Is there a way to isolate x for tanh x = 1?

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[SOLVED] is there a way to isolate x?

Homework Statement


Where does the tangent of tanh x = 1?

Homework Equations


[tex]\\f(x)=\tanh x[/tex]
[tex]f^\prime(x)=\\sech^2x=1-\tanh^2x[/tex]

The Attempt at a Solution


[tex]\\f^\prime(x)=1-\tanh^2x\rightarrow1-\tanh^2x=1\rightarrow\sqrt{\tanh^2x}=\sqrt{0}\rightarrow \tanh x=0[/tex]

Since I have shown that the tangent = 1 when tanh x = 0, I thought it may be sufficient to simply add another line saying x = 0 since we know (based on the definition of tanh) that if tanh x = 0 then x = 0. However, I am wondering if there is a way to approach this to actually isolate x without using this assumption? Does that make sense? Thanks for reading.
 
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what does cosh and sinh equal in terms of exponentials?
 
You can...

as you said, [tex]\\sech^2x = 1 - \\tanh^2x[/tex]
so for [tex]\\sech^2x = 1[/tex] then
[tex]1- \frac{e^x-e^{-x}}{e^x+e^{-x}} = 1[/tex]

[tex]\Rightarrow[/tex] ...
 
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You know that tan(alpha) = 1 when alpha = pi/4. So now just write tanhx in terms of exponentials and solve using simple algebra (tanhx = pi/4).
 
Alright, I think I've got it. Continuing from where I left off:

[tex]$ 0 = \tanh x =\frac{e^{2x}-1}{e^{2x}+1}[/tex]

Therefore, [tex]e^{2x}-1=0[/tex] which gives [tex]x=0[/tex] as the only solution. Looks good yeah? Cheers.
 
So I could, in turn, use a trig identity to show that x = 0 when d/dx tanh x = 1. It should be correct if you read from the beginning. Would you happen to know how to mark posts as [SOLVED] do you? Cheers.