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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).
Solution 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.
Solution 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 `
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