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Chapter 2: Problem 46

Difference Quotient Find \(f(a), f(a+h),\) and the difference quotient \(\frac{f(a+h)-f(a)}{h},\) where \(h \neq 0.\) $$f(x)=\frac{1}{x+1}$$

Short Answer

Expert verified
The difference quotient is \( \frac{-1}{(a+h+1)(a+1)} \).

Step by step solution

01

Evaluate f(a)

First, let's find what \( f(a) \) is for the function \( f(x) = \frac{1}{x+1} \). Substitute \( a \) in place of \( x \): \[ f(a) = \frac{1}{a+1} \]
02

Evaluate f(a+h)

Next, evaluate \( f(a+h) \) by substituting \( a+h \) into the function: \[ f(a+h) = \frac{1}{(a+h) + 1} = \frac{1}{a+h+1} \]
03

Compute the Difference Quotient

Now, compute the difference quotient \( \frac{f(a+h)-f(a)}{h} \) using the expressions we found:\[ \frac{f(a+h)-f(a)}{h} = \frac{\frac{1}{a+h+1} - \frac{1}{a+1}}{h} \]
04

Simplify the Difference Quotient

To simplify, find a common denominator for the fractions in the numerator:The common denominator is \((a+h+1)(a+1)\).Rewrite the fractions:\[ \frac{1}{a+h+1} = \frac{a+1}{(a+h+1)(a+1)} \] and \[ \frac{1}{a+1} = \frac{a+h+1}{(a+h+1)(a+1)} \]Subtract the numerators: \[ \frac{a+1 - (a+h+1)}{(a+h+1)(a+1)} = \frac{-h}{(a+h+1)(a+1)} \]Finally, divide by \( h \): \[ \frac{\frac{-h}{(a+h+1)(a+1)}}{h} = \frac{-1}{(a+h+1)(a+1)} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Functions
Rational functions are mathematical expressions that include fractions where both the numerator and the denominator are polynomials. For example, in the function \( f(x) = \frac{1}{x+1} \), the numerator is 1, which is a constant, and the denominator is \( x+1 \), which is a polynomial of degree 1. These functions can often present unique challenges such as restrictions on the domain.
  • **Polynomials**: In rational functions, polynomials include terms made up of constants, variables, and exponents, all connected by addition, subtraction, and multiplication.
  • **Restrictions**: One must always check where the expression is undefined. For example, here, \( x+1 = 0 \) leads to division by zero, which is undefined, hence the domain of \( f(x) \) excludes \( x = -1 \).
  • **Behavior**: Rational functions can have interesting properties such as vertical asymptotes and defining holes at specific points in their graph due to these restrictions.
Understanding rational functions is crucial when calculating things like the difference quotient, as it involves manipulation and simplification of these functions.
Function Evaluation
Function evaluation is a fundamental concept where specific input values are substituted into a function to determine the output. In the context of our exercise, evaluating the function involves determining \( f(a) \) and \( f(a+h) \).
  • **Substitution**: Replace the variable \( x \) with a specific value, such as \( a \) or \( a+h \), in the function \( f(x) = \frac{1}{x+1} \).
  • **Calculation Steps**: For \( f(a) \), we substitute \( a \) into the function to get \( \frac{1}{a+1} \). Similarly, for \( f(a+h) \), we replace \( x \) with \( a+h \) to obtain \( \frac{1}{a+h+1} \).
  • **Purpose in Difference Quotient**: This evaluation is essential to compute the difference in outputs which forms the numerator of the difference quotient formula.
Function evaluation helps identify changes between two points, which is pivotal in understanding the rate of change in rational functions, as demonstrated by the difference quotient.
Simplification of Expressions
Simplifying expressions is a crucial step in understanding and working with complex equations, such as when computing the difference quotient. A simplified expression is more manageable and often essential for solving. Here鈥檚 how it applies to our exercise:
  • **Common Denominator**: In the difference quotient \( \frac{f(a+h)-f(a)}{h} \), the first step is to find a common denominator for the fractions in the numerator, which is \((a+h+1)(a+1)\).
  • **Combining Fractions**: Convert both fractions to have the same denominator and then combine them. This involves manipulating the numerators, where \( \frac{1}{a+h+1} \) becomes \( \frac{a+1}{(a+h+1)(a+1)} \) and \( \frac{1}{a+1} \) becomes \( \frac{a+h+1}{(a+h+1)(a+1)} \).
  • **Cancel Terms and Simplify**: Once combined, subtract the numerators, which results in \( \frac{-h}{(a+h+1)(a+1)} \). Finally, divide by \( h \) to obtain the simplified form \( \frac{-1}{(a+h+1)(a+1)} \).
Simplification helps to clearly see the underlying patterns or behavior of the function, particularly in how variables interact with one another within the difference quotient.

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