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Minimum value problem

  1. Jul 17, 2007 #1
    The function f(x) is given by: a*cosh(x) + b*sinh(x), where a and b are positive constants. Prove that if (i) a > b, f(x) has a minimum value of [tex] \sqrt{a^2 - b^2}\\ [/tex] (ii) a<b, f(x) has no maximum or minimum values.
    (i) f(x) = a*cosh(x) + b*sinh(x), Therefore f'(x) = a*sinh(x) + b*cosh(x) = 0, for a minimum or a maximum. Therefore f'(x) = [tex] \frac{\exp^x(a+b) + \exp^{-x}(b-a)}{2}=0 \\ [/tex]. Therefore f'(x) = [tex] \exp^x(a + b) = - \exp^{-x}(b - a) \\ [/tex]. Therefore [tex] \exp^{2x}(a + b) = a - b \\ [/tex]. Hence [tex] 2x = \ln(\frac{a - b}{a + b}) \\ [/tex]. Hence [tex] x = \ln(\frac{a-b}{a+b})^\frac{1}{2}\\ [/tex] Substituting for x in f(x) does not give a minimum value of [tex] \sqrt{a^2 - b^2}\\ [/tex]Any help would be appreciated. I was thinking that this thread could be solved like my last thread without the use of calculus to find the minimum, if that is the case, please tell me how to get started with f(x), and I will abandon the calculus approach to this problem. Thanks.
     
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  3. Jul 17, 2007 #2

    Dick

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    Your solution is correct. Notice if a<b then there is no real root. So you have part (ii) already. Substitute your x into the original function and you will get sqrt(a^2-b^2) if you simplify carefully. You could probably do it like the trig case, but this approach doesn't seem to be going so badly.
     
  4. Jul 18, 2007 #3
    When I substitute into the original function, a*cosh(x) +b*sinh(x), I get the following[tex] \frac{a}{2}[\exp^{\ln{(\frac{a-b}{a+b}})^{\frac{1}{2}}} + \exp^{- \ln{(\frac{a-b}{a+b}})^{\frac{1}{2}}}] \\[/tex] [tex] + \frac{b}{2}[ \exp^{\ln{(\frac{a-b}{a+b}})^{\frac{1}{2}}} - \exp^{-\ln{(\frac{a-b}{a+b}})^{\frac{1}{2}}}] \\ [/tex], which seems to me to give the following [tex] \frac{a}{2}[\sqrt{\frac{a-b}{a+b}} - \sqrt{\frac{a-b}{a+b}}] + \frac{b}{2}[\sqrt{\frac{a-b}{a+b}} + \sqrt{\frac{a-b}{a+b}}]\\[/tex] Therefore we get [tex] \frac{a}{2}[0] + \frac{b}{2}[2\sqrt{\frac{a-b}{a+b}} \\[/tex]
     
    Last edited: Jul 18, 2007
  5. Jul 18, 2007 #4

    Dick

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    Noooo. The exponentials of the negative logs invert the fraction in the square root. So your first bracketted quantity should be [sqrt((a-b)/(a+b))+sqrt((a+b)/(a-b))], e.g. exp(-log(a/b))=b/a.
     
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