Use repeatedly the fact that |f(x)| >= a iff f(x) >= a or f(x) <= -a.
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.
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...
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.
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.
x^{2}
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.
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.
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?