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Local extrema

  1. Feb 3, 2010 #1
    how can i find the local extrema of x-coordinates in this equation?

  2. jcsd
  3. Feb 3, 2010 #2


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    Take a first derivative and set it equal to 0. Solve for x.
  4. Feb 4, 2010 #3
    yeah i have to solve for x but how wil i get rid of the derivative 4x^3??? i don't know how to remove it
  5. Feb 4, 2010 #4
    You don't need to remove it.

    Hint: There's a pretty obvious solution, find it and use polynomial division
  6. Feb 4, 2010 #5
    here's my solution i really don't know if this is correct but please correct me so

    take the derivative = 4x^3-6x+2

    Set it equal to zero and begin solving. 4x^3-6x+2 = 0
    4x^3-6x = -2
    2x(2x^2-3) = -2
    2x^2 - 3 = 0 | 2x = 0
    2x^2 = 3 | x = 0
    x^2 = 3/2
    x = plus or minus the square root of 3/2 and x = 0
  7. Feb 4, 2010 #6
    Well, you have a wrong step

    how would 2x(2x^2-3)=-2
    Suggests 2x^2-3=0 or 2x=0? If 2x^2-3=0 or 2x=0 then 2x(2x^2-3) would equal 0, certainly not -2.

    Look at 4x^3-6x+2=0.
    There's a complicated general formula for this. But I say you can at least guess one obvious solution, then your problem reduces to solving a second degree equation.
  8. Feb 4, 2010 #7
    could you tell me at least what is that formula please im running out of time i have to turn this paper in tomorrow
  9. Feb 4, 2010 #8


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    "If ab= 0 then either a= 0 or b= 0". Not ab equal to anything other than 0!

    Did you notice that [itex]4(1)^3- 6(1)+ 2= 0[/itex]? Once you know that 1 is a root, you know that x- 1 is a factor. Divide [itex]4x^2- 6x+ 2[/itex] by x- 1 to reduce to a quadratic equation for the other roots.
  10. Feb 4, 2010 #9


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    Your roots are wrong, but once you get them correctly (plug them back into the equation to make sure it checks out). Then plug them back one by one in the original equation ( x^4 - 3x + 2x) and see which one gives you the highest value - that would be your maximum.
    Last edited: Feb 4, 2010
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