Find the domain of each of the following functions: f(x, y, z) 3x-4y+2z √9-x²-y²-2² = {(x, y, z) = R³ | x² + y² + z² < 8} ○ {(x, y, z) = R³ | x² + y² + z² ≤ 9} ○ {(x, y, z) = R³ | x² + y² + z² <9} {(x, y, z) = R³ | x² + y² + x² > 9}
Find the domain of each of the following functions: f(x, y, z) 3x-4y+2z √9-x²-y²-2² = {(x, y, z) = R³ | x² + y² + z² < 8} ○ {(x, y, z) = R³ | x² + y² + z² ≤ 9} ○ {(x, y, z) = R³ | x² + y² + z² <9} {(x, y, z) = R³ | x² + y² + x² > 9}
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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![### Determining the Domain of a Given Function
In this exercise, you are tasked with finding the domain of the given function:
\[ f(x, y, z) = \frac{3x - 4y + 2z}{\sqrt{9 - x^2 - y^2 - z^2}} \]
To determine the domain, we must identify the set of all points \((x, y, z)\) for which the function is defined. Specifically, we need to ensure that the denominator is not zero and the expression under the square root is non-negative.
The expression under the square root, \(9 - x^2 - y^2 - z^2\), must be positive for the square root to be real and defined. Therefore, we need:
\[ 9 - x^2 - y^2 - z^2 > 0 \]
This simplifies to:
\[ x^2 + y^2 + z^2 < 9 \]
Thus, the domain of the function \(f(x, y, z)\) consists of all points \((x, y, z)\) in \(\mathbb{R}^3\) (3-dimensional real space) such that the sum of their squares is less than 9.
### Options for the Domain
Given this inequality, let's analyze the provided options:
1. \(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 < 8\}\)
2. \(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \leq 9\}\)
3. \(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 < 9\}\)
4. \(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 > 9\}\)
The correct option is:
\(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 < 9\}\)
This correctly represents the set of points \((x,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde5535f4-bfc5-4672-a20d-4797632e4aa9%2Fbfb08200-36cd-4704-8a62-65c39a861843%2Fogaqcxh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining the Domain of a Given Function
In this exercise, you are tasked with finding the domain of the given function:
\[ f(x, y, z) = \frac{3x - 4y + 2z}{\sqrt{9 - x^2 - y^2 - z^2}} \]
To determine the domain, we must identify the set of all points \((x, y, z)\) for which the function is defined. Specifically, we need to ensure that the denominator is not zero and the expression under the square root is non-negative.
The expression under the square root, \(9 - x^2 - y^2 - z^2\), must be positive for the square root to be real and defined. Therefore, we need:
\[ 9 - x^2 - y^2 - z^2 > 0 \]
This simplifies to:
\[ x^2 + y^2 + z^2 < 9 \]
Thus, the domain of the function \(f(x, y, z)\) consists of all points \((x, y, z)\) in \(\mathbb{R}^3\) (3-dimensional real space) such that the sum of their squares is less than 9.
### Options for the Domain
Given this inequality, let's analyze the provided options:
1. \(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 < 8\}\)
2. \(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \leq 9\}\)
3. \(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 < 9\}\)
4. \(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 > 9\}\)
The correct option is:
\(\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 < 9\}\)
This correctly represents the set of points \((x,
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