Inverse Relationship

1. Feb 23, 2006

zombeast

Ok, rookie question.. but I have no one to verify it other than you folks.

Problem: Find the inverse of $$y = \frac{1-e^-x}{e^x+1}$$

The question is... can you do this:

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

I thought this was illegal because the $$e^-x$$ was tied to the 1 through the minus sign. Can the $$e^-x$$ be moved to the bottom of the denominator?

****EDIT****
I cant get tex to work right. it should be e^-x "e to the power of negative x" in the numerator.

Thanks

Last edited: Feb 23, 2006
2. Feb 23, 2006

d_leet

You can't do that.

3. Feb 23, 2006

AKG

No, you can't do that. Besides, you'd end up getting y = 1, which is clearly not equal to the original y, and which also has no inverse.

4. Feb 23, 2006

AKG

Actually, this given y doesn't have an inverse either. It does on $(-\infty,\,\ln(2)]$ and on $[\ln(2),\,\infty)$, but not on both simultaneously.

5. Feb 23, 2006

zombeast

Well ya see, this was a test question tonight on a calculus test I had and I came home racking my brain trying to figure out where I went wrong.

AKG, are you saying that this does not have an inverse unless its in the domain you specified above?

6. Feb 23, 2006

d_leet

What was the test question, because I highly doubt that it was to find the inverse of that function.

7. Feb 23, 2006

zombeast

No, this was a calc 1 exam that covered some review of old stuff as well as lmits and beginning derivatives. This was the problem.

If this was a positve x in the exponent it would be easy to solve.

8. Feb 23, 2006

zombeast

The test question was the original question.

Find the inverse of: $$y = \frac{1-e^-x}{e^x+1}$$

(again, that should be negative x exponent in the numerator)

9. Feb 23, 2006

AKG

Yeah, it has an inverse only if the domain is some subset of $(-\infty,\,\ln(2)]$ or a subset of $[\ln(2),\,\infty)$.

You want to find the inverse of the function g = R o exp, where exp is the exponential function, and R is the rational function defined by:

R(x) = (x-1)/[x(x+1)]

so you know that g-1 = exp-1 o R-1 = ln o R-1. You can find the inverse of R by rearranging the equation above so that it takes the form:

ax² + bx + c = 0

where the coefficients a, b, and c might contain some occurence of R(x), then solve for x using the quadratic formula. This gives you an expression for x in terms of R(x), from which you can figure out the inverse of R.

Last edited: Feb 23, 2006
10. Feb 23, 2006

AKG

To get the LaTeX to work right, click on the image below to see what to do:

$e^{-x}$, not $e^-x[/tex] 11. Feb 23, 2006 zombeast Thank you for the LaTex help. When you say "o R.." what does the 'o' stand for? Are you saying 'of' ? R of exp? I just want to make sure I'm on the same page as you. Thanks again. 12. Feb 23, 2006 zombeast Testing: $$y = \frac{1-e^{-x}}{e^x+1}$$ 13. Feb 23, 2006 arildno You should follow AKG's advice: Rewrite your expression as: $$(e^{x}+1)y=1-e^{-x}\to{e}^{x}(e^{x}+1})y=e^{x}-1$$ Introduce [itex]t=e^{x}$, and rewrite the quadratic equation in t to standard form and solve it, remembering that t>0

14. Feb 23, 2006

AKG

The 'o' denotes function composition. So f o g is the function defined by:

(f o g)(x) = f(g(x))

15. Feb 23, 2006

zombeast

Cool Thanks guys. I'll check this out later when I get home from work. I appreciate it.