equation of inverse of $\displaystyle y= 1/(x+2) -1$

i know you swap the x and y and then solve for y but I get stuck.

Printable View

- Feb 11th 2009, 09:19 PMbrentwoodbcequation of inverse
equation of inverse of $\displaystyle y= 1/(x+2) -1$

i know you swap the x and y and then solve for y but I get stuck. - Feb 11th 2009, 09:38 PMADARSH
-Swap x and y

so

$\displaystyle

x=1/(y+2)-1

$

Next add one to both sides

so

$\displaystyle x+1=\frac{1}{y+2}$

Now Multiply both sides with

$\displaystyle \frac{y+2}{x+1}$

$\displaystyle

y+2 = \frac{1}{x+1}

$

Deduct 2 from both sides

$\displaystyle y= \frac{1}{x+1} - 2$

the thing on RHS is your inverse:) - Feb 11th 2009, 09:43 PMbrentwoodbc
thank you! I got to the point of multiplying and I just multiplied by the recipocal so I got something like xy = -y +2x -1 If that makes any sense (obv.wrong.)

I understand know so thank you. - Feb 11th 2009, 09:50 PMADARSH
I think it will be clear

$\displaystyle (x+1)=\frac{1}{y+2}$

$\displaystyle (x+1)\frac{(y+2)}{(x+1)}=\frac{(y+2)}{(y+2)(x+1)}$

$\displaystyle (y+2)=\frac{1}{(x+1)}$

$\displaystyle (y)=\frac{1}{(x+1)}-2$

$\displaystyle (y)=\frac{-2x-1}{(x+1)}$

$\displaystyle f^{-1}(x)=\frac{-2x-1}{(x+1)}$