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Math Help - Help with finding stationary point (work needs checking)

  1. #1
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    Help with finding stationary point (work needs checking)

    Let f( x ) = 2x^3 - 3x^2 - 12x + 6, x E R (E= element of). Then (give the exact coordinates and the type of each stationary point such as local maximum, local minimum, and stationary point of inflexion)

    Here is my working, please notify me where i've made errors and how to fix them with correct procedure/calculation.
    f( x ) = 2x^3 - 3x^2 - 12x + 6, x E R
    f'(x) = 6x^2 - 2x - 12
    Let f'(x)= 0
    Therefore 6x^2-2x-12=0
    => factorise using quadratic formula
    a = 6 b = -2 c= -12
    Results:
    [see attached image]

    f( result 1 ) =
    f( result 2 ) =
    (inability to use latex, i shall upload my answers shortly, also i'm having difficult onwards after factorise 6x^2-2x-12=0)
    Thanks.
    Attached Thumbnails Attached Thumbnails Help with finding stationary point (work needs checking)-answers.jpg  
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by 99.95 View Post
    Let f( x ) = 2x^3 - 3x^2 - 12x + 6, x E R (E= element of). Then (give the exact coordinates and the type of each stationary point such as local maximum, local minimum, and stationary point of inflexion)

    Here is my working, please notify me where i've made errors and how to fix them with correct procedure/calculation.
    f( x ) = 2x^3 - 3x^2 - 12x + 6, x E R
    f'(x) = 6x^2 - 2x - 12
    Let f'(x)= 0
    Therefore 6x^2-2x-12=0
    => factorise using quadratic formula
    a = 6 b = -2 c= -12
    Results:
    [see attached image]

    f( result 1 ) =
    f( result 2 ) =
    (inability to use latex, i shall upload my answers shortly, also i'm having difficult onwards after factorise 6x^2-2x-12=0)
    Thanks.
     6x^2-6x-12=0

    u can't do this :

     \displaystyle \frac {d}{dx} 3x^2 = 2x ?????


    result will be :
    x_1=-1 \; \; ; x_2=2
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  3. #3
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    silly mistakes will always be my downfall and thank you very much
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