topics_15_1_Functions_of_Several_Variables_15_2_Limits_and_Continuity (1)

# Topics_15_1_Functions_of_Several_Variables_15_2_Limits_and_Continuity (1)

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Unformatted text preview: 15.1 Multivariable Functions Problem 6/page 902. Let f ( x, y ) = ln( x + y- 1) . (a) Evaluate f (1 , 1), f ( e, 1), f (- 4 , 6). (b) Find and sketch the domain of f . (c) Find the range of f . Solution. (a) f (1 , 1) = ln(1 + 1- 1) = ln(1) = 0; f ( e, 1) = ln( e + 1- 1) = ln( e ) = 1; f (- 4 , 6) = ln(- 4 + 6- 1) = ln(1) = 0 . (b) The function z = f ( x, y ) is well defined for x + y- 1 > 0. All points ( x, y ) on the x, y- plane satisfying x + y- 1 > 0 form one of the half-planes determined by the straight line x + y- 1 = 0 . In order to determine which half-plane is the domain of z = ln( x + y- 1) we take a point say (2 , 2) in the upper half-plane and check if it satisfies x + y- 1 > 0. It does hence, graphically the domain is the upper half- plane: { ( x, y ) : x + y- 1 > } . (c) Obviously, the range of z = f ( x, y ) is (-∞ , ∞ ), i.e.,-∞ < z < ∞ . The plot of z = ln( x + y- 1) shows that geometrically, z = ln( x + y- 1) is a surface. The graph of each two-variable function is a surface. 1 Problem 16/page 902. Let f ( x, y ) = √ y + q 25- x 2- y 2 . (a) Find and sketch the domain D of f . (b) Find the range of f . Solution. (a) The function z = f ( x, y ) is well defined for y ≥ 0 and 25- x 2- y 2 ≥ 0. Graphically, y ≥ 0 is the upper half-plane determined by (with a boundary) y = 0. Graphically, 25- x 2- y 2 ≥ 0 is the interior of the circle x 2 + y 2 = 25 including the circle. This is the disk of radius 5, centered at the origin. The intersection of the half-plane and the disk will give the domain...
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## This note was uploaded on 02/05/2010 for the course MATH 264 taught by Professor Dr.d.dryanov during the Fall '09 term at Concordia Canada.

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Topics_15_1_Functions_of_Several_Variables_15_2_Limits_and_Continuity (1)

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