Download Section 9.1 Interval Notation and Business Applications Using

Math 27 Section 9.1 - Page 1
Section 9.1
Interval Notation and Business Applications Using Linear Inequalities
I.
Graphs
A.
The Real Number Line is the line where:
1.
O is the origin.
2.
The negative direction is to the left.
3.
The positive direction is to the right.
4.
Each point corresponds to a different real number called its coordinate.
B.
C.
Graphing Inequalities
1.
If the inequality is < or >, the boundary point will be a ) or (. This means that
the boundary point is not included in the solution set.
2.
If the inequality is < or >, the boundary point will be a ] or [. This means that
the boundary point is included in the solution set.
Examples - Graph each inequality.
1.
x < -3
Since we do not have an "equal" as part of the inequality, the boundary
"punctuation" will be a ).
This inequality is read "x is less than -3", which is all the real numbers to the
left of -3 on the number line. The graph will look like:
Answer:
)
-3
2.
x>7
Since we do have an "equal" as part of this inequality, the boundary
"punctuation" will be [.
This inequality is read "x is greater than or equal to 7", which is all real
numbers to the right of 7 on the number line. The graph will look like:
Answer;
[
7
3.
Now you try one:
Answer:
x<5
]
5
4.
-8 < x < 17
This is a compound inequality since we have more than one inequality in the
problem. So on one side, our boundary will be (, on the other side it will be ].
This inequality is read "x is between -8 and 17, not including -8, but including
17". The graph will look like:
Answer:
(
]
−8
17
II.
Intervals
A.
An interval of the real number line is a set of points on the real number line.
B.
If the boundary point is part of the interval, we use a square bracket.
C.
If the boundary point is not part of the interval, we use a parenthesis.
D.
Writing Intervals from Graphs
1.
Is the boundary point part of the graph?
2.
List the boundary points, with a comma in between them, along with the
appropriate "punctuation".
© Copyright 2008 by John Fetcho. All rights reserved.
Math 27 Section 9.1 - Page 2
E.
Examples - Write the intervals for the above graphs.
1.
x < -3
There is no left hand endpoint, so we will use −∞ for the left, with ( since we
can never reach infinity. Then since -3 is not included on the right, we will use
) on the right. So the interval will be:
Answer:
(− ∞,−3)
2.
x>7
7 is the left-hand endpoint and it is included, so we will use [ on the left.
There is no right-hand endpoint, so we will use ∞ and ) since we can never
reach infinity. So the interval will be:
Answer:
[7, ∞ )
3.
Now you try one:
Answer:
(−∞,5]
x<5
4.
-8 < x < 17
Since we have a left-hand boundary (-8) that is not included, we will use ( on
that end. The right-hand boundary (17) is included, so we will use ] on that
end. The interval will be:
Answer:
(−
−8, 17]
5.
III.
Properties of Inequalities - (Let a, b, and c be real numbers)
A.
Addition (Subtraction) Property
If a < b, then a + c < b + c
B.
IV.
Multiplication (Division) Property
1.
If a < b and c > 0, then ac < bc.
2.
If a < b and c < 0, then ac > bc.
Writing Solutions
A.
Graphically
1.
If you do not have equality, use parentheses.
2.
If you do have equality, use a square bracket.
B.
V.
Note that in all intervals, we always write them from smallest number to
largest.
Interval Notation
1.
If you do not have equality, use parentheses.
2.
If you do have equality, use a square bracket.
Solving Single Inequalities
A.
Solve in the same way you would solve an equality.
B.
Be careful when multiplying/dividing by a negative and change the inequality sign.
C.
When you have solved the problem algebraically, graph the solution set and then
write an interval.
© Copyright 2008 by John Fetcho. All rights reserved.
Math 27 Section 9.1 - Page 3
D.
Examples - Solve, graph, and write an interval.
13
 1
1.
 y + 4  ≤ (y − 6 )
25
 3
Our first step is to distribute on both sides - so dig out your calculators and
warm up the fraction key!
3
1
y +2 ≤ y −2
10
3
Now let’s multiply everything by the LCD to get rid of the fractions. In this
case, the LCD is 30:
9y + 60 < 10y - 60
When solving inequalities, it is best to get the variable terms on the left hand
side. So to do that, we’ll subtract 10y from both sides:
-y + 60 < -60
Now subtract 60 from both sides:
-y < -120
Now divide by -1. Remember to change the inequality sign!
Answer:
y > 120
Since we have equality included, we will use a square bracket when we
graph:
Answer:
[
120
Finally, we need to write the interval:
[120, ∞ )
Answer:
Note that we need all three to have a complete answer.
y 2
+ ≤4
3 5
We begin this one by finding the LCD and multiplying both sides by it to
clear the fractions:
5y + 6 < 60
Now we subtract 6 from both sides to get:
5y < 54
Divide both sides by 5. Since we are dividing by a positive number, we
do not need to change the inequality sign:
54
Answer:
y<
5
Since we have equality, we need to use a square bracket when we
graph:
Answer:
]
54
5
Now we need to write the interval:
54 

Answer:
 − ∞,
5 

2.
3.
Now you try one:
Answer:
b>9
(
9
(9, ∞ )
4b - 6 > 2b + 12
© Copyright 2008 by John Fetcho. All rights reserved.
Math 27 Section 9.1 - Page 4
4.
If the cost function for producing x units of a product is C(x) = 15,000 + 12x
and the revenue function for selling x units is R(x) = 32x, find:
a.
The profit function.
b.
How many units must be produced and sold to have a profit gain?
(Page 614, #34)
a.
To come up with a profit function, we need to realize the basic
relationship between the three functions, namely:
Profit = Revenue − Cost
So in this case, we will have:
P(x) = R(x) - C(x) = 32x - (15,000 + 12x) = 32x - 15,000 - 12x
Answer: P(x) = 20x − 15,000
b.
To determine how many units to produce and sell to have a profit gain
(also called the break-even point), we set the profit function greater
than or equal to zero and solve.
20x - 15,000 > 0
20x > 15,000
x > 750
Answer: The manufacturer must produce and sell 750 units to have a
profit gain.
© Copyright 2008 by John Fetcho. All rights reserved.