1. ## 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 ??
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
right??!

2. 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})\\ \end{aligned}

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!