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Write the interval in the form \(\\{x:|x-c|< r\\}\) or \(\\{x:|x-c| \leq r\\}\). \([3,4 \sqrt{2}]\)

Short Answer

Expert verified
\(\{x: |x - \frac{3+4\sqrt{2}}{2}| \leq \frac{4\sqrt{2} - 3}{2}\}\)

Step by step solution

01

Identify the interval endpoints

The given interval is \([3, 4\sqrt{2}]\). We recognize that the lower endpoint is 3 and the upper endpoint is \(4\sqrt{2}\).
02

Calculate the center (c) of the interval

To find the center \(c\) of the interval, we need to calculate the midpoint. The midpoint is given by the formula: \(c = \frac{a+b}{2}\), where \(a\) and \(b\) are the endpoints of the interval. Here, \(a = 3\) and \(b = 4\sqrt{2}\). Thus, the center \(c = \frac{3 + 4\sqrt{2}}{2}\).
03

Calculate the radius (r) of the interval

The radius \(r\) is the distance from the center of the interval to either endpoint. It is given by the formula: \(r = \frac{b-a}{2}\). With \(a = 3\) and \(b = 4\sqrt{2}\), we have \(r = \frac{4\sqrt{2} - 3}{2}\).
04

Write the interval in set notation

Substitute the calculated center \(c\) and radius \(r\) into the set notation \(\{x:|x-c|\leq r\}\). Thus, the interval \([3, 4\sqrt{2}]\) can be expressed as \(\{x: \left| x - \frac{3+4\sqrt{2}}{2} \right| \leq \frac{4\sqrt{2} - 3}{2} \}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Set Notation
Set notation is a way to describe a set of numbers, particularly useful when dealing with intervals on a number line. It provides a concise way to express the range of values a variable can take, often using logic and symbols. In set notation, the expression \( \{ x : |x - c| \leq r \} \) represents all values of \( x \) that are within a distance of \( r \) from a central point \( c \). This is akin to saying that \( x \) is in the interval between \( c - r \) and \( c + r \), inclusive of the endpoints.

In our given example, the set notation for the interval \([3, 4\sqrt{2}]\) uses the midpoint (or center) \( c \) and the radius \( r \). Set notation transforms an interval into inequality form, which is particularly handy in advanced mathematical contexts like calculus and analysis. This way, it highlights not just the endpoints, but also the relationship of the entire set of numbers within that interval to a specific central value.
  • Central point \( c \)
  • Distance \( r \) to endpoints
Midpoint Calculation
The concept of the midpoint is crucial in understanding intervals and is at the heart of converting intervals to set notation. The midpoint of an interval is simply the average of the endpoints, effectively dividing the interval into two equal parts. It serves as the central point \( c \) in our set notation expression.

To calculate the midpoint \( c \), use the formula:
  • \( c = \frac{a+b}{2} \)
Here, \( a \) and \( b \) are the lower and upper endpoints of the interval, respectively. For the interval \([3, 4\sqrt{2}]\), the calculation becomes:
  • \( c = \frac{3 + 4\sqrt{2}}{2} \)
Finding the midpoint helps in setting up the inequality \( |x-c| \leq r \), which portrays how far any number \( x \) within the interval can stray from this central point.
Radius Calculation
Calculating the radius \( r \) is integral to fully describing an interval in set notation. The radius, analogous to the concept of the radius of a circle, represents the maximum allowable distance from the midpoint to the endpoints of the interval. It tells us how much space there is on each side of the central point \( c \) within the interval.

The formula to determine the radius is:
  • \( r = \frac{b-a}{2} \)
Where \( b \) is the upper endpoint and \( a \) is the lower endpoint of the interval. Applying this to our example interval \([3, 4\sqrt{2}]\), we calculate:
  • \( r = \frac{4\sqrt{2} - 3}{2} \)
This setup ensures that all points in \( \{ x : |x - c| \leq r \} \) will fall within the desired interval limits, providing a clear framework for understanding how the interval behaves relative to its central point. This fundamental concept in mathematics ensures a comprehensive grasp of interval behavior, especially in more sophisticated mathematical applications.

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Most popular questions from this chapter

Let \(P_{0}=\left(x_{0}, y_{0}\right)=(4,16),\left(x_{1}, y_{1}\right)=(1,2),\) and \(\left(x_{2}, y_{2}\right)=\) \((2,6) .\) Obtain an expression for the sum of the squares of the errors \(d_{1}^{2}+d_{2}^{2}\) associated with a line through \(P_{0}\) that has slope \(m .\) With \(m\) measured along the horizontal axis and \(d_{1}^{2}+d_{2}^{2}\) measured along the vertical axis, graph \(d_{1}^{2}+d_{2}^{2}\) in the viewing window \([4.5,5] \times[0,1.2] .\) Use your plot to find the slope of the regression line \(\mathcal{L}\) through \(P_{0}\). Now plot the sum of the absolute errors \(\mathrm{d}_{1}+\mathrm{d}_{2}\) in the viewing window \([3.5,6] \times[0,7] .\) Use your plot to find the value of \(m\) that minimizes \(d_{1}+d_{2}\). Finally, on the same coordinate axes, plot the three data points, the line through \(P_{0}\) that minimizes \(d_{1}+d_{2},\) and \(\mathcal{L}\).

Let \(\mathcal{T}\) be the collection of functions of the form $$ x \mapsto \frac{a x+b}{c x+d} $$ with \(a d-b c=1 .\) (These functions are called linear fractional transformations.) Show that the composition of two functions in \(\mathcal{T}\) is also in \(\mathcal{T}\). By setting \(a=d=1\) and \(b=c=0,\) we see that the identity function \(I(x)=x\) is in \(\mathcal{T}\). Show that every function in \(\mathcal{T}\) has an inverse that is also in \(\mathcal{T}\) and that is obtained by swapping \(a\) and \(d,\) and negating \(b\) and \(c\)

Find a function \(g\) such that \(g \circ f=h\) \(f(x)=x^{2}-9, h(x)=2 x^{2}\)

Find a function \(g\) such that \(g \circ f=h\) \(f(x)=(x-1) / x^{2}, h(x)=x^{2} /(x-1)\)

Let \(P=(s, t)\) be a point in the \(x y\) -plane. Let \(P^{\prime}=(t, s)\) Calculate the slope of the line \(\ell^{\prime}\) that passes through \(P\) and \(P^{\prime}\) Deduce that \(\ell^{\prime}\) is perpendicular to the line \(\ell\) whose equation is \(y=x .\) Let \(Q\) be the point of intersection of \(\ell\) and \(\ell^{\prime}\) Show that \(P\) and \(P^{\prime}\) are equidistant from \(Q\). (As a result, \(P\) and \(P^{\prime}\) are reflections of each other through the line \(y=x\)

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