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Math Help - absolute value

  1. #1
    Super Member dhiab's Avatar
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    absolute value

    Solve in R :
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  2. #2
    MHF Contributor alexmahone's Avatar
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    |2x+1|\geq|3-x|

    Squaring both sides,

    4x^2+4x+1\geq9-6x+x^2

    (x+4)(3x-2)\geq0

    So, x\geq-4 and x\geq\frac{2}{3}
    ie x\geq\frac{2}{3}

    Or x\leq4 and x\leq\frac{2}{3}
    ie x\leq-4
    Last edited by alexmahone; June 13th 2009 at 08:25 AM.
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  3. #3
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    Talking

    You have:

    . . . . . \left|2x\,+\,1\right|\,\leq\,\left|3\,-\,x\right|

    The insides of the absolute values change signs at x = -1/2 and at x = 3, so you have three intervals; namely, (-\infty,\, -1/2),\, (-1/2,\,3),\, and (3,\,\infty).

    Consider each interval separately. If x < -1/2, then 2x + 1 < 0 and 3 - x > 0, so you have:

    . . . . . -(2x\, +\, 1)\, \leq\, 3\, -\, x

    . . . . . -2x\, -\, 1\, \leq\, 3\, -\, x

    . . . . . -1\, \leq\, 3\, +\, x

    . . . . . -4\, \leq\, x

    Since x = -4 is less than -1/2, then this solution is valid: -4\,\leq\,x\,\leq\,-1/2.

    If -1/2 < x < 3, then 2x + 1 > 0 and 3 - x > 0, so:

    . . . . . 2x\, +\, 1\, \leq\, 3\, -\, x

    . . . . . 3x\, \leq\, 2

    . . . . . x\, \leq\, \frac{2}{3}

    Then the inequality works for -1/2\, \leq\, x\, \leq\, 2/3.

    Now check when x > 3:

    . . . . . 2x\, +\, 1\, \leq\, x\, -\, 3

    . . . . . x\, \leq\, -4

    But x cannot be both greater than three and less than minus four, so there is no solution on this interval.

    . . . . . \mbox{solution: }\, -4\, \leq\, x\, \leq\, \frac{2}{3}

    This can be confirmed by graphing, by the way. For further information, try here.
    . . . . .
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by stapel View Post
    You have:

    . . . . . \left|2x\,+\,1\right|\,\leq\,\left|3\,-\,x\right|

    The insides of the absolute values change signs at x = -1/2 and at x = 3, so you have three intervals; namely, (-\infty,\, -1/2),\, (-1/2,\,3),\, and (3,\,\infty).

    Consider each interval separately. If x < -1/2, then 2x + 1 < 0 and 3 - x > 0, so you have:

    . . . . . -(2x\, +\, 1)\, \leq\, 3\, -\, x

    . . . . . -2x\, -\, 1\, \leq\, 3\, -\, x

    . . . . . -1\, \leq\, 3\, +\, x

    . . . . . -4\, \leq\, x

    Since x = -4 is less than -1/2, then this solution is valid: -4\,\leq\,x\,\leq\,-1/2.

    If -1/2 < x < 3, then 2x + 1 > 0 and 3 - x > 0, so:

    . . . . . 2x\, +\, 1\, \leq\, 3\, -\, x

    . . . . . 3x\, \leq\, 2

    . . . . . x\, \leq\, \frac{2}{3}

    Then the inequality works for -1/2\, \leq\, x\, \leq\, 2/3.

    Now check when x > 3:

    . . . . . 2x\, +\, 1\, \leq\, x\, -\, 3

    . . . . . x\, \leq\, -4

    But x cannot be both greater than three and less than minus four, so there is no solution on this interval.

    . . . . . \mbox{solution: }\, -4\, \leq\, x\, \leq\, \frac{2}{3}

    This can be confirmed by graphing, by the way. For further information, try here.
    . . . . .
    Sorry to break your bubble but you've copied the question wrong

    Counterexample: x=0 doesn't satisfy the OP's inequality.
    Last edited by alexmahone; June 13th 2009 at 08:26 AM.
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  5. #5
    Super Member dhiab's Avatar
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    Anathor resolution :
    Attached Files Attached Files
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by dhiab View Post
    Anathor resolution :
    Your first post has the inequality reversed.
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  7. #7
    Super Member dhiab's Avatar
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    Quote Originally Posted by alexmahone View Post
    |2x+1|\geq|3-x|

    Squaring both sides,

    4x^2+4x+1\geq9-6x+x^2

    (x+4)(3x-2)\geq0

    So, x\geq-4 and x\geq\frac{2}{3}
    ie x\geq\frac{2}{3}

    Or x\leq4 and x\leq\frac{2}{3}
    ie x\leq-4
    Hello : your resolution is wrong, a product AB is negative mean that A, B are of opposite sign
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  8. #8
    Super Member dhiab's Avatar
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    hello i'm sorry alexmahone
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  9. #9
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by dhiab View Post
    Hello : your resolution is wrong, a product AB is negative mean that A, B are of opposite sign
    My solution is right. A product AB is positive means that A, B are of same sign.
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