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Thread: Quick question

  1. #1
    Jan 2009

    Quick question

    find and sketch the domain of f(x,y) = sqrt(1+ln(xy))
    it will be:
    graph of y=1/ex is approximately the same as y=1/x
    where should i shade it ??
    it have curve in quadratic 1 and curve in quadratic 3
    i will shade it inside both curves or just the curve in the quadratic 1??

    i think,
    since y > 1/ex
    ---> xy > 1/e
    product is positive
    so I will shade the curve in quadratic 3
    Last edited by TWiX; Dec 22nd 2009 at 07:24 AM.
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  2. #2
    Senior Member
    Dec 2008
    If our function is

    $\displaystyle f(x,y)=\sqrt{1+\ln\,(xy)},$

    then $\displaystyle f$ is defined exactly where $\displaystyle 1+\ln\,(xy)$ is defined and non-negative; that is, where

    $\displaystyle 1+\ln\,(xy)\ge 0.$

    By two reversible operations, we find that this is equivalent to saying

    $\displaystyle \begin{aligned}
    \ln\,(xy)&\ge -1\\
    xy&\ge e^{-1}.\quad\quad (e^x\mbox{ increasing})\\

    Therefore, $\displaystyle xy$ must be positive and $\displaystyle (x,y)$ must lie in the first or third quadrant. Now, if we knew that $\displaystyle x$ were positive, we could divide by $\displaystyle x$ to obtain

    $\displaystyle y\ge \frac{1}{ex},$

    but in the third quadrant, $\displaystyle x<0$, and we can't do this without reversing the inequality. In fact, while your answer is correct in the first quadrant, in the third quadrant it will be

    $\displaystyle y \le \frac{1}{ex}.$

    Hope this helps!
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