Thread: Mathmatics Question?? Absolute Value and Inequalities

Ok, I received some help earlier.
I'm hoping for some kind souls to help again.
Now I'm completely lost on a problem.

Solve and Express in interval notation.
1<=|x|<=4

I came up with this
-4<=x<=4 and x<=-1 and 1<=x

the book says it is
[-4,-1]U[1,4]

I know what interval notation is, but I'm lost on the how they grouped the different x values.
What step am I missing to get the books answer?

[edit1]
Do we default grouping the negative values together and the positive values together?


I'm not sure, but let me try to explain why I see it grouping in this way.

-4 is less then or equal to x which is less then or equal to 4
this covers all values from -4 to 4
x is also less then or equal to -1
x is greater than or equal to 1

-4 to 4 is my outside boundries
so
-4 to -1 is one of the boundries

1 to 4 is the upper side boundry.

Is this the way they are looking at it or is there a fail safe rule that I can use?
[/edit1]

basically, imagine this:

1 <= |x| <= 4

we get the following

1 <= +x <= 4

now, if we put -x and remove the absolute value we get

1 <= -x <= 4

which is(clearing the -)

-4 <= x <= -1

therefore

1 <= |x| <= 4 means

1 <= x <= 4 &&
-4 <= x <= -1

that is where your two intervals come from, sorry if I am being unclear.

on the prior problem I'm not sure why it uses a connecting "or" instead of a connecting "and" in the interval notation.

here is a similar problem.
I believe I did it correctly according to what you have mentioned and what is in the book.
Unfortunately I can not check this one in the book.

0<|x-5|<=1/2

-1/2<=x-5<=1/2
9/2<=x<=11/2

0<x-5
5<x
x-5<0
x<5

therefore in interval notation

[-1/2,5)*and/or*(5,1/2]

*and/or !!not sure which or why i would use "and" over "or" in this problem!!*

>>9/2<=x<=11/2

This looks correct to me, so [9/2,11/2] I believe would be the answer.

Lets see...

0<|9/2-5|<=1/2 = 0<|-1/2|<=1/2 = 0<1/2<=1/2 <--yes

0<|11/2-5|<=1/2 = 0<|1/2|<=1/2 <---yes

Edit: Wait I don't think that's right...

Ok I did it again and got:

[9/2,5)u(5,11/2]

Oh I think thats what Draco got hehe

this is the last college algebra class prior to trig.
these rules are probably used in calculas though.

follow up question:

link referencing the below rule
http://www.wtamu.edu/academic/anns/m...22_linineq.htm

rule: Absolute Value and Inequalities

if
|x|>1
then
x<-1
and
1<x

So does that not throw a wrench into this portion of the problem
"0 < |x - 5| "
wouldn't that disprove your calculations?

case i.
0 < x - 5 <= .5
5 < x <= 5.5

case ii.
0 < 5 - x <= .5
-5 < -x <= -4.5
5 > x >= 4.5

I might not have seen in my book on how the rule you are using works.

[Edited]


As for the problem
0 < |x - 5|
Remember that the absolute value always is possible. The only way this inequality can be false is when x - 5 == 0 which it is when x == 5
The solution is thus:
x != 5

As for the original problem,
1<=|x|<=4
Geometrically, |x| is the distance from the origin to x. The distance is between 1 and 4. Thus, the solution is:
1<=x<=4
OR
-4<=x<=-1

OK, perhaps I'll try to explain this a little better this time.

When you encounter an absolute value, you have to separate the equation of inequality into two different equations where you assume that the expression within || are negative and positive respectively.

Then, the solutions is always given by (i) OR (ii)

I'll post some examples:
(EDIT: litte typo here, x <= 0 should be x >= 0
[-4,1] should be [-4,-1])