Inverse of y = \frac{1-e^-x}{e^x+1}: Solve Here!

[zombeast]

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

Please help me out if you can.

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 can't get tex to work right. it should be e^-x "e to the power of negative x" in the numerator.

Thanks

 

[d_leet]

You can't do that.

 

[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.

 

[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.

 

[zombeast]

Well you 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?

 

[d_leet]

Well you 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?

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

 

[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.

 

[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)

 

[AKG]

Well you 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?

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.

 

[AKG]

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

e^{-x}, not e^-x[/tex]

 

[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.

 

[zombeast]

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

 

[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 t=e^{x}, and rewrite the quadratic equation in t to standard form and solve it, remembering that t>0

 

[AKG]

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

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

 

[zombeast]

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