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3. Determine the domain of the following function, then sketch it: 1 f(x, y) = V² + y? – 9

3. Determine the domain of the following function, then sketch it: 1 f(x, y) = V² + y? – 9

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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Question
**Question 3:** Determine the domain of the following function, then sketch it: \[ f(x, y) = \frac{1}{\sqrt{x^2 + y^2 - 9}} \] **Explanation:** The function \( f(x, y) \) is defined as \( \frac{1}{\sqrt{x^2 + y^2 - 9}} \). The domain of this function is determined by the values of \( (x, y) \) for which the expression inside the square root is positive, since the square root and division by zero are undefined for non-positive numbers. Thus, for \( f(x, y) \) to be defined, the condition is: \[ x^2 + y^2 - 9 > 0 \] This implies: \[ x^2 + y^2 > 9 \] The domain includes all points \((x, y)\) that are outside the circle defined by the equation \( x^2 + y^2 = 9 \), which is a circle with radius 3, centered at the origin \((0,0)\).
Transcribed Image Text:**Question 3:** Determine the domain of the following function, then sketch it: \[ f(x, y) = \frac{1}{\sqrt{x^2 + y^2 - 9}} \] **Explanation:** The function \( f(x, y) \) is defined as \( \frac{1}{\sqrt{x^2 + y^2 - 9}} \). The domain of this function is determined by the values of \( (x, y) \) for which the expression inside the square root is positive, since the square root and division by zero are undefined for non-positive numbers. Thus, for \( f(x, y) \) to be defined, the condition is: \[ x^2 + y^2 - 9 > 0 \] This implies: \[ x^2 + y^2 > 9 \] The domain includes all points \((x, y)\) that are outside the circle defined by the equation \( x^2 + y^2 = 9 \), which is a circle with radius 3, centered at the origin \((0,0)\).
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