# Domain of a function - f(x,y)=root(1-x)+root(1-y)

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## Main Question or Discussion Point

I have came up with a solution for this - in order for this function to be defined, we must have an x and y between negative infinity up to and including the number 1.

If asked to graph this domain, does the domain lie on the x-y plane of three dimensional space, and is it the intersection of the domains of x and y?

I know the only sensible way to my question will likely go over my head - I Just started vector calculus - so no worries, I find it difficult for this to make sense to me.

I am thinking of the domain as the values "under" the z-plane in 3d space, if that makes any sense... at all - or above! or maybe a point in three-d space is ON the plane, when z is zero?

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jedishrfu
Mentor
i think this is simplay a 2D problem in x and y unless you want to make it 3D by stating f(x,y)=z

next what does the root(1-x) mean? I don't understand how to even plot this if I could.

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well it's 1-x all under a square root sign

root(1-x) = sqrt(1-x)

you know?

jedishrfu
Mentor
okay got it so thats why x>1 or y>1 is out of bounds if you're dealing with real numbers but okay for complex numbers.

so I got a sideways parabola from ending at z=0 x=1 y=1

You might google on graph 3D calculator and try your formula. I did but I didnt have the flash plugin installed (I'm on linux) so I had to imagine how it looks rotating the x curve around (1,1,0)

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really..hmm.. I got a rectangular plane - I don't think we are considering z here? hold it constant, and just focus on y and x right?

so we have x and y have both to be less than or equal to negative one. So they intersect on a rectangular plane going back to negative infinity....?

Assuming you are just dealing in real numbers, your domain is simply to do with x and y and it is $\{x,y \in\mathbb R: x \leq 1,\ y \leq 1\}$