Thread: Double Absolute Value Inequalities ???

1. Double Absolute Value Inequalities ???

Hey everyone- can anyone pls assist with how to solve these:

-5< |x|<2

This is nowhere in my book and I am not technically required to know this for my class but I am interested in how this would be solved, I do know how to do it if it were either/or:

i.e. |x|<2 you would just do this: -2< x <2

and obviously |x|> -5 would be all real #s

but to I have no idea how to solve when its already compounded with two different values at once!

Any help greatly appreciated!

2. Re: Double Absolute Value Inequalities ???

Originally Posted by cds333
-5< |x|<2
This is nowhere in my book and I am not technically required to know this for my class but I am interested in how this would be solved, I do know how to do it if it were either/or:
i.e. |x|<2 you would just do this: -2< x <2
and obviously |x|> -5 would be all real #s
Because that is true then the problem is just
$\displaystyle |x|<2$.

3. Double / Compound Absolute Value Inequalities ???

Hey thanks plato!

So what you're saying is the left side gives (-∞,∞), and the right side gives (-2,2), but since the right side is more specific it overrules the left?

What if the problem were something like: 2 ≤ |x| ≤ 5

Thanks again!

4. Re: Double / Compound Absolute Value Inequalities ???

$\displaystyle a\le x\le b$ should be treated as two inequalities: $\displaystyle a\le x$ and $\displaystyle x\le b$. Notice the "and"- they must both be true which means that the solution set of the "double" inequality is the intersection of the two separate inequalities.

Further, $\displaystyle a\le |x|$ itself means $\displaystyle a \le x$ if x is non-negative or $\displaystyle a\le -x$ if x is negative. Here, we have "or" so the solution set is the union of the two separate solution sets.

$\displaystyle 2\le |x|\le 5$ gives the two inequalities $\displaystyle 2\le |x|$ and $\displaystyle |x|\le 5$.

(1)$\displaystyle 2\le |x|$ means
(a) $\displaystyle 2\le x$ as long as x is non-negative which it certainly is if it is larger than or equal to 2: [tex]2\le x.
(b) $\displaystyle 2\le -x$ which is the same as $\displaystyle -2\ge x$ as long as x is negative, which it is as long as x is less than -2: $\displaystyle x\le -2$.
The union of those two sets is the two separate intervals, $\displaystyle [-\infty, -2)\cup[2, \infty)$

(2)$\displaystyle |x|\le 5$ means
(a) $\displaystyle x\le 5$ as long as x is non-negative: $\displaystyle 0\le x\le 5$
(b) $\displaystyle -x\le 5$ which is the same as $\displaystyle x\ge -5$ as long as x is negative: $\displaystyle -5\le x< 0$
The union of those two sets is the single interval, [-5, 5].

Now take the intersection of those two sets. The interval $\displaystyle (-\infty, -2]$ intersect [-5, 5] is [-5, -2]. The interval [2, \infty) intersect [-5, 5] is [2, 5]. The solution set for [tex]2\le x\le 5[/itex] is $\displaystyle [-5, -2]\cup [2, 5]$.

You probably could have done that more simply by reducing the absolute value first: $\displaystyle 2\le x\le 5$ and $\displaystyle 2\le -x\le 5$