Download MTH55_Lec-54_sec_8-5a_PolyNom_InEqual

Chabot Mathematics
§8.5 PolyNom
InEqualities
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
1
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[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Review § 8.4
MTH 55
 Any QUESTIONS About
• §8.4 → Equations in Quadratic Form
 Any QUESTIONS About HomeWork
• §8.3 → HW-40
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PolyNomial InEquality
 PolyNomial InEqualities:
3
2 2
5x  8  3, 3x  2 x  x , x  5x  4  0.
 Second-Degree polynomial inequalities
in one variable are called Quadratic
inequalities.
 To solve polynomial inequalities, focus
attention on where the outputs of a
polynomial function are positive and
where they are negative
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Example  Solve x  2 x  8  0.
 SOLUTION: Consider the related
function y = f(x) = x2 + 2x − 8
y
and its graph.
8
 Since the graph positive
y-values
opens upward,
-5
the y-values are
positive outside
the interval formed
by the x-Intercepts
Chabot College Mathematics
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6
positive
y-values
4
2
-4 -3 -2 -1
-2
-4
-6
-8
1
2 3 4 5
x
y = x2 + 2x – 8
-10
-12
-14
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2
Example  Solve x  2 x  8  0.
 SOLUTION: Thus y > 0 for x-values
OUTSIDE of the region described
y
the x-intercepts
8
 It follows that the
solution set of the
inequality is
x | x  4 or x  2,
or  , 4    2,  .
6
positive
y-values
positive
y-values
4
2
-5 -4 -3 -2 -1
-2
-4
-6
-8
1
2 3 4 5
y = x2 + 2x – 8
-10
-12
-14
Chabot College Mathematics
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x
Bruce Mayer, PE
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StreamLine Solution
 In the next example, simplify the
process by concentrating on the sign of
a polynomial function over each
interval formed by the x-intercepts.
 We will do this by tracking the sign of
each factor.
• By looking at how many positive or
negative factors are being multiplied, we
will be able to determine the sign of the
polynomial function.
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Example  Solve f x   x  9 x  0
3
 SOLUTION - first solve the related
equation
f (x) = 0
x3 − 9x = 0
x(x − 3)(x + 3) = 0
Factoring
Using the principle
x = 0 or x − 3 = 0 or x + 3 = 0 of zero products
x = 0 or x = 3 or x = −3.
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Example  Solve f x   x  9 x  0
3
 The function f(x) has zeros at −3, 0 & 3.
 We will use the factorization
f(x) = x(x − 3)(x + 3).
 The product x(x − 3)(x + 3) is positive or
negative, depending on the signs of x,
x − 3, and x + 3.
 This is easily determined using a chart
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[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve f x   x  9 x  0
3
 The Sign Chart for f(x) = x(x – 3)(x + 3)
Interval:
(, 3)
Sign of x:
Sign of x – 3:
Sign of x + 3:
Sign of product
x(x – 3)(x + 3)
–
–
–
–
(3,0)
(0,3)
(3, )
–
–
+
+
–
+
+
+
+
–3
0
+
–
3
+
 Note that ANY Negative No. < 0
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Example  Solve f x   x  9 x  0
3
 A product is negative when it has an
odd number of negative factors.
 Since the sign allows for EQuality, the
ENDpoints −3, 0, and 3 ARE solutions.
 The chart then displays solution set:
(, 3]  [0, 3], or  x | x  3 or 0  x  3.
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Solve PolyNom InEq by Factoring
1. Add or subtract to get 0 on one side of
a “f(x) Greater/Less Than” InEquality

e.g.; 3x2 < 7 − 5x  3x2 + 5x − 7 < 0
2. Next solve the related f(x) = 0
polynomial equation by factoring.
3. Use the numbers found in step (1) to
divide the number line into intervals.
4. Using a test value (or Test-Pt) from
each interval, determine the sign of
each factor over that interval.
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Solve PolyNom InEq by Factoring
5. Determine the sign of the product of
the factors over each interval.
6. Select the interval(s) for which the
inequality is satisfied and write
set-builder notation or interval notation
for the solution set.
7. Include the endpoints of the intervals
whenever ≤ or ≥ describes the InEquality
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Example Solve x  2x  7.
2
 SOLUTION – Obtain Zero on One Side
the InEquality
x 2  2x ─ 7  2x  7  2x ─ 7
x 2  2x ─ 7  0
 Next Solve the
2
x

2
x

7

0
2
Associated
x  2x  7  0
Equation by
2
Quadratic x   2   2   4 17 
Formula
2 1
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[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example Solve x  2x  7.
2
 Isolating x from Quadratic Formula
x  1.8
2  32
so
x
 1 2 2
x  3.8
2
 This divides the No. line into 3 intervals
0
0
– 4 –3 –2 –1 0
1 2 2
1
2
3 4
5
1 2 2
 Pick “Round Numbers as Test Points
• −3, 0, and 4
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Example Solve x  2x  7.
2
 Constructing the Sign Table: x2−2x−7 > 0
Interval
Point
Value
Result
x 2  2x  7
,1  2 2 
1  2 2,1  2 2 
1  2 2, 
−3
8
+
0
−7
–
4
1
+
x  2x  7  0
2
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[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example Solve x  2x  7.
2
 The Sign Intervals on the Number Line
++++++++ 0 ––––––––– 0+++++
– 4 –3 –2 –1 0
1 2 2
1
2
3 4
5
1 2 2
 Thus for x2 > 2x +7  x2 − 2x − 7 > 0 is
POSITIVE (exceeds Zero) in intervals
,1  2 2 
)
– 4 –3 –2 –1 0
1 2 2
Chabot College Mathematics
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1  2
1
2

2,  .
(
3 4
1 2 2
5
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + x ≤ 12
 Solution: write the related equation
x2 + x − 12 = 0 Set the quadratic expression equal to 0
(x − 3)(x + 4) = 0 Factor
x = 3 or x = −4 Use the zero-products theorem
 Plot the Break-Points” on a number line
to create boundaries between the three
intervals
I
-10 -9
-8
II
-7
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-6
-5
-4
-3
-2
-1
III
0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + x ≤ 12
 Choose a test number (or Test-Pt) from
each interval and substitute that value
into x2 + x − 12 ≤ 0
I
-10 -9
Test

-8
-7
-6
-5
-4
-3
-2
-1
III
0
1
2
3
4
5
6
7
8
9
10
For [−4, 3]
choose 0
For [3, ) we
choose 4
(–5)2 + (–5) – 12 (0)2 + (0) – 12
= 25 – 5 – 12
= 0 – 12
(4)2 + (4) – 12
= 16 + 4 – 12
For (−, −4],
choose −5
=8
= –12
=8
False
True
False
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II
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + x ≤ 12
 The Table Reveals the at the CENTRAL
interval is the Solution
• Last time the solution was the
EXTERIOR Intervals
 The solution is the interval [−4, 3]
[
-10 -9
-8
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-7
-6
-5
-4
-3
-2
-1
0
1
2
]
3
4
5
6
7
8
9
10
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + 5x > 0
 Solution: write & Solve the related
2
equation
x  5x  0
x( x  5)  0
x  0 or x  5
 Plot the Break-Pts on the No. line to
form boundaries between the 3 intervals
I
-10 -9
II
-8
-7
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-6
)
-5
-4
-3
III
-2
-1
(0
1
2
3
4
5
6
7
8
9
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Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + 5x > 0
 Choose a Test-Point from each interval
and substitue that value into The
InEquality x2 + 5x > 0
For (−, −5),
choose −6
For (0, )
choose 1
(–6)2 + 5(–6) > 0 (–1)2 + 5(–1) > 0
36 – 30 > 0
1–5>0
(1)2 + 5(1) > 0
1+5>0
–4 > 0
False
6>0
True
6>0
True
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For (−5, 0)
choose −1
Bruce Mayer, PE
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Example  Solve x2 + 5x > 0
 The Sign Table Reveals that the
EXTERIOR intervals are the Solution
 The solution in interval Notation
(−∞, −5) U (0, ∞)
 The Solution on the Number Line
(
)
-10 -9
-8
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-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  (x + 3)(x + 1)(x – 2) < 0
 Solution: write the related equation
(x + 3)(x + 1)(x – 2) = 0 Write Related Eqn
x + 3 = 0 or x + 1 = 0 or x − 2 = 0
Using Zero-Products Theorem
x = −3 or x = −1 or x = 2 Solving for x
 Plot the Break-Points” on a number line
to create FOUR intervals
-10 -9
-8
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-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  (x + 3)(x + 1)(x – 2) < 0
 Solution: Create Sign/Truth Chart
-10 -9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
(, 3) (3, 1)
Interval
5
6
7
8
9
10
(1, 2)
(2, )
Test Number
4
2
0
3
Test Results
True or False
18 < 0
True
4<0
False
6 < 0
True
24 < 0
False
 Thus the Soln Set: (−∞, −3) U (−1, 2)
• On Number Line:
)
-10 -9
-8
-7
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-6
-5
-4
-3
(
-2
-1
)
0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
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Example  Skid Marks
 A car involved in an accident left skid
marks over 75 feet long. Under the road
conditions at the accident, the distance
d (in feet) it takes a car traveling v miles
per hour to stop is given by the equation
2
d  0.05v  v.
 The accident occurred in a 25 mph zone.
 Was the driver Speeding?
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Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Skid Marks
 SOLUTION: Solve the InEquality:
(stopping distance) > 75 feet, or 0.05v  v  75.
2
0.05v 2  v  75
0.05v 2  v  75  0
0.05v 2  v  75  0
5v 2  100v  7500  0
v 2  20v  1500  0
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v  50 v  30   0
v  50  0 or v  30  0
v  50
or v  30
 The Solns divide
the number line
into 3 intervals
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Skid Marks
 Make Sign-Chart, and Divide No. Line
Interval
Point
Result
– 60
Value
2
0.05v  v
45
(–∞, –50)
(–50, 30)
0
–75
–
(30, ∞)
40
45
+
+
++++++ 0 ––––––––––––––– 0 ++++++
– 60 –50
Chabot College Mathematics
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0
30 40
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Skid Marks
 ++++ on No. Line indicates Speeding
++++++ 0 ––––––––––––––– 0 ++++++
– 60 –50
0
30 40
 For this situation, we look at only the
positive values of v. Note that the
numbers corresponding to speeds
between 0 and 30 mph (i.e., 0 ≤ v ≤ 30 )
are not solutions of 0.05v2+v
 Thus, the car was traveling more than 30
miles per hour  Excessive Speed
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28
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x4 ≤ 1
 SOLUTION: Solve Related Eqn for x
x 1
4
x  1  0, x  1  0, x  1  0
2
x4  1  0
x

x  1x  1x
x  1x  1x
2

1 x 1  0
Chabot College Mathematics
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x  1, x  1
2
 These Break2
 1  0 Points divide the
number line into
2
1  0
3 intervals


Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x4 ≤ 1
 Make Sign Chart & Associated No. Line
Value of

 Result
Interval Point x  1x  1 x 2  1
(−∞, –1)
−2
15
+
(–1, 1)
0
−1
–
(1, ∞)
2
15
+
++++++++0 –––0 +++++++
–3 –2 –1 0
Chabot College Mathematics
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1
2
3 4
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Example  Solve x4 ≤ 1
 The Sign-Regions on No. Line
++++++++0 –––0 +++++++
–3 –2 –1 0
1
2
3 4
 The solution set consists of all x
between −1 & 1, including both −1 & 1.
–2
[
–1
0
]
1
2
{−1 ≤ x ≤ 1} or [−1, 1]
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Bruce Mayer, PE
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ONE-Sign Theorem
 If a polynomial equation has NO
REAL solution, then the polynomial
is either ALWAYS positive or
always negative
 Example  Use One-Sign Theorem to
Solve InEquality:
x – 2x  2  0.
2
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Bruce Mayer, PE
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Example  1-Sign Theorem x 2 – 2x  2  0.
 SOLUTION:
Solve Related Eqn
x – 2x  2  0.
2
 The Expression does NOT factor so use
Quadratic Formula
• First Test Discriminant for Real Solutions
b – 4ac  2   4 12    4
2
2
 Discriminant is NEGATIVE →
NO Real-Number Solution
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33
Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  1-Sign Theorem x 2 – 2x  2  0.
b – 4ac  2   4 12    4
2
2
 Since the discriminant is negative, there
are no real roots.
 Use 0 as a test point, which yields 2.
 The inequality is always positive, the
solution set is ALL REAL Numbers
• Using Interval Notion: (−∞, ∞).
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WhiteBoard Work
 Problems From §8.5 Exercise Set
• 16, 28, 34, 68
 x2 − x − 6 < 0
Graphically
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All Done for Today
Acapulco
Cliff
Diving
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Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
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Bruce Mayer, PE
[email protected] • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
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5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
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4
5
6
5
5
y
4
4
3
3
2
2
1
1
0
-10
-8
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-4
-2
-2
-1
0
2
4
6
-1
0
-3
x
0
1
2
3
4
5
-2
-1
-3
-2
M55_§JBerland_Graphs_0806.xls
-3
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-4
M55_§JBerland_Graphs_0806.xls
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