Sec. 4.6(A) Solving Higher Order Inequalities

1. 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 ______________ open circle closed circle ( , ) [ , ] (2 , 3 ] 2 3 (4 , 4 (- -2 (- R Steps for solving a Higher OrderInequality. 1. 2. 3. 4. 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.

2. Ex1: Solve: x2 - 1 < 0 Ex2: Solve: x2– 4x < 0 Ex3: Solve: x2+ 9 > 0 [-1 , 1] -1 1 (x + 1)(x – 1) = 0 - 1 < 0 4 – 1 < 0 3 < 0 F - 1 < 0 0 – 1 < 0 -1 < 0 T - 1 < 0 4 – 1 < 0 3 < 0 F x + 1 = 0 x – 1 = 0 x = -1 x = 1 (0 , 4) 0 4 x(x – 4) = 0 - 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 x = 0 x – 4 = 0 x = 4 + 9 = 0 = -9 = 2 imaginary answers

3. Ex4: (x – 1)(x – 2)(x – 3) < 0 Ex5: 16x4 – 4x2> 0 Ex6: x3 > 8 0< 0 T 1 2 3 x – 1 = 0 x - 2 = 0 x – 3 = 0 x = 1 x = 2 x = 3 < 0 -0.375 < 0 T (0.5)(-0.5)(-1.5) < 0 0.375 < 0 F 6 < 0 F (- 0 - > 0 16 - 4 > 0 12 > 0 T = 0 (2x + 1)(2x – 1) = 0 = 0 x = 0 (D.R.) - > 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 2 > 8 27 > 8 T > 8 0 > 8 F x – 2 = 0 x = 2 = 0

4. Ex7: x3 + 4x2 + 4x < 0 -2 0 + 4(-3) < 0 -3 < 0 T = 0 x (x + 2)(x + 2) = 0 + 4 + 4(1) < 0 1 + 4 + 4 < 0 9 < 0 F + 4 + 4(-1) < 0 -1 + 4 – 4 < 0 -1 < 0 T x = 0 x + 2 = 0 x = -2 (D.R.) Solve the following using your graphing calculator. Ex8: x2 – 2x – 3 <0 Ex9: Ex10: x3> 4 < 0