Results 1 to 6 of 6

Math Help - Relative and Absolute Extrema? Did I solve correctly?

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    87

    Relative and Absolute Extrema? Did I solve correctly?

    Find all relative and absolute extrema in the function, k(x) = (x^2 - 4)^-2, on the interval [-3, 3].

    My work:
    k'(x) = [(-4x)] / [(x^2 - 4)^3]
    Critical Points: x = 2, x = -2
    Since there are vertical asymptotes at x = +/- 2, there are no relative extrema.
    There is no absolute maximum because the function has no upper bound.
    When x = -3, k(-3) = .04
    When x = 3, k(3) = .04
    Therefore, the function has absolute minimum of k(-3) = k(3) = .04

    Basically there are:
    No relative extrema
    No absolute maximum
    Absolute minimum at k(-3) and k(3)

    Is this correct?
    Any help is appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member BariMutation's Avatar
    Joined
    Dec 2010
    From
    I'm omnipresent
    Posts
    30
    You're almost completely correct. For the very fact that there are vertical asymptotes (which, by definition, go to infinity), there are neither absolute min's or max's.

    However, consider what a relative min/max is compared to a absolute min/max.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    87
    A relative min/max must have lower/higher values on each side, whereas a absolute min/max can be an endpoint. Does this mean that there are no relative or absolute extrema in this function?
    Thank you so much for answering!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member BariMutation's Avatar
    Joined
    Dec 2010
    From
    I'm omnipresent
    Posts
    30
    There are no absolute max points because the peak of the graph occurs at x=2, which is not defined (as you found). There are also no absolute min points because it has a horizontal asymptote, causing the graph to get closer and closer to 0 for infinity. However, since you have the bounds of [-3,3], there is almost certainly either a relative max, min, or both. Here's a hint - you found it already, you just mislabeled it.

    Remember, 95% of the time, if there are bounds given, there will be a relative max, min, or both.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2009
    Posts
    87
    Oh ok,
    So .04 is the relative minimum at x = +/- 3? I didn't think that relative extrema could exist on endpoints?
    Thanks again
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member BariMutation's Avatar
    Joined
    Dec 2010
    From
    I'm omnipresent
    Posts
    30
    Yeah, they can. Think about x^2 on [0,5]. The relative min would be 0, since that is the min of the parabola.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relative and Absolute Extrema
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 5th 2010, 02:07 PM
  2. Relative and Absolute Extrema
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 1st 2010, 10:47 PM
  3. Replies: 1
    Last Post: November 18th 2009, 11:22 PM
  4. Replies: 1
    Last Post: April 21st 2009, 11:32 AM
  5. relative and absolute extrema
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 27th 2006, 03:34 PM

Search Tags


/mathhelpforum @mathhelpforum