Chapter 4: Problem 8
The function, \(f(x)=\sqrt[3]{x}\) is continuous.
1\. How close must \(x\) be to 1 in order to insure that \(f(x)\) is within 0.1 of
\(f(1)=1\) (that is, to insure that \(0.9
Short Answer
Expert verified
1. \(\delta = 0.271\), 2. \(\delta = 0.000025\), 3. \(\delta = 0.001\).
Step by step solution
01
Understanding Continuity and Neighborhoods
We are given a function \( f(x) = \sqrt[3]{x} \) and need to determine how close \( x \) must be to a particular value so that \( f(x) \) stays within a specified range around specific function values. This involves understanding the concept of continuity and neighborhoods - essentially taking a small interval around the given \( x \) values.
02
Find \( \delta \) for \( x=1 \)
The goal is to ensure \(0.9 < \sqrt[3]{x} < 1.1\). We calculate\(0.9 < \sqrt[3]{x} < 1.1\). Cubing each part gives:0.9^3 < x < 1.1^3.\(0.729 < x < 1.331\).This means \( x \) must be within the interval \((0.729, 1.331)\) to ensure the function stays within \((0.9, 1.1)\). The closest this allows \( x \) to be to \(1\) is \(0.271\), meaning we can write \( 1 - \delta = 0.729 \) and \( 1 + \delta = 1.331\).Thus \( \delta = 0.271 \).
03
Find \( \delta \) for \( x=\frac{1}{8} \)
We need \(0.4999 < \sqrt[3]{x} < 0.5001\). Cubing gives:0.4999^3 < x < 0.5001^3.Calculating these values gives:\(0.1249750001 < x < 0.125025001\). Therefore, \( x \) should be in the interval \((0.1249750001, 0.125025001)\). The closest \( x \) can be to \(\frac{1}{8} = \frac{0.125}{1}\) still ensuring \( f(x) \) is within range is \(\delta = 0.000025\) making \( x \) within the range \(\delta = 0.000025\) of \( \frac{1}{8}\).
04
Find \( \delta \) for \( x=0 \)
We aim to have \(-0.1 < \sqrt[3]{x} < 0.1\). Cubing gives:\(-0.1^3 < x < 0.1^3\) \(-0.001 < x < 0.001\).This means \( x \) must be within \(-0.001 < x < 0.001\) to meet the requirement that \( f(x) \) is within 0.1 of 0, the value of \( f(0)\). Therefore, \( \delta = 0.001 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Neighborhoods
In the world of calculus, the concept of a "neighborhood" is pivotal for understanding continuity and limits. When we talk about a neighborhood in mathematics, we're looking at an interval or range around a specific point. Imagine a small circle drawn around a certain point on a graph.The term "neighborhood" refers to the circle's area where the behavior of a function can be closely examined without crossing that boundary.
- A neighborhood is defined by an interval \( (a - ext{Radius}, a + ext{Radius}) \) around a point \( a \).
- In a neighborhood, the function's behavior remains consistent with the behavior at the point itself.
- This concept helps in ensuring the values remain within a desired range as they approach the point.
Cubic Root Function
The cubic root function, represented as \( f(x) = \sqrt[3]{x} \), is a fundamental function in algebra and calculus. It is the inverse operation of cubing a number.Why is the cubic root function important?
- The cubic root function transforms a number into a smaller value, indicating the original number that was cubed to get the input.
- It is a continuous and smooth function, meaning there are no breaks or jumps in its graph.
- The function provides a way to understand how numbers grow when cubed and how to reverse that process.
Delta-Delta Definition
In calculus, the \( \delta-\delta \) definition is a formal method to ensure continuity. It's a way to determine how close values need to be to a point to achieve a specific function behavior. Let's break it down.The process involves:
- Identifying a point of interest, say \( x_0 \).
- Deciding on an acceptable range for the function value, often denoted by \( \epsilon \).
- Determining how close \( x \) must be to \( x_0 \) (within \( \delta \)) so that \( f(x) \) stays within \( \epsilon \) of \( f(x_0) \).