Download 2.7_Polynomials Rat Inequalities

Section 2.7
Polynomials and Rational Inequalities
PAGE 359 – 369 IN TEXTBOOK.
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UP FRONT)
Introduction
Tailgaters beware: If your car is
going 35 miles per hour on dry
pavement, your required
stopping distance is 160 feet, or
the width of a football field.
At 65 mph, the distance required
is 410 feet, or approximately the
length of one and one- tenth
football fields. Figure 2.43 shows
stopping distances for cars at
various speeds on dry roads and
on wet roads.
Introduction Cont.
 A car’s required stopping distance, f(x), in feet, on dry
pavement traveling at mph can be modeled by the
quadratic function f (x) = 0.0875x2 – 0.4x + 66.6
 How can we use this function to determine speeds on dry
pavement requiring stopping distances that exceed the
length of one and one- half football fields, or 540 feet?
 We must solve the inequality 0.0875x2 – 0.4x + 66.6 > 540
Introduction Cont.
The form of this inequality is ax2 + bx + c > 0 which is
called a quadratic inequality.
A polynomial inequality is any inequality that can be
put into one of the forms:
f  x   0, f  x   0
or
f  x   0, f  x   0
Remember, f stands for polynomial function!!!
Solving Polynomial Inequalities
Graphs help us visualize the solution of
polynomial inequalities. For example, the
graph of f(x) = x2 – 7x + 10 is shown in
Figure 2.44.
What are the x-intercepts?
These are going to be referred to as
boundary points.
Where is the graph greater than zero?
Where is the graph less than zero?
Solving Polynomial Inequalities
Locating the x-intercepts of a polynomial function, is an important
step in finding the solution set for polynomial inequalities in the
form f(x) > 0 or f(x) < 0.
We use the x-intercepts of f as boundary points that divide the
real number line into intervals. On each interval, the graph of f is
either above the x-axis [ f(x) > 0 ] or below the x-axis [ f(x) < 0 ].
For this reason, x-intercepts play a fundamental role in solving
polynomial inequalities. The x-intercepts are found by solving the
equation f(x) = 0.
Procedure for Solving Polynomial Inequalities
Express the inequality in the form f(x) > 0 or f(x) < 0 where f is a
polynomial function.
2.
Solve the equation f (x) = 0. The real solutions are the boundary points.
3.
Locate these boundary points on a number line, thereby dividing the
number into intervals.
4.
Choose one representative number, called a test value, within each interval
and evaluate f at that number.
a.
If the value of f is positive, then f(x) > 0 for all numbers, x in the interval.
b.
If the value of f is negative, then f(x) < 0 for all numbers, x in the interval.
5.
Write the solution set, selecting the interval or intervals that satisfy the
given inequality.
This procedure is valid if < is replaced by ≤ or > is replaced by ≥. However, if the
inequality involved ≤ or ≥, include the boundary points [the solutions of
f(x) = 0 ] in the solution set.
1.
Example 1
Solve and graph the solution set on a real numberline:
x 2  x  12
10
-5
0
5
10
Step 1 Express the inequality in the form f(x) < 0 or f(x) > 0. x 2  x  12  0
Step 2 Solve the equation f(x) = 0 x 2  x  12  0
 x  4  x  3  0
Step 3 Locate boundary points on number line & separate into intervals.
x  4,  3
Step 4 Choose test values within each interval & evaluate f(x).
Step 5 Write solution set, selecting the interval(s) that satisfy the function.
 3, 4 
Example 2
Solve and graph the solution set on a real number line:
2 x  x  15
2
10
-5
0
5
Step 1 Express the inequality in the form f(x) < 0 or f(x) > 0.
Step 2 Solve the equation f(x) = 0
10
2x 2  x  15  0
2x 2  x  15  0
2x  5  x  3  0
Step 3 Locate boundary points on number line & separate into intervals.
x 
5
, 3
2
Step 4 Choose test values within each interval & evaluate f(x).
Step 5 Write solution set, selecting the interval(s) that satisfy the function.

5
, 3 U  ,  
2
Example 3
Solve and graph the solution set on a real number line:
3
2
x  x  4x  4
-2
-1
0
1
2
Step 1 Express the inequality in the form f(x) < 0 or f(x) > 0.
Step 2 Solve the equation f(x) = 0 x 3  x 2  4x  4  0
x
2

 4  x  1  0
x 3  x 2  4x  4  0
 x  2 x  2 x  1  0
Step 3 Locate boundary points on number line & separate into intervals.
x  2,  1, 2
Step 4 Choose test values within each interval & evaluate f(x).
Step 5 Write solution set, selecting the interval(s) that satisfy the function.
 ,  2 U 1, 2
Example 4
Solve and graph the solution set on a real number line:
3
2
x  3x  x  3
-4
-2
0
2
4
Step 1 Express the inequality in the form f(x) < 0 or f(x) > 0.
Step 2 Solve the equation f(x) = 0 x 3  3x 2  x  3  0
x
2

 1  x  3  0
x 3  3x 2  x  3  0
 x  1 x  1x  3  0
Step 3 Locate boundary points on number line & separate into intervals.
x  3,  1, 1
Step 4 Choose test values within each interval & evaluate f(x).
Step 5 Write solution set, selecting the interval(s) that satisfy the function.
 3,  1 U 1,  
Try these problems!
Page 366
Problems 2, 12, 18, 22
2.
 x  3 x  5  0
12.
9 x 2  3x  2  0
18.
4x  1  4x
22.
3x 2  5 x  0
2
A rational inequality is an inequality that can be put into one of the forms
f ( x)  0, f ( x)  0,
where f is a rational function.
f ( x)  0
f ( x)  0,
An example of a rational inequality is… 3 x  3
0
2x  4
or
This inequality in the form f (x) > 0, can be compared to the rational function…
f ( x) 
3x  3
2x  4
We start to solve this by first finding the xintercept: set the numerator equal to 0.
We now find the domain of the function.
2x  4  0
2 x  -4
x  -2
3x  3  0
3x  -3
x  -1
These numbers are our boundary points that we use to create intervals on the number line.
-2
-1
0
1
 , 2   1, 
2
y
+


+


x















Example 5


+
5x  2
Solve and graph the solution set
0
3x  1

5x  2  0
5x  2
2
x   0.4
5
+
















-1
 , 0.33   0.4,  

3x  1  0
3x  -1
x


0

1
1
 0.33
3x


Example 6



+


8x  2
Solve and graph the solution set
0
4x  1
8x  2  0
8x  2
1
x    0.25
4
+










4x  1  0
4x  1
1
x   0.25
4x



-1





 0.25,0.25
0
1

x 1
Solve and graph the solution set
2
x3
x 1
x  1 2  x  3
x  1   2x  6
20

0
0
x3
x  3 1 x  3 
x3
Example 7
x  1  2x  6
0
x3
x  5  0
-x  5
x  5
0
x3
x3 0
x  3
x  5
5,
 3
-5
-3
-1
Example 8
A student is standing on the top of a university building which is 80 feet tall.
If he throws a softball up in the air with an initial velocity of 32 feet/sec,
during which time period will the ball's height exceed that of the university
building? s(t )  -16t 2  v0 t  s0
16t  32t  80  80
2
80 ft
16t 2  32t  0
16t  t  2  0
16t  t  2  0
t  0, 2
 0, 2
It will be above 80 feet from 0 to 2 seconds.
Review Time!!!!
Add these problems
to your notes paper to
help you review!
Additional Practice problems can be found in your textbook
1. Express the solution set of the
following inequality in interval
notation: x3 - 9x ≥ 0
(a)  , 3   0,3
(b)  3, 0  3,  
(c)  , 3   0,3
(d)  3, 0  3,  
2.
Express the solution set in
x5
interval notation.
 0.
2x  1
(a)
1 
 , 5
2 
(b)  , 5 
(c) 5,  
(d)
1 
 2 , 5 