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Math Help - Curve Sketching

  1. #1
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    Curve Sketching

    Let f(x) = ax3 + bx2 + cx + 1. Determine the values of a,b, and c so that f(x) has a point of inflection at x = 2, a local minimum at x = -2, and f(1) = 2.

    So, f'(x) = 3ax2 + 2bx + c
    f"(x) = 6ax + 2b

    I know that f'(x) = 0 when x = -2 so: 12a - 4b + c = 0
    and f"(x) = 0 when x = 2 so: 6a + b = 0
    and a + b + c = 1 since f(1) = 2

    I'm not sure if i'm going in the right direction with this question but i'm stuck at this point. Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by steezin View Post
    Let f(x) = ax3 + bx2 + cx + 1. Determine the values of a,b, and c so that f(x) has a point of inflection at x = 2, a local minimum at x = -2, and f(1) = 2.

    So, f'(x) = 3ax2 + 2bx + c
    f"(x) = 6ax + 2b

    I know that f'(x) = 0 when x = -2 so: 12a - 4b + c = 0
    and f"(x) = 0 when x = 2 so: 6a + b = 0
    and a + b + c = 1 since f(1) = 2

    I'm not sure if i'm going in the right direction with this question but i'm stuck at this point. Any help would be appreciated.
    solve the system of equations ...

    12a-4b+c=0<br />

    6a+b=0

    a+b+c=1

    subtract the 3rd equation from the 1st ...

    11a-5b=-1

    from the 2nd equation ... b = -6a

    11a-5(-6a) = -1

    a = -\frac{1}{41}

    b = \frac{6}{41}

    c = \frac{36}{41}

    yes, I know ... some ugly coefficients.
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