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.
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 |
Solution
