equation of inverse

**brentwoodbc** · February 11th 2009, 09:19 PM

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

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

**ADARSH** · February 11th 2009, 09:38 PM

-Swap x and y
so
$<br /> x=1/(y+2)-1<br />$
Next add one to both sides

so

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

Now Multiply both sides with

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

$<br /> y+2 = \frac{1}{x+1}<br />$

Deduct 2 from both sides

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

the thing on RHS is your inverse

**brentwoodbc** · February 11th 2009, 09:43 PM

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.

**ADARSH** · February 11th 2009, 09:50 PM

Originally Posted by brentwoodbc

I understand know so thank you.

I think it will be clear

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

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

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

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

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

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

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