Sec. 4.6(A) Solving Higher Order Inequalities We will use a new notation for writing inequalities. Interval notation looks like the following:
> or < will use ______________ < or > will use ______________
Inequality Notation Graph Interval Notation 2 < x 3
x > 4
x > 2 or x
x 2
3
Steps for solving a Higher Order Inequality.
1.
2.
3.
4.
open circle( , )
closed circle[ , ]
2 3(2 , 3 ]
4 (4 ,
-2 (-
R(-
Get a zero on one side of the equation.
Find the x-int.’s and graph those one a # line.
Pick points in the regions separated by the x-int’s. to test.
Write a solution of the graph drawn using interval notation.
Ex1: Solve: x2 - 1 < 0
Ex2: Solve: x2 – 4x < 0
Ex3: Solve: x2 + 9 > 0
(x + 1)(x – 1) = 0
x + 1 = 0 x – 1 = 0 x = -1 x = 1
-1 1 - 1 < 0 4 – 1 < 0 3 < 0
F
- 1 < 0 0 – 1 < 0 -1 < 0
T
- 1 < 0 4 – 1 < 0 3 < 0
F
[-1 , 1]
x(x – 4) = 0
x = 0 x – 4 = 0 x = 4
0 4 - 4(-1) < 0 1 + 4 < 0 5 < 0
F
- 4(2) < 0 4 – 8 < 0 -4 < 0
T
- 4(5) < 0 25 – 20 < 0 5 < 0
F
(0 , 4)
+ 9 = 0 = -9 =
2 imaginary answers
(−∞ ,∞ )
Ex4: (x – 1)(x – 2)(x – 3) < 0
Ex5: 16x4 – 4x2 > 0
Ex6: x3 > 8
x – 1 = 0 x - 2 = 0 x – 3 = 0 x = 1 x = 2 x = 3
1 2 30 < 0 T
(0.5)(-0.5)(-1.5) < 0 0.375 < 0
F
< 0 -0.375 < 0
T 6 < 0
F (-
= 0 (2x + 1)(2x – 1) = 0 = 0 x = 0 (D.R.)
0 - > 0 16 - 4 > 0 12 > 0
T
- > 0 0.0625 – 0.25 > 0 -0.1875 > 0
F
- > 0 0.0625 – 0.25 > 0 -0.1875 > 0
F - > 0 0 – 0 > 0 0 > 0
T
- > 0 16 – 4 > 0 12 > 0
T
- 8 > 0
= 0x – 2 = 0 x = 2
2 > 8 0 > 8
F
> 8 27 > 8 T
(𝟐 , ∞ )
Solve the following using your graphing calculator. Ex8: x2 – 2x – 3 < 0 Ex9: Ex10: x3 > 4
< 0
Ex7: x3 + 4x2 + 4x < 0
= 0x (x + 2)(x + 2) = 0
x = 0 x + 2 = 0 x = -2 (D.R.)
-2 0+ 4(-3) < 0 -3 < 0
T + 4 + 4(-1) < 0 -1 + 4 – 4 < 0 -1 < 0
T
+ 4 + 4(1) < 0 1 + 4 + 4 < 0 9 < 0
F
(−∞ ,𝟎 ]
[−𝟏 ,𝟑 ] (−𝟐 ,𝟎 )∪ (𝟎 ,𝟐 )
(𝟏 .𝟓𝟖𝟕 , ∞ )
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