Results 1 to 4 of 4

Math Help - Local Maximum and Minimum

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    14

    Local Maximum and Minimum

    I'm trying to find the local maximum and minimum values for the equation f(x)=sin2x-x, 0 < x < \pi but I seem to keep coming up with the wrong answer.

    My working goes like this (in radians):
    f'(x)=2cos2x-1
    f''(x)=-4sin2x

    Local maximums and minimums appear where the gradient is 0, so:

    2cos2x-1=0
    2cos2x=1
    cos2x=\frac{1}{2}
    x=\frac{1}{6}\pi

    f''(\frac{1}{6}\pi ) = -4sin\frac{1}{3}\pi = -4\times 0.866 = -3.464

    Therefore there is a local maximum at x=\frac{1}{6}\pi

    What I can't work out is how to find the local minimum in the same range. I think that, because local maximums and minimums are \pi radians apart on the Cosine graph, they will be x=\frac{1}{2}\pi radians apart on the graph of Cos 2x, which would put the minimum point at x=\frac{2}{3}\pi but the book I'm working from puts it at x=\frac{5}{6}\pi .
    Last edited by djnorris2000; October 5th 2011 at 12:36 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Local Maximum and Minimum

    You have determined the x-coordinate of the maximum correctly, but notice your equation has two solutions ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    14

    Re: Local Maximum and Minimum

    Thanks, I can see there's a second solution, but I make it 2/3 pi while the book I'm working from makes it 5/6 pi and I'm not sure why I'm arriving at a different answer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5

    Re: Local Maximum and Minimum

    Quote Originally Posted by djnorris2000 View Post
    Thanks, I can see there's a second solution, but I make it 2/3 pi while the book I'm working from makes it 5/6 pi and I'm not sure why I'm arriving at a different answer.
    \cos(2x) = \frac{1}{2} \Rightarrow 2x = \frac{\pi}{3}, ~ \frac{5 \pi}{3} \Rightarrow x = ....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: March 21st 2011, 02:23 AM
  2. Replies: 6
    Last Post: January 5th 2011, 03:34 AM
  3. Local Minimum and Maximum; Saddle Point
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 25th 2010, 09:37 PM
  4. local minimum and maximum values
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 16th 2010, 04:05 PM
  5. absolute and local maximum and minimum
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 18th 2009, 03:40 PM

Search Tags


/mathhelpforum @mathhelpforum