Results 1 to 3 of 3

Math Help - question on using derivative to find orginal problem

  1. #1
    Junior Member
    Joined
    Mar 2007
    Posts
    37

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by UMStudent View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,811
    Thanks
    701
    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) .= .ax + bx + cx + d
    Then: .Q'(x) .= .3ax + 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) .= .6x - 6x + c

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

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

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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 21st 2009, 12:25 PM
  2. Find the derivative of...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 4th 2009, 10:52 AM
  3. Replies: 1
    Last Post: October 19th 2008, 10:29 AM
  4. Question on Derivative Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 12th 2008, 07:16 PM
  5. ??? cant find the derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 13th 2008, 01:21 PM

Search Tags


/mathhelpforum @mathhelpforum