57325-0136147054_12

57325-0136147054_12 - Section 12.2 12.2.1 Because f x y = 4 3 x 2 y is defined for all x and y the domain of f is the entire two-dimensional plane

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Section 12.2 12.2.1: Because f ( x, y ) = 4 3 x 2 y is defined for all x and y , the domain of f is the entire two-dimensional plane. 12.2.2: Because x 2 + 2 y 2 = 0 for all x and y , the domain of f ( x, y ) = p x 2 + 2 y 2 is the entire two- dimensional plane. 12.2.3: If either x or y is nonzero, then x 2 + y 2 > 0, and so f ( x, y ) is definedbut not if x = y = 0. Hence the domain of f consists of all points ( x, y ) in the plane other than the origin. 12.2.4: If x 6 = y then the denominator in f ( x, y ) is nonzero, and thus f ( x, y ) is definedbut not if x = y . So the domain of f consists of all those points ( x, y ) in the plane for which y 6 = x . 12.2.5: The real number z has a unique cube root z 1 / 3 regardless of the value of z . Hence the domain of f ( x, y ) = ( y x 2 ) 1 / 3 consists of all points in the xy-plane. 12.2.6: The real number z has a unique cube root z 1 / 3 regardless of the value of z . But 2 x is real if and only if x = 0. Therefore the domain of f ( x, y ) = (2 x ) 1 / 2 + (3 y ) 1 / 3 consists of all those points ( x, y ) for which x = 0. 12.2.7: Because arcsin z is a real number if and only if 1 5 z 5 1, the domain of the given function f ( x, y ) = sin 1 ( x 2 + y 2 ) consists of those points ( x, y ) in the xy-plane for which x 2 + y 2 5 1; that is, the set of all points on and within the unit circle. 12.2.8: Because arctan z is defined for every real number z , the only obstruction to the computation of f ( x ) = arctan( y/x ) is the possibility that x = 0. This obstruction is insurmountable, and therefore the domain of f consists of all those points ( x, y ) in the xy-plane for which x 6 = 0; that is, all points other than those on the y-axis. 12.2.9: For every real number z , exp( z ) is defined and unique. Therefore the domain of the given function f ( x, y ) = exp( x 2 y 2 ) consists of all points ( x, y ) in the entire xy-plane. 12.2.10: Because ln z is a unique real number if and only if z > 0, the domain of f ( x, y ) = ln( x 2 y 2 1) consists of those points ( x, y ) for which x 2 y 2 1 > 0; that is, for which y 2 < x 2 1. This is the region bounded by the hyperbola with equation x 2 y 2 = 1, shown shaded in the following figure; the bounding hyperbola itself is not part of the domain of f . 1441 x y 12.2.11: Because ln z is a unique real number if and only if z > 0, then domain of f ( x, y ) = ln( y x ) consists of those points ( x, y ) for which y > x . This is the region above the graph of the straight line with equation y = x (the line itself is not part of the domain of f ). 12.2.12: Because z is a unique real number if and only if z = 0, the domain of the given function f ( x, y ) = p 4 x 2 y 2 consists of those points ( x, y ) for which x 2 + y 2 5 4. That is, the domain consists of all those points ( x, y ) on and within the circle with center (0 , 0) and radius 2....
View Full Document

This document was uploaded on 11/28/2010.

Page1 / 213

57325-0136147054_12 - Section 12.2 12.2.1 Because f x y = 4 3 x 2 y is defined for all x and y the domain of f is the entire two-dimensional plane

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online