Results 1 to 3 of 3

Math Help - Antiderivative problem.

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    37

    Antiderivative problem.

    Find f if f''(x) = 2 + cos(x), f(0) = -1 and f(pi/2) = 0

    After going through the steps, I found f(x) = x^2 - cos(x) - (pi/2)x

    This follows through correctly if you differentiate it twice, and it satisfies both given substitutions in the question, but my answer key says:

    f(x) = -cos(x) + x^2 - x + (pi/2)(1 - pi/2)

    Did I slip up somewhere?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Tulki View Post
    Find f if f''(x) = 2 + cos(x), f(0) = -1 and f(pi/2) = 0

    After going through the steps, I found f(x) = x^2 - cos(x) - (pi/2)x

    This follows through correctly if you differentiate it twice, and it satisfies both given substitutions in the question, but my answer key says:

    f(x) = -cos(x) + x^2 - x + (pi/2)(1 - pi/2)

    Did I slip up somewhere?
    Does the answer key satisfy f(0) = -1?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    37
    So it seems it does not... ha! Thanks for the heads up, maybe I did get it right after all. :P
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. antiderivative car problem
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 27th 2010, 10:38 PM
  2. Antiderivative Problem
    Posted in the Calculus Forum
    Replies: 7
    Last Post: November 10th 2009, 07:29 PM
  3. antiderivative problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 23rd 2008, 11:41 AM
  4. Antiderivative problem
    Posted in the Calculus Forum
    Replies: 10
    Last Post: May 31st 2008, 09:42 PM
  5. Antiderivative Problem?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 5th 2007, 02:22 PM

Search Tags


/mathhelpforum @mathhelpforum