interval notation and absolute value

CALCULUS 1 – Algebra review

Intervals and Interval Notation



Intervals are sets of real numbers. The notation uses square and

round brackets to show these sets of numbers.





Round bracket – go up to but do not include this number in the set





( 3 , 7 ) - this interval would include all numbers between 3

and 7, but NOT 3 or 7.






Square bracket – include this number in the set








Square bracket – include this number in the set




[ 3 , 7 ] - this interval would include all numbers from 3 to 7..



When working with equations containing an inequality, the symbols

for the inequality determine how you graph and represent the

solution as an interval.

Round bracket - less than ( < ) , greater than ( > )




- open circle on a graph







Square bracket – less than or equal to ( ≤ ), greater than or equal

to ( ≥ )








Square bracket - less than or equal to ( ≤ ), greater than or equal

to ( ≥ )





- closed circle on a graph





to ( ≥ )



EXAMPLE : Solve and graph and show your answer as an interval 753 x





to ( ≥ )




4

123

55

753

x

x

x





to ( ≥ )




4

123

55

753

x

x

x

4 graph





to ( ≥ )




4

123

55

753

x

x

x

4

) ,4 (

graph

interval





to ( ≥ )



EXAMPLE # 2 : Solve and graph and show your answer

as an interval

5123 x





to ( ≥ )




as an interval

5123 x

31

622

11 1

5123

x

x

x





to ( ≥ )




as an interval

5123 x

31

622

11 1

5123

x

x

x

This results in two graphs…

x < 3

x ≥ -1

3 - 1





to ( ≥ )




as an interval

5123 x

31

622

11 1

5123

x

x

x

The solution set is

where the two graphs

overlap ( share )

3 - 1





to ( ≥ )




as an interval

5123 x

31

622

11 1

5123

x

x

x

The solution set is

where the two graphs

overlap ( share )

3 - 1

[ -1 , 3 ) interval





to ( ≥ )




as an interval

01272 xx





to ( ≥ )




as an interval

3,4

034

01272

x

xx

xx

01272 xx





to ( ≥ )




as an interval

3,4

034

01272

x

xx

xx

These are our critical points

01272 xx





to ( ≥ )




as an interval

3,4

034

01272

x

xx

xx


- 3 - 4

01272 xx

Graph the critical points and then

use a test point to find “true/false”





to ( ≥ )




as an interval

3,4

034

01272

x

xx

xx


- 3 - 4

01272 xx



TEST x = 0

TRUE 012

01200

0120702

0

TRUE FALSE TRUE





to ( ≥ )




as an interval

3,4

034

01272

x

xx

xx


- 3 - 4

01272 xx



TEST x = 0

TRUE 012

01200

0120702

0

TRUE FALSE TRUE

,34,interval


Absolute Value Equations

Remember, absolute value equations have two possible answers;

positive and negative. So when solving, drop the absolute value

sign, and set the equation equal to the original answer, and also it’s

negative counterpart.







EXAMPLE # 1 : Solve 752 x







EXAMPLE # 1 : Solve 752 x

16

2

2

2

2

2

12

2212

55 5

7527

752

x

x

x

x

x







EXAMPLE # 2 : Solve 6222

xx








xx

Remember u substitution

from pre-calc ?

023

06

6

2Let

2

2

uu

uu

uu

xu








xx


from pre-calc ?

023

06

6

2Let

2

2

uu

uu

uu

xu

22 and 32

022 and 032

xx

xx








xx


from pre-calc ?

023

06

6

2Let

2

2

uu

uu

uu

xu

22 and 32

022 and 032

xx

xx

Can’t have an absolute value equal

to a negative answer








xx


from pre-calc ?

023

06

6

2Let

2

2

uu

uu

uu

xu

51

323

32

032

x

x

x

x

Now solve the

absolute value

equation …







EXAMPLE # 3 : Solve , and show the solution set as an interval. 532 x








14

2

2

2

2

2

8

228

33 3

5325

x

x

x

x








14

2

2

2

2

2

8

228

33 3

5325

x

x

x

x I like to graph the solution to determine the

interval…

4 1 xx

-1 4








14

2

2

2

2

2

8

228

33 3

5325

x

x

x

x I like to graph the solution to determine the

interval…

4 1 xx

-1 4

)4,1(interval

interval notation and absolute value

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