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    f)(i) Show that the curve y = x3 - 6x2 + 11x - 6 cuts the x-axis at x = 1, 2 and 3.

    (ii) Find the x-coordinates of the two stationary points on the curve.

    (iii) Find d2y/dx2 and establish which is a maximum and which is a minimum.



    i) (x3 - 6x2 + 11x - 6) / (x-1) = (x^2 - 5x + 6)

    x^2 - 5x + 6 = (x-2)(x-3)

    I got this bit correct... but(ii)

    dy/dx = 3x^2 - 12x + 11

    x 33 || +(-12) -> none go in so use other method

    x = -b +- √b^2 - 4ac
    ------------------
    2a

    a = 3
    b = -12
    c = 11


    x = 12 +- √144 - 132
    ---------------------
    6
    x = 2+- √12
    x = 2+- 3√2
    It says its wrong
    A: 2+-√3/3 .... what did I do wrong?

    (iii) is linked to this
    I would of wrote
    2+ 3√2(minimum)
    2- 3√2(maximum)
    A:
    2-√3/3(maximum)
    2+√3/3(minimum)
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    (Original post by ckfeister)

    x = 12 +- √144 - 132
    ---------------------
    6
    x = 2+- √12
    x = 2+- 3√2
    It says its wrong
    A: 2+-√3/3 .... what did I do wrong?
    x=\frac{12\pm\sqrt{144-132}}{6}

    That's fine.

    Then, here's the simplification in detail.

    =\frac{12\pm\sqrt{12}}{6}

    =\frac{12}{6}\pm\frac{\sqrt{12}}  {6}

    =2\pm\frac{\sqrt{12}}{6}

    =2\pm\frac{\sqrt{4\times 3}}{6}

    =2\pm\frac{2\sqrt{3}}{6}

    =2\pm\frac{\sqrt{3}}{3}
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    (Original post by ckfeister)
    f)(i) Show that the curve y = x3 - 6x2 + 11x - 6 cuts the x-axis at x = 1, 2 and 3.

    (ii) Find the x-coordinates of the two stationary points on the curve.

    (iii) Find d2y/dx2 and establish which is a maximum and which is a minimum.



    i) (x3 - 6x2 + 11x - 6) / (x-1) = (x^2 - 5x + 6)

    x^2 - 5x + 6 = (x-2)(x-3)

    I got this bit correct... but(ii)

    dy/dx = 3x^2 - 12x + 11

    x 33 || +(-12) -> none go in so use other method

    x = -b +- √b^2 - 4ac
    ------------------
    2a

    a = 3
    b = -12
    c = 11


    x = 12 +- √144 - 132
    ---------------------
    6
    x = 2+- √12
    x = 2+- 3√2
    It says its wrong
    A: 2+-√3/3 .... what did I do wrong?

    (iii) is linked to this
    I would of wrote
    2+ 3√2(minimum)
    2- 3√2(maximum)
    A:
    2-√3/3(maximum)
    2+√3/3(minimum)
    For the first part, the first thing that went wrong is that  \sqrt {12} \neq 3\sqrt 2 as to have suggested. Also in your answer you should have  x=(1/6)(12\pm \sqrt{12}) but you divided the 12 by 6 but forgot to divide the  \sqrt{12} by 6.
    For the second part remember that If  d^2y/dx^2<0 at a stationary point then it is a maximum point and the opposite is true for if it is a minimum point.
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    (Original post by ghostwalker)
    x=\frac{12\pm\sqrt{144-132}}{6}

    That's fine.

    Then, here's the simplification in detail.

    =\frac{12\pm\sqrt{12}}{6}

    =\frac{12}{6}\pm\frac{\sqrt{12}}  {6}

    =2\pm\frac{\sqrt{12}}{6}

    =2\pm\frac{\sqrt{4\times 3}}{6}

    =2\pm\frac{2\sqrt{3}}{6}

    =2\pm\frac{\sqrt{3}}{3}
    How do you get the signs up you got?
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    (Original post by ckfeister)
    How do you get the signs up you got?
    LaTex - there's a link in the Useful Resources widget.
 
 
 
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