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CALCULUS AB WORKSHEET THREE
WARM-UP AND REVIEW
“Inequalities”
[1.]
Explain, define or illustrate the following terms or phrases: intersection, algebraic
notation, set notation, interval notation, union, inequality, simple inequality,
compound inequality, sign plot, rational inequality, comparison to zero, and
quadratic inequality.
[2.]
Explain the process of solving
a. Simple Inequality
b. Compound Inequality
c. Complex Inequality
SOLVE EACH INEQUALITY BY USING THE CIS METHOD - SHOW ALL OF YOUR
WORK – REPRESENT SOLUTIONS IN INTERVAL NOTATION – VERIFY SOLUTIONS
USING A GRAPHING CALCULATOR
[3.]
4x 11  x 1  3x  7
[4.]
9x  8x  2  3x
[5.]
3x
5
x2
[6.]
x 1
 3
x
[7.]
x2
2
x 1
[8.]
4
2
3 7
x
x
[9.]
3x  8 2

2x  3 3
[10.]
x 2  2x 15  0
[11.]
4  x 2  3x
[12.]
x 10  3x 2
[13.]
x3
0
x2  4
[14.]
4
1

2x  1 x  1
[15.]
[17.]
x2  9
2
x  3x  2
0
[16.]
3x 1
2
x  x6
1
A firm can sell at a price of $100.00 per unit all the “Disraeli Gears” it produces.
If x units are produced each day, the number of dollars in the total cost of each
day’s production is given by the expression:
x 2  20x  700
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CALCULUS AB WORKSHEET THREE
WARM UP AND REVIEW – INEQUALITIES
Determine how many units should be produced each day so that the firm is
guaranteed a profit.
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CALCULUS AB WORKSHEET THREE
WARM UP AND REVIEW – INEQUALITIES
[18.]
A rectangular plot of ground is to be enclosed by a fence and then divided down
the middle by another fence. The fence down the middle costs $3.00 per running
foot and the perimeter fence costs $6.00 per running foot. The area of the plot is to
be exactly 1800 square feet and the cost of the fence is not to exceed $2310.00.
Determine the restrictions on the dimensions of the plot.
SOLUTIONS
[1.]
Intersection refers to what two solutions (sets)
have in common.
Algebraic, Set and Interval Notation are
different ways of writing the solution to an
equation or inequality.
expression compared to zero. Then determine
where the expression is positive or negative
by using a sign plot. The sign plot can then be
used to determine the solution.
[3.]
Union is the combination of two solutions
(sets).
x (2, )
[4.]
9x  8x  2 U 8x  2  3x
x   172
U
x   25
An inequality is any comparison statement
using the symbols <. ≤, > or ≥.
x 
Isolating the x term on one side by addition
and/or multiplication can solve a simple
inequality.
The solution will be the intersection of the
two individual solutions. Since the individual
solutions have no elements in common, the
final solution is the null set.
A compound inequality is any quality
containing two comparison symbols. The
inequality must be writing as two separate
inequalities and solved individually.
[5.]
x  , 5
[6.]
x1
3  0
x
x 1  3x
0
x
4x  1
0
x
A sign plot is a way of illustrating where a
variable expression is positive or negative by
first determining where it is equal to zero.
Complex inequalities (rational or quadratic)
are best solved by comparison to zero and
then using a sign plot to obtain a solution.
[2.]
A simple inequality is solved by algebraically
isolating the variable on the left or right side
of the inequality symbol by addition and/or
multiplication. Once the variable is isolated
the result is the algebraic solution to the
inequality.
A compound inequality of the form
A  B  C can be solved by splitting the
inequality into two distinct inequalities and
solving each separately by then taking the
intersection of the two separate solutions.
2,  
Create a sign plot to determine the intervals
the rational expression is zero and positive.

 x  ,  14
[7.]

(0, )
x  1 , 4 
[8.]
A complex inequality is best solved by
comparison to zero. Rewrite the given
inequality so that the result is a variable
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CALCULUS WORKSHEET THREE
WARM UP AND REVIEW – INEQUALITIES
4
2
3 7
x
x
4
2
 3  7  0
x
x
2
40
x
2  4x
0
x
Create a sign plot to determine the intervals
the rational expression is zero and positive.

 x  ,  12
[9.]

Create a sign plot to determine the intervals
the rational expression is negative.

 x  ,  5
2
[15.] x 3,1
[16.]
(0,)
[10.]
x 2  2x  15  0
(x  5)(x  3)  0
Create a sign plot to determine the intervals
where the product of the factors is zero and
positive.
[11.] x ( ,  4]
[3,)
x (, 2)
[12.]
x 10  3x
2
3x  x 10  0
(3x  5)(x  2)  0
2
Create a sign plot to determine the intervals
where the product of the factors is positive.
[13.] x 2,2 
Create a sign plot to determine the intervals
the rational expression is zero and negative.
[1, )
 x  , 2 
2, 3
3x  1
1
x2  x  6
3x  1
1  0
x2  x  6
3x  1 1x 2  x  6 
0
x2  x  6
x 2  4x  5
0
x2  x  6
1 x 2  4x  5 
0
x2  x  6
1(x  5)(x  1)
0
(x  3)(x  2)
x  23 ,6 
x (, 5]
 –1, 12 
[1, 3) [5,)
[17.] Producing between 10 and 70 units daily
will result in a profit.
[18.] Let x be the horizontal length of the fence
and y vertical length. Therefore the total
amount (T) of fencing can be defined by the
equation
T  x  x  y y y
53 , 
The total cost (C) of the fencing can be
defined by the equation
3, 
C  6x  6x  6y  6y  3y
 12x 15y
[14.]
4
1

2x  1 x  1
4
1

0
2x  1 x  1
4(x  1)  1(2x  1)
0
(2x  1)(x  1)
Since we do not want the cost of the fence to
exceed $2310 we can define the cost
inequality in the following manner
C  2310
12x  15y  2310
2x  5
0
(2x  1)(x  1)
Note that we have two variables. So we must
eliminate either x or y. This is a common
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CALCULUS WORKSHEET THREE
WARM UP AND REVIEW – INEQUALITIES
type of situation. You must be given an
initial condition. In this case the initial
condition is the restriction on the area. The
area (A) can be defined as follows:
A  xy  1800
Solve the area equation for x or y and
substitute into the cost inequality.
x
1800
y
1800 
12 
  15y  2310
 y 
Since you know that y is always a positive
value you can multiply each side by y without
affecting the final outcome. In other words
multiplying an inequality by a positive value
does not effect the direction of the inequality.
12 
1800
 15y  2310
y
15y 2  2310y  21600  0
15  y  144  y  10   0
Create a sign plot to find where the variable
expression is negative. Since y is a positive
value you need not consider in values of y
that are negative. Sign plot should reveal that
if y is a value in the interval [10,144] you
have satisfied the requirements of the
problem. To obtain the x restrictions,
substitute the endpoints of 10 and 144 in for y
into the initial condition and obtain the
interval [12.5,180]
Therefore vertical length that is between 10
feet and 144 feet (inclusive) paired with the
appropriate horizontal length of 12.5 feet to
180 feet (inclusive) will result in a
rectangular plot of 1800 square feet and not
exceed the price of fencing of $2310.
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