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Chapter 9: Problem 11

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

Short Answer

Expert verified
\(\frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}\)

Step by step solution

01

Analyze the Denominator

The rational function to decompose is \( \frac{2}{(x-1)(x+1)} \). The denominator consists of two distinct linear factors: \((x-1)\) and \((x+1)\). This indicates that the decomposition will have two terms.
02

Set Up the Partial Fraction Form

For a rational function with a denominator of distinct linear factors, the partial fraction decomposition takes the form: \[\frac{A}{x-1} + \frac{B}{x+1}\]where \(A\) and \(B\) are constants to be determined.
03

Write the Equation to Clear the Denominator

Multiply both sides of the equation by the common denominator \((x-1)(x+1)\) to eliminate the fractions. This gives:\[2 = A(x+1) + B(x-1)\]
04

Expand and Simplify

Distribute the terms on the right-hand side:\[2 = A(x+1) + B(x-1)\]which simplifies to:\[2 = Ax + A + Bx - B\]Combining like terms yields:\[2 = (A+B)x + (A-B)\]
05

Equate Coefficients

Since the equation must hold for all values of \(x\), equate the coefficients of like terms from both sides. This results in a system of equations:1. \(A + B = 0\)2. \(A - B = 2\)
06

Solve the System of Equations

Solve the system for \(A\) and \(B\). Start by adding the two equations:\[(A + B) + (A - B) = 0 + 2\]This simplifies to:\[2A = 2\]Thus, \(A = 1\).Substitute \(A = 1\) back into equation 1:\[1 + B = 0\]So, \(B = -1\).
07

Write the Partial Fraction Decomposition

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Function
A rational function is basically a fraction composed of two polynomials. Think of it as any standard fraction but with polynomial expressions instead of simple numbers. The function given in the exercise, \( \frac{2}{(x-1)(x+1)} \), is a typical example, where the numerator is the constant 2, and the denominator is the product of two linear polynomials \((x-1)\) and \((x+1)\).
  • A rational function can have zeros and poles. Zeros occur where the numerator equals zero, and poles occur where the denominator equals zero. In this case, the denominator has zeros at \(x = 1\) and \(x = -1\).
  • The degree of the polynomial in the numerator is crucial. If it's less than the degree of the denominator, partial fraction decomposition is possible directly.
As with any rational function, understanding its behavior at specific points helps in breaking it into simpler terms, like we do during partial fraction decomposition. This makes it easier to analyze or integrate further.
Linear Factors
Linear factors are expressions of the form \(ax+b\), where \(a\) and \(b\) are constants, and \(x\) is the variable. For the given rational function, the denominator consists of two distinct linear factors, \((x-1)\) and \((x+1)\).

The concept of linear factors is crucial when setting up the partial fraction decomposition:
  • Each distinct linear factor in the denominator suggests a component of the partial fraction. This means if you have factors like \((x-1)\) and \((x+1)\), you'll have components \(\frac{A}{x-1}\) and \(\frac{B}{x+1}\).
  • This setup allows any rational function to be expressed as a sum of simpler fractions, making calculations such as integration much simpler.
  • When the factors are distinct and linear, the process is straightforward, involving finding values that satisfy the equality formed when the original fraction is equated with the sum of simpler fractions.
Understanding linear factors is a big help in decomposing complex rational functions into manageable parts.
Algebraic Manipulation
Algebraic manipulation involves using algebraic techniques to simplify or rearrange expressions and equations. When performing partial fraction decomposition, we use algebraic manipulation to determine the constants \(A\) and \(B\). Here's how it's done:

First, set up the equation based on the linear factors, resulting in an expression like \[\frac{A}{x-1} + \frac{B}{x+1}\].
  • Multiply both sides by the common denominator, \((x-1)(x+1)\), to clear the fractions. This helps in forming an equation without denominators.
  • Next, expand and simplify the right-hand side by distributing \(A\) and \(B\), simplifying it to form an expression with \(x\).
  • Once simplified, equate the coefficients on both sides with those of the numerator to find the values of \(A\) and \(B\).
This process might seem tedious but it allows the breaking down of the original complex rational function into sums of fractions that are easier to handle or integrate.

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Most popular questions from this chapter

Express the rule in function notation. (For example, the rule "square, then subtract 5 " is expressed as the function \(f(x)=x^{2}-5\). Divide by \(7,\) then subtract 4

Nutrition A nutritionist is studying the effects of the nutrients folic acid, choline, and inositol. He has three types of food available, and each type contains the following amounts of these nutrients per ounce: \begin{array}{|l|ccc|} \hline & \text { Type A } & \text { Type B } & \text { Type C } \\ \hline \text { Folic acid (mg) } & 3 & 1 & 3 \\ \text { Choline (mg) } & 4 & 2 & 4 \\ \text { Inositol (mg) } & 3 & 2 & 4 \\ \hline \end{array} (a) Find the inverse of the matrix $$\left[\begin{array}{lll} 3 & 1 & 3 \\ 4 & 2 & 4 \\ 3 & 2 & 4 \end{array}\right]$$ and use it to solve the remaining parts of this problem. (b) How many ounces of each food should the nutritionist feed his laboratory rats if he wants their daily diet to contain \(10 \mathrm{mg}\) of folic acid, \(14 \mathrm{mg}\) of choline, and \(13 \mathrm{mg}\) of inositol? (c) How much of each food is needed to supply \(9 \mathrm{mg}\) of folic acid, 12 mg of choline, and 10 mg of inositol? (d) Will any combination of these foods supply \(2 \mathrm{mg}\) of folic acid, \(4 \mathrm{mg}\) of choline, and \(11 \mathrm{mg}\) of inositol?

Find the determinant of the matrix. Determine whether the matrix has an inverse, but don't calculate the inverse. $$\left[\begin{array}{rrrr} 1 & 2 & 0 & 2 \\ 3 & -4 & 0 & 4 \\ 0 & 1 & 6 & 0 \\ 1 & 0 & 2 & 0 \end{array}\right]$$

Find the inverse of the matrix. For what value(s) of \(x\) if any, does the matrix have no inverse? $$\left[\begin{array}{ll} 2 & x \\ x & x^{2} \end{array}\right]$$

A biologist is performing an experiment on the effects of various combinations of vitamins. She wishes to feed each of her laboratory rabbits a diet that contains exactly \(9 \mathrm{mg}\) of niacin, \(14 \mathrm{mg}\) of thiamin, and \(32 \mathrm{mg}\) of riboflavin. She has available three different types of commercial rabbit pellets; their vitamin content (per ounce) is given in the table. How many ounces of each type of food should each rabbit be given daily to satisfy the experiment requirements? $$\begin{array}{|l|c|c|c|} \hline & \text { Type A } & \text { Type B } & \text { Type C } \\ \hline \text { Niacin (mg) } & 2 & 3 & 1 \\ \text { Thiamin (mg) } & 3 & 1 & 3 \\ \text { Riboflavin (mg) } & 8 & 5 & 7 \\ \hline \end{array}$$
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