Problem 48 - College Algebra and Calculus: An Applied Approach

Question

Write the partial fraction decomposition for the rational expression. Check your result algebraically. Then assign a value to the constant \(a\) and use a graphing utility to check the result graphically. $$ \frac{1}{x(x+a)} $$

Step-by-Step Solution

Answer

The partial fraction decomposition of \(\frac{1}{x(x+a)}\) is \(\frac{1}{ax} - \frac{1}{ax+a}\). The result has been verified algebraically and can also be confirmed graphically for any specific value assigned to constant a.

Step 1: Write the General Form

Begin with expressing the given rational function as the sum of its partial fraction decomposition. For the fraction \(\frac{1}{x(x+a)}\), the general decomposition is \(\frac{A}{x} + \frac{B}{x+a}\), where A and B are constants we're going to find.

Step 2: Combine the Fractions

Set the expression equal to the original fraction: \(\frac{1}{x(x+a)} = \frac{A}{x} + \frac{B}{x+a}\). Cross-multiply to find a common denominator so that the fractions can be combined: \(\frac{1}{x(x+a)} = \frac{A(x+a) + Bx}{x(x+a)}\). Now, the denominators are the same, so the numerators must also be equal.

Step 3: Equate the Numerators

Set the numerators on both sides equal to each other: \(1 = A(x + a) + Bx\). Now, in order to find the values of A and B, we'll form two equations by choosing suitable values for x. Conveniently, we choose x = 0 and x = -a.

Step 4: Solve the System of Equations

Plug x = 0 into the equation and simplify to get the equation Aa = 1, then solving gives A = \(\frac{1}{a}\). Next, set x = -a to obtain the equation -Ba = 1, which simplifies to B = \(\frac{-1}{a}\). So, the partial fraction decomposition is \(\frac{1}{ax} - \frac{1}{ax+a}\).

Step 5: Validate Algebraically

To check the result, combine the fractions: \(\frac{1}{ax} - \frac{1}{ax+a} = \frac{(ax+a) - ax}{(ax)(ax+a)} = \frac{1}{x(x+a)}\). This is the original function, so the result is correct algebraically.

Step 6: Validate graphically

To verify graphically, use a graphing utility to plot both the original function \(\frac{1}{x(x+a)}\) and the decomposed function \(\frac{1}{ax} - \frac{1}{ax+a}\). Regardless of the specific value of a, both graphs should coincide, confirming the correctness of the result graphically.

Key Concepts

Rational Expressions

Rational expressions are fractions that involve polynomials in their numerator and denominator. They resemble ratios of polynomials and are hence called 'rational'. Simplifying and manipulating these expressions is a common task in algebra, which also serves as a basis for more advanced fields like calculus.

For instance, the expression \( \frac{1}{x(x+a)} \) is a rational expression with a polynomial, \( x(x+a) \) in its denominator. Simplifying rational expressions can involve reducing them to their simplest form, factoring polynomials, and finding common denominators. One popular technique for simplifying complex rational expressions is partial fraction decomposition.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging, simplifying, or rewriting algebraic expressions using the rules and properties of arithmetic and algebra. This includes operations like addition, subtraction, multiplication, division, and the application of exponent rules.

In our exercise, algebraic manipulation involves cross-multiplying to achieve a common denominator and combining like terms. For example, \( \frac{A}{x} + \frac{B}{x+a} \) can be rewritten as \( \frac{A(x+a) + Bx}{x(x+a)} \) after cross-multiplying. Mastery of these skills allows students to solve equations, simplify expressions, and understand how algebraic operations affect the form of an expression.

System of Equations

A system of equations is a set of two or more equations that share a common set of variables. The solution to a system is the set of values that satisfies all equations in the system simultaneously.

In the provided exercise, we encountered a system of equations when determining the constants for the partial fraction decomposition. By setting the numerators equal (\(1 = A(x + a) + Bx\)) and selecting strategic values for \(x\), we created a system of linear equations: \(Aa = 1\) and \(-Ba = 1\). The solution to this system provides the values for \(A\) and \(B\) that make the original and decomposed functions equivalent. Solving systems of equations is a foundational tool in algebra that can be applied to countless real-world problems.

Graphing Utilities

Graphing utilities are tools—either software or hardware like graphing calculators—that allow users to visually represent equations and functions. They offer an intuitive understanding of mathematical concepts and provide a graphical validation of algebraic manipulation.

In validating our solution from the exercise, a graphing utility can plot both the original function \(\frac{1}{x(x+a)}\) and the decomposed function \(\frac{1}{ax} - \frac{1}{ax+a}\). If both graphs overlap, it confirms the algebraic solution visually. Students can learn a lot from graphing utilities as they can reveal asymptotes, intercepts, and the general shape of the function, which can be insightful not just for homework but also for more advanced studies.