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5. Sketch the domain of f. (a). (Q24) f(x, y) = √√√x² + y² - 4. (b). (Q28(b)) f(x, y) = ln(y - 2x).

5. Sketch the domain of f. (a). (Q24) f(x, y) = √√√x² + y² - 4. (b). (Q28(b)) f(x, y) = ln(y - 2x).

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**5. Sketch the domain of \( f \).** (a). (Q24) \( f(x, y) = \sqrt{x^2 + y^2 - 4} \). **Graph Explanation:** - The graph is a coordinate plane with horizontal \( x \)-axis and vertical \( y \)-axis. - The origin is labeled as \( O \). - The function involves a square root, implying that the expression inside the square root, \( x^2 + y^2 - 4 \), must be greater than or equal to zero for the domain to be real numbers. This represents a region outside or on a circle centered at the origin with radius 2. (b). (Q28(b)) \( f(x, y) = \ln(y - 2x) \). **Graph Explanation:** - This graph is similar to the first one, with \( x \) and \( y \)-axes and origin labeled \( O \). - The function involves a natural logarithm, implying that the argument \( y - 2x \) must be greater than zero for the function to be defined. Thus, the domain is the region of the plane above the line \( y = 2x \).
Transcribed Image Text:**5. Sketch the domain of \( f \).** (a). (Q24) \( f(x, y) = \sqrt{x^2 + y^2 - 4} \). **Graph Explanation:** - The graph is a coordinate plane with horizontal \( x \)-axis and vertical \( y \)-axis. - The origin is labeled as \( O \). - The function involves a square root, implying that the expression inside the square root, \( x^2 + y^2 - 4 \), must be greater than or equal to zero for the domain to be real numbers. This represents a region outside or on a circle centered at the origin with radius 2. (b). (Q28(b)) \( f(x, y) = \ln(y - 2x) \). **Graph Explanation:** - This graph is similar to the first one, with \( x \) and \( y \)-axes and origin labeled \( O \). - The function involves a natural logarithm, implying that the argument \( y - 2x \) must be greater than zero for the function to be defined. Thus, the domain is the region of the plane above the line \( y = 2x \).
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given function f left parenthesis x comma y right parenthesis equals square root of x squared plus y squared minus 4 end root

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