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
Find the indicated function and write its domain in interval notation.
![### Problem Statement:
Find the indicated function and write its domain.
Given:
\[ p(x) = x^2 + 6x \]
\[ q(x) = \sqrt{1 - x} \]
Required:
\[ \left(\frac{q}{p}\right)(x) = ? \]
### Solution:
1. **Calculate \(\frac{q}{p}(x)\):**
\[
\frac{q(x)}{p(x)} = \frac{\sqrt{1 - x}}{x^2 + 6x}
\]
2. **Domain of \( \frac{q}{p}(x) \):**
- For \( q(x) = \sqrt{1 - x} \) to be real, \( 1 - x \geq 0 \). Therefore, \( x \leq 1 \).
- For \( \frac{q}{p}(x) \) to be defined, \( x^2 + 6x \neq 0 \).
- Solve \( x^2 + 6x = 0 \):
\[
x(x + 6) = 0 \implies x = 0 \text{ or } x = -6
\]
Hence, the domain of \( \frac{q}{p}(x) \) is all values of \( x \) such that \( x \leq 1 \) and \( x \neq 0 \) and \( x \neq -6 \).
### Graph Explanation:
In the document, there is a small graph shown:
- The graph represents a function with a visible curve.
- The vertical axis ranges from \(-5\) to \(5\).
- The horizontal axis ranges from \(-5\) to \(5\).
(Note: Since this is an educational website, teachers and students should use graphing tools to get an intuitive understanding of the function's behavior and constraints, adhering to the domain restrictions.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feec97995-04ea-480a-b226-22e319b4bb5a%2Fb27e2cde-8943-4a40-81de-65a204578cba%2Fs79s2kp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Find the indicated function and write its domain.
Given:
\[ p(x) = x^2 + 6x \]
\[ q(x) = \sqrt{1 - x} \]
Required:
\[ \left(\frac{q}{p}\right)(x) = ? \]
### Solution:
1. **Calculate \(\frac{q}{p}(x)\):**
\[
\frac{q(x)}{p(x)} = \frac{\sqrt{1 - x}}{x^2 + 6x}
\]
2. **Domain of \( \frac{q}{p}(x) \):**
- For \( q(x) = \sqrt{1 - x} \) to be real, \( 1 - x \geq 0 \). Therefore, \( x \leq 1 \).
- For \( \frac{q}{p}(x) \) to be defined, \( x^2 + 6x \neq 0 \).
- Solve \( x^2 + 6x = 0 \):
\[
x(x + 6) = 0 \implies x = 0 \text{ or } x = -6
\]
Hence, the domain of \( \frac{q}{p}(x) \) is all values of \( x \) such that \( x \leq 1 \) and \( x \neq 0 \) and \( x \neq -6 \).
### Graph Explanation:
In the document, there is a small graph shown:
- The graph represents a function with a visible curve.
- The vertical axis ranges from \(-5\) to \(5\).
- The horizontal axis ranges from \(-5\) to \(5\).
(Note: Since this is an educational website, teachers and students should use graphing tools to get an intuitive understanding of the function's behavior and constraints, adhering to the domain restrictions.)
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