/*! 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 10 Exercises \(5-10\) refer to the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 10

Exercises \(5-10\) refer to the function $$f(x)=\left\\{\begin{array}{ccc}{x^{2}-1,} & {-1 \leq x < 0} \\ {2 x,} & {0 < x < 1} \\ {1,} & {x=1} \\ {-2 x+4,} & {1 < x < 2} \\ {0,} & {2 < x<3}\end{array}\right.$$ graphed in the accompanying figure. To what new value should \(f(1)\) be changed to remove the discontinuity?

Short Answer

Expert verified
Change \(f(1)\) to 2 to remove the discontinuity.

Step by step solution

01

Identify the Discontinuity at x = 1

The problem states that there is a discontinuity at \(x = 1\). To find the discontinuity, we examine \(f(x)\) at \(x = 1\) and its neighboring intervals.
02

Evaluate the Limit from the Left

To find the left-hand limit as \(x\) approaches 1, we consider the interval \(0 < x < 1\) where \(f(x) = 2x\). The left-hand limit can be written as:\[\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} 2x = 2(1) = 2.\]
03

Evaluate the Limit from the Right

To find the right-hand limit as \(x\) approaches 1, we consider the interval \(1 < x < 2\) where \(f(x) = -2x + 4\). The right-hand limit can be written as:\[\lim_{{x \to 1^+}} f(x) = \lim_{{x \to 1^+}} (-2x + 4) = -2(1) + 4 = 2.\]
04

Compare Limits and Function Value at x = 1

We have established that both the left-hand limit and right-hand limit as \(x\) approaches 1 are 2. However, currently, \(f(1) = 1\). In order for the function to be continuous at \(x = 1\), \(f(1)\) must equal both the left-hand and right-hand limits.
05

Determine the New Value for f(1)

To remove the discontinuity at \(x = 1\), change \(f(1)\) to 2, matching the left-hand and right-hand limits, so that the function is continuous at that point.

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.

Piecewise Functions Explained
A piecewise function is a type of function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. Each piece or segment has its own unique rule. This kind of function is common in scenarios where different conditions apply to different sections of a real-life scenario, such as tax brackets or temperature settings.

Understanding piecewise functions involves:
  • Defining segments: Each part of the function \(f(x)\) is defined for a specific range of \(x\). For example, for the given function \(f(x)\), \(x\) values are divided into multiple ranges with corresponding expressions like \(x^2 - 1\), \2x\, \1\, etc.
  • Checking endpoints: Attention must be given to where each segment starts and ends, and whether it is inclusive or exclusive. In our case, notice carefully how some parts use \(-1 \leq x < 0\) while others like \(1 < x < 2\) rely on exclusivity.
  • Smooth transitions: The goal is often to have transitions at the endpoints be as seamless as possible, eliminating any sudden jumps which create discontinuities.
Piecewise functions are vital for modeling situations with distinct changes in behavior at different intervals, often requiring close evaluation of continuity and limits.
Understanding Limits
Limits are a foundational concept in calculus used to understand the behavior of functions as they approach a specific point. They allow us to investigate what happens near a particular value and are key when analyzing function continuity.

Here鈥檚 how limits work:
  • Left-hand limit: This represents the value that \(f(x)\) approaches as \(x\) comes close to a particular point from the left (denoted as \(x \to c^-\)). For the problem, \(\lim_{{x \to 1^-}} 2x=2\).
  • Right-hand limit: Conversely, this is the value that \(f(x)\) approaches as \(x\) approaches from the right (denoted as \(x \to c^+\)). Example: \(\lim_{{x \to 1^+}} (-2x + 4) = 2\).
  • Overall limit: For a limit at a point to exist, both the left-hand limit and right-hand limit should be the same. If they are equal, that common value is the limit of the function at \(x = c\).
In the context of the exercise, determining the limits as \(x\) approaches 1 from both sides helps us identify if there is a discontinuity at that point, guiding us to correct \(f(1)\) for continuity.
Dealing with Discontinuous Functions
Discontinuous functions exhibit breaks or holes at certain points within their domain, where the function is not continuous. These breaks occur when the limits from the left and right do not match or the function at that point doesn't match those limits.

To address discontinuities:
  • Identifying the break: Breaks are located where the piecewise function rules change, or the function changes abruptly at a point. In our exercise, this was identified at \(x = 1\).
  • Matching limits: As seen in the problem, making adjustments to ensure the left-hand and right-hand limits are equal can sometimes be enough. E.g., both limits at \(x = 1\) needed to equal 2 to remove the discontinuity.
  • Adjusting the function value: Often, we fix discontinuities by modifying the value the function takes at that specific point. For the exercise, \(f(1)\) was adjusted from 1 to 2 to achieve continuity.
Understanding and fixing discontinuities is crucial for functions that model physical phenomena, ensuring predictions and analyses based on these models are accurate and reliable.

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

In Exercises \(47-50,\) graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0\) .If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$ f(x)=(1+2 x)^{1 / x} $$

Find the limits in Exercises \(80-86\). $$\lim _{x \rightarrow-\infty}\left(\sqrt{x^{2}+3}+x\right)$$

A function value Show that the function \(F(x)=(x-a)^{2}\) . \((x-b)^{2}+x\) takes on the value \((a+b) / 2\) for some value of \(x .\)

Use formal definitions to prove the limit statements in Exercises \(89-92 .\) $$\lim _{x \rightarrow 0} \frac{1}{|x|}=\infty$$

You will find a graphing calculator useful. Let $$h(x)=\left(x^{2}-2 x-3\right) /\left(x^{2}-4 x+3\right)$$ a. Make a table of the values of \(h\) at \(x=2.9,2.99,2.999,\) and so on. Then estimate lim_{x} \rightarrow 3 \(h(x)\) . What estimate do you arrive at if you evaluate \(h\) at \(x=3.1,3.01,3.001, \ldots\) instead? b. Support your conclusions in part (a) by graphing \(h\) near \(x_{0}=3\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow 3\) . c. Find lim_{x\rightarrow3} \(h(x)\) algebraically.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

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.