Results 1 to 3 of 3

Math Help - Finding the maximum and minimum

  1. #1
    Member
    Joined
    Oct 2007
    Posts
    178

    Finding the maximum and minimum

    Use analytic methods to find the extreme values of the function on the interval and where they occur.

    Extreme values includes local and absolute max and min.

    Analytic methods meaning I can't graph anything.

    g(x)= \sin (x + \frac{\pi}{4})
    To find the critical points i need to find when y'= 0.
    g'(x)=\cos(x + \frac{\pi}{4})

    I got the points \frac{\pi}{4}, \frac{5\pi}{4}
    When plugged into g(x) they yield 1 and -1, respectively.

    These are only 2 critical points; I need 4.

    Thanks for the help!
    Last edited by Truthbetold; January 9th 2008 at 05:59 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2007
    Posts
    178

    Another Problem

    Derivative is x^{2/3} + \frac{2(x+2)}{3x^(1/3)}

    The original function is x^{2/3}(x+2)
    The answer says the critical points are 0 and -4/5.
    The points appear to be 0, and -2.

    What did I do wrong?
    Last edited by Truthbetold; January 9th 2008 at 07:31 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1
    Quote Originally Posted by Truthbetold View Post
    Derivative is x^[2/3] + \frac{2(x+2)}{3x^(1/3)}

    The original function is x^[2/3](x+2)
    The answer says the critical points are 0 and -4/5.
    The points appear to be 0, and -2.

    What did I do wrong?
    x^[2/3] + \frac{2(x+2)}{3x^(1/3)} = 0

    Multiply both sides by x^{1/3}. (Note that x = 0 is not in the domain of the derivative, so we lose nothing by doing this.)

    x + \frac{2}{3}(x + 2) = 0

    So I'm getting that x = -4/5 is a critical point.

    x = 0 is defined to be a critical point since the derivative doesn't exist there.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 12
    Last Post: May 20th 2011, 11:56 AM
  2. Replies: 1
    Last Post: March 20th 2010, 04:32 AM
  3. Finding the Minimum & Maximum
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: September 11th 2009, 07:24 PM
  4. Finding Relative Maximum and Minimum
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 5th 2009, 07:45 PM
  5. Replies: 0
    Last Post: November 19th 2007, 12:51 AM

Search Tags


/mathhelpforum @mathhelpforum