Problem 6 - 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 \(x, 0<|x-c|<\delta \Rightarrow a < x < b\) $$ a=2.7591, \quad b=3.2391, \quad c=3 $$

Step-by-Step Solution

Answer

Choose \(\delta = 0.2391\). This ensures \(a < x < b\) for all \(x\) with \(0 < |x-c| < \delta\).

Step 1: Understand the Problem

You are given an interval \((a, b)\) on the real number line where \(a = 2.7591\) and \(b = 3.2391\). The point \(c = 3\) lies inside this interval. The goal is to find a value of \(\delta > 0\) such that for all \(x\) satisfying \(0 < |x-c| < \delta\), it follows that \(a < x < b\).

Step 2: Sketch the Given Interval and Point

On the real number line, sketch the interval from \(a = 2.7591\) to \(b = 3.2391\). Mark the point \(c = 3\) inside this interval. This visually confirms that \(c\) is equidistant from both boundaries and helps in visualizing where \(x\) can lie.

Step 3: Determine Limits for \(x\)

The condition \(a < x < b\) translates into \(2.7591 < x < 3.2391\). For \(x\) to satisfy \(0 < |x - 3| < \delta\), \(x\) must be within the bounds around \(c\).

Step 4: Calculate \(\delta\) from the Interval Edges

To ensure \(|x-3| < \delta\) keeps \(x\) within the interval, calculate the distance from \(c\) to each edge: \(c-a=3-2.7591=0.2409\) and \(b-c=3.2391-3=0.2391\). The smaller distance is \(0.2391\), which will be the maximal possible \(\delta\).

Step 5: Confirm \(\delta\) Ensure Condition

Choose \(\delta = 0.2391\). Then \(0<|x-3|<0.2391\) implies \(a

Key Concepts

Interval Notation

Interval notation is a way of representing a range of numbers on the real number line. It provides a clear picture of which numbers are included or excluded from the range. For example, the interval \((a, b)\) means that all numbers \(x\) such that \(a < x < b\) are included. Here, the parentheses indicate that the endpoints \(a\) and \(b\) themselves are not included in the interval. This is known as an "open interval."
An open interval differs from a closed interval, which would include the endpoints, denoted as \([a, b]\). There are also half-open or half-closed intervals, like \((a, b]\) or \([a, b)\), including only one of the endpoints.
In this problem, you have an open interval \((2.7591, 3.2391)\) where the point \(c = 3\) is inside. We need to ensure that any value \(x\) within a specified distance of \(c\) remains in this interval.

Real Number Line

The real number line is a visual representation of all real numbers arranged in a linear format. Each point on the line corresponds to a real number. In this context, it allows us to place intervals and specific points. For instance, the interval \((2.7591, 3.2391)\) defines all points \(x\) on the number line that are strictly between these two endpoints.
By marking this interval on the real number line along with the point \(c = 3\), you can easily see that \(c\) lies between the endpoints. The visual aid helps in understanding the range and position of \(x\) concerning the given interval.
The real number line is useful for illustrating the concept of inequalities as well. Here, it visually supports the requirement that every \(x\) within this interval agrees with \(a < x < b\).

Distance Calculation

Distance calculation is crucial in determining how far a point is from another, which in this context is used to find how close \(x\) can be to \(c = 3\), while still remaining in the interval \((2.7591, 3.2391)\).
We calculate the distance from the center point \(c\) to the bounds \(a\) and \(b\). In this exercise, we'll do two calculations: one for \(c-a\), which equals \(3 - 2.7591 = 0.2409\), and the other for \(b-c\), resulting in \(3.2391 - 3 = 0.2391\).
These simple distance calculations show how far \(c\) is from each endpoint of the interval. Here, the smaller value, \(0.2391\), is chosen as \(\delta\) to ensure that \(x\) does not reach any interval edge. By setting \(0 < |x - 3| < 0.2391\), we ensure \(x\) remains securely within the defined interval boundaries.

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Key Concepts

Interval Notation

Interval notation is a way of representing a range of numbers on the real number line. It provides a clear picture of which numbers are included or excluded from the range. For example, the interval \((a, b)\) means that all numbers \(x\) such that \(a < x < b\) are included. Here, the parentheses indicate that the endpoints \(a\) and \(b\) themselves are not included in the interval. This is known as an "open interval."
An open interval differs from a closed interval, which would include the endpoints, denoted as \([a, b]\). There are also half-open or half-closed intervals, like \((a, b]\) or \([a, b)\), including only one of the endpoints.
In this problem, you have an open interval \((2.7591, 3.2391)\) where the point \(c = 3\) is inside. We need to ensure that any value \(x\) within a specified distance of \(c\) remains in this interval.

Real Number Line

The real number line is a visual representation of all real numbers arranged in a linear format. Each point on the line corresponds to a real number. In this context, it allows us to place intervals and specific points. For instance, the interval \((2.7591, 3.2391)\) defines all points \(x\) on the number line that are strictly between these two endpoints.
By marking this interval on the real number line along with the point \(c = 3\), you can easily see that \(c\) lies between the endpoints. The visual aid helps in understanding the range and position of \(x\) concerning the given interval.
The real number line is useful for illustrating the concept of inequalities as well. Here, it visually supports the requirement that every \(x\) within this interval agrees with \(a < x < b\).

Distance Calculation

Distance calculation is crucial in determining how far a point is from another, which in this context is used to find how close \(x\) can be to \(c = 3\), while still remaining in the interval \((2.7591, 3.2391)\).
We calculate the distance from the center point \(c\) to the bounds \(a\) and \(b\). In this exercise, we'll do two calculations: one for \(c-a\), which equals \(3 - 2.7591 = 0.2409\), and the other for \(b-c\), resulting in \(3.2391 - 3 = 0.2391\).
These simple distance calculations show how far \(c\) is from each endpoint of the interval. Here, the smaller value, \(0.2391\), is chosen as \(\delta\) to ensure that \(x\) does not reach any interval edge. By setting \(0 < |x - 3| < 0.2391\), we ensure \(x\) remains securely within the defined interval boundaries.