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Math Help - quick Calc Equation question

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    quick Calc Equation question

    Hello,
    I am stuck with this one question, I do not even know how to begin. Any help would be greatly appreciated.

    Determine a, b, c and d so that the graph of ax^3+bx^2+cx+d has a point of inflection at the origin and a relative maximum at the point (2,4).

    I know the point of inflection is when the second derivative equals 0 and the max is when the first derivative equals 0, going from negative y values to positive y values. but how do i use that?

    Thanks for all your help!
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  2. #2
    Member TheMasterMind's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    Hello,
    I am stuck with this one question, I do not even know how to begin. Any help would be greatly appreciated.

    Determine a, b, c and d so that the graph of ax^3+bx^2+cx+d has a point of inflection at the origin and a relative maximum at the point (2,4).

    I know the point of inflection is when the second derivative equals 0 and the max is when the first derivative equals 0, going from negative y values to positive y values. but how do i use that?

    Thanks for all your help!


    Let f(x) = ax3 + bx2 + cx + d
    f'(x) = 3ax2 + 2bx + c
    f''(x) = 6ax + 2b

    f(2) = 8a + 4b + 2c + d = 8a + 2c = 4
    f'(2) = 12a + 4b + c = 12a + c = 0

    Solve

    8a + 2c - 2(12a + c) = 4 - 0
    -16a = 4
    a = -1/4

    12(-1/4) + c = 0
    c = 3

    Therefore, f(x) = -1/4 x3 + 3x
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  3. #3
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    This question was answered yesterday here: http://www.mathhelpforum.com/math-he...213-b-c-d.html
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