Chapter 1.1, Problem 64E

To determine

To graph: the given set.

Explanation of Solution

Given information:

An interval is given as

  $\left(-\infty ,6\right]\cap \left(2,10\right)$

Concept used:

An interval $\left(a,b\right)$ in terms of an inequality can be expressed as $a<x<b$ .

An interval $\left[a,b\right]$ in terms of an inequality can be expressed as $a\le x\le b$ .

An interval $\left[a,b\right)$ in terms of an inequality can be expressed as $a\le x<b$ .

An interval $\left(a,b\right]$ in terms of an inequality can be expressed as $a<x\le b$ .

An interval $\left(-\infty ,b\right]$ in terms of an inequality can be expressed as $x\le b$ .

An interval $\left(-\infty ,b\right)$ in terms of an inequality can be expressed as $x<b$ .

An interval $\left[a,\infty \right)$ in terms of an inequality can be expressed as $x\ge a$ .

An interval $\left(a,\infty \right)$ in terms of an inequality can be expressed as $x>a$ .

For any two or more sets, intersection of the sets gives the elements which are the only common elements of the sets. Intersection is represented by the symbol $\cap$ .

Calculation:

Consider the given interval.

  $\left(-\infty ,6\right]\cap \left(2,10\right)$

Let us consider a variable x o express the inequality.

So, the interval can be expressed as an inequality as $x\le 6\cap 2<x<10$ .

Combining the interval will be $2<x\le 6=\left(2,6\right]$

Graph:

Now, graph the interval on number line as shown:

  

Here, $6$ is included to the interval so, closed ball is placed for $6$ on the number line.

Here, $2$ is excluded from the interval so, open ball is placed for $2$ on the number line.