1. inverse functions

Ok need to find the inverse and i need to show that
f ^(-1) o f= x= f o f^(-1)

p(x) =

x
-----
x - 1

i did

y=
x
---
y -1 = (y-1) x = y then xy -1x = y -xy
then

-1x = y(-x)
----- -------
-x -x

which means x = y ?

then i guess i need to find the inverse of that? i dont get it?
the inverse of x =y is y=x ?? i'm lost???

2. $\displaystyle y=\frac{x}{x-1}\Rightarrow yx-y=x\Rightarrow x=\frac{y}{y-1}$

Then $\displaystyle f^{-1}(y)=\frac{y}{y-1}$

3. Hello, b_austintx!

Your algebra is off . . .

Find the inverse and verify that: .$\displaystyle f\circ f^{-1} \,=\,x$

. . $\displaystyle p(x) \:=\:\frac{x}{x-1}$

We have: .$\displaystyle y \:=\:\frac{x}{x-1}$

Switch variables: .$\displaystyle x \:=\:\frac{y}{y-1}$

Solve for $\displaystyle y\!:\;\;x(y-1) \:=\:y \quad\Rightarrow\quad xy - x \:=\:y \quad\Rightarrow\quad xy-y \:=\:x$

. . Factor: .$\displaystyle y(x-1) \:=\:x \quad\Rightarrow\quad y \:=\: \frac{x}{x-1}$

Therefore: .$\displaystyle p^{-1}(x) \:=\:\frac{x}{x-1}$

. .
Interesting . . . p(x) is its own inverse!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We will verify that: .$\displaystyle p\circ p^{-1} \:=\:x$

$\displaystyle p\circ p^{-1} \;=\;p\left(p^{-1}\right) \;= \;p\left(\frac{x}{x-1}\right) \;=\; \frac{\dfrac{x}{x-1}}{\dfrac{x}{x-1} - 1}$

Multiply by $\displaystyle \frac{x-1}{x-1}\!:\quad \frac{(x-1)\left(\dfrac{x}{x-1}\right)} {(x-1)\left(\dfrac{x}{x-1} - 1\right)} \;=\;\frac{x}{x - (x-1)} \;=\; \frac{x}{1} \;=\;x$
. . Done!