Problem 20 Each of Exercises $15-30$ give... [FREE SOLUTION]

Chapter 2: Problem 20

Each of Exercises $15-30$ gives a function $f(x)$ and numbers $L, x_{0}$ and $\epsilon > 0 .$ In each case, find an open interval about $x_{0}$ on which the inequality $|f(x)-L| < \epsilon$ holds. Then give a value for $\delta > 0$ such that for all $x$ satisfying $0 < \left|x-x_{0}\right| < \delta$ the inequality $|f(x)-L| < \epsilon$ holds. $$ f(x)=\sqrt{x-7}, \quad L=4, \quad x_{0}=23, \quad \epsilon=1 $$

Short Answer

Expert verified

The open interval is $16 < x < 32$, and $\delta = 7$.

Step by step solution

Understanding the Problem

We need to find an open interval around $x_0 = 23$ where the inequality $|f(x) - L| < \epsilon$ holds. Here, $f(x) = \sqrt{x-7}$, $L = 4$, and $\epsilon = 1$. We must then determine $\delta > 0$ such that for all $x$ satisfying $0 < |x - x_0| < \delta$, the inequality $|f(x) - L| < \epsilon$ is guaranteed.

Express the Absolute Inequality

Start by expressing $|f(x) - L| < \epsilon$ for $f(x) = \sqrt{x-7}$ and $L = 4$. This gives us: \[|\sqrt{x-7} - 4| < 1.\]

Solve the Inequality

Solve the inequality $|\sqrt{x-7} - 4| < 1$. This results in two inequalities: $\sqrt{x-7} - 4 < 1$ and $\sqrt{x-7} - 4 > -1$.1. For $\sqrt{x-7} - 4 < 1$, solve $\sqrt{x-7} < 5$ which implies $x-7 < 25$, or $x < 32$.2. For $\sqrt{x-7} - 4 > -1$, solve $\sqrt{x-7} > 3$ which implies $x-7 > 9$, or $x > 16$. Hence, the interval for $x$ is $16 < x < 32$.

Find the Open Interval

Determine the open interval around $x_0 = 23$ where both inequalities hold: $16 < x < 32$. Since $23$ is within this range, this is the valid interval on which $|f(x) - L| < \epsilon$.

Determine $\delta$

The interval $16 < x < 32$ gives bounds centered around $x_0 = 23$. Calculate the distance from $x_0$ to the nearest boundary:- From $23$ to $16$, the distance is $23 - 16 = 7$.- From $23$ to $32$, the distance is $32 - 23 = 9$.The smaller distance, $7$, is the value for $\delta$. Therefore, we choose $\delta = 7$.

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Key Concepts

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

Understanding the Epsilon-Delta Definition

The epsilon-delta ($ \epsilon-\delta $) definition is central to understanding limits in calculus. It's used to formally define the limit of a function at a point. The definition states that for a given limit $ L $, there should exist a $ \delta > 0 $ such that if $ x $ is within the distance of $ \delta $ from $ x_0 $ (except possibly at $ x_0 $ itself), then the function $ f(x) $ is within $ \epsilon $ of $ L $. In simpler terms:

Think of $ \epsilon $ as a tolerance for how close $ f(x) $ can get to the limit $ L $.
$ \delta $ is a measure of how close $ x $ has to be to $ x_0 $ to achieve that goal.

Understanding this concept helps in determining the open interval within which $ f(x) $ stays close to $ L $, which is crucial for solving inequality constraints in calculus problems.

Decoding Function Intervals

Function intervals describe where a function behaves or fulfills certain conditions. In the exercise, we need to locate an open interval around $ x_0 = 23 $ on which $ |\sqrt{x-7} - 4| < 1 $ holds. Start by isolating the function expression:

Solve $ |\sqrt{x-7} - 4| < 1 $ to break it into two separate inequalities, creating a range for $ x $.
The resulting interval is from $ 16 < x < 32 $.

This outcome reflects the values of $ x $ that make the function $ f(x) = \sqrt{x-7} $ stay within 1 unit of 4. Identifying such an interval helps in many calculus problems, especially those involving function behaviors and limits.

Approaching Absolute Inequality Solving

Solving absolute inequalities such as $ |\sqrt{x-7} - 4| < 1 $ involves breaking down the problem into manageable parts. Here's how:

Break the inequality into two as follows: $ \sqrt{x-7} - 4 < 1 $ and $ \sqrt{x-7} - 4 > -1 $.
Solve these linear inequalities separately:
- For $ \sqrt{x-7} < 5 $, deduce $ x < 32 $ by squaring both sides and adding 7.
- For $ \sqrt{x-7} > 3 $, deduce $ x > 16 $ similarly.

The intersection of these solutions gives the interval where $ x $ satisfies the original inequality. Thus, handling absolute inequalities efficiently demands careful separation and solving each separately, leading to a comprehensive understanding of the function's behavior.

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91影视

Short Answer

Step by step solution

Understanding the Problem

Express the Absolute Inequality

Solve the Inequality

Find the Open Interval

Determine \(\delta\)

Key Concepts

Understanding the Epsilon-Delta Definition

Decoding Function Intervals

Approaching Absolute Inequality Solving

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