Download Section 1.7 Inequalities

Section 1.7 Inequalities
• Properties of Inequalities:
– Trichotomy Property: For all real numbers a and b, one of the following statements is true: a < b or
a = b or a > b.
– Transitive Property: If a, b, and c are real numbers, then
if a < b and b < c, then a < c.
if a > b and b > c, then a > c.
– Addition/Subtraction Properties: Let a, b, and c be real numbers.
∗ If a < b, then a + c < b + c
∗ If a < b, then a − c < b − c.
∗ Similarly for ≤, ≥, >
– Multiplication/Division Properties: Let a, b, and c be real numbers.
1. If a < b and c > 0, then ca < cb.
a b
If a < b and c > 0, then < .
c c
2. If a < b and c < 0, then ca > cb.
a b
If a < b and c > 0, then > .
c c
NOTE: If we multiply or divide by a negative number, we MUST flip the inequality.
• Solving Linear Inequalities:
Ex: Solve 5(x − 4) > 25.
5(x − 4) > 25
5x − 20 > 25
5x > 45
x>9
We can write this solution in interval notation: (9, ∞).
Ex: Solve −4(x + 3) ≥ 16.
−4x − 12 ≥ 16
−4x ≥ 28
We want the coefficient of x to be 1, so we must divide by −4. When we do so, we MUST flip the inequality
sign!!
−4x
28
≤
−4
−4
x ≤ −7
In interval notation, this is (−∞, −7].
• Compound Inequalities:
Ex: Solve −5 ≤ 2x + 1 < 9
Let’s get x by itself. Remember that if we do something to one side of this compound inequality, that we must
do it to all three sides:
−5 − 1 ≤ 2x + 1 − 1 < 9 − 1
−6 ≤ 2x < 8
−6 2x 8
≤
<
2
2
2
−3 ≤ x < 4
In interval notation, this is [−3, 4).
Ex: Solve x + 1 < 2x − 3 ≤ 3x − 5.
Here, we cannot isolate x because there are x’s on all three sides. We must treat these as two separate inequalities instead of as a compound inequality.
x + 1 < 2x − 3
AND
2x − 3 ≤ 3x − 5
1 < x−3
AND
−3 ≤ x−5
4<x
2≤x
AND
In interval notation, we have (4, ∞) and [2, ∞).
These intervals overlap. We can combine them into one interval, (4, ∞).
• Quadratic Inequalities: We solve the quadratic equation (replace the inequality sign with an = sign). Then we
must check each interval by using a test point.
Ex: Solve x2 + 5x + 6 < 0.
Solve x2 + 5x + 6 = 0. Factor:
(x + 2)(x + 3) = 0
x+2 = 0
OR
x+3 = 0
x = −2
OR
x = −3
We now must check three intervals: (−∞, −3), (−3, −2), (−2, ∞).
Check (−∞, −3): Pick any number in this interval and see if it satisfies the original inequality: −5 ∈ (−∞, −3).
(−5)2 + 5(5) + 6? <?0
25 + 25 + 6? <?0
56? <?0
False. So there are no solutions in this interval.
Check (−3, −2): Pick any number in this interval and see if it satisfies the original inequality: −2.5 ∈
(−3, −2).
(−2.5)2 + 5(−2.5) + 6? <?0
6.25 − 12.5 + 6? <?0
−.25? <?0
2
True. There are solutions in this interval.
Check (−2, ∞): Pick any number in this interval and see if it satisfies the original inequality: 0 ∈ (−2, ∞).
02 + 5(0) + 6? <?0
6? <?0
False. There are no solutions in this interval.
Therefore, the solution set for this inequality is (−2, ∞).
Ex: Solve 9x2 ≤ 24x − 16.
Solve 9x2 = 24x − 16 first.
9x2 − 24x + 16 = 0
(3x − 4)(3x − 4) = 0
3x − 4 = 0
3x = 4
4
x=
3
There are two intervals to check: (−∞, 43 ) and ( 43 , ∞). Note, we know x =
only need to check the open intervals.
Check (−∞, 34 ). Test point: 0 ∈ (−∞, 34 ).
4
3
satisfies this inequality, so we
9(0)2 ? ≤?24(0) − 16
0? ≤? − 16
False. There are no solutions in this interval.
Check ( 43 , ∞). Test point: 2 ∈ ( 43 , ∞).
9(2)2 ? ≤?24(2) − 16
9(4)? ≤?48 − 16
36? ≤?32
False. There are no solutions in this interval.
Therefore the only solution for this inequality is x = 43 . In interval notation, this is [ 43 , 43 ].
• Rational Inequalities
x2 + 2x − 3
> 0.
x2 + 4x + 3
We first solve this as an equation rather than an inequality. Factor the top and bottom.
Ex: Solve
(x − 1)(x + 3)
=0
(x + 1)(x + 3)
The solutions to the numerator are:
x=1
OR
x = −3
The solutions to the denominator are:
x = −1
OR
x = −3
We now have to check four intervals: (−∞, −3), (−3, −1), (−1, 1), (1, ∞)
3