/*! 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 22 Sketch a possible graph for a fu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 22

Sketch a possible graph for a function \(f\) with the specified properties. (Many different solutions are possible.) (i) the domain of \(f\) is \([-2,1]\) (ii) \(f(-2)=f(0)=f(1)=0\) (iii) \(\lim _{x \rightarrow-2^{+}} f(x)=2, \lim _{x \rightarrow 0} f(x)=0\), and \(\lim _{x \rightarrow 1^{-}} f(x)=1\)

Short Answer

Expert verified
Graph is continuous except at \(x = -2\) and \(x = 1\) with specified limits. Passes through \((0,0)\).

Step by step solution

01

Understand the Domain and Intercepts of the Function

The function is defined for all values of \(x\) from \(-2\) to \(1\), inclusive. This means the graph must exist between these two points on the \(x\)-axis. Additionally, the points \((-2, 0)\), \((0, 0)\), and \((1, 0)\) must all lie on the graph because these are given that \(f(-2) = 0\), \(f(0) = 0\), and \(f(1) = 0\).
02

Behavior near x = -2

Approach \(x = -2\) from the right ([x \rightarrow -2^{+}]). The function approaches the value 2, implying the graph should approach the point \((-2, 2)\) as \(x\) approaches \(-2\) from the right. However, the function has a point \((-2, 0)\), indicating a vertical drop at \(x = -2\).
03

Behavior around x = 0

The function approaches the value 0 as \(x\) approaches 0 from either direction \((\lim_{x \rightarrow 0} f(x) = 0)\). This means the graph should smoothly go through the point \((0, 0)\) without any discontinuity or jump.
04

Behavior near x = 1

Approach \(x = 1\) from the left ([x \rightarrow 1^{-}]). The function approaches the value 1 at this limit. Therefore, the graph should approach the point \((1, 1)\) as it approaches \(x = 1\) from the left. However, since \(f(1) = 0\), there is a jump down to this point at \(x = 1\).
05

Sketching the Graph

Combine all the information: draw a curve starting at \((2, 0)\), shooting up to \((-2, 2)\) with a right-open discontinuity, then smoothly dropping to \((0, 0)\) along the \(x\)-axis, and continue to approach \((1, 1)\) from the left but drop to \((1, 0)\) at the point. Make sure to depict the open points as necessary for the described limits.

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.

Understanding Function Behavior
The behavior of a function refers to how the function's graph behaves as the input, or \( x \), approaches certain values or intervals. In this exercise, we focus on how the function \( f \) behaves around specific points on the graph, based on given conditions.
  • At \( x = -2 \), the graph should approach the point \( (-2, 2) \) as \( x \) approaches \(-2^+\) (from the right). However, at \( x = -2 \), the function value is \( 0 \), indicating a sharp drop, or discontinuity, down to \( (-2, 0) \).
  • As \( x \) approaches \( 0 \) from either side, the function smoothly passes through the origin \( (0, 0) \), with no jumps or gaps. This showcases continuous behavior at \( x = 0 \).
  • Towards \( x = 1 \), the function tends towards the point \( (1, 1) \) as \( x \) approaches from the left, \( x \rightarrow 1^- \). However, the graph jumps down to \( (1, 0) \) when it actually reaches \( x = 1 \). Thus, there's a noticeable discontinuity at this point as well.
Function behavior allows us to understand how the graph transitions and changes direction based on the given conditions.
Exploring Limits in Calculus
Limits are crucial in calculus for understanding the behavior of functions as they approach certain points, but not necessarily at those points. They describe what value a function is tending towards, providing insight into its potential behavior even if the function doesn't touch those values.
For this graph:
  • The limit \( \lim_{x \rightarrow -2^+} f(x) = 2 \) informs us that as \( x \) gets very close to \(-2\) from the right, \( f(x) \) gets close to 2. However, the function value itself at \( x = -2 \) is \( 0 \), displaying a vertical discontinuity.
  • \( \lim_{x \rightarrow 0} f(x) = 0 \) means the function smoothly approaches 0 as \( x \) nears \( 0 \) from both sides, resulting in continuity at \( x = 0 \).
  • Finally, \( \lim_{x \rightarrow 1^-} f(x) = 1 \) tells us that the graph approaches \( (1, 1) \) as \( x \) nears \( 1 \) from the left. Yet, the function itself jumps to \( (1, 0) \), again showcasing a distinctive discontinuity.
Knowing limits helps us sketch graphs accurately, especially when dealing with potential jumps or holes.
Grasping Domain and Range
The domain and range of a function are fundamental concepts that dictate where the function is defined and what values it can take.
For the given function \( f \):
  • The domain is the set of all possible \( x \)-values for which the function is defined. Here, the domain is \([-2, 1]\), meaning the graph only exists between \( x = -2 \) and \( x = 1 \), including these endpoints.
  • The range refers to all possible \( y \)-values that the function can output given the domain. With the conditions \( f(-2)=f(0)=f(1)=0 \), and the limits set, the range includes values from the lowest point \( 0 \) up to the peak it approaches when near \((1, 1)\) from the left. However, due to the jumps seen, certain values may not lie exactly on the graph, like \( 2 \) at \( x = -2 \).
To sketch a graph accurately, understanding both the domain and range provides a structure to plot how the function appears visually between the specified points. Both elements guide where the graph can and cannot go, based on defined mathematical conditions.

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 some population models it is assumed that a given ecological system possesses a carrying capacity \(L .\) Populations greater than the carrying capacity tend to decline toward \(L\), while populations less than the carrying capacity tend to increase toward \(L\). Explain why these assumptions are reasonable, and discuss how the concepts of this section apply to such a model.

Suppose that \(f\) is an invertible function, \(f(0)=0, f\) is continuous at 0 , and \(\lim _{x \rightarrow 0}(f(x) / x)\) exists. Given that \(L=\lim _{x \rightarrow 0}(f(x) / x)\), show $$ \lim _{x \rightarrow 0} \frac{x}{f^{-1}(x)}=L $$

Sketch the graphs of the curves \(y=1-x^{2}, y=\cos x\) and \(y=f(x)\), where \(f\) is a function that satisfies the inequalities $$ 1-x^{2} \leq f(x) \leq \cos x $$ for all \(x\) in the interval \((-\pi / 2, \pi / 2)\). What can you say about the limit of \(f(x)\) as \(x \rightarrow 0 ?\) Explain.

True-False Determine whether the statement is true or false. Explain your answer. If \(\lim _{x \rightarrow a} f(x)\) and \(\lim _{x \rightarrow a} g(x)\) exist, then so does \(\lim _{x \rightarrow a}[f(x)+g(x)]\)

Find the limits. $$ \lim _{x \rightarrow 0} \frac{\sin 6 x}{\sin 8 x} $$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

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.