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Homework Help: Is there a way to isolate x?

  1. Mar 9, 2008 #1
    [SOLVED] is there a way to isolate x?

    1. The problem statement, all variables and given/known data
    Where does the tangent of tanh x = 1?

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

    3. 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.
     
    Last edited: Mar 9, 2008
  2. jcsd
  3. Mar 9, 2008 #2
    what does cosh and sinh equal in terms of exponentials?
     
  4. Mar 9, 2008 #3
    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] ...
     
    Last edited: Mar 9, 2008
  5. Mar 9, 2008 #4
    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).
     
  6. Mar 9, 2008 #5
    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.
     
  7. Mar 10, 2008 #6
    why are you setting tanhx = 0?...
     
  8. Mar 10, 2008 #7
    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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