Inequalities, trigonometric and x exponent.

  • Thread starter Mutaja
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  • #1
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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).

Homework Statement


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

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


Homework Equations





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.
 

Answers and Replies

  • #2
35,129
6,876
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).

Homework Statement


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

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


Homework Equations





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.

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.
 
  • #3
239
0
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.

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.
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.

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.
 
  • #4
35,129
6,876
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.
 
  • #5
35,129
6,876
I'm sorry, I didn't know it mattered as I can't see any visual difference, but I'll keep that in mind.
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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