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Cosh x > 1

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data
    True or false?

    Cosh x > 1 for all x

    3. The attempt at a solution

    the answer is true, and I'm not sure why. if h is a coefficient, then I'm pretty sure I can think of some values of x and h that would be < 1.
    For instance, if h=1 and x=[tex]\pi[/tex]/3

    What does the question even mean?
     
  2. jcsd
  3. Nov 23, 2009 #2

    Mark44

    Staff: Mentor

    cosh is short for hyperbolic cosine. h is not a parameter.

    This function is defined as
    [tex]cosh(x)~=~\frac{e^{x} + e^{-x}}{2}[/tex]

    Edit: no i in exponents

    You might need to write the right side of the equation above as a Maclaurin series to show that cosh(x) >= 1 for all x.
     
    Last edited: Nov 23, 2009
  4. Nov 23, 2009 #3
    cosh x = (Exp[x] + Exp[-x])/2
     
  5. Nov 23, 2009 #4
    jakncoke, by Exp[x] do you mean ex where e is the constant that's approx. 2.718 ?
     
  6. Nov 23, 2009 #5
    yes.
     
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