1. ## equation 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.

2. -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

3. 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.

4. Originally Posted by brentwoodbc
I understand know so thank you.
I think it will be clear

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

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

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

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

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

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