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Math Help - Critical Numbers

  1. #1
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    Critical Numbers

    Hello,
    I am trying to find the critical numbers for
    f'(x) = 36x^3 - 72x^2 - 72

    I got it to:
    36(x^3 - 2x^2 - 2)

    However, I am confused to as if I am supposed to factor it more?

    Or how would I solve for
    36(x^3 - 2x^2 - 2) = 0

    Thanks very much in advance.
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  2. #2
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    What you are doing will find the zeros if any.

    For the critical numbers take the derivative equal to zero and solve for x, these are the critical numbers I think you are after.
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  3. #3
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    Try using synthetic division to see if you can break the polynomial down.
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  4. #4
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    Quote Originally Posted by advancedfunctions2010 View Post
    Hello,
    I am trying to find the critical numbers for
    f'(x) = 36x^3 - 72x^2 - 72

    I got it to:
    36(x^3 - 2x^2 - 2)

    However, I am confused to as if I am supposed to factor it more?

    Or how would I solve for
    36(x^3 - 2x^2 - 2) = 0

    Thanks very much in advance.
    If you get f'(x) = 36x^3 - 72x^2 - 72 as part of finding the critical values of f(x) then you're expected to solve x^3 - 2x^2 - 2 = 0. Since there is no simple solution I assume you're expected to use technology such as a CAS: solve x^3 - 2x^2 - 2 = 0 - Wolfram|Alpha

    On the other hand, if you have to find the critical values of f'(x) = 36x^3 - 72x^2 - 72, then you need to solve f''(x) = 0.
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  5. #5
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    The orignal question is
    determine the local maximum and minimum values of f(x) = (3x^2 -6x)(3x^2+-6x)

    I then got f'(x) = 36x^3 - 72x^2 - 72

    but I am not supposed to use technology to solve.. very confused
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  6. #6
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    1: \displaystyle f'(x)=2(3x^2-6x)(6x-6)

    2: \displaystyle f'(x)=(6x-6)(3x^2+6x)+(3x^2-6x)(6x+6)=(3x^2+6x)(6x+6x+12)=(3x^2+6x)(12x+12)

    You have a +- 6 so I did both derivatives since I wasn't sure what was a typo.

    Critical points of 1: 1, 0, 2

    2: -1, 0, -2
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  7. #7
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    sorry, it was supposed to be "+6"
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  8. #8
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    Number 2 then.
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  9. #9
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    thanks so much!
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  10. #10
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    Quote Originally Posted by advancedfunctions2010 View Post
    The orignal question is
    determine the local maximum and minimum values of f(x) = (3x^2 -6x)(3x^2+-6x)

    I then got f'(x) = 36x^3 - 72x^2 - 72

    but I am not supposed to use technology to solve.. very confused
    You have a difference times a sum, which gives a difference of squares!

    (3x^2 -6x)(3x^2+-6x)=\left(3x^2\right)^2-\left(6x\right)^2=9x^4-72x^2.

    The derivative will have only two terms.

    The following has been edited.
    Should have been:

    (3x^2 -6x)(3x^2+6x)=\left(3x^2\right)^2-\left(6x\right)^2=9x^4-72x^2.
    Last edited by SammyS; December 16th 2010 at 06:50 PM. Reason: Dumb typo.
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  11. #11
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    Quote Originally Posted by SammyS View Post
    You have a difference times a sum, which gives a difference of squares!

    (3x^2 -6x)(3x^2+-6x)=\left(3x^2\right)^2-\left(6x\right)^2=9x^4-72x^2.

    The derivative will have only two terms.
    (3x^2+6x)(12x+12)=\frac{d}{dx}[9x^4-72x^2]
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