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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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Have you tried to draw a picture in the $xOy$ plane? –  Jack Mar 26 '12 at 3:34
    
Of the domain you mean? –  Jonas Mar 26 '12 at 3:37
    
I mean a picture for the two inequalities in your post. –  Jack Mar 26 '12 at 5:31

2 Answers 2

up vote 1 down vote accepted

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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Both inequalities are necessary, since we are not give that $y\geq 0$. –  Brett Frankel Mar 26 '12 at 3:41
    
What a silly mistake. Thanks for pointing it out. –  Isaac Solomon Mar 26 '12 at 3:43
    
I've tried to understand how you've resoned regarding the inverse function. So far i know that we need to have an injective function in order for a inverse to exist, but when you say image, are you refering to the range? That can't be right though since the range don't include any negative coordinates...please bear with me, really want to understand it –  Jonas Mar 26 '12 at 3:49
    
In order to have an inverse, you need your function to be both surjective and injective. However, this function is not surjective, as you will never get, say, $(-1,-1)$ in the range (also known as the image). –  Isaac Solomon Mar 26 '12 at 3:53
    
Thanks a lot, finally got it now :) –  Jonas Mar 26 '12 at 4:06

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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Thanks, great answer :) –  Jonas Mar 26 '12 at 3:34
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You can use the arrows to vote for helpful questions, Jonas! –  The Chaz 2.0 Mar 26 '12 at 3:55

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