Problem 87 - Algebra and Trigonometry

Question

Simplify the expression. (This type of expression arises in calculus when using the “quotient rule.”) $$ \frac{3(1+x)^{1 / 3}-x(1+x)^{-2 / 1}}{(1+x)^{2 / 2}} $$

Step-by-Step Solution

Answer

The expression simplifies to \( \frac{3(1+x)^{7/3} - x}{(1+x)^3} \).

Step 1: Simplify the Exponent Terms

Recall that \( (1+x)^{-2} = rac{1}{(1+x)^{2}} \) and that \( (1+x)^{1/3} \) is a cube root of \( (1+x) \). Also, \( (1+x)^{2/2} \) simplifies to \( (1+x)^1 = (1+x) \). Substitute these back into the expression, getting: \[ \frac{3(1+x)^{1/3} - x\cdot\frac{1}{(1+x)^{2}}}{(1+x)} \]

Step 2: Factor and Simplify the Numerator

Factor out \( (1+x)^{-2} \) from the numerator: \[ \left( 3(1+x)^{1/3}(1+x)^2 - x \right)(1+x)^{-2} \] This results in: \[ \left( 3(1+x)^{7/3} - x \right)(1+x)^{-2} \]

Step 3: Simplify the Division

Divide the entire expression by \( (1+x) \), which simplifies with the \( (1+x)^{-2} \) term: \[ \frac{3(1+x)^{7/3} - x}{(1+x)^3} \]

Step 4: Final Simplification Review

Simplification resulted in \( \frac{3(1+x)^{7/3} - x}{(1+x)^3} \). Ensure each term is as simplified as possible considering standard operations.

Key Concepts

Understanding Exponent Rules

Exponent rules are fundamental in simplifying expressions involving powers. They help rearrange and simplify expressions initially more complex.

• The most basic exponent rule is that any term raised to the power of zero is equal to one: \( a^0 = 1 \).
• When multiplying like bases with exponents, you add the exponents: \( a^m \times a^n = a^{m+n} \).
• When dividing, you subtract the exponents: \( \frac{a^m}{a^n} = a^{m-n} \).
• When raising a power to another power, you multiply the exponents: \( (a^m)^n = a^{m\cdot n} \).

In our exercise, these rules are applied to simplify the terms. For example, using the rule that negative exponents mean taking the reciprocal, \( (1+x)^{-2} = \frac{1}{(1+x)^2} \). The division with same-base exponents factors out the term to simplify the overall expression.
Additionally, understanding these rules allows one to recognize that \( (1+x)^{2/2} \) simplifies directly to \( (1+x)^1 \) or simply \( 1+x \). This conversion is crucial for achieving a simplified form.

Applying the Quotient Rule

The quotient rule is specifically used when dividing two differentiable functions. In simpler terms, it helps you solve expressions like \( \frac{f(x)}{g(x)} \) where both \( f(x) \) and \( g(x) \) are polynomials or functions. Although the exercise does not involve derivatives, the structuring mimics a quotient rule scenario.

• The quotient rule formula is \( \frac{d}{dx}\left[ \frac{u}{v} \right] = \frac{v \cdot du - u \cdot dv}{v^2} \), where \( u \) and \( v \) are functions of \( x \).

In this exercise's expression, none of the terms require differentiation, but they require manipulation to simplify the numerator before dividing. Simplifying before directly applying expressions in calculus helpsin ensuring accurate mathematics without unnecessary complexity.
These steps show the methodology similar to algebraic fraction manipulations and are essential in calculus integrationwhere similar element sense occurs during optimization problems.

Simplifying Rational Expressions

A rational expression is a ratio of two polynomials. Simplifying these types of expressions involves the same principles used in simplifying fractions. The process requires simplifying both the numerator and the denominator as far as possible before performing any operations between them.

• Simplify terms within each polynomial using factorization.
• Cancel any common factors in the numerator and the denominator.

In the given problem, the expression is simplified by first addressing the terms in the numerator and rearranging them using exponent rules. Then, further simplification is achieved by rewriting the division with like terms in the denominator effectively reducing the expression. This results in an expression where the terms are easier to handle mathematically and better represent calculations involving polynomial degrees and roots.
Understanding rational expressions is crucial for progressing in higher mathematics, as it forms the foundation for handling complex algebraic equations and functions. Identifying and simplifying these parts makes tackling more advanced topics like calculus significantly easier.