Problem 1 - Thomas Calculus

Question

In Exercises \(1-6,\) sketch the interval \((a, b)\) on the \(x\) -axis with the point \(c\) inside. Then find a value of \(\delta>0\) such that for all \(x, 0<|x-c|<\delta \Rightarrow a < x < b\) $$ a=1, \quad b=7, \quad c=5 $$

Step-by-Step Solution

Answer

Choose \( \delta = 2 \) so that \(x\) remains within the interval \((1, 7)\) for \(3 < x < 7\).

Step 1: Understand the Problem

We need to sketch the interval \((a, b)\), which is \((1, 7)\), on the \(x\)-axis. Additionally, we need to place the point \(c = 5\) inside this interval. Our goal is to find \(\delta > 0\) such that any \(x\) satisfying \(0 < |x - c| < \delta\) also satisfies \(a < x < b\).

Step 2: Interpret the Condition

We are given that \(0 < |x - c| < \delta\). This means the distance from \(x\) to \(c\) is less than \(\delta\), but \(x\) is not equal to \(c\). We need \(x\) to be strictly within the interval \((1, 7)\).

Step 3: Visualize the Interval

On the \(x\)-axis, sketch the interval from \(a = 1\) to \(b = 7\). Place the point \(c = 5\) inside this interval: - [1, 5, 7] This indicates that \(c = 5\) is the midpoint or closer to the middle of \(a\) and \(b\).

Step 4: Calculate Distance from c to a and b

Calculate the distance from \(c\) to \(a\) and \(b\):- Distance to \(a = 1\): \(5 - 1 = 4\).- Distance to \(b = 7\): \(7 - 5 = 2\).The smaller distance will be important for choosing \(\delta\).

Step 5: Determine \(\delta\)

Choose \(\delta\) to be the smallest of the two distances calculated:- Since the distance to \(b\) is smaller, we choose \(\delta = 2\).This ensures that \(x\) remains within the interval no matter how close \(x\) gets to \(c\).

Step 6: Verify and Interpret Result

If \(0 < |x - 5| < 2\), it implies that \(3 < x < 7\). Since \(3 > a\) (which is 1), it's ensured that \(a < x < b\) (i.e., \(1 < x < 7\)). Thus, \(\delta = 2\) satisfies the required condition.

Key Concepts

Open Interval

In calculus, an open interval is a crucial concept often encountered in problems involving functions and limits. An open interval

• is denoted as \((a, b)\).
• includes all real numbers \(x\) such that \(a < x < b\).
• excludes the endpoints \(a\) and \(b\).

Open intervals allow us to discuss values in between two points without including the points themselves.
This is important in calculus when discussing behaviors around certain values without considering the values exactly.
In the exercise provided, we're dealing with the open interval \((1, 7)\).Here, the interval includes all numbers greater than 1 but less than 7,therefore excluding 1 and 7 themselves.
This is represented on a number line as a pair of parentheses markers at points 1 and 7, with the range in between fully covered.

Epsilon-Delta Definition

The epsilon-delta definition is a formal way of describing the concept of limits in calculus.
It's used to specify how close a function gets to a limit near a particular point. In simpler terms,

• for a limit \(L\) of function \(f(x)\) as \(x\) approaches \(c\), given any \(\epsilon > 0\),
• there exists a \(\delta > 0\) such that for all \(x\),
• \(0 < |x - c| < \delta\) implies \(|f(x) - L| < \epsilon\).

In the problem you're working with, we're simplifying this context.
Instead of a function approaching a limit, we're focusing on a specific point \(c = 5\) within the interval \((1, 7)\).
The goal is to find \(\delta\) such that any \(x\) is within our interval if it lies less than \(\delta\) away from \(c\).
This ensures that \(x\) is close enough to \(c\) but remains in \((1, 7)\), maintaining the required continuity around \(c\).
This is an excellent example of how the concept of limits permeates calculus.

Inequality Solving

Inequality solving involves finding the set of values that satisfy a given inequality condition.
It's a fundamental skill in calculus and mathematics in general.
For inequalities like \(0 < |x - c| < \delta\),

• we're interested in the distance from \(x\) to a point \(c\).
• This ensures \(x\) isn't exactly \(c\),
• yet lies within a determined range \(\delta\).

The inequality is broken down into

• \(-\delta < x - c < \delta\)

which further simplifies to

• \(c - \delta < x < c + \delta\).

By solving inequality problems, we can determine the range of \(x\) that satisfies particular conditions.
In this exercise, we determined that \(\delta = 2\) is the smallest distance from \(c = 5\) to the endpoints \(1\) and \(7\).
This choice of \(\delta\) guarantees maintainence of \(x\) within the open interval. Solving such inequalities is essential in establishing logical bounds in calculus problems.