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CALCULUS 1 – Algebra review
Intervals and Interval Notation
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.
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
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.
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
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
[ 3 , 7 ] - this interval would include all numbers from 3 to 7..
CALCULUS 1 – Algebra review
Intervals and Interval Notation
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 ( > )
CALCULUS 1 – Algebra review
Intervals and Interval Notation
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
CALCULUS 1 – Algebra review
Intervals and Interval Notation
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 ( ≥ )
CALCULUS 1 – Algebra review
Intervals and Interval Notation
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 ( ≥ )
- closed circle on a graph
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE : Solve and graph
3x  5  7
and show your answer as an interval
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE : Solve and graph
3x  5  7
 5  5
3x  12
x4
3x  5  7
and show your answer as an interval
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE : Solve and graph
3x  5  7
 5  5
3x  12
x4
3x  5  7
and show your answer as an interval
4
graph
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE : Solve and graph
3x  5  7
 5  5
3x  12
x4
3x  5  7
and show your answer as an interval
4
graph
( 4,  )
interval
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
as an interval
 3  2 x 1  5
and show your answer
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
 3  2 x 1  5
as an interval
 3  2x 1  5
 1   1  1
 2  2x  6
1  x  3
and show your answer
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
 3  2 x 1  5
and show your answer
as an interval
 3  2x 1  5
 1   1  1
 2  2x  6
1  x  3
-1
3
This results in two graphs…
x<3
x ≥ -1
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
 3  2 x 1  5
and show your answer
as an interval
 3  2x 1  5
 1   1  1
 2  2x  6
1  x  3
-1
3
The solution set is
where the two graphs
overlap ( share )
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
 3  2 x 1  5
and show your answer
as an interval
 3  2x 1  5
 1   1  1
 2  2x  6
1  x  3
-1
3
The solution set is
where the two graphs
overlap ( share )
[ -1 , 3 )
interval
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 3 : Solve and graph
as an interval
x 2  7 x  12  0
and show your answer
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
x 2  7 x  12  0
as an interval
x 2  7 x  12  0
x  4x  3  0
x  4,3
and show your answer
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
x 2  7 x  12  0
as an interval
x 2  7 x  12  0
x  4x  3  0
x  4,3
These are our critical points
and show your answer
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
x 2  7 x  12  0
and show your answer
as an interval
x 2  7 x  12  0
x  4x  3  0
x  4,3
These are our critical points
Graph the critical points and then
use a test point to find “true/false”
-4
-3
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
x 2  7 x  12  0
and show your answer
as an interval
x  7 x  12  0
2
x  4x  3  0
x  4,3
These are our critical points
Graph the critical points and then
use a test point to find “true/false”
TRUE
FALSE
-4
TRUE
-3
TEST x = 0
02  70  12  0
0  0  12  0
12  0
TRUE
0
CALCULUS 1 – Algebra review
Intervals and Interval Notation
Round bracket - less than ( < ) , greater than ( > )
- open circle on a graph
Square bracket - less than or equal to ( ≤ ), greater than or equal
to ( ≥ )
- closed circle on a graph
EXAMPLE # 2 : Solve and graph
x 2  7 x  12  0
and show your answer
as an interval
x  7 x  12  0
2
x  4x  3  0
x  4,3
These are our critical points
Graph the critical points and then
use a test point to find “true/false”
TRUE
FALSE
-4
TRUE
-3
TEST x = 0
02  70  12  0
0  0  12  0
12  0
TRUE
0
 ,4   3,
interval
CALCULUS 1 – Algebra review
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.
CALCULUS 1 – Algebra review
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
2x  5  7
CALCULUS 1 – Algebra review
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
2x  5  7
2x  5  7
 7  2x  5  7
5 
 5  5
 12  2 x  2
 12 2 x 2


2
2 2
6  x 1
CALCULUS 1 – Algebra review
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 # 2 : Solve
x2  x2 6
2
CALCULUS 1 – Algebra review
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 # 2 : Solve
x2  x2 6
2
Let u  x  2
u2  u  6
u2  u  6  0
u  3u  2  0
Remember u substitution
from pre-calc ?
CALCULUS 1 – Algebra review
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 # 2 : Solve
x2  x2 6
2
Let u  x  2
u2  u  6
u2  u  6  0
u  3u  2  0
Remember u substitution
from pre-calc ?
x 2 3  0
and
x2 2 0
x2 3
and
x  2  2
CALCULUS 1 – Algebra review
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 # 2 : Solve
x2  x2 6
2
Let u  x  2
u2  u  6
u2  u  6  0
u  3u  2  0
Remember u substitution
from pre-calc ?
x 2 3  0
and
x2 2 0
x2 3
and
x  2  2
Can’t have an absolute value equal
to a negative answer
CALCULUS 1 – Algebra review
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 # 3 : Solve
 2x  3  5
, and show the solution set as an interval.
CALCULUS 1 – Algebra review
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 # 3 : Solve
 2x  3  5
 5  2 x  3  5
3 
 3  3
 8  2 x  2
 8  2x
2


2 2 2
4  x  1
, and show the solution set as an interval.
CALCULUS 1 – Algebra review
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 # 3 : Solve
 2x  3  5
 5  2 x  3  5
3 
 3  3
 8  2 x  2
 8  2x
2


2 2 2
4  x  1
, and show the solution set as an interval.
I like to graph the solution to determine the
interval…
x  1
x4
-1
4
CALCULUS 1 – Algebra review
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 # 3 : Solve
 2x  3  5
 5  2 x  3  5
3 
 3  3
 8  2 x  2
 8  2x
2


2 2 2
4  x  1
, and show the solution set as an interval.
I like to graph the solution to determine the
interval…
x  1
x4
4
-1
(1,4)
interval