Problem 109 - Precalculus : Building Concepts and Connections

Question

In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=\frac{1}{x-3}, x \neq 3$$

Step-by-Step Solution

Answer

The difference quotient of the function \( f(x) = \frac{1}{{x-3}} \) is \( \frac{-1}{{(x+h-3) \times (x-3)}} \).

Step 1: Substitute the function into the difference quotient

Let's fill in the formula for the difference quotient replacing \( f(x) \) and \( f(x+h) \) by the function \( \frac{1}{{x-3}} \) which leads to: \( \frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{{(x+h)-3}} - \frac{1}{{x-3}}}{h} = \frac{\frac{1}{{x+h-3}} - \frac{1}{{x-3}}}{h} \)

Step 2: Simplify the difference in the numerator

The numerators are fractions themselves, hence you have to calculate with fractions in the numerator. Use a common denominator, which is \( (x+h-3) \times (x-3) \), to simplify this: \( \frac{\frac{1}{{x+h-3}} - \frac{1}{{x-3}}}{h} = \frac{(x-3)-(x+h-3)}{{h \times (x+h-3) \times (x-3)}} = \frac{-h}{{h \times (x+h-3) \times (x-3)}} \)

Step 3: Simplify the denominator

Cancel out the \( h \) from the numerator with \( h \) in the denominator: \( \frac{-h}{{h \times (x+h-3) \times (x-3)}} = \frac{-1}{{(x+h-3) \times (x-3)}} \)

Key Concepts

Understanding Rational Functions

Rational functions are an essential concept in mathematics, particularly in calculus and algebra. A rational function is any function that can be expressed as the quotient of two polynomials. More specifically, a rational function has the form \( f(x) = \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). In the given exercise, the function \( f(x) = \frac{1}{x-3} \) is a simple example of a rational function.

• The numerator in this function is \(1\), which is actually a constant polynomial.
• The denominator \(x-3\) is a linear polynomial (first degree).

Understanding rational functions involves recognizing their behavior when \( x \) approaches values that make the denominator zero. These values create vertical asymptotes in the graph of the function. In our case, \( x = 3 \) is such a point, commonly resulting in undefined behavior if not properly examined with limits.

Algebraic Simplification

Algebraic simplification is a crucial skill necessary for evaluating difference quotients or solving complex expressions. The main goal here is to simplify the terms to make calculations more manageable. In the context of rational functions, simplification often involves common techniques such as factoring and finding common denominators.In the provided solution, algebraic simplification is applied by finding a common denominator for the fractions in the numerator:

• The expression \( \frac{1}{x+h-3} - \frac{1}{x-3} \) is tackled by rewriting it with a common denominator: \((x+h-3)(x-3)\).
• This allows combining the fractions into a single fraction.
• The resultant numerator \((x-3) - (x+h-3)\) simplifies to \(-h\), with a crucial cancellation step removing \(h\) for the final simplification process.

These simplification steps are vital in calculating expressions like the difference quotient, allowing us to evaluate limits more efficiently.

The Role of Limits in Difference Quotients

The concept of limits is integral to the understanding of difference quotients. A limit examines what happens to a function as the input approaches a specific value. In calculus, "limits" underpin the foundational concepts such as derivatives and integrals.When we calculate the difference quotient \( \frac{f(x+h)-f(x)}{h} \), we are essentially investigating how the function behaves as \( h \) approaches 0. This is pivotal in determining the slope of the tangent line to the function at a given point and ultimately contributes to finding the derivative.During the simplification process in the original solution, after the term \( -h \) is divided out, the result is a new expression: \( \frac{-1}{{(x+h-3) \times (x-3)}} \). For limits,

• We then consider the expression as \( h \) approaches 0.
• This evaluation results in a smoother and more accessible form for calculating instantaneous rates of change.

Understanding limits in this context helps students conceptually grasp why derivatives work, making it a powerful tool in the realm of calculus.