AP Calc

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Solve the following absolute value inequalities:

|x+1|<=|x-3|


I promise I'm not posting my whole homework assignment. These are questions that I skipped and have no clue how to do out of a huge packet.

  • AP Calc -

    The key to absolute value problems is to remember the definition of |n|.

    |n| = n if n >= 0
    |n| = -n if n < 0

    So, here we have

    |x+1|<=|x-3|

    if (x-3) >= 0, |x-3| = x-3
    Also, since that means x > 3, x+1 >= 0, so |x+1| = x+1 and we have

    x+1 <= x-3
    1 < -3
    not possible

    If (x-3) < 0, (that is, x<3) we have
    |x+1| <= -(x-3)
    Now, if x >= -1, x+1 >= 0, and we have
    x+1 <= -x+3
    2x <= 2
    x <= 1
    But, we required above that x >= -1, so we get a solution set -1 <= x <= 1

    If (x-3) < 0 and (x+1) < 0 we have
    -(x+1) <= -(x-3)
    -x-1 <= -x+3
    -1 < 3
    So x < -1 is also a solution.

    Combining the solution sets, we see that

    x <= 1 is the complete solution set

    To see how this works graphically, visit

    http://rechneronline.de/function-graphs/

    and enter
    abs(x+1)
    abs(x-3)
    as your two functions, and set the domain for x from -5 to 5, and the range y from 0 to 5

    This is an excellent place to try out viewing function graphs to confirm your algebra.

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