Problem 71 - Applied Calculus

Question

For each function, find and simplify \(\frac{f(x+h)-f(x)}{h} .\) (Assume \(\left.h \neq 0 .\right)\) (See instructions on previous page.) $$ f(x)=\frac{2}{x} $$

Step-by-Step Solution

Answer

\( \frac{-2}{x(x+h)} \)

Step 1: Understand the Function

We are given the function \( f(x) = \frac{2}{x} \). We need to find the expression \( \frac{f(x+h) - f(x)}{h} \) and simplify it.

Step 2: Substitute into the Difference Quotient

First, find \( f(x+h) \). This is \( f(x+h) = \frac{2}{x+h} \). Substitute both \( f(x+h) \) and \( f(x) \) into the difference quotient: \( \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \).

Step 3: Find a Common Denominator

To simplify, find a common denominator for the fractions in the numerator. The common denominator of \( x+h \) and \( x \) is \( x(x+h) \). Rewrite the fractions: \( \frac{2x - 2(x+h)}{x(x+h)h} \).

Step 4: Simplify the Numerator

Simplify the numerator of the expression: \( 2x - 2x - 2h = -2h \). The expression is now \( \frac{-2h}{x(x+h)h} \).

Step 5: Cancel \( h \) from the Numerator and Denominator

We can cancel \( h \) in the numerator and the denominator, as \( h eq 0 \). The expression simplifies to \( \frac{-2}{x(x+h)} \).

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus, important for finding the derivative of a function. It is defined as \[\frac{f(x+h) - f(x)}{h},\]where \( h eq 0 \). This formula represents the average rate of change of the function \( f(x) \) over the interval \( [x, x+h] \). By calculating the limit as \( h \) approaches zero, it gives the exact rate of change at a single point, known as the derivative.

• The numerator \( f(x+h) - f(x) \) calculates the change in the function's value.
• The denominator \( h \) represents the change in \( x \).

For the function \( f(x) = \frac{2}{x} \), the difference quotient helps us understand how the function behaves around any point \( x \). Finding \( f(x+h) \) involves substituting \( x+h \) in place of \( x \), leading to \( f(x+h) = \frac{2}{x+h} \). This substitution is the first step toward evaluating the difference quotient for this function. Let's delve deeper into simplifying the expression.

Function Simplification

In calculus, simplifying functions is crucial because it makes it easier to analyze complex expressions. Simplification often involves algebraic techniques such as factoring, distributing, and combining like terms. In the context of the difference quotient, simplification focuses on ensuring the expression is in its simplest form without any removable discontinuities, such as unnecessary terms that can be canceled.

• The first step in simplifying the difference quotient \( \frac{f(x+h) - f(x)}{h} \) for \( f(x) = \frac{2}{x} \) involves combining the terms \( \frac{2}{x+h} \) and \( \frac{2}{x} \).
• To do this, a common denominator \( x(x+h) \) is used.
• This results in a single fraction in the numerator: \( \frac{2x - 2(x+h)}{x(x+h)} \), which simplifies to \( -2h \) after distributing and subtracting.

The final step of the simplification is to cancel \( h \) from both the numerator and denominator since \( h eq 0 \). This leaves us with the simplified expression \( \frac{-2}{x(x+h)} \). By simplifying the function, we avoid potential mistakes and gain a clearer insight into the problem.

Rational Functions

Rational functions are functions represented as the ratio of two polynomials. In our example, the function \( f(x) = \frac{2}{x} \) is a simple rational function. Rational functions are crucial in calculus and other areas of mathematics due to their properties.

• The denominator cannot be zero, which determines the function's domain.
• They can exhibit vertical asymptotes, holes, and horizontal asymptotes based on the degree of the numerator and the denominator.

For our function, since the denominator is \( x \), it highlights that \( x eq 0 \) to avoid division by zero. As we simplify the difference quotient of \( f(x) \), we see the rational function concept in action.
The original difference quotient expression involves a rational expression where simplifying leads us from a more complex structure to \( \frac{-2}{x(x+h)} \). This final expression is simpler and reflects our understanding of rational functions' behavior. Mastering rational functions is essential for solving many calculus problems, as they frequently appear in limits, derivatives, and integrals.