Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

For \(g(u)=3 /(u-2)\), find and simplify \([g(x+h)-\) \(g(x)] / h\)

Step-by-Step Solution

Answer

The simplified form is \(\frac{-3}{(x+h-2)(x-2)}\).

Step 1: Identify the Function Terms

We are given the function \(g(u) = \frac{3}{u-2}\). We need to find and simplify \(\frac{g(x+h) - g(x)}{h}\). First, identify the function terms: \(g(x) = \frac{3}{x-2}\) and \(g(x+h) = \frac{3}{(x+h)-2}\).

Step 2: Substitute and Expand

Substitute the expressions for \(g(x+h)\) and \(g(x)\) into the difference quotient: \(\frac{g(x+h) - g(x)}{h} = \frac{\frac{3}{x+h-2} - \frac{3}{x-2}}{h}\).

Step 3: Find a Common Denominator

To simplify the expression, find a common denominator for \(\frac{3}{x+h-2}\) and \(\frac{3}{x-2}\). The common denominator is \((x+h-2)(x-2)\).

Step 4: Combine the Fractions

Combine the fractions: \(\frac{3(x-2) - 3(x+h-2)}{(x+h-2)(x-2)} = \frac{3x - 6 - (3x + 3h - 6)}{(x+h-2)(x-2)} \). Simplify the numerator.

Step 5: Simplify the Numerator

Simplify the numerator: \(3x - 6 - 3x - 3h + 6 = -3h\). Substitute back: \(\frac{-3h}{(x+h-2)(x-2)}\).

Step 6: Simplify the Total Expression

Divide through by \(h\): \(\frac{-3h}{h(x+h-2)(x-2)} = \frac{-3}{(x+h-2)(x-2)}\). Thus, the simplified form of the difference quotient is \(\frac{-3}{(x+h-2)(x-2)}\).

Key Concepts

Function Simplification

Function simplification is a critical step in calculus and algebra, allowing us to break down complex expressions into more manageable parts.
This is essential when working with difference quotients, rational functions, and calculus limits.
In our exercise, we started with the function \(g(u) = \frac{3}{u-2}\). To find the difference quotient \(\frac{g(x+h) - g(x)}{h}\), we substituted \(g(x) = \frac{3}{x-2}\) and \(g(x+h) = \frac{3}{x+h-2}\).
The goal was to simplify the expression by reducing the complexity of the terms involved:

• Substitution helps us express how functions change over a small increment \(h\), a fundamental idea in calculus.
• We found a common denominator \((x+h-2)(x-2)\) to combine the terms in the difference quotient efficiently.
• Through simplifying like terms, we arrived at the simplified form \(\frac{-3}{(x+h-2)(x-2)}\), showcasing the step-by-step reduction to a clearer expression.

Simplifying expressions not only makes handling calculations easier but also boils down the problem to its essence, helping us better understand the underlying mathematical principles.
Next, let's explore rational functions and their significance.

Rational Functions

Rational functions, like our given \(g(u) = \frac{3}{u-2}\), are functions made up of polynomial ratios.
These functions are defined where the denominator is not zero.
Key characteristics of rational functions include:

• Each rational function has the form \(\frac{p(x)}{q(x)}\), where \(p(x)\) and \(q(x)\) are polynomials and \(q(x) eq 0\).
• These functions can exhibit behaviors such as vertical asymptotes (where the function is undefined) and horizontal asymptotes (behavior as \(x\) approaches infinity).
• When we simplify rational function expressions, we're often looking to cancel factors in the numerator and denominator to clarify the function's structure.

In our exercise, handling the difference quotient involved careful treatment of rational function terms. Combining and reducing these fractions required finding a common denominator. This practice is common in calculus when dealing with rational expressions, emphasizing their importance not only in ease of computation but also in the in-depth understanding of function behavior.
Understanding rational functions is crucial since it forms the basis for solving complex calculus problems involving limits, derivatives, and integrals.

Limits in Calculus

Limits are a fundamental concept in calculus, dealing with the behavior of functions as they approach specific points or infinity.
Understanding limits helps us grasp concepts like continuity, the definition of derivatives, and integration.
In the context of our exercise, limits play a role in interpreting the difference quotient \(\frac{g(x+h) - g(x)}{h}\):

• A difference quotient represents the average rate of change of a function over an interval \(h\), and by taking the limit as \(h \to 0\), we find the instantaneous rate of change or the derivative.
• The simplification to \(\frac{-3}{(x+h-2)(x-2)}\) prepares the expression for limit evaluation, crucial in verifying continuity or finding derivatives.
• This approach underlines the method behind deriving functions, providing a foundational tool for further calculus studies.

Understanding limits, therefore, is not just about finding where a function heads to as it nears a point but also about facilitating calculations that describe changes in functions under various conditions.
Mastering this concept is fundamental to advancing in calculus and related mathematical fields.