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A Vanishing Polynomial Problem (Posted on 2007-01-17) Difficulty: 2 of 5
A cubic polynomial M(x) is such that M(-2)=0, and has relative extremes at x=-1 and x=1/3.

Determine M(x) given that ∫-11M(x)dx= 14/3.

See The Solution Submitted by K Sengupta    
Rating: 1.5000 (2 votes)

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Solution Comment 2 of 2 |

Let M(x) = Ax^3 + Bx^2 + Cx + D. then
    M'(x) = 3Ax^2 + 2Bx + C
Setting M'(x) = 0 gives
       B     sqrt(B^2 -3AC)
    - --- - ---------------- = -1      (1)
       3A         3A
       B     sqrt(B^2 -3AC)     1
    - --- + ---------------- = ---     (2)
       3A         3A            3
Adding (1) and (2) gives
    B = A
Subtracting (1) from (2) gives
    A(A+C) = 0
Since M(x) is to be cubic,
    C = -A
Therefore,
    M(x) = Ax^3 + Ax^2 - Ax + D
Since M(-2) = 0, D = 2A. Thus,
    M(x) = A(x^3 + x^2 - x + 2)
           |- 1
     14    |
    --- =  |  M(x)dx
     3     |
          -| -1
           |- 1
           |
        = A| (x^3 + x^2 - x + 2)dx
           |
          -| -1
              x^4   x^3   x^2        | 1
        = A ( --- + --- - --- + 2x ) |
               4     3     2         | -1
             14
        = A ---
             3
Thus, A = 1 and
    M(x) = x^3 + x^2 - x + 2

 

  Posted by Bractals on 2007-01-17 12:52:16
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