91Ó°ÊÓ

A rational function is a type of mathematical expression that is made by dividing one polynomial by another. Just like fractions in arithmetic, these expressions are indicated by two separate polynomial functions: a numerator and a denominator. In our exercise, the rational function is represented as \[\frac{2}{(x-1)(x+1)}\]which means the numerator is 2, and the denominator is \[ (x-1)(x+1). \]In the wider scope, rational functions are pivotal because they embody the idea of division in polynomials, enabling complex expressions to be simplified or analyzed more conveniently. They play a crucial role in calculus and algebra for solving equations, doing curve sketching, and analyzing asymptotic behavior. Without a good understanding of rational functions, moving forward with advanced algebraic concepts can be challenging.

The concept of a common denominator is essential when dealing with fractions, including rational functions. The purpose of finding a common denominator is to create equivalent fractions that can easily be added or subtracted. In our example, we're decomposing the rational function \[\frac{2}{(x-1)(x+1)}\]into partial fractions:\[\frac{A}{x-1} + \frac{B}{x+1}\]To do this, we multiply each term by the common denominator \[(x-1)(x+1)\] to eliminate the fractions:\[2 = A(x+1) + B(x-1).\] By clearing the fractions, it simplifies the equation and helps in setting up a straightforward path to solve for unknown coefficients like \(A\) and \(B\). The common denominator is a vital step not only in partial fraction decomposition but also in many arithmetic operations involving fractions.

Equating coefficients is a technique used to find the constants when decomposing a rational function. Once you multiply both sides by the common denominator, as in:\[2 = A(x+1) + B(x-1),\]you expand and combine like terms to form a polynomial equation. This results in \[(A + B)x + (A - B).\]To solve for \(A\) and \(B\), you equate the coefficients of corresponding terms on both sides of the equation. Since there was no \(x\) term in \(2\), we assume:

• \(A + B = 0\)
• \(A - B = 2\)

By equating coefficients, problems which initially seem complicated become simple algebra. For example, these two equations lead to the conclusion \(A = 1\) and \(B = -1\). This step is not only crucial in partial fraction decomposition but also in many areas of algebra and calculus, where clarity in understanding variable relationships is required.