- #1
zombeast
- 9
- 0
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 [tex]y = \frac{1-e^-x}{e^x+1}[/tex]
The question is... can you do this:
[tex]y = \frac{1-e^-x}{e^x+1} = y = \frac{1}{e^x + 1 - e^x} = \frac{1}{1} = 1 [/tex]
I thought this was illegal because the [tex]e^-x[/tex] was tied to the 1 through the minus sign. Can the [tex]e^-x[/tex] 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
Please help me out if you can.
Problem: Find the inverse of [tex]y = \frac{1-e^-x}{e^x+1}[/tex]
The question is... can you do this:
[tex]y = \frac{1-e^-x}{e^x+1} = y = \frac{1}{e^x + 1 - e^x} = \frac{1}{1} = 1 [/tex]
I thought this was illegal because the [tex]e^-x[/tex] was tied to the 1 through the minus sign. Can the [tex]e^-x[/tex] 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
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