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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![### Analysis of the Function and Graph Sketching
#### Problem Statement:
Analyze and sketch a graph of the function. Find any intercepts, relative extrema, points of inflection, and asymptotes.
#### Function Given:
\[ y = \frac{x}{x^2 + 4} \]
### Step-by-Step Analysis:
1. **Intercepts:**
- **X-intercept:** Set \( y = 0 \)
\[ \frac{x}{x^2 + 4} = 0 \]
Solving for \( x \), we get \( x = 0 \). So, the x-intercept is at (0, 0).
- **Y-intercept:** Set \( x = 0 \)
\[ y = \frac{0}{0^2 + 4} = 0 \]
So, the y-intercept is also at (0, 0).
2. **Relative Extrema:**
- To find relative maxima or minima, first find the derivative of \( y \):
\[
y = \frac{x}{x^2 + 4}
\]
Differentiate using the quotient rule:
\[
y' = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{x^2 + 4 - 2x^2}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}
\]
Set \( y' = 0 \):
\[
\frac{4 - x^2}{(x^2 + 4)^2} = 0 \implies 4 - x^2 = 0 \implies x^2 = 4 \implies x = \pm2
\]
To classify these critical points, use the second derivative test.
3. **Points of Inflection:**
- Assume the second derivative (\( y'' \)) is found and analyze possible points where concavity changes.
4. **Asymptotes:**
- **Vertical Asymptotes:** The denominator \( x^2 + 4 \neq 0 \) for any real \( x \), thus there are no vertical asymptotes.
- **Horizontal](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6e1a3f1-b538-4464-8269-7761edea80d0%2F350d7364-1224-41b0-95b1-209848105248%2F5b7dd33_processed.png&w=3840&q=75)
Transcribed Image Text:### Analysis of the Function and Graph Sketching
#### Problem Statement:
Analyze and sketch a graph of the function. Find any intercepts, relative extrema, points of inflection, and asymptotes.
#### Function Given:
\[ y = \frac{x}{x^2 + 4} \]
### Step-by-Step Analysis:
1. **Intercepts:**
- **X-intercept:** Set \( y = 0 \)
\[ \frac{x}{x^2 + 4} = 0 \]
Solving for \( x \), we get \( x = 0 \). So, the x-intercept is at (0, 0).
- **Y-intercept:** Set \( x = 0 \)
\[ y = \frac{0}{0^2 + 4} = 0 \]
So, the y-intercept is also at (0, 0).
2. **Relative Extrema:**
- To find relative maxima or minima, first find the derivative of \( y \):
\[
y = \frac{x}{x^2 + 4}
\]
Differentiate using the quotient rule:
\[
y' = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{x^2 + 4 - 2x^2}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}
\]
Set \( y' = 0 \):
\[
\frac{4 - x^2}{(x^2 + 4)^2} = 0 \implies 4 - x^2 = 0 \implies x^2 = 4 \implies x = \pm2
\]
To classify these critical points, use the second derivative test.
3. **Points of Inflection:**
- Assume the second derivative (\( y'' \)) is found and analyze possible points where concavity changes.
4. **Asymptotes:**
- **Vertical Asymptotes:** The denominator \( x^2 + 4 \neq 0 \) for any real \( x \), thus there are no vertical asymptotes.
- **Horizontal
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