1. Essential Question: What is one important difference between solving equations and solving inequalities?

2.  Interval Notation ◦ [c, d] → all real numbers x such that c < x < d ◦ (c, d) → all real numbers x such that c < x < d ◦ [c, d) → all real numbers x such that c < x < d ◦ (c, d] → all real numbers x such that c < x < d  Brackets “[“ & “]” represent that the endpoints of an inequality are included in the solution  Parenthesis “(“ & “)” represent that the endpoints of an inequality are not included  The above examples only cover compound inequalities. Any ideas how to represent simple inequalities?

3.  For lines going to the right of b ◦ [b, ∞) → all real numbers x such that x > b ◦ (b, ∞) → all real numbers x such that x > b  For lines going to the left of b ◦ (-∞, b] → all real numbers x such that x < b ◦ (-∞, b) → all real numbers x such that x < b  For the entire number line ◦ (-∞, ∞) → represents all real numbers  Why don’t we use brackets around ∞?

4.  Basic Principles for Solving Inequalities 1.Add or subtract the same quantity on both sides of the inequality 2.Multiply or divide both sides of the inequality by the same positive quantity 3.Multiply or divide both sides of the inequality by the same negative quantity, and reverse the direction of the inequality  The third rule is the only change from solving traditional equalities

5.  Ex. 1 – Solving an Extended Linear Inequality ◦ ◦ Written in interval notation, our solution is -1 < x < 6 [-1, 6]

6.  Ex. 2 – Solving an Extended Linear Inequality ◦ In interval notation, our solution is (-3, -1 / 5 )

7.  Assignment ◦ Page 124 ◦ Problems 5-39, odd exercises ◦ Show work ◦ Answers in interval notation

8. Essential Question: What is one important difference between solving equations and solving inequalities?

9.  Solving Other Inequalities ◦ The solutions of an inequality of the form f(x) < g(x) consist of intervals on the x-axis where the graph of f is below the graph of g ◦ To solve inequalities 1)Write the inequality in one of the following forms f(x) > 0f(x) > 0f(x) < 0f(x) < 0 2)Determine the zeros of f, both real and extraneous, exactly if possible, approximately otherwise. 3)Determine the interval, or intervals, on the x-axis where the graph of f is above (in the case of > 0) or below (< 0) the x-axis, using +∞ and the zeros you found in step 2 as endpoints.

10.  Ex 3: Solving an Inequality ◦ Solve x 4 + 10x 3 + 21x 2 > 40x + 80  Get the inequality equal to 0  Subtract 40x and subtract 80 from both sides ◦ x 4 + 10x 3 + 21x 2 – 40x – 80 > 0  We don’t know how to get exact solutions for a 4 th degree polynomial, so we need to get approximate solutions by graphing  The zeros are at -1.53 and 1.89  Because it’s >, we’re looking for where the line is above the x-axis ◦ (-∞,-1.53) and (1.89, ∞)  Note that parenthesis are used because the original problem was > and not >

11.  Ex 4: Solving a Quadratic Inequality ◦ Solve 2x 2 + 3x – 4 < 0  We can find exact solutions with quadratic equations. This one cannot be factored, so the quadratic function must be used.

12.  Ex 4: Continued ◦ We know the zeros are and ◦ They are about -2.35 and 0.85 in decimal form ◦ There are two options used to determine the interval, a) graph or b)Select a point between zeros and test  2x 2 + 3x – 4 < 0  x < -2.35 [let’s use -3]: 2(-3) 2 + 3(-3) – 4 = 5  -2.35 < x < 0.85 [we’ll use 0]: 2(0) 2 + 3(0) – 4 = -4  x > 0.85 [we’ll use 1]: 2(1) 2 + 3(1) – 4 = 1  The first and last tests give us a value > 0, only the test in the middle satisfied the original inequality.  is our interval solution

13.  Ex 5: Solving an Inequality ◦ Solve (x + 5)(x – 2) 6 (x – 8) < 0 ◦ The zeros of the function are -5, 2, and 8 a)Graphing  Graphing confirms that the graph is below x during -5 < x < 8 b)Testing intervals  x < -5 [test -6, result: 3,670,016]  -5 < x < 2 [test 0, result: -2560]  2 < x < 8 [test 3, result: -40]  x > 8 [test 9, result: 1,647,086]  Testing confirms the graphing conclusion  Interval solution is [-5, 8]

14.  Bonus example, solving fractional inequalities ◦ Problem #58 ◦ The equation has one real solution, 2x-1 = 0, x=½ ◦ The equation has one extraneous solution  5x + 3 = 0, x = -3 / 5 ◦ Graphing  Graph is positive between -∞ and -3 / 5 and then ½ to ∞ ◦ Testing intervals  x < -3 / 5 [test: -1, result: 3 / 2 ]  -3 / 5 < x < ½ [test: 0, result: -1 / 3 ]  x > ½ [test: 1, result: 1 / 8 ] ◦ Both confirm an interval solution of [-∞, -3 / 5 ) and [½, ∞]

15.  Assignment ◦ Page 124-125 ◦ Problems 41-69, odd exercises