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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![### Topic: Partial Fraction Decomposition
This set of notes demonstrates the process of partial fraction decomposition with a specific example. The goal is to find the constants \(A\), \(B\), and \(C\).
#### Given Equation:
\[
\frac{3x + 1}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3}
\]
#### Steps:
1. **Express the Equation:**
\[
3x + 1 = A(x + 3) + B(x - 2)
\]
2. **Expand and Combine Like Terms:**
\[
3x + 1 = (A + B)x + (3A - 2B)
\]
3. **Setup the System of Equations:**
- By comparing coefficients, we have:
- \( A + B = 3 \)
- \( 3A - 2B = 1 \)
4. **Solve for Constants:**
- **Using \( x = 0 \):**
Plugging \( x = 0 \) into the equation:
\[
1 = 3A + 2C \implies 1 = 3A + 2C = 2
\]
- Simplifying gives \( C \).
- **Using \( x = 2 \):**
Plugging \( x = 2 \) into the equation:
\[
7 = A(7) \implies A = 1
\]
- **Using \( x = 1 \):**
Plugging \( x = 1 \) into the equation:
\[
4 = 5 - B \implies B = 1
\]
#### Final Solution:
The constants are determined as \( A = 1 \), \( B = 1 \), and \( C \) is found from the equation system.
These values are boxed and highlighted in the solution as follows:
- \( A = 1 \)
- \( B = 1 \)
- Appropriate calculations are shown, with equations and simplifications.
This example demonstrates how to decompose a rational function into partial fractions by solving a system of equations derived from equating coefficients.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd524c353-1214-4780-ae6a-4868b2895696%2F598cbe4d-cb95-4e3a-a511-385b796e0c09%2Fdrh435_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Topic: Partial Fraction Decomposition
This set of notes demonstrates the process of partial fraction decomposition with a specific example. The goal is to find the constants \(A\), \(B\), and \(C\).
#### Given Equation:
\[
\frac{3x + 1}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3}
\]
#### Steps:
1. **Express the Equation:**
\[
3x + 1 = A(x + 3) + B(x - 2)
\]
2. **Expand and Combine Like Terms:**
\[
3x + 1 = (A + B)x + (3A - 2B)
\]
3. **Setup the System of Equations:**
- By comparing coefficients, we have:
- \( A + B = 3 \)
- \( 3A - 2B = 1 \)
4. **Solve for Constants:**
- **Using \( x = 0 \):**
Plugging \( x = 0 \) into the equation:
\[
1 = 3A + 2C \implies 1 = 3A + 2C = 2
\]
- Simplifying gives \( C \).
- **Using \( x = 2 \):**
Plugging \( x = 2 \) into the equation:
\[
7 = A(7) \implies A = 1
\]
- **Using \( x = 1 \):**
Plugging \( x = 1 \) into the equation:
\[
4 = 5 - B \implies B = 1
\]
#### Final Solution:
The constants are determined as \( A = 1 \), \( B = 1 \), and \( C \) is found from the equation system.
These values are boxed and highlighted in the solution as follows:
- \( A = 1 \)
- \( B = 1 \)
- Appropriate calculations are shown, with equations and simplifications.
This example demonstrates how to decompose a rational function into partial fractions by solving a system of equations derived from equating coefficients.
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