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Math Help - Find MINs, MAXs, POI and concavity...HELP!

  1. #1
    Newbie maksim378's Avatar
    Joined
    Nov 2007
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    Exclamation Find MINs, MAXs, POI and concavity...HELP!

    I already asked for help in another Thread I posted. Please help me out.

    I'm having problems with solving for X inhere...

    y'=
    - 2cos(x/(x-2))

    (x-2)^2


    now I have to set it equal to zero and solve for X.

    0=- 2cos(x/(x-2))
    (x-2)^2

    what are the possible values for X? by the way, my domain is from 0 to 2pi.
    Quote Originally Posted by maksim378 View Post
    After I did the 1st derivative test, I had to find MINs, MAXs, POI as I said before.
    to find the critical value, I have to find the answer for X in the derivative (that was my last question).
    let's say that the answer for X was 2, lets imagine.
    then we'd have to do the line test where I'd put the point (2) on the line and choose any other points around:

    1 2 3

    then I'd plug those other 2 numbers (1 and 3) into the derivative. the answer would give me an idea of how the graph looks like. when I plug in 1 into the derivate, let's sa taht the answer is positive and for 3, it's positive also
    so those signs help me out on the number line
    +++ +++
    1 2 3

    if the line was -----+++++ then it's a MIN, if it was ++++----, then it's a MAX, if the line is --- --- or +++ +++ then that's a POI.
    then I'd plug the answer for X that we've found in the derivative (2) into the original equation and that woul give me the Y coordinate. so now I have the X which is 2 and the Y coordinate. now I now the coordinates of a point which happened to be our Poin Of Inflection.



    that's what I'd do with an easy problem. in the one I have, we had to put N in order to solve for X. N is any number, then the answer for X will be different everytime... then I won't be able to find MIN or MAX or POI coordinates... or is there a way to do that?


    P.S. btw, the answer for X in my real derivative -2cos(x/(x-2))/(x-2)^2 can't be 0, I figure that out. I hope Im at least right in this case.
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
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    So you have
    0 = -2cos[x / (x-2)] / (x-2)^2

    Continue simplifying that.
    0 = -2cos[x / (x-2)]
    0 = cos[x / (x-2)]
    angle [x / (x-2)] = arccos(0) = pi/2 or 3pi/2 ---***

    When [x /(x-2)] = pi/2,
    Cross multiply,
    2x = (x-2)pi
    2x = pi(x) -2pi
    x(2 -pi) = -2pi
    x = 2pi / (pi-2) = 5.50387679 radians -------answer.

    When [x /(x-2)] = 3pi/2,
    Cross multiply,
    2x = (x-2)3pi
    2x = 3pi(x) -6pi
    x(2 -3pi) = -6pi
    x = 6pi / (3pi-2) = 2.5387366 radians -------answer also.
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