Essential Question: What is one important difference between solving equations and

solving inequalities?

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?

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 ∞?

Basic Principles for Solving Inequalities1. Add or subtract the same quantity on both sides

of the inequality2. Multiply or divide both sides of the inequality by

the same positive quantity3. 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

Ex. 1 – Solving an Extended Linear Inequality

◦ ◦ Written in interval notation, our solution is

2 3 5 2 11x x 2 3 5 and 3 5 2 11x x x

3 3 2 5 52x xx 1 and 6x x

-1 < x < 6[-1, 6]

Ex. 2 – Solving an Extended Linear Inequality

◦ In interval notation, our solution is

4 3 5 18x 5

1 5

3 3

1

3

5

x

x

1 13 is the same a

5

s

5

5

5

35

x x

(-3, -1/5)

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

Essential Question: What is one important difference between solving equations and

solving inequalities?

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 inequalities1) Write the inequality in one of the following forms• f(x) > 0 f(x) > 0 f(x) < 0

f(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.

Ex 3: Solving an Inequality◦ Solve x4 + 10x3 + 21x2 > 40x + 80

Get the inequality equal to 0 Subtract 40x and subtract 80 from both sides

◦ x4 + 10x3 + 21x2 – 40x – 80 > 0 We don’t know how to get exact solutions for a 4th

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 >

Ex 4: Solving a Quadratic Inequality◦ Solve 2x2 + 3x – 4 < 0

We can find exact solutions with quadratic equations. This one cannot be factored, so the quadratic function must be used.

22 (3) (3) 4(2)( 4)4

2 2(2)

3 9 32 3 41

4 4

b b ac

a

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 orb) Select a point between zeros and test

2x2 + 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

3 41

4

3 41

4

3 41 3 41,

4 4

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]

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, ∞]

2 10

5 3

x

x

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