)^2) -1)/3`

= `((3x +1) -1)/3`

= `(3x)/3`

= x

= IX, Identity function

Since, fog(y) and gof(x) are identity function.

Thus, f is invertible.

`So , f^-1(x)= g (x) = (sqrt(x+6)-1)/3`

Now,

`f^-1 (43)=(sqrt(43+6)-1)/3 = (sqrt(49)-1)/3 = (7-1)/3 = 6/3 =2 `

And `f^-1 (163) = (sqrt (163+6) -1)/3 = (sqrt(169)-1)/3 = (13- 1)/3 = 12/3 =4 `