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

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Homework Help Overview

The discussion revolves around isolating the variable x in the equation tanh x = 1. Participants explore the properties of the hyperbolic tangent function and its relationship to exponential functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants attempt to manipulate the derivative of the hyperbolic tangent function to find solutions, while others question the validity of setting tanh x to zero. There are discussions about expressing tanh in terms of exponentials and using trigonometric identities.

Discussion Status

The conversation includes various attempts to isolate x, with some participants suggesting algebraic manipulations and others exploring the implications of specific values. There is no explicit consensus on a singular approach, but multiple lines of reasoning are being explored.

Contextual Notes

Participants are navigating assumptions about the properties of hyperbolic functions and their definitions, as well as the implications of setting certain values in the context of the problem.

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[SOLVED] is there a way to isolate x?

Homework Statement


Where does the tangent of tanh x = 1?

Homework Equations


\\f(x)=\tanh x
f^\prime(x)=\\sech^2x=1-\tanh^2x

The Attempt at a Solution


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

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.
 
Last edited:
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what does cosh and sinh equal in terms of exponentials?
 
You can...

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

\Rightarrow ...
 
Last edited:
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:

$ 0 = \tanh x =\frac{e^{2x}-1}{e^{2x}+1}

Therefore, e^{2x}-1=0 which gives x=0 as the only solution. Looks good yeah? Cheers.
 
why are you setting tanhx = 0?...
 
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.
 

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