Problem 28 - Precalculus: Mathematical for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{x-4}{(2 x-5)^{2}}$$

Step-by-Step Solution

Answer

\(\frac{x - 4}{(2x-5)^2} = \frac{1/2}{2x-5} - \frac{3/2}{(2x-5)^2}\)

Step 1: Simplify the Rational Function

The given rational function is \(\frac{x-4}{(2x-5)^{2}}\). Notice that the denominator \((2x-5)^{2}\) is a repeated linear factor. Therefore, the partial fraction decomposition will require terms corresponding to this repeated factor.

Step 2: Setup the Partial Fraction Decomposition

For a rational function with a repeated linear factor \((2x-5)^{2}\), the decomposition takes the form:\[ \frac{x-4}{(2x-5)^2} = \frac{A}{2x-5} + \frac{B}{(2x-5)^2} \]where \(A\) and \(B\) are constants to be determined.

Step 3: Multiply Through by the Denominator

Multiply both sides of the equation by \((2x-5)^2\) to eliminate the denominators:\[ x-4 = A(2x-5) + B \]This step allows us to find values for \(A\) and \(B\) by eliminating the common denominator.

Step 4: Expand and Simplify the Equation

Expand the right-hand side of the equation:\[ x-4 = A(2x-5) + B = 2Ax - 5A + B \]Now, we have:\[ x - 4 = 2Ax - 5A + B \]

Step 5: Equate Coefficients

Equate the coefficients from both sides:1. For the coefficient of \(x\), we find: \[ 1 = 2A \]2. For the constant term, we find:\[ -4 = -5A + B \]

Step 6: Solve the Equations for Constants

First, solve for \(A\):\[ 1 = 2A \Rightarrow A = \frac{1}{2} \]Next, substitute \(A = \frac{1}{2}\) into the second equation:\[ -4 = -5\left(\frac{1}{2}\right) + B \Rightarrow -4 = -\frac{5}{2} + B \Rightarrow B = -4 + \frac{5}{2} \Rightarrow B = -\frac{8}{2} + \frac{5}{2} \Rightarrow B = -\frac{3}{2} \]

Step 7: Write the Partial Fraction Decomposition

Plug the values of \(A\) and \(B\) back into the partial fraction form:\[ \frac{x - 4}{(2x-5)^2} = \frac{\frac{1}{2}}{2x-5} + \frac{-\frac{3}{2}}{(2x-5)^2} \]The decomposition is complete.

Key Concepts

Rational Function

A rational function is a type of function that involves a ratio of two polynomials. In simpler terms, think of it like a fraction where both the numerator and the denominator are polynomials. These functions are quite common in algebra and calculus and can represent a wide array of real-world situations.

For example, in this exercise, \( \frac{x-4}{(2x-5)^{2}} \) is a rational function. The numerator is a polynomial (\(x-4\)), and the denominator is also a polynomial (\((2x-5)^{2}\)).

The beauty of rational functions lies in their ability to provide detailed information about the relationships between different variables. They are used to model behaviors such as ratios of quantities, rates of change, and can even extend to more complex systems.

• They provide insights into asymptotic behavior and discontinuities.
• They can have horizontal, vertical, or slant asymptotes depending on the degree of the numerator and denominator.
• Partial fraction decomposition is frequently used with rational functions to simplify integrals.

Understanding rational functions and their behavior is essential for tackling many different kinds of mathematical problems.

Repeated Linear Factor

A repeated linear factor appears when a linear term in the denominator is raised to a power greater than one. In our exercise, this concept comes into play because the denominator of our rational function, \( (2x-5)^2 \), is exactly such a case. Rather than being just \(2x-5\), it is squared, indicating repetition.

When decomposing a rational function that contains a repeated linear factor, each power of the factor must be represented in the partial fraction decomposition. This ensures that the decomposition accounts for the multiplicity of the roots. This is why we write the decomposition as:
\( \frac{x-4}{(2x-5)^{2}} = \frac{A}{2x-5} + \frac{B}{(2x-5)^{2}} \)

By setting it up this way, it allows for all parts of the denominator to be considered:

• Each repeated factor requires a separate term with potentially different constants.
• This incremental approach helps in solving the coefficients needed for the decomposition.

The proper setup using these repeated factors is crucial in solving more advanced algebraic expressions as they ensure every component is addressed.

Denominator

The denominator of a rational function can tell us many things about the function itself. It's the part of the fraction that lies below the line. The nature of the denominator affects how the entire function behaves. It determines where the function is undefined and the potential for vertical asymptotes.

In our example, the denominator is \( (2x-5)^{2} \), which is not just a linear factor but a repeated one. This affects the partial fraction decomposition we can perform.

Why does the denominator matter in partial fraction decomposition?

• The form and factors of the denominator guide how we set up the decomposition.
• It helps us determine the types of fractions we need, as each root and multiplicity impacts the terms in decomposition.

Understanding the peculiarities of the denominator ensures that any transformation we do, such as partial fraction decomposition, maintains the integrity of the original rational expression.