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.
-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
I think it will be clear
$\displaystyle (x+1)=\frac{1}{y+2}$
$\displaystyle (x+1)\frac{(y+2)}{(x+1)}=\frac{(y+2)}{(y+2)(x+1)}$
$\displaystyle (y+2)=\frac{1}{(x+1)}$
$\displaystyle (y)=\frac{1}{(x+1)}-2$
$\displaystyle (y)=\frac{-2x-1}{(x+1)}$
$\displaystyle f^{-1}(x)=\frac{-2x-1}{(x+1)}$