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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![**Analyzing and Sketching the Graph of the Function**
**Problem Statement:**
Analyze and sketch a graph of the function. Find any intercepts, relative extrema, points of inflection, and asymptotes.
\[ y = \frac{x} {x^2 + 4} \]
### Steps to Analyze and Sketch the Graph:
1. **Identify Intercepts:**
- **X-intercepts:** Set \( y = 0 \) and solve for \( x \).
\[ 0 = \frac{x} {x^2 + 4} \]
Thus, \( x = 0 \) is the x-intercept.
- **Y-intercepts:** Set \( x = 0 \) and solve for \( y \).
\[ y = \frac{0} {0^2 + 4} = 0 \]
Thus, \( y = 0 \) is the y-intercept.
2. **Determine Relative Extrema:**
- First, find the first derivative of \( y \) to locate critical points.
\[ y' = \frac{(x^2 + 4)(1) - x(2x)} {(x^2 + 4)^2} = \frac{x^2 + 4 - 2x^2} {(x^2 + 4)^2} = \frac{4 - x^2} {(x^2 + 4)^2} \]
- Set \( y' = 0 \) to find critical points.
\[ \frac{4 - x^2} {(x^2 + 4)^2} = 0 \]
So, \( x = \pm 2 \).
- Determine if these points are relative maxima or minima by evaluating the second derivative.
3. **Points of Inflection and Concavity:**
- Find the second derivative \( y'' \).
\[ y'' = \frac{d} {dx} \left( \frac{4 - x^2} {(x^2 + 4)^2} \right) \]
This requires further differentiation which can be complex. Analyze the concavity based on the sign of \( y'' \).
4. **Asymptotes:**
- **Vertical Asymptotes:** Occur where the denominator is zero. Here, \( x^2 +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6e1a3f1-b538-4464-8269-7761edea80d0%2Fae1871ea-5a67-4b19-b481-cbea4f7d1ad3%2Fy7mn2hs_processed.png&w=3840&q=75)
Transcribed Image Text:**Analyzing and Sketching the Graph of the Function**
**Problem Statement:**
Analyze and sketch a graph of the function. Find any intercepts, relative extrema, points of inflection, and asymptotes.
\[ y = \frac{x} {x^2 + 4} \]
### Steps to Analyze and Sketch the Graph:
1. **Identify Intercepts:**
- **X-intercepts:** Set \( y = 0 \) and solve for \( x \).
\[ 0 = \frac{x} {x^2 + 4} \]
Thus, \( x = 0 \) is the x-intercept.
- **Y-intercepts:** Set \( x = 0 \) and solve for \( y \).
\[ y = \frac{0} {0^2 + 4} = 0 \]
Thus, \( y = 0 \) is the y-intercept.
2. **Determine Relative Extrema:**
- First, find the first derivative of \( y \) to locate critical points.
\[ y' = \frac{(x^2 + 4)(1) - x(2x)} {(x^2 + 4)^2} = \frac{x^2 + 4 - 2x^2} {(x^2 + 4)^2} = \frac{4 - x^2} {(x^2 + 4)^2} \]
- Set \( y' = 0 \) to find critical points.
\[ \frac{4 - x^2} {(x^2 + 4)^2} = 0 \]
So, \( x = \pm 2 \).
- Determine if these points are relative maxima or minima by evaluating the second derivative.
3. **Points of Inflection and Concavity:**
- Find the second derivative \( y'' \).
\[ y'' = \frac{d} {dx} \left( \frac{4 - x^2} {(x^2 + 4)^2} \right) \]
This requires further differentiation which can be complex. Analyze the concavity based on the sign of \( y'' \).
4. **Asymptotes:**
- **Vertical Asymptotes:** Occur where the denominator is zero. Here, \( x^2 +
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