Results 1 to 8 of 8

Math Help - Extrema of a closed interval??

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    23

    Extrema of a closed interval??

    How do I approach this problem?
    Find the extrema of f(x)=x^(1/3) on the closed interval of [-1,1]

    Help! Please and thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by painterchica16 View Post
    How do I approach this problem?
    Find the extrema of f(x)=x^(1/3) on the closed interval of [-1,1]

    Help! Please and thank you!
    the same way you find extrema anywhere. find the derivative, set it equal to zero, solve for x. then just pick the x's that fall in that interval. then classify each as max, min, or inflection point
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    23
    I tried that, but when I set the derivative (1/3x^-2/3) equal to 0, I get x=0, and then I plug in -1 and 1 in the derivative and get one for both, which doesnt seem right. What am I doing wrong?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by painterchica16 View Post
    I tried that, but when I set the derivative (1/3x^-2/3) equal to 0, I get x=0, and then I plug in -1 and 1 in the derivative and get one for both, which doesnt seem right. What am I doing wrong?
    if you got x = 0, then x = 0 would be the extreme point in [-1,1]. however, x = 0 is wrong. \frac 13 x^{-2/3} = \frac 1{3 \sqrt[3]{x^2}} is never zero.

    if this is a problem that asks you about the absolute extrema, then it remains for you to test the end points. the one that gives the largest value is your absolute max, the one that gives the smallest value is the absolute min

    and to get the points, you do not plug into the derivative, you plug into the original function. you want the value of the function at that point
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    Posts
    23
    I'm trying this over and over again, but I still get x=0. I know you've already helped me so much, but could take it back from the derivative and go from there? This topic kinda confuses me more than it should and I'm still lost. Thank you so much!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by painterchica16 View Post
    I'm trying this over and over again, but I still get x=0. I know you've already helped me so much, but could take it back from the derivative and go from there? This topic kinda confuses me more than it should and I'm still lost. Thank you so much!
    your derivative is correct. if you set it equal to zero, there is no solution. you have x in the denominator, it cannot be zero!

    now, test the end points. that is, find f(-1) and f(1). the larger one of those values is you absolute max, and the smaller is your absolute min
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    be sure to familiarize yourself with what extreme points are and what absolute extreme points are. i think that's one of your problems here.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2008
    Posts
    23
    Thank you so much! I really appreciate the help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Extrema on interval?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 22nd 2011, 03:52 PM
  2. Extrema on an Interval
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 5th 2010, 02:54 AM
  3. Replies: 1
    Last Post: June 9th 2009, 09:29 PM
  4. Extrema on an interval find critical numbers
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 9th 2009, 03:13 PM
  5. Extrema on a closed interval
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 13th 2008, 06:08 PM

Search Tags


/mathhelpforum @mathhelpforum