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Im having some problem figuring out the domain for a function:

$f(x,y)=(\sqrt {x+y},\sqrt {x-y})$

Since the square root is involved, i must have the expression within them >= 0, therefore i set up the following inequlities:

  1. $x-y\geqslant0$

  2. $x+y\geqslant0$ <--- is this one neccesary?

So my main problem, besides deciding whether or not it has an inverse, is that im unsure if i need to include the $x+y\geq 0$ constrain? or is $x-y\geq 0$ simply enough in this case.

Best Regards

Joe

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    I mean a picture for the two inequalities in your post.2012-03-26

2 Answers 2

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You have two constraints $x + y \geq 0$ and $x - y \geq 0$. These are equivalent to $x \geq - y$ and $x \geq y$. Since, as Brett pointed out, $y$ may be negative, both inequalities are necessary.

If this is a function from $\mathbb{R}^2$ to $\mathbb{R}^2$, it is not invertible, as its image does not contain elements with negative coordinates.

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    Thanks a lot, finally got it now :)2012-03-26
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You just need to know that $x+y$ and $x-y$ are $\geq 0$, since this is all you need to make sense of the square root function for real numbers.

Adding the two relations, we see that $2x=x+y+x-y\geq 0$, so we in fact see that $x\geq 0$. Since we must be able to subtract both $y$ and $-y$ from $x$ and get a positive number, we see that $x-|y|\geq 0$ This last relation is also sufficient, since we can deduce the original relations $x-y\geq 0$ and $x+y\geq 0$ from it.

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    You can use the arrows to vote for helpful questions, Jonas!2012-03-26