Transcript Polynomial Inequalities

Polynomial
Inequalities
2.7
Definition of a Polynomial
Inequality
A polynomial inequality is any inequality
that can be put in one of the forms
f(x) < 0, f(x) > 0, f(x) < 0, or f(x)> 0
where f is a polynomial function.
Procedure for Solving Polynomial
Inequalities
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Express the inequality in the standard form
f(x) < 0 or
f(x) > 0.
Find the zeros of f. The real zeros are the boundary points.
Locate these boundary points on a number line, thereby dividing
the number line into intervals.
Choose one representative number within each interval and
evaluate f at that number. If the value of f is positive, then f(x)>0
for all numbers, x, in the interval. If the value of f is negative, then
f(x)<0 for all numbers, x, in the interval.
Write the solution set; selecting the interval(s) that satisfy the
given inequality.
Example
Solve and graph the solution set on a real number line:
2x2 – 3x > 2.
Solution
Step 1 Write the inequality in standard form. We can write by subtracting
2 from both sides to get zero on the right.
2x2 – 3x – 2
2x2 – 3x – 2
2–2
0
>
>
Step 2 Solve the related quadratic equation. Replace the inequality sign with an
equal sign. Thus, we will solve.
2x2 – 3x – 2
=
(2x + 1)(x – 2) =
2x + 1 = 0
or
x–2
x = -1/2 or
x
=
The boundary points are –1/2 and 2.
0
This is the related quadratic equation.
0
=
0
2
Factor.
Set each factor equal to 0.
Solve for x.
Example cont.
Solve and graph the solution set on a real number line:
2x2 – 3x > 2.
Solution
Step 3 Locate the boundary points on a number line. The
number line with the boundary points is shown as follows:
-1/2
2
x
0
The boundary points divide the number line into three test
intervals. Including the boundary points (because of the given
greater than or equal to sign), the intervals are (-ºº, -1/2], [-1/2, 2],
[2, -ºº).
Example cont.
Solve and graph the solution set on a real number line:
2x2 – 3x > 2.
Solution
Step 4 Take one representative number within each test interval and
substitute that number into the original inequality.
Test Interval
Representative
Number
Substitute into
2x∆ – 3x > 2
Conclusion
(-ºº, -1/2]
-1
2(-1) 2 – 3(-1) > 2
5 > 2, True
(-ºº, -1/2] belongs
to the solution set.
[-1/2, 2]
0
2(0) 2 – 3(0) > 2
0 > 2, False
[2, ºº)
3
2(3)2 – 3(3) > 2
9 > 2, True
[-1/2, 2] does not
belong to the
solution set.
[2, ºº) belongs to
the solution set.
Example cont.
Solve and graph the solution set on a real number line:
2x2 – 3x > 2.
Solution
Step 5 The solution set are the intervals that produced a true statement.
Our analysis shows that the solution set is
(-ºº, -1/2] or [2, ºº).
The graph of the solution set on a number line is shown as follows:
-5
-2
1
4
-1/2
2
)
(
-4
-1
2
5
x
-3
0
3
YU TRY!

Solve the inequality:
x  x  30
2
A : (,3)  (1, )
B : (,2.30)  (1.30, )
C : (2.30,1.30)
D : (3,1)
Senteo Question
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Text Example
Solve and graph the solution set:
x 1
2
x3
Solution
Step 1 Express the inequality so that one side is zero and the other side is
a single quotient. We subtract 2 from both sides to obtain zero on the right.
x 1
 2 This is the given inequality.
x3
x 1
2 0
x3
Subtract 2 from both sides, obtaining 0 on the right.
x  1 2(x  3)

 0 The least common denominator is x + 3. Express 2
x3
x3
in terms of this denominator.
x  1  2(x  3)
0
x3
x  1  2x  6
0
x3
x  5
0
x3
Subtract rational expressions.
Apply the distributive property.
Simplify.
Text Example cont.
Solve and graph the solution set:
x 1
2
x3
Step 3 Locate boundary points on a number line.
The boundary points divide the number line into three test intervals, namely
A: [-ºº, -5), (-5, -3), (-3, ºº].
B: (-ºº, -5], [-5, -3), (-3, ºº).
C: (-ºº, -5], [-5, -3], [-3, ºº).
YU PARTICIPATE!
Senteo Question
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Text Example cont.
Solve and graph the solution set:
x 1
2
x3
Step 4 Take one representative number within each test interval and
substitute that number into the original equality.
Test
Interval
(-ºº, -5)
(-5,-3)
x  5
Substitute into x  3
f (x) 
Representative
Number
-6
-4
f (6) 
(6)  5
6  3
 1/ 3
f (4) 
(4)  5
4  3
1
(-3, ºº)
0
f (0) 
 5/ 3
0  5
03
Conclusion
f(x )< 0 for all x in
(-ºº, -5)
f(x) > 0 for all x in
(-5,-3)
f(x) < 0 for all x in
(-3, ºº)
Text Example cont.
Solve and graph the solution set:
x 1
2
x3
Step 5 The solution set are the intervals that produced a true statement.
Our analysis shows that the solution set is:
A: [-1, -3]
B: [-1, -3)
C: [-5, -3]
D: [-5, -3)
Senteo Question
YU CAN DO IT!
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Senteo Question Object->Properties...
The Position Formula for a Free-Falling
Object Near Earth’s Surface
• An object that is falling or vertically projected into the
air has its height in feet above the ground given by
•
s = -16 t 2 + v0 t + s0
• where s is the height in feet, v0 is the original velocity
(initial velocity) of the object in feet per second, t is
the time that the object is in motion in seconds, and s0
is the original height (initial height) of the object in
feet.
Example
An object is propelled straight up from ground level with an initial
velocity of 80 fps. Its height at time t is described by s = -16 t 2 + 80 t
where the height, s, is measured in feet and the time, t, is measured in
seconds. In which time interval will the object be more than 64 feet
above the ground?
Solution
-16 t 2 + 80 t > 64
-16 t 2 + 80 t – 64 > 0
-16 t 2 + 80 t – 64 = 0
-16 (t – 1)(t – 4) = 0
(t – 1)(t – 4) = 0
t – 1 = 0 or t – 4 = 0
t=1
t=4
This is the inequality implied by the problem’s
question. We must find t.
Subtract 64 from both sides.
Solve the related quadratic equation.
Factor.
Divide each side by -16.
Set each factor equal to 0.
Solve for t. The boundary points are 1 and 4.
Example cont.
An object is propelled straight up from ground level with an initial velocity of
80 fps. Its height at time t is described by s = -16 t ∆ + 80 t where the height, s,
is measured in feet and the time, t, is measured in seconds. In which time
interval will the object be more than 64 feet above the ground?
Solution
-16 t 2 + 80 t > 64
t=1
t=4
This is the inequality implied by the problem’s
question. We must find t.
The boundary points are 1 and 4.
Since neither boundary point satisfy the
inequality, 1 and 4 are not part of the solution.
1
4
x
With test intervals (-ºº, 1), (1, 4), and (4, ºº), we could use 0, 2, and 5 as test
points for our analysis.
Example cont.
An object is propelled straight up from ground level with an initial velocity of
80 fps. Its height at time t is described by s = -16 t 2 + 80 t where the height, s,
is measured in feet and the time, t, is measured in seconds. In which time
interval will the object be more than 64 feet above the ground?
Solution
Test
Interval
Representative
Number
Substitute into
(x – 1)(x – 4) < 0
Conclusion
(-ºº, 1) 0
(0 – 1)(0 – 4) < 0
4 < 0, False
(-ºº, 1) does not belong to
the solution set.
(1, 4)
2
(2 – 1)(2 – 4) < 0
-2 < 0, True
(1, 4) belongs to the
solution set.
(4, ºº)
5
(5 – 1)(5 – 4) < 0
4 < 0, False
(4, ºº) does not belong to
the solution set.
The object will be above 64 feet between 1 and 4 seconds.