Problem 72
Question
Write the partial fraction decomposition for the rational expression. Check your result algebraically by combining fractions, and check your result graphically by using a graphing utility to graph the rational expression and the partial fractions in the same viewing window. $$\frac{4 x^{4}}{(2 x-1)^{3}}$$
Step-by-Step Solution
Verified Answer
The partial fraction decomposition of \(\frac{4x^4}{(2x-1)^3}\) is \(\frac{3/4}{2x-1} + \frac{0}{(2x-1)^2} + \frac{1/2}{(2x-1)^3}
1Step 1: Identify the Fractions and Form a General Statement
The fraction is already an improper fraction, so we don't have to do a long division. The denominator \((2x-1)^3\) has the common factor \(2x-1\) repeating three times. So, we can form a general statement:\[\frac{4x^4}{(2x-1)^3} = \frac{A}{2x-1} + \frac{B}{(2x-1)^2} + \frac{C}{(2x-1)^3}\]where A, B, and C are constants that we need to find.
2Step 2: Clear the Denominator
We multiply through by \((2x - 1)^3\) to clear the denominator:\[4x^4 = A(2x - 1)^2 + B(2x - 1)+ C\]
3Step 3: Substitute Suitable Values for x
Now, we choose suitable values for x to eliminate A, B, and C in turn. A suitable value would be x = 1/2, giving:\[4(1/2)^4 = A(0) + B(0) + C\]which simplifies to\[\frac{1}{2} = C\]Now that we have C, we need to find A and B.
4Step 4: Differentiate to Find Further Coefficients
Differentiating both sides of the equation (using chain rule for A term) gives:\[16x^3 = 2A(2x-1)(2) + B\]And plugging in the value x = 1/2 again gives B = 0. Now we differentiate this equation again, obtaining: \[48x^2 = 4A \]And using x = 1/2, we get:\[3 = 4A\]Therefore, \(A = 3/4\).
5Step 5: Write the Decomposed Fraction
Substitute the calculated values of A, B, and C back into the general decomposition we originally declared. This gives:\[\frac{4x^4}{(2x-1)^3} = \frac{3/4}{2x-1} + \frac{0}{(2x-1)^2} + \frac{1/2}{(2x-1)^3}\]
Key Concepts
Rational Expression DecompositionAlgebraic TechniquesGraphing Utility
Rational Expression Decomposition
When dealing with rational expressions, especially complex ones with higher degree polynomials, we often turn to partial fraction decomposition. This method breaks down a complicated fraction into simpler 'partial' fractions that are easier to integrate, differentiate, or otherwise work with.
Consider the given exercise \[\frac{4 x^{4}}{(2 x-1)^{3}}\]. Our goal is to express this as a sum of simpler rational expressions. We anticipate a decomposition involving terms of the form \[\frac{A}{2x-1}, \frac{B}{(2x-1)^2},\] and \[\frac{C}{(2x-1)^3}\], where A, B, and C are constants to be determined. This decomposition is based on the fact that \[(2x-1)^3\] can be thought of as a product of same factors \[(2x-1)(2x-1)(2x-1)\].
Consider the given exercise \[\frac{4 x^{4}}{(2 x-1)^{3}}\]. Our goal is to express this as a sum of simpler rational expressions. We anticipate a decomposition involving terms of the form \[\frac{A}{2x-1}, \frac{B}{(2x-1)^2},\] and \[\frac{C}{(2x-1)^3}\], where A, B, and C are constants to be determined. This decomposition is based on the fact that \[(2x-1)^3\] can be thought of as a product of same factors \[(2x-1)(2x-1)(2x-1)\].
- First, we equate the original rational expression with our proposed partial fractions.
- Next, we clear the denominators to derive a polynomial equation.
- Then, we solve this equation for A, B, and C by strategically choosing values for x or using differentiation.
- Once these constants are found, we substitute them back to get our decomposed expression.
Algebraic Techniques
The exercise employs several algebraic techniques that are crucial for solving a range of mathematical problems. These methods include manipulating equations, differentiating to find coefficients, and strategically substituting values for variables to simplify expressions.
We use these techniques in the solution as follows:
We use these techniques in the solution as follows:
- Multiplication of both sides by the common denominator to eliminate fractions.
- Substitution of specific values of x to simplify the equation and solve for the unknown constants one at a time.
- Differentiation, which takes advantage of the repeated differentiation rule (chain rule) to find coefficients by reducing the degree of the terms.
Graphing Utility
Using a graphing utility is an exceedingly helpful way to visualize the problem and verify solutions in algebra and calculus. Graphing the original rational expression alongside its decomposed partial fractions serves as a check; both graphs should be identical since they represent the same mathematical relationship.
In our example, we could graph \[\frac{4 x^{4}}{(2 x-1)^{3}}\] using a graphing calculator or software, and then graph \[\frac{3/4}{2x-1} + \frac{0}{(2x-1)^2} + \frac{1/2}{(2x-1)^3}\] separately. If the decomposed fractions have been found correctly, the two graphs will lie on top of one another.
In our example, we could graph \[\frac{4 x^{4}}{(2 x-1)^{3}}\] using a graphing calculator or software, and then graph \[\frac{3/4}{2x-1} + \frac{0}{(2x-1)^2} + \frac{1/2}{(2x-1)^3}\] separately. If the decomposed fractions have been found correctly, the two graphs will lie on top of one another.
- Graphing the expressions provides a visual confirmation of the algebraic work.
- It can also help identify errors or discrepancies in the algebraic solution.
- Furthermore, visualization aids in understanding the behavior of the function, such as asymptotes and intercepts.
Other exercises in this chapter
Problem 72
Solve the system graphically or algebraically. Explain your choice of method. $$\left\\{\begin{array}{l} y=x^{3}-2 x^{2}+x-1 \\ y=-x^{2}+3 x-1 \end{array}\right
View solution Problem 72
Operations with Matrices Use a graphing utility to perform the operations for the matrices \(A, B,\) and \(C\) and the scalar \(c .\) Write a brief statement co
View solution Problem 72
Use matrices to solve the system of equations, if possible. Use Gauss-Jordan elimination. $$\left\\{\begin{array}{c} x+y+4 z=5 \\ 2 x+y-z=9 \end{array}\right.$$
View solution Problem 72
If \(A\) is a \(2 \times 2\) matrix given by \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right],\) then \(A\) is invertible if and only if \(a d-b c \n
View solution