Problem 85 - A Graphical Approach to Precalculus with Limits

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) j 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: Write the Expression for f(x)

The given function is \( f(x) = \frac{1}{2x} \). We need to find \( f(x + h) \) first.

Step 2: Calculate f(x + h)

To find \( f(x + h) \), substitute \( x + h \) into \( f(x) \): \[ f(x + h) = \frac{1}{2(x + h)} \]

Step 3: Set Up the Difference Quotient

Now, substitute \( f(x) \) and \( f(x + h) \) into the difference quotient formula:\[ \frac{f(x + h) - f(x)}{h} = \frac{\frac{1}{2(x + h)} - \frac{1}{2x}}{h} \]

Step 4: Combine the Fractions

To combine the fractions in the numerator, find a common denominator, which is \( 2x(x + h) \). This gives:\[ \frac{1}{2(x + h)} - \frac{1}{2x} = \frac{x - (x + h)}{2x(x + h)} \]This simplifies to:\[ \frac{-h}{2x(x + h)} \]

Step 5: Simplify the Difference Quotient

Substitute the simplified expression from Step 4 into the difference quotient:\[ \frac{\frac{-h}{2x(x + h)}}{h} \]Simplify by canceling \( h \) from the numerator and denominator:\[ = \frac{-1}{2x(x + h)} \]

Step 6: Final Simplified Expression

The simplified form of the difference quotient is:\[ \frac{-1}{2x(x + h)} \]

Key Concepts

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. In this exercise, the function is given by \( f(x) = \frac{1}{2x} \). This is a specific type of rational function where the numerator is a constant and the denominator is a simple monomial. Understanding the behavior of rational functions is important because they often appear in various mathematical analyses and problems.

Here are some key points about rational functions:

• They can be expressed in the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials.
• The domain of a rational function excludes values that make the denominator zero.
• Rational functions can have vertical and horizontal asymptotes, which are lines that the graph of the function approaches but never actually touches.

In our case, for \( f(x) = \frac{1}{2x} \), the function is defined for all \( x eq 0 \), as \( x = 0 \) would make the denominator zero.

Limit Process

The limit process is a fundamental concept in calculus, and it is essential for understanding how functions behave as they approach a particular point. This concept is closely tied to finding the difference quotient, which is a step towards determining the derivative of a function.

For the function \( f(x) = \frac{1}{2x} \), we begin by considering \( f(x + h) \), giving us \( f(x + h) = \frac{1}{2(x + h)} \). The difference quotient is then formulated as:

• \( \frac{f(x + h) - f(x)}{h} \)
• Substituting the expressions, it becomes \( \frac{\frac{1}{2(x + h)} - \frac{1}{2x}}{h} \)

The key here is to simplify this expression as \( h \) approaches zero. This involves algebraic manipulation, typically requiring common denominators or other strategies to simplify the expression. This process reveals how the function behaves near a certain point, allowing us to evaluate the derivative, which is the core of differential calculus.

Simplifying Expressions

Simplifying expressions is a crucial skill in mathematics, allowing for easier interpretation and solution of problems. In this exercise, simplifying the difference quotient is important for obtaining the most manageable form of the expression.

Here’s a breakdown of the simplification process in this exercise:

• Initially, combine the fractions by finding a common denominator in \( \frac{f(x + h) - f(x)}{h} \). For \( \frac{1}{2(x + h)} - \frac{1}{2x} \), the common denominator is \( 2x(x + h) \).
• Rewrite the expression: \( \frac{x - (x + h)}{2x(x + h)} \).
• Simply this to \( \frac{-h}{2x(x + h)} \), by canceling terms in the numerator.
• Finally, the \( h \) in the numerator cancels with \( h \) in the denominator, leaving us with \( \frac{-1}{2x(x + h)} \).

Mastering the simplification of expressions not only improves computational accuracy but also enhances your ability to solve more complex algebraic equations and calculus problems in the future.