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anonymous

  • 5 years ago

Absolute and local extrema: Determine where the following functions have absolute and local extrema, if anywhere. f(x)=1/(xsinx), defined on the interval (o,pi)

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  1. anonymous
    • 5 years ago
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    Extrema are known by a different name, minima/maxima. Do you know how to find the minimum/maximum of an equation?

  2. anonymous
    • 5 years ago
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    We take the derivative of the original equation and equate it to zero right?

  3. anonymous
    • 5 years ago
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    Correct.

  4. anonymous
    • 5 years ago
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    i got x=0, x=-tanx

  5. anonymous
    • 5 years ago
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    and then i don't know how to continue:(

  6. anonymous
    • 5 years ago
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    One second. x = -tan x doesn't look right to me, but I'm doing the derivative now.

  7. anonymous
    • 5 years ago
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    What was the derivative you arrived at?

  8. anonymous
    • 5 years ago
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    f'(x)= [-(xsinx)^-2] (sinx+xcosx)

  9. anonymous
    • 5 years ago
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    Alright, so to evaluate this poses an interesting problem. If x=0, both the numerator and the denominator go to 0. L'hopital's rule may help us with this issue. Alternatively, is there any way (other than x=0) to get xcosx - sinx to equal zero?

  10. anonymous
    • 5 years ago
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    ummm, i don't know, i'm confused too.

  11. anonymous
    • 5 years ago
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    From examination, this function does not appear to have any extrema. L'hopitals rule doesn't help eliminate the 0/0 problem, and xcosx - sinx doesnt equal 0 at any other point in (0, pi).

  12. anonymous
    • 5 years ago
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    so does the function have no absoulte nor local extrema?

  13. anonymous
    • 5 years ago
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    None that are so easily found, at least. If this is for calc 1 level, I would answer that it does not.

  14. anonymous
    • 5 years ago
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    i was think about that too, thank you so much! :)

  15. anonymous
    • 5 years ago
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    thinking*

  16. anonymous
    • 5 years ago
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    My pleasure. Anything else I can help you with today?

  17. anonymous
    • 5 years ago
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    yes please! if you are willing to help :D Absolute Maximum: The word "threat" was originally pronounced "threet". In linguistics, this shift in pronunciation is called a language change. It is theorized that any given language change spreads through a population according to a logistic differential equation such as the one described below. Consider a population of 1000. Let y(t) denote the number of people who have, as of time t, adopted the lanyage change. Then y'(t) = ky(t) [1- (y(t))/1000] where k is a positive constant. What is y(t) when y'(t) is maximal?

  18. anonymous
    • 5 years ago
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    i equated the function to zero, and i got y(t)=0, y(t)=1000

  19. anonymous
    • 5 years ago
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    So your equation is \[y'(t) = k y(t) [1 - \frac{y(t)}{1000}]\] correct?

  20. anonymous
    • 5 years ago
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    yes :)

  21. anonymous
    • 5 years ago
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    So, I'm solving it now, but what I'm doing is taking the derivative of both sides, and then subbing in the given value of y'(t) when you see it. Then I will set that equal to zero.

  22. anonymous
    • 5 years ago
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    By following your instructions, i still got y(t)=0 and y(t)=1000. Is that correct?

  23. anonymous
    • 5 years ago
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    That isn't what I'm getting. So, after taking the derivative and substituting in, I end up with: \[y''(t) = k ^{2}y(t) - \frac{k ^{2}[y(t)]^{2}}{1000} - \frac{2k ^{2}[y(t)]^{2} - \frac{2k ^{2}[y(t)]^{3}}{1000}}{1000} = 0\] Simplifying further, I end up at: \[[y(t)]^{2} - 1500 y(t) + 500000 = 0\] Could you solve from here?

  24. anonymous
    • 5 years ago
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    can i do y(t) [y(t) -1500] = -500000 ? or do i need to do the quadratic formula?

  25. anonymous
    • 5 years ago
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    You can factor it pretty easily, actually. And since we don't care what the expression for y(t) is, just what it equates to, you can treat it as a variable when doing so.

  26. anonymous
    • 5 years ago
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    [y(t) - 500][y(t) - 1000] = 0

  27. anonymous
    • 5 years ago
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    oh yeah i just got help by using the calculator:P Thanks for your help! your really a lifesaver!

  28. anonymous
    • 5 years ago
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    My pleasure :) Just keep in mind that calculus is easy. Algebra is the "hard" part of math. Take your time doing the algebraic portions, and you'll be just fine.

  29. anonymous
    • 5 years ago
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    our prof just simply don't teach anything and we need to read the textbook and do our assignment by ourselves. Thats really hard!

  30. anonymous
    • 5 years ago
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    Indeed. As a graduate student, I feel your pain. You'll have a lot of bad professors, those that either don't care to teach, or just don't know how. Finding places like this can help. More importantly, check around for alternative texts (the university library almost always has a few for any subject) to help make things a bit more clear for you. Of course, for math, there is Paul's Online Math Notes :)

  31. anonymous
    • 5 years ago
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    Anyway, good luck!

  32. anonymous
    • 5 years ago
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    Thank you so much! :D bye!

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