Problem 40 - Introductory Algebra for College Students

Question

Find and simplify. \(\frac{f(x)-f(h)}{x-h}\) $$f(x)=x^{3}-1$$

Step-by-Step Solution

Answer

The simplified version of the original expression is \(x^{2}+ox+h^{2}\)

Step 1: Rewrite the Expression

Replace \(f(x)\) and \(f(h)\) with \(x^{3}-1\) and \(h^{3}-1\) respectively in the expression \(\frac{f(x)-f(h)}{x-h}\) to get \(\frac{x^{3}-1 - (h^{3}-1)}{x-h}\)

Step 2: Simplify the Numerator

The next step is to simplify the numerator to get \(\frac{x^{3}-h^{3}}{x-h}\)

Step 3: Apply Difference of Cubes formula

The phrase \(x^{3}-h^{3}\) is known as the difference of cubes, which can be factored using the formula \(a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})\). Applying this formula gives \(\frac{(x-h)(x^{2}+hx+h^{2})}{x-h}\)

Step 4: Simplify Expression

The terms \(x-h\) in the numerator and denominator cancel each other, simplifying the expression to \(x^{2}+xh+h^{2}\)

Key Concepts

Simplifying Expressions

Simplifying algebraic expressions is a fundamental skill in mathematics, allowing one to condense and streamline computations. It may involve combining like terms, factoring, expanding polynomials, and simplifying fractions. Here are some general steps to simplify an expression:

• First, perform any necessary substitutions. For instance, if a function is defined such as f(x) = x³ - 1, and you need to work with f(x) and f(h), you'll replace these with their respective expressions.
• Next, combine like terms and apply arithmetic operations. In the example f(x) - f(h), we combine like terms after the substitution to get a new form of the expression.
• Afterward, look for common factors that can be simplified. In the given exercise, the difference of cubes in the numerator and a common linear factor in the denominator allowed us to simplify the expression further.
• Finally, cancel out any common terms. In the step-by-step solution, we noticed an x-h in both the numerator and denominator, which were eliminated to simplify the expression to x² + xh + h².

Remember, while simplifying, always ensure that each step follows mathematical laws and that no crucial terms or factors are omitted, as this affords clarity and correctness in the solution.

Factoring Polynomials

Factoring polynomials is a technique used to express a polynomial as the product of its factors. It simplifies many algebraic procedures, including division and finding roots. The difference of cubes formula, a³ - b³ = (a - b)(a² + ab + b²), is an essential tool applied when a polynomial is in the form of the difference between two cubes, much like x³ - h³ in the provided problem.

• Identify the structure: Recognize the pattern of the cubes in the expression.
• Apply the formula: Use the difference of cubes formula to break down the polynomial into factors. In the exercise, substituting a = x and b = h into the formula results in (x - h)(x² + xh + h²).
• Simplify further: After factoring, there may be more simplification possible by canceling out common terms or factoring further, if applicable.

Understanding and applying factoring techniques efficiently can save time and enhance problem-solving skills in algebra.

Algebraic Functions

Algebraic functions are relations where each input value is associated with only one output value, often represented by an equation involving variables like x and y. The function given in the exercise, f(x) = x³ - 1, describes a cube relation where input x determines the output after executing the operations indicated.

• Function Notation: In algebra, functions are denoted by names such as f, followed by parentheses enclosing the input variable, as in f(x).
• Function Evaluation: To find the value of a function at a specific input, simply replace the input variable with the given number or expression and compute accordingly.
• Function Behavior: Understanding how to manipulate functions is crucial. When the input changes, as it does from x to h in this problem, we must carefully evaluate the function for each input separately before proceeding with operations like subtraction.

In the context of this exercise, we're evaluating the function for two inputs and then studying how the function's output changes as the input changes—a concept known as the difference quotient, prevalent in calculus as a precursor to derivatives.