Problem 89 - A Graphical Approach to College Algebra

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) ) for each function \(f\). Simplify completely. $$f(x)=\frac{1}{2 x}$$

Step-by-Step Solution

Answer

The difference quotient is \( \frac{-1}{2x(x+h)} \).

Step 1: Substitute into the Difference Quotient Formula

The difference quotient is given by \( \frac{f(x+h)-f(x)}{h} \). For the function \( f(x) = \frac{1}{2x} \), calculate \( f(x+h) \) as \( \frac{1}{2(x+h)} \). Now, the difference quotient becomes \( \frac{\frac{1}{2(x+h)} - \frac{1}{2x}}{h} \).

Step 2: Simplify the Numerator

To simplify \( \frac{1}{2(x+h)} - \frac{1}{2x} \), find a common denominator. The common denominator is \( 2x(x+h) \). Rewrite the expression as \( \frac{x - (x+h)}{2x(x+h)} = \frac{x-x-h}{2x(x+h)} \). This simplifies to \( \frac{-h}{2x(x+h)} \).

Step 3: Simplify the Entire Expression

Insert the simplified numerator back into the original difference quotient: \( \frac{-h}{2x(x+h)h} \). Cancel \( h \) from the numerator and the denominator, yielding \( \frac{-1}{2x(x+h)} \).

Step 4: Final Simplification

The difference quotient simplifies completely to \( \frac{-1}{2x(x+h)} \). There are no more terms or factors that can be simplified further.

Key Concepts

Function Simplification

Function simplification is a crucial process in both algebra and calculus. It involves reducing a complex expression to its simplest form, which can make it easier to analyze and understand. When dealing with the difference quotient, the goal is to simplify the expression so that we can derive deeper insights or further apply calculus concepts like derivation.

In our specific case, we're looking at the function \( f(x) = \frac{1}{2x} \). The first step involves substituting this function into the difference quotient formula, which yields \( \frac{\frac{1}{2(x+h)} - \frac{1}{2x}}{h} \). Simplifying this requires carefully handling the fractions within the numerator.

To simplify \( \frac{1}{2(x+h)} - \frac{1}{2x} \), we need to identify a common denominator, which means multiplying respective denominators together, resulting in \( 2x(x+h) \). By converting each term to share this common denominator, we can combine the fractions and simplify the numerator. This is a key skill in both algebra and calculus.

• Identify common denominators in algebraic fractions.
• Subtract numerators after rewriting the fractions with common denominators.
• Cancel out common factors when possible to achieve the simplest form.

Algebraic Expressions

Algebraic expressions are the building blocks of many mathematical concepts. They consist of numbers, variables, and arithmetic operations. Manipulating these expressions allows us to solve equations, work with functions, and understand calculus concepts.

In our exercise, we began with algebraic expressions in the form of fractions. Handling fractions requires careful attention to detail, especially when manipulating the numerators and denominators to simplify an expression or solve problems like the difference quotient. Understanding how to find and use a common denominator for simplifying algebraic fractions is key here.

Let’s break down the steps:

• Write fractions with a common denominator like \( 2x(x+h) \).
• Subtract the numerators: \( x - (x+h) = -h \).
• Simplify the result: \( \frac{-h}{2x(x+h)} \).

By mastering these steps, you lay a strong foundation in algebra, which is essential for learning more advanced mathematical concepts in calculus.

Calculus Concepts

Calculus focuses on the study of change and motion, using foundations built from algebraic principles. One of the first concepts introduced in calculus is the difference quotient, which helps us approximate the rate of change of functions.

The difference quotient itself, \( \frac{f(x+h)-f(x)}{h} \), is a critical concept because it forms the basis of the derivative. Calculating and simplifying this expression involves several steps that reinforce foundational algebra skills. Understanding how to simplify and apply these concepts is crucial when you begin moving on to derivatives and other calculus topics.

• Simplifying the difference quotient is akin to finding a function's "instinctual" rate of change.
• The final form \( \frac{-1}{2x(x+h)} \) is easy to analyze and shows how changes in \( x \) affect the overall function.
• This simplification process paves the way for more advanced topics in calculus, such as limits and derivatives.

By thoroughly understanding and simplifying these expressions, students develop a strong calculus foundation, enabling them to tackle more complex challenges as they progress in their studies.