math 31 lessons precalculus 6. absolute values and inequalities
Post on 04-Jan-2016
215 Views
Preview:
TRANSCRIPT
MATH 31 LESSONS
PreCalculus
6. Absolute Values and Inequalities
A. Absolute Values
The absolute value of a number a, denoted by |a| , is the
distance from a to 0 on the real number line.
Distances are always positive or zero.
The absolute value of a is also called the magnitude of a.
Properties of absolute values
0if, aaa If a is positive, then it remains positive.
e.g.
0if, aaa If a is positive, then it remains positive.
55
041.012since,1212
Properties of absolute values
0if, aaa If a is negative, then multiply it by 1 to make it positive.
e.g.
0if, aaa If a is negative, then multiply it by 1 to make it positive.
777
2
032.12since,22
Ex. 1 Simplify
Try this example on your own first.Then, check out the solution.
217314
since
217314
2]173[14
014.12
01.1173
217314
2]173[14
21734
217314
2]173[14
21734
517
B. Absolute Value Equalities
e.g.
If |x| = 5, then x can be 5 or 5
axax ifonlyandif
Alternative Perspective
When |x| = 5, the argument x could be either positive or negative.
So, we have to deal with the two cases separately.
Alternative Perspective
When |x| = 5, the argument x could be either positive or negative.
So, we have to deal with the two cases separately.
Case 1: x is positive
|x| = 5
x = 5
Simply drop the absolute values and solve
Alternative Perspective
When |x| = 5, the argument x could be either positive or negative.
So, we have to deal with the two cases separately.
Case 1: x is positive Case 2: x is negative
|x| = 5 |x| = 5
x = 5 x = 5
x = 5Simply drop the absolute values and solve
Multiply the argument (inside the absolute value) by 1 and solve
The alternative perspective leads to a straight-forward
method to solve any absolute value equation:
If | f(x) | = g(x) , then:
f(x) = g(x) or f(x) = g(x)
Simply drop the absolute values and solve
Multiply the argument (inside the absolute value) by 1 and solve
e.g. Solve 1172 x
1172 x
1172 x or 1172 x
Simply drop the absolute values and solve
Multiply the argument (inside the absolute value) by 1 and solve
1172 x
1172 x or 1172 x
182 x 1172 x
1172 x
1172 x or 1172 x
182 x
9x
1172 x
42 x
2x
Ex. 2 Solve
Try this example on your own first.Then, check out the solution.
2252 xx
2252 xx
2252 xx 2252 xxor
Simply drop the absolute values and solve
Multiply the argument (inside the absolute value) by 1 and solve
2252 xx
2252 xx 2252 xxor
052 xx
05 xx
5,0x
2252 xx
2252 xx 2252 xxor
052 xx 2252 xx
05 xx
5,0x
0452 xx
014 xx
4,1x
C. Absolute Value Inequalities
As we will see, the method used for absolute value
equalities will also help us when we do absolute
value inequalities.
We consider 2 cases.
Case 1: |x| a
Consider the inequality |x| 5
Again, the argument x could be positive or negative.
So, we will have to consider each case separately.
|x| 5
5x or 5 x
If x is positive, then simply drop the absolute value
If x is negative, then multiply the argument by 1 to make it positive
|x| 5
5x or 5 x
Remember: When you multiply both sides by a negative, you must flip the inequality
5x
|x| 5
5x or 5x
,55,
5or5 xx
-5 5
All values less than -5 and greater than 5 have a magnitude greater than 5.
Case 1: |x| a
Consider the inequality |x| 7
Again, the argument x could be positive or negative.
So, we will have to consider each case separately.
|x| 7
7x and 7 x
If x is positive, then simply drop the absolute value
If x is negative, then multiply the argument by 1 to make it positive
|x| 7
7x and 7x
7,7
77 x
7 7
All values between -7 and 7 have a magnitude smaller than 7
Note:
When the absolute value is “greater than”,
you use “or” (union) in the solution
For the inequality |x| > 3,
the solution is x < -3 or x > 3
“Great-or”
When the absolute value is “less than”,
you use “and” (intersection) in the solution
For the inequality |x| < 11,
the solution is x > -11 and x < 11
(i.e. -11 < x < 11 )
“Less-and”
Ex. 3 Solve
Express your answer in interval notation.
Try this example on your own first.Then, check out the solution.
423 y
423 y
423 y 423 yor
“Great-or”
If x is positive, then simply drop the absolute value
If x is negative, then multiply the argument by 1 to make it positive
423 y
423 y 423 y
23 y 423 y
or
423 y
423 y 423 y
23 y
3
2y
423 y
63 y
2y
or
3
2y 2yor
,3
22,
-2 2/3
Ex. 4 Solve
Express your answer in interval notation.
Try this example on your own first.Then, check out the solution.
242 xx
242 xx
242 xx 242 xxand
“Less-and”
If x is positive, then simply drop the absolute value
If x is negative, then multiply the argument by 1 to make it positive
242 xx
242 xx 242 xx
062 xx 242 xx
022 xx
and
We must solve each inequality separately.
The overall solution must be the intersection between the two solutions sets.
Inequality 1: 062 xx
023 xx
Inequality 1:
Critical values:
062 xx
023 xx
2,3x
Recall, the CV’s are where f(x) = 0
-3 2
023 xx
Interval x + 3 x - 2 (x + 3) (x - 2)
x < -3
-3 < x < 2
x > 2
The CV’s become the “endpoints” of the intervals.
Set up an interval test.
-3 2
023 xx
Interval x + 3 x - 2 (x + 3) (x - 2)
x < -3
-3 < x < 2
x > 2
x = -4 +
+
+ + +
x = 0
x = 3
Inequality 2:
012 xx
022 xx
Inequality 2:
Critical values:
012 xx
1,2x
Recall, the CV’s are where f(x) = 0
022 xx
-2 1
012 xx
Interval x + 2 x - 1 (x + 2) (x - 1)
x < -2
-2 < x < 1
x > 1
The CV’s become the “endpoints” of the intervals.
Set up an interval test.
-2 1
012 xx
Interval x + 2 x - 1 (x + 2) (x - 1)
x < -2
-2 < x < 1
x > 1
x = -3 +
+
+ + +
x = 0
x = 2
Remember, the overall solutions must be intersection between
the two solutions. It must satisfy both.
A straightforward way to find the intersection is to draw
both solutions on the same number line:
-3 1-2 2
The intersection is the overlap between the two solutions.
i.e.
-2-3 21
2123 xorx
2,12,3
top related