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Math Help - Critical point question

  1. #1
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    Critical point question

    hi everyone

    having problems finding critical points for this two equation to fined the maximum,minimum or saddle points.

    a) x^4+y^4-4xy+1
    f_x=4x^3-4y, f_y=4y^3-4x

    b) f(x,y)= x^4-2x^2y+2y
    f_x=4x^3-4xy, f_y=2-2x^2

    need help to find the critical points for this two equations,really hope someone can help me. really appreciate all your help & support.
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  2. #2
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    How far have you come in your attempts? Where do critical points for a given function occur?
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  3. #3
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    thank you for replying.

    a) x^3-y=0, y^3-x=0

    x^3=y
    y^3=x
    im stuck here, unable to find the critical point
    b) 4x^3-4xy=0
    4x^3=4xy
    x^2=y

    2-2x^2=0
    2=2x^2
    x=1 & -1

    critical point, = (1,1) & (-1,1) ..is this right???

    realyl hope someone can help, appreciate all help & support.
    Last edited by anderson; March 12th 2010 at 06:03 AM.
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  4. #4
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    a: Remember that the two derivatives form a system of equations:
    x^3-y = 0
    y^3-x = 0

    imply that

    x^9=x or x(x^8-1) = 0 (see why?)

    What values for critical points does this give you?

    b: Seems right to me, so far, however, you are being a little hasty. It's quite right that
    4x^3-4xy = 0 implies that
    x^2 = y
    but it also implies that
    x(x^2-y) = 0. See what you missed?
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  5. #5
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    thank you for replying,

    a)
    (x^3)^3-x=0
    x(x^8-1)
    x= -1,0 & 1
    critical points=(-1,-1)(0,0) & (1,1)
    b)you are right, but is my final answer for critical point correct?
    critical point=(1,1) & (-1,1) .. i think it has to be correct..not sure,a bit confused though

    is this correct?

    thank you again for all help & support.
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