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Hyperbolic Graph Problem

  1. May 30, 2010 #1
    please find attached the problem and the short and sweet Answer.
    I can't understand the last step of the answer.

    Attached Files:

  2. jcsd
  3. May 31, 2010 #2


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    Given that (k, -1) is on the graph of y= cosh(x)- 3sinh(x), show that
    [tex]e^{2k}- e^k- 2= 0[/tex]

    First the part you say you understand, but I'll write it out so others can follow:

    By definition
    [tex]cosh(x)= \frac{e^x+ e^{-x}}{2}[/itex]
    [tex]sinh(x)= \frac{e^x- e^{-x}}{2}[/itex]

    [tex]cosh(x)- 3sinh(x)= \frac{e^x+ e^{-x}}{2}- \frac{3e^x- 3e^{-x}}{2}[/tex]
    [tex]= \frac{-2e^x+ 4e^{-x}}{2}[/tex]

    and the fact that (k, -1) is on the graph means that
    [tex]cosh(k)- 3sinh(k)= \frac{-2e^k+ 4e^{-k}}{2}= -1[/tex]

    Multiplying through by 2 gives
    [tex]-2e^k+ 4e^{-k}= -2[/tex]

    Now, for the step you say you don't understand: Multiply through by [itex]e^k[/itex] to get:
    [tex]-2e^{2k}+ 4= -2e^{k}[/tex]
    amd divide by -2 to get
    [tex]e^{2k}- 2= e^{k}[/itex]

    Finally, subtract [itex]e^k[/itex] from both sides:
    [tex]e^{2k}- e^k- 2= 0[/tex]
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