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Lecture

# Appendixes A-C - Winter 2013

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School
University of Waterloo
Department
Mathematics
Course
MATH 127
Professor
Andrew Beltaos
Semester
Winter

Description
Math 127 - Winter 2013 Appendix A Notation: ℤ is the set of all integers - To say x is an integer, we write x ⋲ ℤ, ℤ = { 0, ± 1, ± 2, ... ) - The natural numbers are the set of all the integers: N, N = { 1, 2, 3, ... } - The set of all rational numbers is denoted ℚ = { p/q | p ⋲ ℤ, q ⋲ ℤ, and q ≠ 0 } - the set of all real numbers is ℝ If we have 2 sets, A and B, the intersection of A and B is denoted as A∩B - the set of everything which is in A and B. A∩B = { x | x ⋲ A and x ⋲ B } The union of A and B, denoted as A∪B, is the set of everything in A and everything in B. A∪B = { x | x ⋲ A or x ⋲ B } Example: remove duplicates Let A = { 1, 2, 3 } and B = { 2, 4, 6 }, then A∩B = { 2 } and A∪B = {1, 2, 3, 4, 6} Linear Inequalities Example 1: Solve the inequality 2 + x < 5x - 3 and sketch the solution on the real number line. 1. Subtract x from both sides 2 < 4x - 3 2. Add 3 to both sides 2 + 3 < 4x → 5 < 4x 3. Divide both sides by 4 5/4 5/4 } 5/4 Example 2: Solve: x² - 2x - 15 ≤ 0 Note: [ -3, 5 ] = [ -3, ∞) ∩ (-∞, 5 ] (x - 5)(x +3) ≤ 0 Note: the LHS will be zero when: We make a table, breaking up the ▯ x - 5 = 0 or x + 3 = 0 (i.e x = 5 or x = -3) real number line at these points 1 x is in x - 5 x + 3 product (- ∞, - 3) - - + (-3, 5) - + - (5, ∞) + + + We know the LHS is negative when x ⋲ (-3, 5) and is zero when x = -3 or 5, therefore, LHS is ≤ 0 when x ⋲ [-3, 5]. Alternate notation: { x ⋲ ℝ | x ≥ -3 and x ≤ 5 } or { x ⋲ ℝ | -3 ≤ x ≤ 5 } -3 5 Example 3: 2 inequalities that need to be Solve the inequality 2 ≤ -x + 1 < 3 satisﬁed at the same time! 2 ≤ -x + 1 -x + 1 < 3 1 ≤ - x x < 2 - 1 ≥ x x > - 2 Absolute Value ( non-negative) Properties of Absolute Values The absolute value of a real number x, denoted |x|, is deﬁned as the distance 1. |ab| = |a| |b| between x 2. |a/b| =|a|/|b| (b≠0) 3. |a | = |a| |a| = a if a ≥ 0 2 |a| = -a if a < 0 √a = |a| Example 1: (pg A7) Express |3x - 2| without using the absolute value symbol ▯ ▯ 3x - 2 ▯▯ if 3x-2 ≥ 0▯ ▯ 3x - 2▯ if x ≥ 2/3 |3x - 2| = { ▯ ▯ ▯ ▯ = { - (3x - 2)▯ if 3x - 2 < 0▯ ▯ 2 - 3x▯ if x < 2/3 2 Math 127 - Winter 2013 Example 2: Solve |2x - 5| |2x - 5| is equivalent to: ▯ ▯ 2x - 5 = 3▯ or▯ 2x - 5 = -3 ▯ ▯ ▯ ▯ ▯ 2x = 8▯ ▯ 2x = 2 ▯ ▯ ▯ ▯ ▯ x = 4▯ ▯ x = 1 Therefore, x = 4 or x = 1. Example 3: (pg A8) Solve |x - 5| < 2 |x - 5| < 2 is equivalent to: ▯▯ -2 < x - 5 < 2 ▯ ▯ ▯ ▯ ▯ 3 < x < 7▯ ▯ ▯ add 5 to both sides Therefore the solution is on the open interval (3,7).
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