Problem 4 - Precalculus Mathematics for Calculus

Question

Consider the expression \(\frac{1}{x}-\frac{2}{x+1}-\frac{x}{(x+1)^{2}}\). (a) How many terms does this expression have? (b) Find the least common denominator of all the terms. (c) Perform the addition and simplify.

Step-by-Step Solution

Answer

(a) 3 terms. (b) LCD is \( x(x+1)^{2} \). (c) Simplified: \( \frac{1-x}{x(x+1)^{2}} \).

Step 1: Identify the Terms

The expression given is \( \frac{1}{x} - \frac{2}{x+1} - \frac{x}{(x+1)^{2}} \). To determine how many terms there are, count each fractional component separated by a plus or minus sign. Here, we see three distinct terms: \( \frac{1}{x} \), \( \frac{2}{x+1} \), and \( \frac{x}{(x+1)^{2}} \).

Step 2: Determine the Least Common Denominator

To find the least common denominator (LCD) of \( \frac{1}{x} - \frac{2}{x+1} - \frac{x}{(x+1)^{2}} \), identify the denominators: \( x \), \( x+1 \), and \( (x+1)^{2}\). The LCD will be the least common multiple of these denominators. Since \( (x+1)^{2} \) includes \( x+1 \), the LCD is \( x \cdot (x+1)^{2} \).

Step 3: Rewrite Each Term with the LCD

Convert each term to have the common denominator \( x(x+1)^{2} \). For \( \frac{1}{x} \), multiply the numerator and denominator by \( (x+1)^{2} \) to get \( \frac{(x+1)^{2}}{x(x+1)^{2}} \). For \( \frac{2}{x+1} \), multiply by \( x(x+1) \) to obtain \( \frac{2x(x+1)}{x(x+1)^{2}} \). For \( \frac{x}{(x+1)^{2}} \), multiply by \( x \) to get \( \frac{x^{2}}{x(x+1)^{2}} \).

Step 4: Combine the Fractions

Now that all terms have the same denominator, combine them into a single fraction: \( \frac{(x+1)^{2} - 2x(x+1) - x^{2}}{x(x+1)^{2}} \).

Step 5: Simplify the Numerator

Expand each part of the numerator: \((x+1)^{2} = x^{2} + 2x + 1\), \(2x(x+1) = 2x^2 + 2x\). Substituting these gives the numerator as \(x^{2} + 2x + 1 - 2x^{2} - 2x - x^{2}\). Simplify to get \(-x + 1\).

Step 6: Write the Simplified Fraction

The expression simplifies to \( \frac{-x+1}{x(x+1)^{2}} \) or, to match the original form: \( \frac{1-x}{x(x+1)^{2}} \).

Key Concepts

Least Common Denominator

When working with rational expressions, identifying the least common denominator (LCD) is crucial. It allows us to combine the fractions by expressing them with a common base.
The least common denominator is essentially the smallest expression that all individual denominators of your fractions can divide into without remainder. To find it, you typically determine the most comprehensive product of all unique factors from each denominator.
For instance, in the given expression \[\frac{1}{x} - \frac{2}{x+1} - \frac{x}{(x+1)^{2}}\], you need a common baseline across different terms:- Denominator 1: \(x\)- Denominator 2: \(x+1\)- Denominator 3: \((x+1)^{2}\) Using these, the LCD is created by choosing the highest power of each unique factor. The LCD here becomes \(x(x+1)^{2}\), as it covers all denominators thoroughly.
Finding the LCD streamlines succeeding steps by consolidating the denominators into a uniform form, crucial for proper fraction addition.

Fraction Addition

Adding fractions, particularly with different denominators, demands precise alignment under a common denominator. This principle is equally applicable to rational expressions, simplifying the terms allows for unencumbered addition.
Once you have determined the least common denominator (LCD), each term needs to be expressed in relation to this unified base. Each original fraction in the expression \[\frac{1}{x}, \frac{2}{x+1}, \frac{x}{(x+1)^{2}}\] requires conversion:

• Multiply \(\frac{1}{x}\) by \((x+1)^{2}\).
• Multiply \(\frac{2}{x+1}\) by \(x(x+1)\).
• Multiply \(\frac{x}{(x+1)^{2}}\) by \(x\).

This results in their nominators being adjusted as well, resulting in:

• \(\frac{(x+1)^{2}}{x(x+1)^{2}}\)
• \(\frac{2x(x+1)}{x(x+1)^{2}}\)
• \(\frac{x^{2}}{x(x+1)^{2}}\)

With a common denominator in place, proceed to sum the numerators directly since they share the same base value. This leads to a consolidated expression that frames the entirety of the addition process cleanly.

Expression Simplification

Simplifying rational expressions is a methodical process that reduces complexity without altering the intrinsic value of the expression. Once rational expressions share a common denominator, you focus on simplifying the numerator as the principal means to expression refinement.
Take the converted addition of the given fractions. Upon achieving a single fraction form \[\frac{-x+1}{x(x+1)^{2}}\], you must break down each part of the numerator:

• Expand \((x+1)^{2}\) as \(x^2 + 2x + 1\).
• Determine \(2x(x+1)\) as \(2x^2 + 2x\).

After gathering the expanded and simplified terms, all that remains is to execute algebraic reduction within the numerator:

• Start with \(x^2 + 2x + 1 - 2x^2 - 2x - x^2\)
• Condense to form \(-x + 1\)

Finally, translate this result into a single simplified fraction: \(\frac{1-x}{x(x+1)^{2}}\). This transformation epitomizes expression simplification; clear, efficient, and readily applicable.