Maximization: find max value of (x^4-x^2)/(x^6+2x^3-1) for x > 1

daniellionyang said:
Determine the maximum value attained by (x^4-x^2)/(x^6+2x^3-1) all x in (1,infinity).
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You posted this to "Algebra" instead of to "Calculus", so I'll guess that you're supposed to use some sort of algebraic method. But I can't guess which, so please reply with that information. (It'll be whatever method they gave you in class, in the section of the book that generated this exercise.)

When you reply, please include a clear listing of your efforts so far. Thank you! ;)
 
stapel said:
You posted this to "Algebra" instead of to "Calculus", so I'll guess that you're supposed to use some sort of algebraic method. But I can't guess which, so please reply with that information. (It'll be whatever method they gave you in class, in the section of the book that generated this exercise.)

When you reply, please include a clear listing of your efforts so far. Thank you! ;)
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I think you can't solve it with calculus, since calculators aren't allowed. I tried factorization, but nothing seems to work! IDK what to do other than factor.
 
From your response I can conclude that you haven't yet studied calculus. But without that knowledge, I know of no way to get an exact value. The question for you then becomes what Stapel asked you earlier: What methods have you used in your class on similar problems? One method you might have used is getting an approximate answer by graphing the function. When I do that, I see that the curve reaches its peak somewhere around 1.5 or 1.6. So, plug in various values around there and see if you can get a feel for where the maximum might be. When doing approximations of this sort, I like to stop at 3 decimal places, but your book and/or teacher might tell you otherwise.
 
I actually have studied calculus. But again, calculators aren't allowed. I will take a calculus solution if found. However, I'm pretty sure this is not possible without calculators. Also, my teacher said the value can be found exactly. The answer is specified to be of the form (a+sqrt{b})/c
 
Is it helpful that the supernal is infinity and the infinitum is -infinity? I think you might be able to bound the equation and use the extreme principle. Or maybe you need RMS-AM-GM-HM or cauchy-schwartz inequalities.
 
daniellionyang said:
I actually have studied calculus. But again, calculators aren't allowed. I will take a calculus solution if found. However, I'm pretty sure this is not possible without calculators. Also, my teacher said the value can be found exactly. The answer is specified to be of the form (a+sqrt{b})/c
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Use of CALCULUS, generally does not require calculators.

At which value of x(>1) we get x^6 +2x^3 -1 to be minimum.

As Ishuda indicated, the above transforms into a quadratic equation (parabola) and it has a minimum (vertex of the parabola). Where is it?
 
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Well, since you do know calculus, why not use the standard method of finding maxima and minima? You know that maxima and minima are found when the first derivative is 0. So, what is the derivative of the given function? You can definitely work that out by hand, no calculator required. Then when is that derivative 0? As a hint, the derivative will be a fraction, so you'll probably only need to consider the numerator (unless the denominator is also 0, in which case the value would be undefined).
 
ksdhart said:
why not use the standard method of finding maxima

when is that derivative 0? As a hint, the derivative will be a fraction, so you'll probably only need to consider the numerator
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That numerator is a 9th-degree polynomial. Can you factor it by hand? ;)
 
Subhotosh Khan said:
Use of CALCULUS, generally does not require calculators.

At which value of x(>1) we get x^6 +2x^3 -1 to be minimum.

As Ishuda indicated, the above transforms into a quadratic equation (parabola) and it has a minimum (vertex of the parabola). Where is it?
Click to expand...

What would finding the vertex of the denominator do? Also, I'm saying it requires a calc since by the quotient rule, you have (f'g-g'f)/g^2, which leads to an 8th degree on the top, unsolvable by hand.
 
Yes. I already figured it out, and typing it into the calculator proves this is right. The question is why... I think the solution involves a clever factorization.​
 
Bob Brown MSEE said:
p(phi)=1/6
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Golden Ratio defined several ways
\(\displaystyle \frac{a\, +\, b}{a}\, = \, \frac{a}{b}\); a & b positive
or
\(\displaystyle \phi\, =\, \frac{\sqrt{5}\, -\, 1}{2}\) ~ 0.618033988749895
or ...

Actually the max appears to be at 1 + \(\displaystyle \phi\)
 
Ref

Ishuda said:
Golden Ratio defined several ways
\(\displaystyle \frac{a\, +\, b}{a}\, = \, \frac{a}{b}\); a & b positive
or
\(\displaystyle \phi\, =\, \frac{\sqrt{5}\, -\, 1}{2}\) ~ 0.618033988749895
or ...

Actually the max appears to be at 1 + \(\displaystyle \phi\)
Click to expand...
google
 
I understand what the golden ratio is :). No need of that. But I think I solved it! Factoring out x^3 in the top and bottom and simplifying under the constraints gives x/(x^2+3x+2)! Then applying weighted AMGM gives that this is greater than 1/(6*6th root(x^3*x*x*x*1*1))=1/6! Thanks people! (BTW, the source of this problem was HMMT)
 
Ishuda said:
\(\displaystyle \phi\, =\, \frac{\sqrt{5}\, -\, 1}{2}\) ~ 0.618033988749895
Click to expand...

That value is actually the reciprocal of the Golden Ratio.

This WolframAlpha page provides an explanation for setting up an equation to solve for phi; scroll down to the yellow (similar) rectangles.

:cool:
 
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