# Help analyzing inverse of an abs function

• Dec 6th 2009, 09:55 AM
mikedwd
Help analyzing inverse of an abs function
I am trying to find an expression for $\displaystyle f^{-1}$

$\displaystyle f(x)=ln\lvert\frac{x}{x-1}\rvert$

And then the algebra

$\displaystyle x=ln\lvert\frac{y}{y-1}\rvert$

$\displaystyle e^{x}=\lvert\frac{y}{y-1}\rvert$

First solution:
$\displaystyle e^{x}=\frac{y}{y-1}$
$\displaystyle ye^{x}-e^{x}=y$
$\displaystyle ye^{x}-y=e^{x}$
$\displaystyle y(e^{x}-1)=e^{x}$
$\displaystyle y=\frac{e^{x}}{e^{x}-1}$

Second Solution:
$\displaystyle -e^{x}=\frac{y}{y-1}$
$\displaystyle -ye^{x}+e^{x}=y$
$\displaystyle e^{x}=y+e^{x}y$
$\displaystyle e^{x}=(1+e^{x})y$
$\displaystyle y=\frac{e^x}{e^{x}+1}$

I think my calculator is telling me that the first solution I did is true for $\displaystyle y<0$ and $\displaystyle y\ge1$ and the second solution is true for all reals.

If I am understanding my calculator correctly, how do I analytically determine that the second solution is the inverse of f for all real values of x? And if this is incorrect, how do I analyze the functions I found for $\displaystyle f^{-1}$?

Thanks a lot if you can help!
• Dec 6th 2009, 12:16 PM
Amer
Quote:

Originally Posted by mikedwd
I am trying to find an expression for $\displaystyle f^{-1}$

$\displaystyle f(x)=ln\lvert\frac{x}{x-1}\rvert$

And then the algebra

$\displaystyle x=ln\lvert\frac{y}{y-1}\rvert$

$\displaystyle e^{x}=\lvert\frac{y}{y-1}\rvert$

First solution:
$\displaystyle e^{x}=\frac{y}{y-1}$
$\displaystyle ye^{x}-e^{x}=y$
$\displaystyle ye^{x}-y=e^{x}$
$\displaystyle y(e^{x}-1)=e^{x}$
$\displaystyle y=\frac{e^{x}}{e^{x}-1}$

Second Solution:
$\displaystyle -e^{x}=\frac{y}{y-1}$
$\displaystyle -ye^{x}+e^{x}=y$
$\displaystyle e^{x}=y+e^{x}y$
$\displaystyle e^{x}=(1+e^{x})y$
$\displaystyle y=\frac{e^x}{e^{x}+1}$

I think my calculator is telling me that the first solution I did is true for $\displaystyle y<0$ and $\displaystyle y\ge1$ and the second solution is true for all reals.

If I am understanding my calculator correctly, how do I analytically determine that the second solution is the inverse of f for all real values of x? And if this is incorrect, how do I analyze the functions I found for $\displaystyle f^{-1}$?

Thanks a lot if you can help!

we can divide the function into two functions if x<0 and x>1

$\displaystyle f(x) = ln \frac{x}{x-1}$

$\displaystyle e^x = \frac{y}{y-1} \Rightarrow y = \frac{e^x}{e^x-1}$ this for y<0 and y>1 this is the range, the domain here is all real numbers

if 0<x<1

$\displaystyle f(x) = ln \frac{-x}{x-1}$

$\displaystyle e^x = \frac{-y}{y-1} \Rightarrow y= \frac{-e^x}{e^x-1}$

this when 0<y<1 this is the range, that means x will take the values when 1-e^x is larger than e^x

$\displaystyle \frac{e^x}{1-e^x}$

$\displaystyle 0<2e^x <1 \Rightarrow 0<e^x < \frac{1}{2}$

$\displaystyle \ln 0 < x < \ln \frac{1}{2} \Rightarrow -\infty < x< \ln 1/2$

0,1 the function is not defined so these points are not included in the domain of f(x)( the range of $\displaystyle f^{-1}$ )
• Dec 6th 2009, 01:10 PM
mikedwd
Quote:

Originally Posted by Amer
we can divide the function into two functions if x<0 and x>1

$\displaystyle f(x) = ln \frac{x}{x-1}$

$\displaystyle e^x = \frac{y}{y-1} \Rightarrow y = \frac{e^x}{e^x-1}$ this for y<0 and y>1 this is the range, the domain here is all real numbers

if 0<x<1

$\displaystyle f(x) = ln \frac{-x}{x-1}$

$\displaystyle e^x = \frac{-y}{y-1} \Rightarrow y= \frac{-e^x}{e^x-1}$

this when 0<y<1 this is the range, that means x will take the values when 1-e^x is larger than e^x

$\displaystyle \frac{e^x}{1-e^x}$

$\displaystyle 0<2e^x <1 \Rightarrow 0<e^x < \frac{1}{2}$

$\displaystyle \ln 0 < x < \ln \frac{1}{2} \Rightarrow -\infty < x< \ln 1/2$

0,1 the function is not defined so these points are not included in the domain of f(x)( the range of $\displaystyle f^{-1}$ )

Yes, except that $\displaystyle e^{x}=\frac{-y}{y-1} \Rightarrow y=\frac{e^{x}}{e^{x}+1}$

So then $\displaystyle e^{x}+1>e^{x} \Rightarrow 1>0$ ?? So does this then mean that the inverse is $\displaystyle y=\frac{e^{x}}{e^{x}+1}$ for all values (because this cannot equal 0 or 1)?

But the output of that is never negative, while the original function can certainly have negative input...I'm still working on this...all help is appreciated!

When I tried that inverse out, indeed it only worked for y values between 0 and 1 (I just tested to see if it would undo the original function).

Hmm...

I wrote a piecewise:

denominator e^x-1 for x<0 and x>1
denominator e^x+1 for 0<x<1

Is that correct?
• Dec 6th 2009, 07:48 PM
Amer
the problem is

we can write the function f(x) like this

$\displaystyle ln \frac{x}{x-1} \;\;\; \; 1<x , x<0$ the range of this peace is all real number

$\displaystyle ln \frac{-x}{x-1} \;\;\;\; 0<x<1$ the range is all real numbers

note that:-f(x) is not one-one function

and the inverses

$\displaystyle y =\frac{e^x}{e^x -1 }$ the inverse of the first function the domain is all real numbers the range is 1<y, y<0

$\displaystyle y= \frac{e^x }{1+e^x}$ the inverse of the second one the domain is all real numbers the range 0<y<1

there exist a real number have a two different images and this contradict with function conditions

as you know the inverse function take the domain of the original function as a range and take the range as a domain if the function we have is one-one

see this is the image of f(x)

Attachment 14305
• Dec 7th 2009, 12:48 PM
mikedwd
Thanks a lot for your help, Amer.

I understand what you're saying and I really appreciate it!

-Mike