Results 1 to 6 of 6
Like Tree2Thanks
  • 1 Post By agentmulder
  • 1 Post By agentmulder

Math Help - Applying function to sides of a disequality

  1. #1
    Newbie
    Joined
    Apr 2013
    From
    Boston
    Posts
    11

    Applying function to sides of a disequality

    A disequation like this

    -1 <= sinx <= 1

    of course is always true, and is equivalent to

    -1 <= sinx
    sinx <= 1

    What if you apply f(x) = x^-1 to both sides? It's a monotically decreasing function for x != 0, so you should just
    reverse the disequality... but!

    -1 >= 1/sinx
    1 <= 1/sinx

    which is nonsense. Am I wrong somewhere?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member agentmulder's Avatar
    Joined
    Aug 2011
    From
    42nd parallel, North America
    Posts
    105
    Thanks
    33

    Re: Applying function to sides of a disequality

    It's probably not total nonsense. There are issues arising because of the asymptotes but at least 1/sinx never gives a value BETWEEN -1 and 1 as the graph shows.

    plot(1/(sinx)) - Wolfram|Alpha



    Note*

     -1 \le sinx \le 1

    is equivalent to

     -1 \le sinx \ \ AND \ \ sinx \le 1

    and in interval notation

     [-1, \infty) \cap (- \infty, 1] = [-1,1]

    The INTERSECTION of the 2 intervals

    but the other one is OR if you interpret it as AND there is no solution.

     -1 \ge \frac{1}{sinx}

    OR

     1 \le \frac{1}{sinx}

    and in interval notation

     (- \infty , -1] \cup [1 , \infty)

    The UNION of the 2 intervals.

    Last edited by agentmulder; April 21st 2013 at 01:12 AM.
    Thanks from MarkFL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2013
    From
    Boston
    Posts
    11

    Re: Applying function to sides of a disequality

    Thanks a lot!

    But then, a chained disequality is an and-equivalent. Why applying a function should change that?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member agentmulder's Avatar
    Joined
    Aug 2011
    From
    42nd parallel, North America
    Posts
    105
    Thanks
    33

    Re: Applying function to sides of a disequality

    Quote Originally Posted by AndrewTahoe View Post
    Thanks a lot!

    But then, a chained disequality is an and-equivalent. Why applying a function should change that?
    I edited my post to include interval notation but if that doesn't help with your new question consider...

     -2 \le 2 \le 4

    Now apply the inverse to everything

     -\frac{1}{2} ?? \frac{1}{2} ?? \frac{1}{4 }

    What can we put for the question marks ??

    we see that

     -\frac{1}{2} \le \frac{1}{2} \ \ OR \ \ \frac{1}{2} \ge \frac{1}{4}

    Will help us with the dilemmas of -??-
    Last edited by agentmulder; April 21st 2013 at 01:37 AM.
    Thanks from MarkFL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    708
    Thanks
    292

    Re: Applying function to sides of a disequality

    Hi,
    A real valued function on a set D of reals is decreasing provided for all a and b in D with a < b, f(a) > f(b). You said f(x)=1/x is decreasing everywhere but 0. In a sense this is right. That is, f is decreasing on (-\infty,0) and decreasing on (0,\infty). But this does not mean f is decreasing on (-\infty,0)\cup(0,\infty). From -2 < 1 it certainly does not follow that -1/2 > 1/1.

    So for your original inequality, if sin(x) > 0 then 1 and sin(x) are in a set D where f(x)=1/x is decreasing and so 1\geq 1/sin(x). But if sin(x) < 0, you can't conclude 1/sin(x)\geq1.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2013
    From
    Boston
    Posts
    11

    Re: Applying function to sides of a disequality

    Ok, that was clear. In fact, the definition of monotonically decreasing make sense when we consider an interval that is a subset of the domain of the function. No "holes" allowed!

    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 15th 2012, 01:54 AM
  2. Help solving disequality and functions domain
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 18th 2011, 06:12 AM
  3. Applying the Law of Cosines
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: December 4th 2010, 04:39 PM
  4. limit from sides in complex function..
    Posted in the Calculus Forum
    Replies: 14
    Last Post: July 12th 2010, 02:16 PM
  5. Replies: 6
    Last Post: October 15th 2007, 05:38 AM

Search Tags


/mathhelpforum @mathhelpforum