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Chapter 1: Problem 61

Simplify the difference quotients\(\frac{f(x+h)-f(x)}{h}\) and \(\frac{f(x)-f(a)}{x-a}\) for the following functions. $$f(x)=\frac{x}{x+1}$$

Short Answer

Expert verified
Answer: The simplified difference quotients are: 1. \(\frac{f(x+h)-f(x)}{h} = \frac{-h(3x + h + 1)}{(x + h + 1)(x + 1)}\) 2. \(\frac{f(x)-f(a)}{x-a} = \frac{1}{(x + 1)(a + 1)}\)

Step by step solution

01

1. f(x+h)

Calculate f(x+h) by replacing x with (x+h) in the given function: $$f(x + h) = \frac{x + h}{(x + h) + 1}$$
02

2. Simplify f(x+h)

Simplify the expression obtained in step 1: $$f(x + h) = \frac{x + h}{x + h + 1}$$
03

3. Calculate the first difference quotient \(\frac{f(x+h)-f(x)}{h}\)

Substitute the expressions for f(x+h) and f(x) into the first difference quotient and simplify: $$\frac{f(x+h)-f(x)}{h} = \frac{\frac{x + h}{x + h + 1} - \frac{x}{x + 1}}{h}$$
04

4. Find a common denominator for the difference

The common denominator for \(\frac{x + h}{x + h + 1}\) and \(\frac{x}{x + 1}\) is \((x + h + 1)(x + 1)\). Rewrite both fractions with this common denominator and simplify: $$\frac{f(x+h)-f(x)}{h} = \frac{\frac{x(x + 1) - (x + h)(x + h + 1)}{h(x + h + 1)(x + 1)}}{h}$$
05

5. Cancel h in the denominator

Cancel the h in the numerator and denominator: $$\frac{f(x+h)-f(x)}{h} = \frac{x(x + 1) - (x + h)(x + h + 1)}{(x + h + 1)(x + 1)}$$
06

6. Expand and simplify the numerator

Expand the numerator and simplify: $$\frac{f(x+h)-f(x)}{h} = \frac{x^2 + x - x^2 - hx - 2hx - h^2}{(x + h + 1)(x + 1)}$$ $$\frac{f(x+h)-f(x)}{h} = \frac{-(hx + 2hx + h^2 + x)}{(x + h + 1)(x + 1)}$$ $$\frac{f(x+h)-f(x)}{h} = \frac{-h(3x + h + 1)}{(x + h + 1)(x + 1)}$$ Now, let's calculate the second difference quotient.
07

7. Calculate the second difference quotient \(\frac{f(x)-f(a)}{x-a}\)

Substitute the expressions for f(x) and f(a) into the second difference quotient and simplify: $$\frac{f(x)-f(a)}{x-a} = \frac{\frac{x}{x + 1} - \frac{a}{a + 1}}{x-a}$$
08

8. Find a common denominator for the difference

The common denominator for \(\frac{x}{x + 1}\) and \(\frac{a}{a + 1}\) is \((x + 1)(a + 1)\). Rewrite both fractions with this common denominator and simplify: $$\frac{f(x)-f(a)}{x-a} = \frac{\frac{x(a + 1) - a(x + 1)}{(x + 1)(a + 1)}}{x-a}$$
09

9. Expand and simplify the numerator

Expand the numerator and simplify: $$\frac{f(x)-f(a)}{x-a} = \frac{ax + x - ax - a}{(x + 1)(a + 1)}$$ $$\frac{f(x)-f(a)}{x-a} = \frac{x - a}{(x + 1)(a + 1)}$$
10

10. Cancel x-a in the numerator and denominator

Rewrite the second difference quotient, canceling the (x-a) in the numerator and denominator: $$\frac{f(x)-f(a)}{x-a} = \frac{1}{(x + 1)(a + 1)}$$ In conclusion, the simplified difference quotients are: $$\frac{f(x+h)-f(x)}{h} = \frac{-h(3x + h + 1)}{(x + h + 1)(x + 1)}$$ $$\frac{f(x)-f(a)}{x-a} = \frac{1}{(x + 1)(a + 1)}$$

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Key Concepts

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

Understanding the Difference Quotient
The difference quotient is a key concept in calculus. It helps us find the derivative of a function, which tells us the rate at which something changes. Essentially, it measures the slope of the secant line between two points on a function.

To calculate the difference quotient, we use the formula \( \frac{f(x+h) - f(x)}{h} \) or \( \frac{f(x) - f(a)}{x-a} \). For functions like rational functions, where you have fractions, it can get a bit tricky at first, but understanding this process is crucial.
  • Step-by-step approach: Start with finding \( f(x+h) \) by replacing every \( x \) in the function with \( x+h \).
  • Simplify: Simplify any complex fractions that arise after substituting.
  • Find common denominators: This helps in combining terms effectively.
By mastering the difference quotient, you get closer to understanding what derivatives are and their significance in calculus.
Limits and Their Role in Calculus
Limits are foundational in the study of calculus. They allow us to understand what happens as a variable approaches a certain value. It is an essential concept when we're dealing with gradients, derivatives, and continuity.

Think of a limit as asking the question: "What does a function approach as the input gets closer and closer to a particular point?" The difference quotient formula is heavily reliant on limits, especially when finding derivatives. Here's why:
  • In the formula \( \frac{f(x+h)-f(x)}{h} \), as \( h \) approaches 0, the difference quotient becomes the derivative.
  • Similarly, in \( \frac{f(x)-f(a)}{x-a} \), as \( x \) approaches \( a \), you determine the instantaneous rate of change.
  • Using limits, you transform the average rate of change into an instantaneous one, which is crucial in calculus.
By getting comfortable with limits, you'll unlock a deeper understanding of calculus topics and solve complex problems more effectively.
Rational Functions in Calculus
Rational functions are a class of functions written as the quotient of two polynomials. They typically take the form \( f(x) = \frac{P(x)}{Q(x)} \). These functions often appear in calculus problems because they have interesting properties and behaviors.

Understanding rational functions involves:
  • Identifying domain restrictions: The denominator \( Q(x) \) should not be zero, as this makes the function undefined.
  • Simplification: Often, terms in the numerator and denominator can be simplified to aid calculations.
  • Analyzing behavior: Asymptotes, which are lines that the graph of the function approaches but never touches, are common with rational functions.
In the context of our difference quotient problems, rational functions require careful handling due to their fraction nature. When simplifying such functions, looking for common denominators and canceling terms becomes particularly important. Mastering these techniques will help you in tackling rational function problems confidently.

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