Thread: question on using derivative to find orginal problem

If you have a third degree polynomial such that Q(1) = 1 Q'(1)=3 Q''(1) = 6 and Q'''(1)=12 what is the original polynomial? Is there some kind of method or forumala for this? If so if you just gave me that i'm sure I could solve the problem i've tried some guess and checking but that takes forever and its not a very wise way to solve something especially if it became a test problem. Thanks.

Originally Posted by UMStudent

If you have a third degree polynomial such that Q(1) = 1 Q'(1)=3 Q''(1) = 6 and Q'''(1)=12 what is the original polynomial? Is there some kind of method or forumala for this? If so if you just gave me that i'm sure I could solve the problem i've tried some guess and checking but that takes forever and its not a very wise way to solve something especially if it became a test problem. Thanks.

This problem is a 4x4 linear sytem.

It leads to a Vandermonde Determinant (for which there is a formula for in this special case (the superfactorial)).

But then you need to use Cramer's rule several times.
Which is a slow algorithm. Thus, there is no efficient way of solving this problem

Hello, UMStudent!

If you have a third degree polynomial such that:
[1] . .Q(1) = 1
[2] . Q'(1) = 3
[3] .Q''(1) = 6
[4] Q'''(1) = 12

What is the original polynomial?

Let the cubic polynomial be:
. . . . . .Q(x) .= .ax³ + bx² + cx + d
Then: .Q'(x) .= .3ax² + 2bc + c
. . . . .Q''(x) .= .6ax + 2b
. . . . Q'''(x) .= .6a


From [4]: .6a = 12 . → . a = 2
. . .Then: .Q''(x) .= .12x + 2b

From [3]: .Q''(1) .= .12 + 2b .= .6 . → . b = -3
. . .Then: .Q'(x) .= .6x² - 6x + c

From [2]: .Q'(1) .= .6 - 6 + c .= .3 . → . c = 3
. . .Then: .Q(x) .= .2x³ - 3x² + 3x + d

From[1]: .Q(1) .= .2 - 3 + 3 + d .= .1 . → . d = -1

Therefore: .Q(x) .= .2x³ - 3x² + 3x - 1