anonymous
  • anonymous
Find and sketch the domains of the following functions.? a) f(x,y)=sqrt((x+y)/(x-y)) b) f(x,y)=sqrt(x+y)/sqrt(x-y)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
By sketching the domain of the function, we are sketch a region on the xy-plane. A. f(x, y) = sqrt((x+y)/(x-y)) Well it should be pretty obvious that x - y = 0 is a domain restriction because if we let x - y = 0 we'd be dividing by zero. So, x = y is one of our domain restrictions. Probably draw a dotted y = x line on the coordinate plane. If we let y = -x, then we would get a 0 inside the square root, provided (0, 0) is not the point chosen. So, draw a solid line y = -x to show that values on the line y = -x are in the domain. However, it may be good to put a hole on (0, 0) to show that value is not in the domain. Then, we have to look at possibilities. We want the quotient under the square root to be positive. For that to happen, the numerator and denominator have to both be positive or both be negative. So... x + y > 0 --> y > -x x - y > 0 --> y < x The region -x < y < x is the region containing the positive x-axis. Shade this region in because we are allowed to pick (x, y) points inside this region. x + y < 0 --> y < -x x - y < 0 --> y > x The region x < y < -x is the region containing the negative x-axis. Shade this region in because we are allowed to pick (x, y) points inside this region.
anonymous
  • anonymous
The only difference between part b and part a is that in part b, we are not allowed to have either function be negative. So, we are only allowed to shade where both x + y and x - y are positive. x + y > 0 --> y > -x x - y > 0 --> y < x So, we shade the region -x < y < x, which is the region containing the positive x-axis.
anonymous
  • anonymous
Thanks

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