Problem 15 - Precalculus Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{5}{(x-1)(x+4)}$$

Step-by-Step Solution

Answer

\(\frac{5}{(x-1)(x+4)} = \frac{1}{x-1} - \frac{1}{x+4}\).

Step 1: Identify the Denominator Factors

The given rational function is \( \frac{5}{(x-1)(x+4)} \). The denominator \((x-1)(x+4)\) is already factored into linear factors.

Step 2: Set Up the Partial Fraction Form

Since the denominator is the product of two distinct linear factors, the partial fraction decomposition will take the form: \( \frac{A}{x-1} + \frac{B}{x+4} \), where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides of the equation \( \frac{5}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4} \) by \((x-1)(x+4)\) to eliminate the fractions, resulting in: \[ 5 = A(x+4) + B(x-1) \]

Step 4: Expand the Right-Hand Side

Distribute \(A\) and \(B\) to get: \[ 5 = Ax + 4A + Bx - B \] Combine like terms: \[ 5 = (A+B)x + (4A-B) \]

Step 5: Solve for Constants A and B

For the equation \(5 = (A+B)x + (4A-B)\) to be an identity for all \(x\), the coefficients of corresponding terms on both sides must be equal.So, equate coefficients:1. \(A + B = 0\) (for the \(x\) term)2. \(4A - B = 5\) (constant term) From the first equation, \(A = -B\). Substitute into the second equation:\[ 4(-B) - B = 5 \]\[ -4B - B = 5 \]\[ -5B = 5 \]\[ B = -1 \]Now, substitute back to find \(A\):\[ A = -(-1) = 1 \]

Step 6: Write the Final Decomposition

Substitute \(A\) and \(B\) back into the partial fractions to obtain the decomposition: \( \frac{5}{(x-1)(x+4)} = \frac{1}{x-1} - \frac{1}{x+4} \).

Key Concepts

Rational Functions

A rational function is a type of function that can be expressed as the ratio of two polynomials. In simpler terms, it's a fraction where both the numerator and the denominator are polynomials. An example of a rational function is \( \frac{5}{(x-1)(x+4)} \). The numerator is "5," a constant polynomial, and the denominator is "\((x-1)(x+4)\)," a polynomial of degree two.

Rational functions can be quite complex when the polynomials in the numerator and denominator have higher degrees. However, understanding them is crucial because they appear frequently in calculus and algebra.

• These functions can exhibit behavior like vertical and horizontal asymptotes, which describe how the function behaves as \(x\) approaches certain values or infinity.
• The degree of these polynomials often dictates this behavior.
• Partial fraction decomposition is a technique used to simplify rational functions, which in turn, makes them easier to integrate or analyze.

Linear Factors

When dealing with rational functions, the denominators can often be broken down into simpler expressions called factors. These factors can either be linear, like \(x - 1\), or non-linear (such as a quadratic term).

Linear factors are polynomials of the first degree, meaning they have the form \(ax + b\), where \(a\) and \(b\) are constants. They play a crucial role in partial fraction decomposition because they allow the original function to be expressed as a sum of simpler fractions.

• Each distinct linear factor corresponds to a term in the partial fraction decomposition.
• If the denominator of a rational function is composed entirely of linear factors, as in our example, the decomposition is more straightforward.
• Repeated linear factors are handled differently, which can introduce additional constants in the decomposition process.

This breakdown helps us to perform operations like integration and differentiation more easily.

Fraction Decomposition

Partial fraction decomposition is a process by which a complex rational function is expressed as the sum of simpler fractions. This is particularly useful in calculus, making integration more accessible and often necessary.

To start with partial fraction decomposition, the denominator must be fully factored, as seen with \((x-1)(x+4)\). Once we identify these factors, we set up our partial fractions: each factor becomes the denominator of a new fraction, with constants as numerators.

• You write a separate fraction for each linear factor, like \( \frac{A}{x-1} \) and \( \frac{B}{x+4} \).
• The goal is to find the values of constants \(A\) and \(B\) that make the equations equivalent.
• This often involves clearing the fractions by multiplying through by the common denominator and equating coefficients of like terms to solve for unknowns.

Once the constants are found, substitute them back to provide a cleaner, easier-to-work-with expression of the original rational function.