Mathmatics Question?? Absolute Value and Inequalities

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  1. #1
    essence of digital xddxogm3's Avatar
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    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]
    Last edited by xviddivxoggmp3; 09-26-2004 at 11:04 PM.
    "Hence to fight and conquer in all your battles is not supreme excellence;
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  2. #2
    'AlHamdulillah
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    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.
    there used to be something here, but not anymore

  3. #3
    Registered User Draco's Avatar
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    You have to remember that you must turn it into two inequalities because of the absolute value sign. Because of the value sign you end up with that one inequality equaling both 1<=x<=4 and 1<=-x<=4. What you then do is flip the signs for the second one, making -4<=x<=-1

  4. #4
    essence of digital xddxogm3's Avatar
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    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!!*
    Last edited by xviddivxoggmp3; 09-26-2004 at 11:48 PM.
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    Registered User Draco's Avatar
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    I'm not sure how you got to the line -1/2<=x-5<=1/2. For the first inequality I got 0<x-5<=1/2 then add five to each side for 5<x<=5.5. For the second I got 0<-(x-5)<=1/2 then change the sign and add five so 4.5<x<=5. I'm not 100% sure on this second one though, I havent done inequalities in a couple of years.

    Post 400!

  6. #6
    carry on JaWiB's Avatar
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    >>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
    Last edited by JaWiB; 09-26-2004 at 11:42 PM.
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  7. #7
    Toaster Zach L.'s Avatar
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    Concisely,

    0 < abs(x - 5) <= .5

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

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

    x in [4.5, 5) U (5, 5.5]

    An easy way to check is to insert a value and watch what happens:

    x = 4.7
    0 < (4.7 - 5) <= .5
    0 < abs(-.3) <= .5
    0 < .3 <= .5 (true)

    x = 5.2
    0 < abs(5.2 - 5) <= .5
    0 < abs(.2) <= .5
    0 < .2 <= .5 (true)

    x = 5
    0 < abs(5 - 5) <= .5
    0 < 0 <= .5 (false)

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    Software Developer jverkoey's Avatar
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    Just wondering, is this calculus? I haven't taken it yet...but this stuff seems really interesting and I'm finding I'm actually understanding it...which is really cool.

  9. #9
    Registered User Draco's Avatar
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    this is intermediate to advanced algebra, more of a pre-calc level

  10. #10
    Software Developer jverkoey's Avatar
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    Oh, I kinda thought it was a bit simple and was wondering why I was actually understanding it, lol. I've just never seen these examples in class before so they seemed odd to me I guess.

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    Registered User Draco's Avatar
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    yeah, I understand you. If this stuff is intersting for you you're going to love calculus.

  12. #12
    essence of digital xddxogm3's Avatar
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    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.
    Last edited by xviddivxoggmp3; 09-27-2004 at 10:42 AM.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  13. #13
    Cheesy Poofs! PJYelton's Avatar
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    rule: Absolute Value and Inequalities

    if
    |x|>1
    then
    x<-1
    and
    1<x
    Actually it says OR instead of AND. Think of it this way, it can't be an AND because a number can't be between 4.5 and 5 AND also be between 5 and 5.5 at the same time, it has to be OR. For example, 4.8 works when you do the math, but if it was AND then it wouldn't because when you say AND it must fit both inequalities which it doesn't. It only fits one of the inequalities. No different than if you were programming with &&'s and ||'s.

  14. #14
    S Sang-drax's Avatar
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    [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
    Last edited by Sang-drax; 09-27-2004 at 04:53 PM. Reason: I was confusing :)
    Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling

  15. #15
    S Sang-drax's Avatar
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    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])
    Attached Images Attached Images  
    Last edited by Sang-drax; 09-27-2004 at 05:42 PM.
    Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling

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