Thread: Solving inequalities with absolute values within absolute values

Solve the inequality:



So I can find the first solution:




I can't figure out how to get the other solution correctly. From looking it up I know it should be . I have never had to deal with absolute values like this before so I don't know where to put all the negative signs.

Use repeatedly the fact that |f(x)| >= a iff f(x) >= a or f(x) <= -a.

I have attempted that a few times and usually I come up with which is incorrect. I guess I just don't know how to properly write it out and I end up missing a step or something every time. Would it be possible to get a step by step of how it is correctly done?

|||x - 1| - 1| - 1| >= 4 iff (a1) ||x - 1| - 1| - 1 >= 4 or (a2) ||x - 1| - 1| - 1 <= -4.

Case (a1). ||x - 1| - 1| - 1 >= 4 iff ||x - 1| - 1| >= 5 iff (b1) |x - 1| - 1 >= 5 or (b2) |x - 1| - 1 <= -5.

Case (b1) |x - 1| - 1 >= 5 iff |x - 1| >= 6 iff (c1) x - 1 >= 6 or (c2) x - 1 <= -6.

Finish analyzing (a2), (b2), (c1) and (c2).

Sorry, this is really confusing now. I think I see where you go from a1 to b1/b2 and then from that to c1/c2 and from the c1/c2 cases I can solve and find that and , which is the correct answer for the question. I'm not really following what happens with a2 here. If I do with a2 as you did with a1 I end up with and , which obviously means I did something wrong there...

Originally Posted by BobRoss

If I do with a2 as you did with a1 I end up with and , which obviously means I did something wrong there...

Note that (a2) has the form |f(n)| <= a, which is equivalent to -a <= f(n) <= a (and not to a choice between f(n) >= a and f(n) <= -a). In this particular case, (a2) is impossible because it is equivalent to ||x - 1| - 1| <= -3, and an absolute value can't be negative. The case (b2) is also impossible.

Okay I think I see now what you've done here. That helps a lot, thank you!

I have another absolute value inequality question if you don't mind. I have read the inequality document that is stickied in this forum but it didn't help.

x²

I can't find any examples like this in my text or any similar examples online so I am lost as to how to properly solve this one too. Another quick walkthrough of this problem would be awesome if you are able.

We have |ab| = |a| |b| for all a, b. So, |x^2 - x| = |x| |x - 1|. Thus, |x - 1| - |x^2 - x| = |x - 1| (1 - |x|). When is a product negative?

It is negative when you multiply a positive and a negative. How are you arriving at this last part:

Originally Posted by emakarov

|x - 1| (1 - |x|).

?

Originally Posted by BobRoss

It is negative when you multiply a positive and a negative.

Yes.

Originally Posted by BobRoss

How are you arriving at this last part

By factoring out |x - 1|.

Perhaps think of it in terms of distances.
Let a = ||x-1| - 1|. Then, via |||x-1| - 1| - 1| >=4 (meaning |a-1|>= 4), the distance a to 1 is >=4
Thus a >=5 or a <= -3.
But a is an absolute value, hence positive.
Thus a >=5.
Let b = |x-1|.
Then the distance from b to 1 is a (a = ||x-1|-1| = |b-1|), and a >=5.
Thus the distance from b to 1 is >=5.
Thus b>= 6 or b <= -4.
But b is an absolute value, hence positive.
Thus b >= 6.
But b = |x-1| = distance from x to 1, and b >=6.
Thus the distance from x to 1 is >=6.
Thus x >=7, or x <= -5.

Originally Posted by emakarov

Yes.

By factoring out |x - 1|.

Gah, I'm terrible at this. I can't figure out how you get to that by factoring out |x-1|.

Originally Posted by BobRoss

I can't figure out how you get to that by factoring out |x-1|.

|x - 1| - |x² - x| < 0 <=>
|x - 1| - |x(x - 1)| < 0 <=>
|x - 1| - |x| |x - 1| < 0 <=>
|x - 1|(1 - |x|) < 0 <=>
|x - 1| > 0 and 1 - |x| < 0.

Okay I should have seen that before. So now I need to analyze the cases

1.) |x-1| is positive and 1-|x| is negative

and the other case

2.) |x-1| is negative and 1-|x| is positive

And since the absolute value cannot be negative, case 2 doesn't exist. And when I solve case 1 I get x>1 and x<-1. Is that correct?