/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Find the average rate of change ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 16

Find the average rate of change of the given function between the following pairs of \(x\) -values. [Hint: See pages 95-96.] a. \(x=3\) and \(x=5\) b. \(x=3\) and \(x=4\) c. \(x=3\) and \(x=3.5\) d. \(x=3\) and \(x=3.1\) e. \(x=3\) and \(x=3.01\) f. What number do your answers seem to be approaching? $$ f(x)=\frac{4}{x} $$

Short Answer

Expert verified
The average rates of change approach \(-\frac{4}{3^2} = -\frac{4}{9} \).

Step by step solution

01

Understand the Formula

The average rate of change of a function \( f(x) \) between two \( x \)-values, \( x_1 \) and \( x_2 \), is given by the formula \( \frac{f(x_2) - f(x_1)}{x_2 - x_1} \). Use this to find the average rate of change for each pair of \( x \)-values.
02

Calculate for x = 3 and x = 5

First, evaluate \( f(x) \) at \( x = 3 \) and \( x = 5 \). \( f(3) = \frac{4}{3} \) and \( f(5) = \frac{4}{5} \). Calculate the average rate of change: \[ \frac{f(5) - f(3)}{5 - 3} = \frac{\frac{4}{5} - \frac{4}{3}}{2} = \frac{-\frac{8}{15}}{2} = -\frac{4}{15}. \]
03

Calculate for x = 3 and x = 4

Evaluate \( f(x) \) at \( x = 3 \) and \( x = 4 \). \( f(3) = \frac{4}{3} \) and \( f(4) = 1 \). Calculate the average rate of change: \[ \frac{f(4) - f(3)}{4 - 3} = \frac{1 - \frac{4}{3}}{1} = -\frac{1}{3}. \]
04

Calculate for x = 3 and x = 3.5

Evaluate \( f(x) \) at \( x = 3 \) and \( x = 3.5 \). \( f(3) = \frac{4}{3} \) and \( f(3.5) = \frac{8}{7} \). Calculate the average rate of change: \[ \frac{f(3.5) - f(3)}{3.5 - 3} = \frac{\frac{8}{7} - \frac{4}{3}}{0.5} = \frac{-\frac{4}{21}}{0.5} = -\frac{8}{21}. \]
05

Calculate for x = 3 and x = 3.1

Evaluate \( f(x) \) at \( x = 3 \) and \( x = 3.1 \). \( f(3) = \frac{4}{3} \) and \( f(3.1) \approx 1.2903 \). Calculate the average rate of change: \[ \frac{f(3.1) - f(3)}{3.1 - 3} \approx \frac{1.2903 - \frac{4}{3}}{0.1} = \frac{-0.0436}{0.1} = -0.436. \]
06

Calculate for x = 3 and x = 3.01

Evaluate \( f(x) \) at \( x = 3 \) and \( x = 3.01 \). \( f(3) = \frac{4}{3} \) and \( f(3.01) \approx 1.3289 \). Calculate the average rate of change: \[ \frac{f(3.01) - f(3)}{3.01 - 3} \approx \frac{1.3289 - \frac{4}{3}}{0.01} = \frac{-0.004299}{0.01} = -0.4299. \]
07

Look for Pattern

Observe the obtained average rates of change for the decreasing intervals: \(-\frac{4}{15}, -\frac{1}{3}, -\frac{8}{21}, -0.436, -0.4299\). These values tend towards the derivative of \( f(x) = \frac{4}{x} \) at \( x = 3 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Calculus
Calculus is a branch of mathematics that focuses on changes. It allows us to explore how things change with respect to each other. One important component of calculus is the concept of rates of change. The average rate of change is used to understand how a quantity changes over a specific interval within a function.
In the given problem, we looked at the function \( f(x) = \frac{4}{x} \) and calculated how it changes between various \( x \)-values. The formula to determine this is \( \frac{f(x_2) - f(x_1)}{x_2 - x_1} \). This helps us find the slope of the line connecting two points on the graph of the function.
By using these calculations, we gain insight into the behavior of a function over a specified range. Calculating the average rate of change becomes essential when dealing with real-world phenomena that require understanding how quickly or slowly something evolves, which is a foundational principle of calculus.
Function Analysis
Function analysis involves studying the properties and behaviors of functions, which are mathematical relationships between sets of numbers. Understanding how functions behave is crucial, especially in calculus, as it tells us about trends and patterns.
When analyzing a function, we look at aspects like continuity, limits, and rates of change. The average rate of change, specifically, reveals the overall behavior of a function across an interval, showing how the value of the function progresses or regresses between two points.
For the function \( f(x) = \frac{4}{x} \), analyzing how its output changes between different \( x \)-values allowed us to gauge its behavior more thoroughly. This kind of analysis helps us determine critical points and understand how functions change in intervals, driving a deeper insight into the mathematical phenomenon they represent.
Mathematical Derivatives
Derivatives are a central concept in calculus, representing the instant rate of change of a function. They are like taking the idea of an average rate of change to its smallest possible interval, approaching zero. This gives a precise snapshot of how a function behaves at any given point.
Finding the derivative of a function can be done through several techniques, one of which comes from noticing patterns in the average rate of change as the interval becomes very small. For \( f(x) = \frac{4}{x} \), by examining rates over smaller intervals around \( x = 3 \), we notice that the average rates approach a certain number, indicating the derivative at that point.
  • This number, when calculated through differentiation, formally defines the slope of the tangent line to the function's graph at \( x = 3 \).
  • In essence, this gives a comprehensive picture of the function鈥檚 behavior at a specific instant.
Understanding derivatives and how they relate to average changes prepares us for deeper concepts in calculus and their application in various fields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=x^{2} \sqrt{1+x^{2}} $$

Derive the Quotient Rule from the Product Rule as follows. a. Define the quotient to be a single function, $$ Q(x)=\frac{f(x)}{g(x)} $$ b. Multiply both sides by \(g(x)\) to obtain the equation \(Q(x) \cdot g(x)=f(x) .\) c. Differentiate each side, using the Product Rule on the left side. d. Solve the resulting formula for the derivative \(Q^{\prime}(x) .\) e. Replace \(Q(x)\) by \(\frac{f(x)}{g(x)}\) and show that the resulting formula for \(Q^{\prime}(x)\) is the same as the Quotient Rule. Note that in this derivation when we differentiated \(Q(x)\) we assumed that the derivative of the quotient exists, whereas in the derivation on pages \(135-136\) we proved that the derivative exists.

By canceling the common factor, \(\frac{(x-1)(x+2)}{x-1}\) simplifies to \(x+2\). At \(x=1\), however, the function \(\frac{(x-1)(x+2)}{x-1}\) is discontinuous (since it is undefined where the denominator is zero), whereas \(x+2\) is continuous. Are these two functions, one obtained from the other by simplification, equal to each other? Explain.

The Chain Rule (page 155 ) states that the derivative of \(f(g(x))\) is \(f^{\prime}(g(x)) \cdot g^{\prime}(x)\). Use Carath茅odory's definition of the derivative to prove the Chain Rule by giving reasons for the following steps. a. Since \(g\) is differentiable at \(x\), there is a function \(G\) that is continuous at 0 and such that \(g(x+h)-g(x)=G(h) \cdot h\), and \(G(0)=g^{\prime}(x)\). b. Since \(f\) is differentiable at \(g(x)\), there is a function \(F\) that is continuous at 0 and such that \(f(g(x)+h)-f(g(x))=F(h) \cdot h\), and \(F(0)=f^{\prime}(g(x))\) c. For the function \(f(g(x))\) we have \(f(g(x+h))-f(g(x))\) \(\quad=f(g(x)+g(x+h)-g(x))-f(g(x))\) \(\quad=f(g(x)+(g(x+h)-g(x)))-f(g(x))\) \(=F(g(x+h)-g(x)) \cdot(g(x+h)-g(x))\) \(=F(g(x+h)-g(x)) \cdot G(h) \cdot h\) Therefore, the derivative of \(f(g(x))\) is \(\begin{aligned} F(g(x+0)-g(x)) \cdot G(0) &=F(0) \cdot G(0) \\ &=f^{\prime}(g(x)) \cdot g^{\prime}(x) \end{aligned}\) as was to be proved.

True or False: If \(\lim _{x \rightarrow 2} f(x)=7\), then \(\lim _{x \rightarrow 2^{+}} f(x)=7\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.