a) Find the function's domain . b) Find the function's range. c) Find the boundary of the function's domain. d) Determine if the domain is an open region, a closed region , or neither. e) Decide if the domain is bounded or unbounded. for: f(x,y) = 1/((16- x^2 -y^2)^(1/2))

Rivka Thorpe 2020-11-02 Answered
a) Find the functions
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Expert Answer

Arham Warner
Answered 2020-11-03 Author has 102 answers

a)Determine the domain of the multivariable function.
f(x,y)=116x2y2
16x2y20
x2y216
1×(x2y2)1×16
x2+y216
Also, 16x2y2>0
16x2y2>0
x2+y2<16
From, x2+y216andx2+y2<16
The domain for f(x,y) is (x,y) R2:x2+y2<16.
b) Evaluate the range for the given function.
Range for 16x2y2is(0,16].
Therefore, range for f(x,y) is
f(x,y)=116x2y2
As 16x2y20,f(x,y)
As 16x2y216,f(x,y)14
Therefore, range for f(x,y)[14,).
c)Determine the boundary of the domain using the domain for f(x, y).
The domain for f(x,y) is (x,y) R2:x2+y2<16.
The equation x2+y2<16 is a circle.
The boundary for the domain is points lying in the circle x2+y2=16.
d) The domain for the function f(x, y) do not contain the boundary points. Therefore, the domain for the function is an open region.
e) The domain for the function is enclosed in the circle.
Hence, the domain is bounded in the circle of equation x2+y2<16.

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