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Math Help - Relative Extrema

  1. #1
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    Relative Extrema

    I have to find a, b, c, and d such that the function defined by :
    f(x) = ax^3+bx^2+cx+d
    will have a relative extrema at points (1,2) and (2,3).

    From the given critical points, I am able to know that when x=1 or x=2, f'(x)= 3ax^2+2bx+c should be equal to zero.

    Therefore f'(x) should have factors like (x-1) and (x-2) or in simplified form x^2-3x+2. Now my problem is that how should I relate the two equations of f'(x) for me to be able to solve for the said unknowns?
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  2. #2
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    You the coefficients of f'(x) should have the ratio 1:-3:2. So 3a:2b:c = 1:-3:2, which gives \frac{3a}{2b} = \frac{1}{-3} and \frac{2b}{c} = \frac{-3}{2}. You get two more equations by plugging the points (1,2) and (2,3) into the function. So you get a system of 4 linear equations with 4 unknowns.
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  3. #3
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    Quote Originally Posted by deltaX View Post
    I have to find a, b, c, and d such that the function defined by :
    f(x) = ax^3+bx^2+cx+d
    will have a relative extrema at points (1,2) and (2,3).

    From the given critical points, I am able to know that when x=1 or x=2, f'(x)= 3ax^2+2bx+c should be equal to zero.

    Therefore f'(x) should have factors like (x-1) and (x-2) or in simplified form x^2-3x+2. Now my problem is that how should I relate the two equations of f'(x) for me to be able to solve for the said unknowns?
    note that f(1) = 2

    a(1^3) + b(1^2) + c(1) + d = 2

    and f(2) = 3

    a(2^3) + b(2^2) + c(2) + d = 3


    f'(1) = 0

    3a(1^2) + 2b(1) + c = 0

    f'(2) = 0

    3a(2^2) + 2b(2) + c = 0


    four equations ... four unknown coefficients.

    solve the system.
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