Problem 14 - Precalculus Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{2 x}{(x-1)(x+1)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{1}{x-1} + \frac{1}{x+1} \).

Step 1: Identify Denominator Factors

The denominator of the function \( \frac{2x}{(x-1)(x+1)} \) is \((x-1)(x+1)\). These are linear factors, meaning they can potentially be used as denominators in the partial fractions.

Step 2: Set Up the Partial Fraction Decomposition

For linear factors \((x-1)\) and \((x+1)\), we set up the partial fraction decomposition as follows: \( \frac{2x}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \). Here, \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides by \((x-1)(x+1)\) to eliminate the denominators: \[ 2x = A(x+1) + B(x-1) \]

Step 4: Expand and Simplify

Expand the right side: \[ 2x = Ax + A + Bx - B \] Combine like terms to get: \[ 2x = (A + B)x + (A - B) \]

Step 5: Equate Coefficients

By equating the coefficients of corresponding terms, we get two equations: 1. \( A + B = 2 \)2. \( A - B = 0 \)

Step 6: Solve the System of Equations

Solve the system of equations:1. From \( A - B = 0 \), we have \( A = B \).2. Substitute \( A = B \) into \( A + B = 2 \): \( A + A = 2 \Rightarrow 2A = 2 \Rightarrow A = 1 \). Hence, \( B = 1 \) as well.

Step 7: Write the Decomposition

Substitute \( A = 1 \) and \( B = 1 \) back into the partial fractions: \[ \frac{2x}{(x-1)(x+1)} = \frac{1}{x-1} + \frac{1}{x+1} \]

Key Concepts

Linear Factors

In algebra, linear factors are essential components, especially when dealing with polynomial and rational functions. A linear factor is simply a factor of the form \(ax + b\), where "a" and "b" are constants. For example, in the rational function \( \frac{2x}{(x-1)(x+1)} \), the linear factors are \((x-1)\) and \((x+1)\). These are the building blocks that help to break down complex expressions into simpler, manageable pieces.

Recognizing linear factors is the first step in the process of partial fraction decomposition. This step ensures that we correctly identify the pieces needed to set up the equation for decomposition. Each factor in the denominator of a rational function represents a potential separate fraction in the decomposition process. Linear factors are straightforward since they are non-repeating and do not have exponents higher than one.

• Linear factors: Basic component forms like \( (x-1) \), \( (x+1) \)
• No exponents larger than one, simplifying setup for partial fractions
• Used as denominators in partial fraction decomposition

Rational Functions

A rational function is a ratio of two polynomials. The general form of a rational function is \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x) eq 0\). The function \( \frac{2x}{(x-1)(x+1)} \) is a perfect example of a rational function where \(P(x) = 2x\) and \(Q(x) = (x-1)(x+1)\). Understanding the characteristics of rational functions is essential as they appear frequently in calculus and when solving real-world problems.

When working with rational functions, it's often helpful to simplify them into smaller, manageable parts. Partial fraction decomposition is a useful technique for breaking down rational functions into a sum of simpler fractions. This method not only simplifies calculations but also provides an easier approach to integration and analysis.

• Comprised of a numerator and a non-zero polynomial denominator
• Partial fraction decomposition simplifies complex rational functions
• Useful in calculus for integration and solving differential equations

System of Equations

Solving a system of equations is a key part in finding the coefficients in partial fraction decomposition. This means figuring out the values of unknowns in the decomposed fractions. For example, after setting up the equation \(2x = A(x+1) + B(x-1)\), we expanded and rearranged it to get a system of linear equations: \(A + B = 2 \) and \(A - B = 0\).

This system is straightforward to solve because it's made up of two linear equations. The solution process typically involves substitution or elimination methods, and in this example, we determined that \(A = 1\) and \(B = 1\). Solving systems of equations is a valuable skill in algebra since it allows us to find precise solutions for unknown values based on given conditions.

• Key step: Equate coefficients to form a solvable system
• Methods: Substitution or elimination offers clear solutions
• Outcome: Determine exact values for partial fraction coefficients