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.(adsbygoogle = window.adsbygoogle || []).push({});

(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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# Homework Help: Minimum value problem

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