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Inequalities, trigonometric and x exponent.

  1. Oct 30, 2013 #1
    These are the two last problems I'll bother you with for a short while (I love this forum, I'll definitely stay on and hopefully be able to contribute in the future).

    1. The problem statement, all variables and given/known data
    Problem 1:
    (##-x^2##-1)sin2x > 0 , xe[0,2[itex]\pi[/itex]]

    Problem 2:
    ##2^{-x^2+x+2}## < 4


    2. Relevant equations



    3. The attempt at a solution

    For the first problem:

    (##-x^2##-1)sin2x > 0 , xe[0,2[itex]\pi[/itex]]

    Since x is an element within a positive boundary, can I write the problem as:

    sin(2x) ≤ 0

    Why, or why not?

    Other than that, I need to refresh my knowledge on solving inequalities.

    Any help or input is appreciated while I attempt to gather more info.
     
  2. jcsd
  3. Oct 30, 2013 #2

    Mark44

    Staff: Mentor

    Your first inequality can be rewritten as -(x2 + 1)sin(2x) ≥ 0, or
    (x2 + 1)sin(2x) ≤ 0

    Since x2 + 1 ≥ 1 for all real x, it can never be zero or negative. If you divide both sides of the inequality by x2 + 1, and the direction of the inequality won't change. That gets you to the inequality you wrote.

    BTW, your mix of LaTeX and HTML tags threw me off for a bit. In LaTeX use leq for ≤. Using the U (underscore) HTML tag under a < symbol confused me for a bit.

    You can also click Go Advanced to see a table of quick symbols off to the right. It has Greek letters and symbols such as ≤, ≥, ≠, ∞, and several others.
     
  4. Oct 30, 2013 #3
    But how do I go from sin2x ≤ 0 to something like x = [itex]\pi[/itex] + n [itex]\pi[/itex]/2? This is just something random I wrote down, but hopefully you get the idea.

    Where do I even look to find the solution for this?

    Also, in respect to my 2nd inequality. I don't know where I should look for tips on how to solve it. My book says absolutely nothing about inequality in this regard. It might sound lazy, but I've looked. I suspect I'm supposed to remember this from pre-calculus math but it's as unfortunate as it's obvious - I don't.

    Sorry I'm not able to make more progress before posting again. Again, thanks for all your help and any input is greatly appreciated.
    I'm sorry, I didn't know it mattered as I can't see any visual difference, but I'll keep that in mind. Thanks.
     
  5. Oct 30, 2013 #4

    Mark44

    Staff: Mentor

    You're not going to get x = ... as the solution to sin(2x) ≤ 0. The simplest way is to sketch a quick graph of y = sin(2x) and note the intervals where the graph touches or goes below the x-axis.

    For your second problem, which is
    $$2^{-x^2 + x + 2} < 4$$

    Note the 4 = 22, so then you have
    $$2^{-x^2 + x + 2} < 2^2$$

    Since y = 2x is a strictly increasing function, if 2A < 2B, then A < B.
     
  6. Oct 30, 2013 #5

    Mark44

    Staff: Mentor

    This was in reference to what I said about using the canned symbols ≤ rather than [ U]<[/ U]; i.e., coming up with your own way to write ≤. A lot of us here at PF will "quote" your post, which means that we're looking at the LaTeX and/or HTML markup, instead of how the post actually appears in the browser, so your ≤ actually seemed to me at first to be <, which was why I was confused.
     
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