Problem 53 - Calculus and its applications

Question

Find the simplified difference quotient for each function listed. $$ f(x)=\frac{1}{1-x} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Difference Quotient

The difference quotient for a function \( f(x) \) is given by the formula: \[\frac{f(x+h) - f(x)}{h}\]where \( h \) is not equal to zero. This represents the average rate of change of the function \( f(x) \) over the interval from \( x \) to \( x+h \).

Step 2: Substitute into the Difference Quotient

First, find \( f(x+h) \) for the function \( f(x) = \frac{1}{1-x} \). Replace \( x \) with \( x+h \) in the function: \[f(x+h) = \frac{1}{1-(x+h)} = \frac{1}{1-x-h}\]Now, substitute \( f(x) \) and \( f(x+h) \) into the difference quotient:\[\frac{\frac{1}{1-x-h} - \frac{1}{1-x}}{h}\]

Step 3: Simplify the Numerator

To simplify \( \frac{1}{1-x-h} - \frac{1}{1-x} \), find a common denominator. The common denominator is \((1-x-h)(1-x)\).Express each fraction with the common denominator and simplify:\[\frac{(1-x) - (1-x-h)}{(1-x-h)(1-x)} = \frac{-h}{(1-x-h)(1-x)}\]

Step 4: Simplify the Difference Quotient

Substitute back the simplified numerator into the difference quotient:\[\frac{-h}{(1-x-h)(1-x)} \times \frac{1}{h} = \frac{-1}{(1-x-h)(1-x)}\]This is the simplified form of the difference quotient.

Key Concepts

Average Rate of Change

The average rate of change is a concept that helps us understand how a function's value changes over a certain interval. Imagine driving a car. The distance you cover over a period of time gives you an average speed. In the case of functions, this is represented through the difference quotient formula:

• \( \frac{f(x+h) - f(x)}{h} \)

Here, \( h \) is the change in the input value (think of an increase in time, like one more hour of driving), and \( f(x+h) - f(x) \) is the change in the function value (similar to the change in the distance traveled). The simplest way to understand this is to think about how the function "moves" from one point to another. If our function is \( \frac{1}{1-x} \), then calculating the average rate of change over a small interval \( h \) gives us insight into the function's behavior between the points \( x \) and \( x+h \). It tells us, on average, how much the function's value is changing per unit of \( x \). This concept is fundamental in calculus as it leads us to the derivative, which provides the exact rate of change at any given point.

Function Simplification

Function simplification makes complex algebraic expressions easier to handle. Consider the expression you had earlier: the difference quotient with \( f(x) = \frac{1}{1-x} \) and \( f(x+h) = \frac{1}{1-x-h} \).First, you find the common denominator, which in this case is \((1-x-h)(1-x)\). It allows you to combine two fractions into one by aligning their denominators, simplifying what initially might appear as a messy algebraic expression.

• For \( \frac{1}{1-x-h} - \frac{1}{1-x} \), you express both fractions in terms of this common denominator.
• This grants you \( \frac{-h}{(1-x-h)(1-x)} \) as the simplified numerator.

Function simplification is often a critical step in analyzing and understanding rational functions. It turns a knotty problem into something more manageable, paving the way for further analysis like finding derivatives or solving for critical points. This is a useful tool in your math toolkit when handling functions that could otherwise be too complicated.

Rational Functions

Rational functions are expressions formed by the ratio of two polynomials. In our example, the function \( f(x) = \frac{1}{1-x} \) is a simple rational function.

• The numerator is \(1\), and the denominator is \(1-x\).
• They typically have vertical asymptotes, places where the function isn't defined, like \( x = 1 \) for this function.

An important element of rational functions is their behavior around these asymptotes. As \( x \) approaches the value that makes the denominator zero, the function value can increase or decrease without bound. For example, at \( x = 1 \), \( f(x) \) becomes undefined since division by zero is not possible.Understanding rational functions involves analyzing these behaviors, finding asymptotes, intersections with axes, and using simplification techniques to resolve their difference quotients. Identifying characteristics like whether a function is increasing, decreasing, or having asymptotic behavior aids deeper comprehension of how these functions behave across different values of \( x \). By dissecting these details, you gain a clearer picture of the function's nature and it frames foundational knowledge for more advanced calculus concepts.