Problem 28 - Calculus Single Variable

Question

Write the interval in the form \(\\{x:|x-c|< r\\}\) or \(\\{x:|x-c| \leq r\\}\). \((\pi-\sqrt{2}, \pi)\)

Step-by-Step Solution

Answer

The interval is \( \{x: |x - \frac{2\pi-\sqrt{2}}{2}| < \frac{\sqrt{2}}{2} \} \).

Step 1: Identify Interval Boundaries

We have the open interval \( (\pi-\sqrt{2}, \pi) \), which means that the interval starts just above \( \pi-\sqrt{2} \) and ends at \( \pi \). This denotes that the interval does not include the end points.

Step 2: Calculate the Midpoint

The midpoint of an interval \( (a, b) \) is given by the formula \( c = \frac{a+b}{2} \). For this interval, \( a = \pi-\sqrt{2} \) and \( b = \pi \). Calculating gives: \[ c = \frac{(\pi-\sqrt{2})+\pi}{2} = \frac{2\pi-\sqrt{2}}{2} \].

Step 3: Calculate the Radius

The radius \( r \) of the interval is half of the difference between the endpoints. We calculate it as follows: \[ r = \frac{b-a}{2} = \frac{\pi - (\pi-\sqrt{2})}{2} = \frac{\sqrt{2}}{2} \].

Step 4: Write the Interval in Absolute Value Form

Using the results from previous steps, the interval can now be expressed as: \[ \{x: |x - \frac{2\pi-\sqrt{2}}{2}| < \frac{\sqrt{2}}{2} \} \]. This notation represents all \( x \) values within \( \frac{\sqrt{2}}{2} \) distance from the center \( \frac{2\pi-\sqrt{2}}{2} \).

Key Concepts

Midpoint Calculation

Calculating the midpoint of an interval is like finding the center point that lies exactly in between two numbers that define your interval's boundaries.
For instance, consider the open interval \((\pi-\sqrt{2}, \pi)\). We need to find the point that sits exactly in the middle of \(\pi-\sqrt{2}\) and \(\pi\). This is crucial as it acts like the balancing point.
The formula used for finding the midpoint, \(c\), is:

• \(c = \frac{a+b}{2}\)

where \(a\) and \(b\) are the interval boundaries.
Upon plugging in our values:

• \(a = \pi-\sqrt{2}\)
• \(b = \pi\)

The calculation becomes:\[ c = \frac{(\pi-\sqrt{2})+\pi}{2} = \frac{2\pi-\sqrt{2}}{2} \]
This midpoint, \(\frac{2\pi-\sqrt{2}}{2}\), will serve as the center point from which we will measure the radius outward in both directions.

Radius of an Interval

The radius of an interval in mathematics represents how far you can stretch out from the midpoint, reaching toward either boundary of the interval.
For our interval, the radius can be understood as simply half the length of the whole interval.
Given the interval is \((\pi-\sqrt{2}, \pi)\), the radius \(r\) is calculated using:

• \(r = \frac{b-a}{2}\)

where \(a\) is \(\pi-\sqrt{2}\) and \(b\) is \(\pi\).
After substituting:\[ r = \frac{\pi - (\pi-\sqrt{2})}{2} = \frac{\sqrt{2}}{2} \]
This result tells us that, starting from the midpoint \(\frac{2\pi-\sqrt{2}}{2}\), you can move \(\frac{\sqrt{2}}{2}\) units to either side to stay within the interval.

Absolute Value Representation

The absolute value notation is a powerful way to express the distance from a number on a number line, always as a positive quantity.
In the context of intervals, it helps us describe the set of all points \(x\) that are within a certain distance from a central point \(c\).
For our specific interval, the absolute value representation is:

• \(\{x: |x - \frac{2\pi-\sqrt{2}}{2}| < \frac{\sqrt{2}}{2}\}\)

This represents all points \(x\) such that the distance from \(\frac{2\pi-\sqrt{2}}{2}\) is less than \(\frac{\sqrt{2}}{2}\).
The absolute value bars, \(|...|\), ensure that we're considering only the magnitude of the difference, disregarding direction.
Thus, absolute value notation efficiently conveys both the center \(c\) and the allowable deviation, \(r\), from it within the interval.