Chapter 1.1, Problem 57E

a)

To determine

To express: the given graph in interval notation.

Answer to Problem 57E

In terms of interval the setcan be expressed as $\left[-3,5\right]$ .

Explanation of Solution

Given information:

Agraphis given as

  

Concept used:

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

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

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

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

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

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

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

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

If the terminal number or point is included in the interval it can be represented by closed ball on the number line otherwise it will be represented by a open ball.

Calculation:

Consider the given graph.

  

Here, for both $-3$ and 5closed ball are placed on the number line, so both will be included to the interval.

So, the set can be expressed as aninterval as $\left[-3,5\right]$ .

b)

To determine

To express: the given graph in interval notation.

Answer to Problem 57E

In terms of interval the set can be expressed as $\left(-3,5\right]$ .

Explanation of Solution

Given information:

A graph is given as

  

Concept used:

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

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

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

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

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

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

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

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

If the terminal number or point is included in the interval it can be represented by closed ball on the number line otherwise it will be represented by a open ball.

Calculation:

Consider the given graph.

  

Here, for $-3$ open ball is placed on the number line, for 5 closed ball is placed, so5 will be included to the interval and $-3$ is not included in the set.

So, the set can be expressed as aninterval as $\left(-3,5\right]$ .