inverse functions

**b_austintx** · April 27th 2009, 06:20 AM

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???

**red_dog** · April 27th 2009, 07:24 AM

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

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

**Soroban** · April 27th 2009, 07:54 AM

Hello, b_austintx!

Your algebra is off . . .

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

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

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

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

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

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

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

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

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

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

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

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

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