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INT
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f(x,y)=sqrt(yx)
Find the boundary of the function's domain.
 2 years ago
 2 years ago
INT Group Title
f(x,y)=sqrt(yx) Find the boundary of the function's domain.
 2 years ago
 2 years ago

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remnant Group TitleBest ResponseYou've already chosen the best response.1
what have you tried so far?
 2 years ago

INT Group TitleBest ResponseYou've already chosen the best response.1
I know it has something to do with interior points and boundary points, but I dont get it. My current understanding is that interior points are arbitrary points in the function's range that if we make them the center of a disk with positive radius, all the points in that disk are also in R. Then boundary points are the same except that the disk has some points inside R and some outside R.  This doesnt make sense to me because wouldn't it depend on how big the radius of the disk is? how do you know how big the radius is?
 2 years ago

remnant Group TitleBest ResponseYou've already chosen the best response.1
Okay, so you have,\[\large f(x, y) = \sqrt{yx}\] and you want to try and find the boundary of the domain. So, you want to find out when this function is valid. So when is it? Well, we know that for the square root of anything, that anything must be greater to or equal to zero so we have,\[y  x \ge 0\] so,\[y \ge x\]
 2 years ago

INT Group TitleBest ResponseYou've already chosen the best response.1
yeah, I've gotten that so far.
 2 years ago

remnant Group TitleBest ResponseYou've already chosen the best response.1
dw:1352071930007:dw
 2 years ago

remnant Group TitleBest ResponseYou've already chosen the best response.1
that is where the function would be valid, whcih means that the boundary of the func's domain would be the line y=x
 2 years ago

INT Group TitleBest ResponseYou've already chosen the best response.1
so basically it's just graphing the domain inequality.
 2 years ago

remnant Group TitleBest ResponseYou've already chosen the best response.1
Seems to be that way, yes
 2 years ago
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