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I do not know where to start! Help me, please!
I have problem finding the solution to these inequalities:
1) |x² - x| - |x| < 1
2) |x| / (x 2)² ≥ 1
3) ||x + 1| - |x - 1|| < 1
4) (√x² + 1) + 2x 1 > 0
5) |x² -1| + |2 - x| < 2
6) 1 + |x - 1| / 1 - |x - 1| ≤ 1
Thanks!
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Hi mmoadi
I think that you must redefine these inequalities without using the absolute value notation , after that you deal with usual inequaliteis containing functions of more than one rule.
Best Wishes
Riad Zaidan
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Hi mmoadi;
At least this one appears easy:
4)
Last edited by bobbym (2009-09-23 10:39:40)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym
When taking the square root for x² you must take the absolute value of x and so you will have two inequalities, and youmust solve these two inequalities
-x+ 1+ 2x 1 > 0 when x<0 ⇒ x>0 which is empty sunce x<0 and
x+ 1+ 2x 1 > 0 when x>0⇒3x>0 ⇒x>0 therefore the solution of this inequality is {x∈R: x>0} mainly the positive reals.
Best Regards
Riad Zaidan
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Last edited by JaneFairfax (2009-09-24 23:34:12)
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Hi rzaidan;
That is exactly how I did it. By splitting it into 2 inequalities as you indicate.
I just did not show the work of solving the first inequality. I just did the second one.
Last edited by bobbym (2009-09-25 00:02:34)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Never mind, forget what I said. Sorry.
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Nothing to forget. I have great respect for you.
Last edited by bobbym (2009-09-25 00:07:45)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thanks.
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Thanks for all
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