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Math Help - Inequalities

  1. #1
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    Inequalities

    The equation of a curve is x^2y^2-x^2+y^2=0
    a) find the equation of the tangents at the origin
    b) find the equations of the real asymptotes
    c) show that the numerical value of y is never greater than the corresponding value of x
    d)show that the numerical value of y is always less than unity

    My problem is with (c) and (d). I've done (a) and (b) already.
    For (c), x-y\geq 0
    y=\pm\frac{x}{\sqrt{x^2+1}}
    Substitute
    x-\frac{x}{\sqrt{x^2+1}}=\frac{x\sqrt{x^2+1}-x}{\sqrt{x^2+1}}
    Now I don't know how to continue to make it such that it is greater or equal to 0, which then i can show that y is always lesser of equal to x.
    Same problem with the next one. The \sqrt{x^2+1} is what stumble me.
    Thanks!
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  2. #2
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    I think you should rewrite this as

    y = \left|\frac{x}{\sqrt{x^2 + 1}}\right|.


    Then to show that y \leq x, you would have

    \left|\frac{x}{\sqrt{x^2 + 1}}\right| \leq x

    -x \leq \frac{x}{\sqrt{x^2 + 1}} \leq x

    -1 \leq \frac{1}{\sqrt{x^2 + 1}}\leq 1.


    This would only ever not be true if the denominator was smaller than the numerator. So it would not be true if \sqrt{x^2 + 1} < 1

    If x^2 + 1 < 1 then

    x^2 < 0.

    Since this is never true, that means that

    -1 \leq \frac{1}{\sqrt{x^2 + 1}} \leq 1 IS true.

    Therefore y \leq x for all x.


    To answer d) you want to show that

    y = \left|\frac{x}{\sqrt{x^2 + 1}}\right| < 1.


    Therefore -1 < \frac{x}{\sqrt{x^2 + 1}} < 1

    -\sqrt{x^2 + 1} < x < \sqrt{x^2 + 1}


    This should be obvious, since x = \sqrt{x^2}.

    \sqrt{x^2} < \sqrt{x^2 + 1}.
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