/*! 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 15 Gives a function \(f(x)\) and nu... [FREE SOLUTION] | 91Ó°ÊÓ

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Gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon \geq 0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. \(f(x)=x+1, \quad L=5, \quad c=4, \quad \epsilon=0.01\)

Short Answer

Expert verified
An open interval is (3.99, 4.01), and \(\delta = 0.01\).

Step by step solution

01

Express the Problem in Inequality

We need to find an interval around \(c\) where the inequality \(|f(x) - L| < \epsilon\) holds. Given that \(f(x) = x + 1\), \(L = 5\), so we can rewrite \(|f(x) - 5| < 0.01\) as \(|x + 1 - 5| < 0.01\). Simplifying, this becomes \(|x - 4| < 0.01\).
02

Determine the Interval

The inequality \(|x - 4| < 0.01\) can be broken into two inequalities: \(-0.01 < x - 4 < 0.01\). Solving these inequalities gives the interval: \(3.99 < x < 4.01\). Thus, the open interval about \(c = 4\) on which the inequality holds is \((3.99, 4.01)\).
03

Determine the Value for \(\delta\)

We need to find \(\delta > 0\) such that for all \(x\) satisfying \(0 < |x - c| < \delta\), the inequality \(|f(x) - L| < \epsilon\) holds. From Step 2, \(|x - 4| < 0.01\) gives us that \(\delta = 0.01\). Thus, \(\delta = 0.01\) is the value for which the condition \(|f(x) - L| < \epsilon\) is satisfied whenever \(0 < |x - 4| < \delta\).

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Key Concepts

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

Epsilon-Delta Definition
The epsilon-delta definition is one of the foundational building blocks of calculus, particularly when we talk about limits. It's a formal way to define what it means when we say a function approaches a certain limit as its input gets closer to a particular value. Let's break it down:
  • Epsilon (\( \epsilon \)): This is a small positive number that represents how close we want the function's output \( f(x) \) to be to a specific limit \( L \). It's a measure of allowed error.
  • Delta (\( \delta \)): This small positive number signifies how close \( x \) needs to be to \( c \), such that \( f(x) \) remains within the epsilon distance of the limit \( L \).
The goal is to show that for every epsilon, however small, there is a delta such that when \( x \) is within delta of \( c \), \( f(x) \) is within epsilon of \( L \). This means:\[ |f(x) - L| < \epsilon \]holds whenever:\[ 0 < |x - c| < \delta \]In our problem, we showed that \( \delta = 0.01 \) ensures \( |x - 4| < 0.01 \), making the function \( x + 1 \) approach 5 closely within an error of 0.01.
Inequalities
Inequalities are a fundamental concept when it comes to understanding the behavior of functions, particularly when you're dealing with limits. Inequalities allow us to describe ranges and margins within which a function must behave.When solving the exercise, we began with the inequality:\[ |f(x) - L| < \epsilon \]This was translated using the function \( f(x) = x+1 \) and the limit \( L = 5 \) into the specific inequality:\[ |x + 1 - 5| < 0.01 \]Which simplifies to:\[ |x - 4| < 0.01 \]This means \( x \) is squeezed between two values, not perfectly hitting \( c \), but close enough within \( (3.99, 4.01) \). Interpreting inequalities gives us a glimpse into the interval that assures the continuity and the limit of functions, wrapping the function’s behavior in a nice bound. It's a protective margin set both around the input \( x \) and the output \( f(x) \).
Continuity
Continuity is an essential property of functions that tells us about the smoothness and unbroken nature of a graph. When a function is continuous at a point \( c \), it means several things:
  • The function is defined at point \( c \).
  • The limit of the function as \( x \) approaches \( c \) exists.
  • The limit of the function as \( x \) approaches \( c \) equals the function’s value at \( c \).
In our exercise example, the function \( f(x) = x + 1 \) is continuous at \( c = 4 \). Here's why:- As \( x \) approaches 4, \( f(x) \) smoothly approaches \( 5 \).- There are no breaks, jumps, or holes in \( f(x) \).For any \( x \) near 4, no matter how small the epsilon, we can find a delta ensuring the function stays as close as we want to the value 5, guaranteeing both the notion of limits and continuity. This seamless transition from point close to \( c \) reflects a continuous nature—a key aspect of studying calculus limits.

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