Problem 29 - Thomas Calculus

Question

Each of Exercises \(15-30\) gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon>0 .\) 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. $$ \begin{array}{l}{f(x)=m x+b, \quad m>0, \quad L=(m / 2)+b} \\ {c=1 / 2, \quad \epsilon=c>0}\end{array} $$

Step-by-Step Solution

Answer

The open interval is \(\left( \frac{1}{2} - \frac{1}{2m}, \frac{1}{2} + \frac{1}{2m} \right)\) with \(\delta = \frac{1}{2m}\).

Step 1: Identify the problem

We need to find an open interval about \(c = \frac{1}{2}\) where \(|f(x) - L| < \epsilon\) holds, and determine \(\delta > 0\) for which \(0 < |x - c| < \delta\) implies \(|f(x) - L| < \epsilon\). The function is given as \(f(x) = mx + b\), \(L = \frac{m}{2} + b\), and \(\epsilon = \frac{1}{2}\).

Step 2: Simplify the inequality

First, express \(|f(x) - L| < \epsilon\): \(|mx + b - (\frac{m}{2} + b)| < \frac{1}{2}\). Simplify this to obtain \(|mx - \frac{m}{2}| < \frac{1}{2}\).

Step 3: Factor out m

Factor out \(m\) to express the absolute inequality: \(m|x - \frac{1}{2}| < \frac{1}{2}\).

Step 4: Solve for x

Divide through by \(m\) (since \(m > 0\)) to get: \(|x - \frac{1}{2}| < \frac{1}{2m}\).

Step 5: Determine delta

The value of \(\delta\) must satisfy \(0 < |x - \frac{1}{2}| < \frac{1}{2m}\). So, choose \(\delta = \frac{1}{2m}\).

Step 6: Conclusion for the interval

The open interval for \(x\) around \(c\) is \(\left( \frac{1}{2} - \frac{1}{2m}, \frac{1}{2} + \frac{1}{2m} \right)\) where \(|f(x) - L| < \epsilon\).

Key Concepts

Epsilon-Delta Definition

One of the foundational concepts in calculus related to limits is the epsilon-delta definition. This definition provides a rigorous way to define what it means for a function to approach a limit.
In simple terms, it asserts that for every possible distance from the limit (called \(\epsilon\)), it is possible to find a corresponding distance from a point \(c\) (called \(\delta\)) such that the function will be closer to the limit than the specified distance \(\epsilon\) whenever the input is within the distance \(\delta\) of \(c\).

• The given inequality \(|f(x) - L| < \epsilon\) describes the closeness of \(f(x)\) to \(L\).
• The inequality \(0 < |x - c| < \delta\) ensures that \(x\) is close enough to \(c\) without being exactly equal.

A typical problem involves finding such a \(\delta\) given a specific \(\epsilon\). This assures that, within that interval, the function behaves predictably.

Linear Functions

Linear functions are among the simplest forms of functions in mathematics. They can be written in the form \(f(x) = mx + b\), where \(m\) and \(b\) are constants.

• The coefficient \(m\) represents the slope of the line, showing how much \(f(x)\) changes for a unit increase in \(x\).
• The constant \(b\) is the y-intercept, where the function crosses the y-axis.

For the given problem, you're handling a linear function in terms of identifying a point \(c\) and evaluating how the function behaves near it.
The stepwise manipulation uses the properties of linearity, such as factoring \(m\) and simplifying inequalities, which follows naturally due to the straightforward nature of linear equations.

Interval Notation

Interval notation is a mathematical notation used to describe a set of numbers between two endpoints. It's a concise way to represent variables usually in conditions or constraints.
In this calculus problem, determining the open interval around point \(c\) is crucial:

• Open intervals, notated like \( (a, b) \), include all numbers between \(a\) and \(b\) but not the endpoints \(a\) and \(b\) themselves.
• In this scenario, \(x\) is within the interval \(\left( \frac{1}{2} - \frac{1}{2m}, \frac{1}{2} + \frac{1}{2m} \right)\).

Interval notation helps easily visualize and understand which values of \(x\) meet the conditions created by the \(\epsilon\) and \(\delta\) definitions, by specifying clear boundaries where the defined inequalities hold.