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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![### Domain of a Function of Two Variables
To determine the domain of the function of two variables:
\[ g(x, y) = \frac{9}{2y + 6x^2} \]
The function \(g(x, y)\) is defined for:
\[ \{ (x, y) \mid y \neq \boxed{\phantom{0}} \} \]
#### Explanation:
- The domain of \(g(x, y)\) includes all pairs \((x, y)\) except those that make the denominator zero.
- To avoid division by zero, the condition \(2y + 6x^2 \neq 0\) must be satisfied.
- Solving \(2y + 6x^2 = 0\) results in \(y \neq -3x^2\).
- Therefore, the function \(g(x, y)\) is defined for all real numbers \(x\) and \(y\) except where \(y = -3x^2\).
### Summary
The set notation for the domain reflects this constraint:
\[ \{ (x, y) \mid y \neq -3x^2 \} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Febbbc1cc-ba98-4671-a28a-276fb59e8560%2F6df97f61-9536-4de0-86ff-574bba1ccb9e%2F6oqu7b_processed.png&w=3840&q=75)
Transcribed Image Text:### Domain of a Function of Two Variables
To determine the domain of the function of two variables:
\[ g(x, y) = \frac{9}{2y + 6x^2} \]
The function \(g(x, y)\) is defined for:
\[ \{ (x, y) \mid y \neq \boxed{\phantom{0}} \} \]
#### Explanation:
- The domain of \(g(x, y)\) includes all pairs \((x, y)\) except those that make the denominator zero.
- To avoid division by zero, the condition \(2y + 6x^2 \neq 0\) must be satisfied.
- Solving \(2y + 6x^2 = 0\) results in \(y \neq -3x^2\).
- Therefore, the function \(g(x, y)\) is defined for all real numbers \(x\) and \(y\) except where \(y = -3x^2\).
### Summary
The set notation for the domain reflects this constraint:
\[ \{ (x, y) \mid y \neq -3x^2 \} \]
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