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Math Help - Quick question

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    236

    Quick question

    find and sketch the domain of f(x,y) = sqrt(1+ln(xy))
    it will be:
    y>1/(ex)
    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
    right??!
    Last edited by TWiX; December 22nd 2009 at 07:24 AM.
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  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    If our function is

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

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

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

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

    \begin{aligned}<br />
\ln\,(xy)&\ge -1\\<br />
xy&\ge e^{-1}.\quad\quad (e^x\mbox{ increasing})\\<br />
\end{aligned}

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

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

    but in the third quadrant, 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

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

    Hope this helps!
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