# Thread: Quick question

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 ??
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??!

2. 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}
\ln\,(xy)&\ge -1\\
xy&\ge e^{-1}.\quad\quad (e^x\mbox{ increasing})\\
\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!