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Section 0.2 (5/31/07)
Set notation and solving inequalities
Overview: Inequalities are almost as important as equations in calculus. Many functions’ domains are
intervals, which are defined by inequalities. Inequalities are needed to study where functions have positive
and negative values. They are also used in the definitions of limits and with derivatives to study where
functions are increasing and decreasing and where their graphs are concave up and concave down. In this
section we describe notation and terminoogy for intervals and other sets and discuss the rules for solving
inequalities. These rules are similar to those for solving equations but are somewhat more difficult to
apply.
Topics:
• Intervals and other sets of numbers
• The absolute value function
• Working with inequalities
Intervals and other sets of numbers
Intervals can be defined, as in the last section, by giving their defining inequalities. With this approach
the interval in Figure 1 is called the interval 0 < x ≤ 2. The heavy line in the drawing indicates that the
points x with 0 < x < 2 are in the interval; the dot shows that x = 2 is in the interval; and the open
circle at x = 0 indicates that the point x = 0 is not.
We can also define intervals with set-builder notation: The symbols {x : P} designate the set
of numbers x that satisfy condition P. With this notation the interval in Figure 1 can be defined by
{x : 0 < x ≤ 2}, which reads “the set of those numbers x such that 0 < x ≤ 2.” We also refer to this
interval as (0;2], where the parenthesis at the left indicates that the point x = 0 is not in the interval and
the square bracket at the right indicates that the point x = 2 is in the interval. Similarly, the interval in
Figure 2 can be referred to either as the interval x ≤ 2, as the interval {x : x ≤ 2}, or as the interval
(−∞;2]; and the interval in Figure 3 can be given by x > 1, by {x : x > 1}, or by (1;∞).
−1 0 1 2 3 x −1 0 1 2 3 x −1 0 1 2 3 x
The interval 0 < x ≤ 2 or The interval x ≤ 2 or The interval x > 1 or
{x : 0 < x ≤ 2} = (0;2] {x : x ≤ 2} = (−∞;2] {x : x > 1} = (1;∞)
FIGURE1 FIGURE2 FIGURE3
To describe a set that consists of two intervals, we use the union symbol ∪ with the convention
†
that A∪B designates the set consisting of the points in set A combined with the points in set B:
A∪B={x:xisinAorxisinB}:
Example 1 (a) Draw on an x-axis the set of points {x : −5 ≤ x < 2 or x > 4}.
(b) Express the set in part (a) as a union of intervals.
Solution (a) We show the intervals on an x-axis by drawing solid lines from x = −5 to x = 2
and to the right of x = 4 and by putting a solid dot at x = −5 and small open circles
at x = 2 and x = 4, as in Figure 4.
(b) Since {x : −5 ≤ x < 2 or x > 4} consists of the interval −5 ≤ x < 2 and the
interval x > 4, it is their union [−5;2) ∪ (4;∞).
†The symbol ∩ is used for the intersection A ∩ B = {x : x is in A and x is in B} of two sets A and B.
1
p. 2 (5/31/07) Section 0.2, Set notation and solving inequalities
[−5;2) (4;∞)
FIGURE4 −5 2 4 x
The absolute value function
Whenwewanttoconsider the size of a number x without reference to whether it is positive or negative,
we use its absolute value, denoted |x|. The absolute value of x equals x if x is ≥ 0 and can obtained
by multiplying x by −1 if it is negative, so that |5| = 5 and |−5| = −(−5) = 5. Thus, the absolutevalue
function y = |x| with variable x can be defined by
|x| = n x for x≥0 (1)
−x for x < 0:
Definition (1) shows that the graph of the absolute value function consists of the line y = x for
x≥0andtheline y =−x for x <0, as shown in Figure 5.
y y = |x|
4
2
FIGURE5 −4 −2 2 4 x
The absolute value of a number x can also be obtained with the formula
√ 2
|x| = x (2)
since, by convention, √x2 denotes the nonnegative square root of x2. For example, if x = −5, then
p 2 √
(−5) = 25=5and this is the absolute value of −5.
Frequently it is convenient to think of the absolute value |x| of a number x as its distance from
the origin on an x-axis (Figure 6). We can also think of |a −b| as the distance between the points a and
b on an x-axis (Figure 7).
|x2| = −x2 |x1| = x1 |a − b|
x2 0 x1 x a b
FIGURE6 FIGURE7
Section 0.2, Set notation and solving inequalities p. 3 (5/31/07)
Example 2 Solve the equation |2x − 8| = 6 for x.
Solution One solution: |2x−8| = 6 if 2x−8 = 6 or if 2x−8 = −6. We solve these equations
by adding 8 to both sides of each and then dividing both sides of each by 2:
2x−8=6 2x−8=−6
2x = 14 2x = 2
x = 7 x=1
The solutions are x = 7 and x = 1. This is illustrated in Figure 8 by the curve
y = |2x −8| and the line y = 6, which intersect at x = 1 and x = 7.
Alternate solution: Dividing both sides of |2x − 8| = 6 by 2 gives |x − 4| = 3, so
the solutions are the points x = 1 and x = 7 that are a distance 3 from x = 4, as can
be seen in Figure 8.
y y = |2x −8|
8
y = 6
4
2
FIGURE8
1 4 7 x
Question 1 Verify the solutions in Example 2 by substituting the values of x in the formula |2x−8|.
The set consisting of a finite number of real numbers x1;x2;:::;xn is denoted {x1;x2;:::;xn}.
Thus, the solution set (the set of solutions) of the equation in Example 2 is the set {1;7}.
Working with inequalities
Aninequalityis a statement such as x3−1 < 4x3−4 that involves an inequality symbol <;>;≤, or
≥. Two inequalities involving the variable x are equivalent if they are satisfied by the same values of x.
We say that an inequality has been solved if it has been replaced by an equivalent inequality in which
the variable appears alone on one side of the inequality sign and not on the other, as in the statement
x>−1.
How does solving inequalities differ from solving equations? If we want to solve the equation
x3−1=4x3+2,wecanadd1tobothsides toobtain x3 =4x3+3;subtract 4x from both sides to have
−3x3 = 3; divide both sides by −3 to have x3 = −1; and finally take cube roots of both sides to obtain
the solution x = −1. We can also perform such procedures on inequalities, but with some modifications.
Here are some basic rules:
Theorem1 (a) Addinganumbertobothsidesofaninequality,subtracting anumberfromboth sides,
or multiplying or dividing both sides by a positive number yields an equivalent inequality.
(b) Multiplying or dividing both sides of an inequality by a negative number and reversing the
direction of the inequality sign yields an equivalent inequality.
(c) Taking an odd power or odd root of both sides of an inequality yields an equivalent inequality.
(d) If the numbers on both sides of an inequality are nonnegative, then taking an even power or
even root of both sides yields an equivalent inequality.
p. 4 (5/31/07) Section 0.2, Set notation and solving inequalities
Example 3 Find the solution set of the inequality 6 + 2x ≤ 10.
Solution Subtracting 6 from both sides of 6 + 2x ≤ 10 gives the equivalent inequality 2x ≤ 4.
Then dividing both sides by the positive number 2 yields x ≤ 2. The solution set is the
interval (−∞;2].
CQuestion 2 Illustrate the result of Example 3 by generating y = 6 + 2x and y = 10 together on
†
your calculator or computer in the window −5 ≤ x ≤ 5;−5 ≤ y ≤ 15.
Example 4 Solve the inequality x3 −1 < 4x3 +2.
Solution Adding 1 to both sides of the inequality gives x3 < 4x3+3. Subtracting 4x3 from both
sides yields −3x2 < 3. Dividing both sides by the negative number −3 and reversing
the direction of the inequality sign gives x3 > −1. Taking cube roots of both sides gives
x>−1.
CQuestion 3 Generate y = x3 −1 and y = 4x3 +2 together in the window −2 ≤ x ≤ 2;−8 ≤ y ≤ 8
and use the curves to explain the solution of Example 4.
Example 5 The smallest and largest modern Mexican coins are a centavo coined in the 1970’s
(1)
and a ten peso piece from the 1950’s (Figure 9). The radius of the centavo is 0.65
centimeters and the radius of the ten peso piece is 2 centimeters. Consequently, the
radii of all modern Mexican coins lie in the interval 0:65 ≤ r ≤ 2 with r measured in
centimeters. What is the smallest interval that contains the areas of all modern Mexican
coins?
FIGURE9
Solution Squaringthethreepositive numbersintheinequalities 0:65 ≤ r ≤ 2 givesthe equivalent
2 2 2
inequalities (0:65) ≤ r ≤ 2 . Then multiplying the new numbers by the positive
number π yields the equivalent inequalities (0:65)2π ≤ πr2 ≤ 4π. Since the area of a
circle of radius r is A = πr2, the smallest interval containing the areas A of all modern
Mexican coins is (0:65)2π ≤ A ≤ 4π.
†Whenever you are asked in this text to generate a graph, generate it on a graphing calculator or computer.
(1)Data from A Guide Book of Mexican CoinsbyT. Buttrey,Jr.andC. Hubbard,Racine, Wisconsin:WesternPublishing
Company, Inc., 1969, pp. 173 and 208.
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