Problem 84 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the difference quotient and simplify your answer. $$g(x)=3 x-1, \frac{g(x+h)-g(x)}{h}, \quad h \neq 0$$

Step-by-Step Solution

Answer

The simplified difference quotient is 3.

Step 1: Substitute \( x + h \) into the Function

Replace \( x \) with \( x + h \) in the function \( g(x) \). The function becomes \( g(x + h) = 3(x + h) - 1 \).

Step 2: Simplify the Function

Apply the distributive property to simplify the function, \( g(x + h) = 3x + 3h - 1 \).

Step 3: Substitute the Functions into the Difference Quotient

Substitute \( g(x + h) \) and \( g(x) \) into the difference quotient equation \( \frac{g(x+h)-g(x)}{h} \). The expression becomes \( \frac{3x + 3h - 1 - (3x - 1)}{h} \).

Step 4: Simplify the Difference Quotient

Simplify the expression in the numerator by cancelling out the common terms \( 3x \) and \( -1 \). The expression becomes \( \frac{3h}{h} \).

Step 5: Cancel out the h Terms

The \( h \) terms cancel in the final expression giving the simplified difference quotient as 3.

Key Concepts

Function Simplification

Function simplification is a fundamental step when working with complicated mathematical expressions. It involves breaking down and reducing a function to its simplest form while keeping its original value intact. In the context of the difference quotient, simplification makes the expression easier to understand and work with.

When you substitute variables as part of solving equations, terms may combine or cancel each other out. In the given exercise, after applying substitution, the expression for the function becomes:

• Expand the expression: \( g(x + h) = 3(x + h) - 1 = 3x + 3h - 1 \)
• Identify common terms: Notice the terms you can simplify such as \(3x\) and \(-1\) in the next steps.

Thus, simplifying expressions is a straightforward method to ensure complex functions become more manageable.

Distributive Property

The distributive property is a key algebraic property used when simplifying expressions, especially when dealing with parentheses. It states that multiplying a single term by a sum is the same as multiplying each addend individually and then adding the results.

In the function \( g(x + h) = 3(x + h) - 1 \), the distributive property allows us to distribute the "3" across the terms inside the parentheses, resulting in:

• Multiply the terms: \( 3(x + h) = 3x + 3h \)

This ensures every term inside the parentheses is considered, maintaining the equivalency of the expression. The distributive property simplifies the process of expanding algebraic expressions and is indispensable when dealing with difference quotients.

Function Substitution

Function substitution involves replacing a variable within a function with another expression in order to evaluate or simplify it. It is especially useful in operations like those involving difference quotients where you evaluate changes in function values.

In the exercise, you first substituted \( x + h \) for \( x \) in the function \( g(x) \) to find \( g(x + h) \). This step is crucial because it helps determine how the function behaves as \( x \) changes to \( x + h \).

Substituting correctly is essential because it sets up the right expression for further simplification:

• Replace and evaluate: \( g(x+h) = 3(x + h) - 1 \) results in \( 3x + 3h - 1 \)

Function substitution is a stepping stone that lays the groundwork for applying other algebraic operations.

Algebraic Fractions

Algebraic fractions refer to expressions that are in the form of a fraction, where the numerator and the denominator contain algebraic expressions. Simplifying these fractions is a skill you often use when working with calculus concepts like difference quotients.

After replacing and simplifying the functions in the quotient formula, you begin to rewrite:

• Difference Quotient: \( \frac{3x + 3h - 1 - (3x - 1)}{h} \)
  
• Simplifying the numerator: \( \frac{3h}{h} \)
• Cancel out common terms: Removing the \( h \) results in \( 3 \)

Understanding how to cancel out terms in algebraic fractions helps you find the behavior of the function as \( h \) approaches zero, which is the essence of evaluating difference quotients.