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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 2 - LimitsProblem 2.7.2
Chapter 2, Problem 2.7.2

Suppose f(x) lies in the interval (2, 6). What is the smallest value of ε such that |f (x)−4|<ε?

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Step 1: Understand the problem. We are given that f(x) is in the interval (2, 6), which means 2 < f(x) < 6. We need to find the smallest value of \( \varepsilon \) such that \( |f(x) - 4| < \varepsilon \).
Step 2: Consider the expression \( |f(x) - 4| \). This represents the distance between f(x) and 4 on the number line.
Step 3: Since f(x) is between 2 and 6, the distance from 4 to the nearest endpoint of the interval (2, 6) will determine the smallest \( \varepsilon \).
Step 4: Calculate the distance from 4 to the endpoints of the interval. The distance from 4 to 2 is \( |4 - 2| = 2 \), and the distance from 4 to 6 is \( |6 - 4| = 2 \).
Step 5: The smallest \( \varepsilon \) is the minimum of these distances, which is 2. Therefore, \( \varepsilon = 2 \) ensures that \( |f(x) - 4| < \varepsilon \) for all f(x) in the interval (2, 6).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Absolute Value Inequality

The expression |f(x) - 4| < ε represents an absolute value inequality, which measures the distance between f(x) and the number 4. This inequality states that the value of f(x) must be within ε units of 4, meaning f(x) can vary but must remain close to this central value.
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Interval Notation

The interval (2, 6) indicates that the function f(x) takes values strictly between 2 and 6. Understanding this interval is crucial because it helps determine the possible values of f(x) and how they relate to the target value of 4, which is central to the absolute value inequality.
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Finding ε

To find the smallest value of ε such that |f(x) - 4| < ε, we need to consider the maximum deviation of f(x) from 4 within the given interval. Since f(x) lies between 2 and 6, the closest points to 4 are 2 and 6, leading to the calculation of ε as the minimum distance from 4 to these endpoints, which is 2.
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