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Math Help - Turning Points

  1. #1
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    Turning Points

    Having real trouble putting these together.

    Yor help would be appreciated

    Let f(x) = x + 3bx + 3cx + d

    a) Show that y =f (x) has two distinct turning points if and only if b > c


    b) If b > c, show that the vertical distance between the turning points is 4(b - c) to the power of 3/2
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  2. #2
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    Hi.

    Quote Originally Posted by Joel View Post
    Having real trouble putting these together.

    Yor help would be appreciated

    Let f(x) = x + 3bx + 3cx + d

    a) Show that y =f (x) has two distinct turning points if and only if b > c
    First you need to find f'(x)

    f'(x) = 3x^2+3*2bx+3c

    Then you need to find the zeros of f'(x),

    so

    3x^2+3*2bx+3c = 0

    Now solve for x and you get the turning points.

    This is a quadratic equation, you should be able to find them.

    Kind regards
    Rapha
    Last edited by Rapha; June 15th 2009 at 02:25 AM.
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  3. #3
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    Quote Originally Posted by Joel View Post
    Having real trouble putting these together.

    Yor help would be appreciated

    Let f(x) = x + 3bx + 3cx + d

    a) Show that y =f (x) has two distinct turning points if and only if b > c


    b) If b > c, show that the vertical distance between the turning points is 4(b - c) to the power of 3/2

    (a) f(x)=x^3+3bx^2+3cx+d

    f'(x)=3x^2+6bx+3c

    x^2+2bx+c=0

    Find the discriminant ...

    (2b)^2-4(1)(c)

    =4b^2-4c

    =4(b^2-c)>0 as we know that b^2>c

    Hence , y=f(x) has 2 distinct turning points .
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