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Math Help - What does this mean? ( Differentiating Functions/Equalities )

  1. #1
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    What does this mean? ( Differentiating Functions/Equalities )

    I have to get the null-point(s) of a curve/function. The original function goes like this:
    (i tried to use the MATH BBCode tags but I don't know how to use them so they caused errors)

    f(x) = -x⁴ / 2 + 2x

    To get the null-points (not sure if it's called that in English) I have to set the above to be equal to zero:

    -x⁴ / 2 + 2x = 0

    Next, I do 2 on the equation:

    -x⁴ + 4x = 0

    Then I say that x = u so that I can use a formula later

    -u + 4u = 0

    Now I use this formula to get u₁ and u₂
    (I think it's from Vieta)

    x + px + q = 0

    ₁x₂ = -p / 2 √((p / 2) - q)


    In my equation I don't have a q so I don't know if that conflicts with the proper use of the formula. Anyway, I went on to replace the variables in the formula with the ones from my equation (leaving out the - q part inside the square-root section):

    ₁u₂ = -2 √4

    That leaves me with these results for ₁u₂

    u₁ = 0
    u₂ = -4


    Since u is actually x we get ₁x₂ from the square-root of u₁

    x₁ = 0
    x₂ = 0


    Now my first question is if the above is correct. Is x₂ zero as well, if u₁ was zero? ( ₁x₂ = √u₁ )

    Next, I'd do the same with u₂ but since u₂ is -4 and you can't get the square-root of negative numbers (without resorting to Complex Numbers) this leaves me with no answer to x₃ and x₄

    Is this true, or what am I doing/understanding wrong?

    Please help me out quickly, I need this homework done today!
    Thanks so much!!
    Last edited by temhawk; February 25th 2009 at 09:59 AM.
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  2. #2
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    Quote Originally Posted by temhawk View Post
    I have to get the null-point(s) of a curve/function. The original function goes like this:
    (i tried to use the MATH BBCode tags but I don't know how to use them so they caused errors)

    f(x) = -x⁴ / 2 + 2x

    To get the null-points (not sure if it's called that in English) I have to set the above to be equal to zero:

    -x⁴ / 2 + 2x = 0

    Next, I do 2 on the equation:

    -x⁴ + 4x = 0

    ...
    All your calculations up to this point are correct. But with your problem there is a much easier way:

    -x^2(x^2-4)=0

    A product equals zero if one factor equals zero. Therefore you have:

    -x^2=0~\vee~x^2-4=0~\implies~\boxed{x = 0~\vee~x = -2~\vee~x = 2}
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  3. #3
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    Oooh, thank you very much!

    I totally forgot about this method. Thanks for the quick help!
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  4. #4
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    I'm stuck again.

    This time I have to get the extreme values. So I set the first derivative of x of the function to be equal to zero.

    f(x) = -x⁴ / 2 + 2x

    f'(x) = 4x / 2 + 4x = 2x + 4x

    2x + 4x = 0


    I continued the same way as before:

    2x (x + 2) = 0

    So x₁ = 0 but I can't get x₂ or x₃
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  5. #5
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    Quote Originally Posted by temhawk View Post
    I'm stuck again.

    This time I have to get the extreme values. So I set the first derivative of x of the function to be equal to zero.

    f(x) = -x⁴ / 2 + 2x

    f'(x) = - 4x / 2 + 4x = 2x + 4x <<<<<< You forgot the negative sign

    2x + 4x = 0


    I continued the same way as before:

    2x (x + 2) = 0

    So x₁ = 0 but I can't get x₂ or x₃
    You've made a tiny but very effective mistake. (See above):

    -2x^3+4x = 0~\implies~-2x(x^2-2)=0

    I leave the rest for you.
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  6. #6
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    Thanks,

    I hate when I forget about the symbols
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