Problem 11 - Calculus for Biology and Medicine

Question

In Problems , write out the partial-fraction decomposition of the function \(f(x)\). $$ f(x)=\frac{4 x+1}{x^{2}-3 x-10} $$

Step-by-Step Solution

Answer

The partial fraction decomposition of \(f(x)\) is \(\frac{3}{x-5} + \frac{1}{x+2}\).

Step 1: Factor the Denominator

First, we need to factor the quadratic denominator \(x^2 - 3x - 10\). We look for two numbers whose product is \(-10\) and whose sum is \(-3\). These numbers are \(-5\) and \(2\), so we can factor the expression as \((x - 5)(x + 2)\).

Step 2: Set Up Partial Fractions

Since the denominator is factored into two distinct linear factors \((x - 5)\) and \((x + 2)\), the partial fraction decomposition will be of the form: \[ \frac{4x + 1}{x^2 - 3x - 10} = \frac{A}{x - 5} + \frac{B}{x + 2} \] where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

To find \(A\) and \(B\), multiply both sides of the equation by the denominator \((x - 5)(x + 2)\): \[ 4x + 1 = A(x + 2) + B(x - 5) \]

Step 4: Expand and Collect Terms

Expand the right side of the equation: \[ A(x + 2) + B(x - 5) = Ax + 2A + Bx - 5B = (A + B)x + (2A - 5B) \] Match the coefficients from both sides: 1. For the \(x\) terms: \(A + B = 4\)2. For the constant terms: \(2A - 5B = 1\)

Step 5: Solve the Systems of Equations

We have the system:1. \(A + B = 4\)2. \(2A - 5B = 1\)From equation (1), \(A = 4 - B\). Substitute \(A = 4 - B\) into equation (2): \[ 2(4 - B) - 5B = 1 \] \[ 8 - 2B - 5B = 1 \]\[ 8 - 7B = 1 \] \[ -7B = -7 \]\[ B = 1 \]Substitute \(B = 1\) back into \(A = 4 - B\):\[ A = 4 - 1 = 3 \]

Step 6: Write the Partial Fraction Decomposition

With \(A = 3\) and \(B = 1\), we can write the partial fraction decomposition:\[ \frac{4x + 1}{x^2 - 3x - 10} = \frac{3}{x - 5} + \frac{1}{x + 2} \]

Key Concepts

Factorization

Factorization is the process of breaking down an expression into a product of its factors. For the expression $x^2 - 3x - 10$, factorization involves finding two numbers that multiply to give the constant term, -10, and add to give the coefficient of the middle term, -3. In this problem, the numbers are -5 and 2, so the quadratic expression can be factored as $(x - 5)(x + 2)$. Factorization is crucial in partial fraction decomposition because it allows us to rewrite the denominator in a form that can be split into simpler fractions. Identifying the correct factors helps in setting up the partial fractions correctly, which is essential for solving the equation.

System of Equations

A system of equations is a set of two or more equations with the same variables. In partial fraction decomposition, you use a system of equations to solve for unknown constants, which in this case are represented by $A$ and $B$. After setting up the partial fractions, we get the equation $4x + 1 = A(x + 2) + B(x - 5)$. When expanded, this leads to two separate equations based on the coefficients: - $A + B = 4$ (for the $x$ terms) - $2A - 5B = 1$ (for the constants) Solving this system helps us find the values of $A$ and $B$, making it a key step in determining the partial fraction decomposition.

Quadratic Expressions

Quadratic expressions are polynomials of degree two, generally written in the form $ax^2 + bx + c$. In this exercise, we work with the quadratic expression $x^2 - 3x - 10$. Understanding quadratic expressions is essential in both factorization and partial fraction decomposition. Key properties of quadratics: - They can be factored into two binomials. - The roots can be found using the quadratic formula if factorization is not straightforward. Quadratic expressions often appear as denominators in rational functions, where decomposition helps in integrating or simplifying the expression. Recognizing and handling these expressions efficiently is a valuable skill in algebra.