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Math Help - linear approximation

  1. #1
    Super Member bigwave's Avatar
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    linear approximation

    verify the given linear approximation at a=0, then determine the values of x for which the linear approximation is accurate to within 0.1

    \sqrt{1+x}\approx1+\frac{1}{2}x

    which assuming means
    \sqrt{1+x}-0.1<1+\frac{1}{2}x<\sqrt{1+x}+0.1

    but couldn't get the given answer of
    -0.69<x<1.09
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  2. #2
    Senior Member
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    First, we notice that the domain is x \ge -1.


    1+\frac{x}{2}<\sqrt{1+x}+0.1

    \implies~ 0.9+0.5x<\sqrt{1+x}

    Since both sides of the inequality are always non-negative over the domain x \ge -1:

    \implies~ (0.9+0.5x)^2<1+x

    \implies~ 0.81+0.9x+0.25x^2<1+x

    \implies~ 0.25x^2 - 0.1x - 0.19<0

    Solving this gives -0.694427<x<1.09443 which matches the solution you gave.



    By the way, the other side doesn't actually give any restriction:

    \sqrt{1+x}-0.1<1+\frac{x}{2}

    \implies~ \sqrt{1+x}<1.1+0.5x

    \implies~ 1+x < (1.1+0.5x)^2

    \implies~ 1+x < 1.21 + 1.1x + 0.25x^2

    \implies~ 0 < 0.25x^2 + 0.1x + 0.21

    Which is always true, so there is no additional restriction here.
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