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Math Help - Linear approximation

  1. #1
    DBA
    DBA is offline
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    Linear approximation

    For what values of x is the linear approximation sqrt(x+3) ~ 7/4 + x/4
    accurate to within 0.5?
    "~" stands for about

    Solution: Accuracy to within 0.5 means that the function should differ by less than 0.5:

    |sqrt(x+3) - ( 7/4 + x/4) | > 0.5

    Equivalently we could write

    sqrt(x+3) - 0.5 < ( 7/4 + x/4) < sqrt(x+3) + 0.5

    I have two questions:
    First, how do they get from "linear approximation sqrt(x+3) ~ 7/4 + x/4 accurate to within 0.5" to the expression "|sqrt(x+3) - ( 7/4 + x/4) | > 0.5"

    Second, how do they get the equvalent term from
    |sqrt(x+3) - ( 7/4 + x/4) | > 0.5
    to
    sqrt(x+3) - 0.5 < ( 7/4 + x/4) < sqrt(x+3) + 0.5

    I thought |u| < a --> -a < u < a
    So, I get -0.5 < sqrt(x+3) - ( 7/4 + x/4) < 0.5

    What do I have to do to get to
    sqrt(x+3) - 0.5 < ( 7/4 + x/4) < sqrt(x+3) + 0.5

    Thanks for any help!

    Additionally I want to add this information I have:
    F(x)= sqrt(x+3)
    Linear approximation f(x) ~ f(a)+f'(a)(x-a)
    a=1
    Last edited by DBA; October 5th 2009 at 08:39 PM. Reason: additional information
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  2. #2
    MHF Contributor
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    If the desired error is less than .5, the starting equation should be |sqrt(x+3) - ( 7/4 + x/4) | < 0.5. You wrote > instead of <.
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