Proving Minimum Value of f(x) = a*cosh(x) + b*sinh(x)

In summary, the function f(x) is given by a*cosh(x) + b*sinh(x), where a and b are positive constants. If a > b, f(x) has a minimum value of sqrt(a^2 - b^2). If a < b, f(x) has no maximum or minimum values. The minimum value can be found by finding the root of the derivative of f(x) and substituting the value back into the original function. Alternatively, the minimum value can be found by simplifying the substitution of the root into the original function.
  • #1
John O' Meara
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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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  • #2
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.
 
  • #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]
 
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  • #4
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.
 

FAQ: Proving Minimum Value of f(x) = a*cosh(x) + b*sinh(x)

What is a minimum value problem?

A minimum value problem is a mathematical problem that involves finding the smallest possible value of a function, typically within a given range of inputs or constraints.

How is a minimum value problem solved?

A minimum value problem is solved using a variety of techniques, including calculus, optimization algorithms, and mathematical modeling. The specific method used depends on the complexity of the problem and the available tools.

What is the importance of minimum value problems?

Minimum value problems are important in many areas of science, engineering, and economics. They allow us to optimize processes and find the most efficient solutions to problems, leading to cost savings, improved performance, and better decision-making.

Can minimum value problems have multiple solutions?

Yes, depending on the constraints and complexity of the problem, there can be multiple solutions that satisfy the minimum value criteria. In some cases, there may even be an infinite number of solutions.

What are some real-world applications of minimum value problems?

Minimum value problems are used in a wide range of fields, including finance, physics, chemistry, and biology. Some examples include finding the shortest path for transportation, minimizing energy consumption in engineering systems, and optimizing drug dosages in medicine.

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