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 the equation be ax^2 + bx^2 + cx + d.

Substituting x=-2,

-8a + 4b - 2c + d = 0

The derivative is

3ax^2 + 2bx + c

and substituting x=-1 and x=1/3 to find the extrema, where the derivative vanishes, gives

3a - 2b + c = 0
a + 2b + 3c = 0

leading to c = -a and b = a

Integrating the original cubic:

ax^4/4 + bx^3/3 + cx^2/2 + dx

and evaluating from -1 to 1 and equating to 14/3, give

b + 3d = 7 or
3d = 7 - a

so

b = a = 7 - 3d
c = 3d - 7

Going back to the original cubic evaluated at -2:

4b - 2c + d = 8a
28 - 12d - 6d + 14 + d = 56 - 24d
7d = 14
d = 2

So going back to where b, a and c were defined in terms of d:
c = -1; a = 1; b = 1

So M(x) = x^3 + x^2 - x + 2

Posted by Charlie on 2007-01-17 10:52:41