Calculus 1 Notes PDF
Document Details
Uploaded by Deleted User
Amman Arab University
Mohammad Al-Smadi
Tags
Summary
These notes cover fundamental concepts from Calculus 1, including sets, intervals, and properties of inequalities. The examples provided demonstrate applications of these concepts to solve problems.
Full Transcript
# Calculus 1 ## Mohammad Al-Smadi ### College of Arts and Sciences ### Department of Mathematics ## Chapter (0): Revision ### Sets A set is a collection of elements. #### Examples 1. The set of natural numbers: IN = {1, 2, 3, ...} 2. The set of integers: Z = {-2, -1, 0, 1, 2, 3, ...} 3. The set of...
# Calculus 1 ## Mohammad Al-Smadi ### College of Arts and Sciences ### Department of Mathematics ## Chapter (0): Revision ### Sets A set is a collection of elements. #### Examples 1. The set of natural numbers: IN = {1, 2, 3, ...} 2. The set of integers: Z = {-2, -1, 0, 1, 2, 3, ...} 3. The set of rational numbers: Q = { a / b, a, b ∈ Z, b ≠ 0} For example: 2 / 3, -3 / 2, √25 = 5, -1/2, 0, 3 √2 ∉ Q. 4. The set of irrational numbers. I, For example: √2, π, e ≈ 2.7, x = 3.14. 5. The set of real numbers: IR = IN ∪ Z ∪ Q ∪ I Note: IN ⊂ Z ⊂ Q ⊂ IR 6. The set of whole numbers: W = {0, 1, 2, ...} # Intervals * (a, b): a < x < b open * (a, b]: a < x ≤ b half-open * [a, b): a ≤ x < b half-open * [a, b]: a ≤ x ≤ b closed * (-∞, b]: x ≤ b * (-∞, b): x < b * [a, ∞): x ≥ a * (a, ∞): x > a * (-∞, ∞) = IR ### Properties of inequalities Let a, b, c, d ∈ IR * a < b ⇒ a + c < b + c * a < b ⇒ a - c < b - c * a < b ⇒ ac < bc if c > 0 (positive) * a < b ⇒ ac > bc if c < 0 (negative) * a < b ⇒ a / c < b / c if c > 0 * a < b ⇒ a / c > b / c if c < 0 * If a < b ⇒ |a| > |b|, where a & b are both positive or both negative. * If a < b and b < c then a < c. * If a < b and c <d then a + c < b + d ### Examples 1. 7 ≤ 2 - 5x ≤ 9 * Solution: 5 ≤ -5x ≤ 7 -1 ≥ x > -7/5 x ∈ (-7/5, -1] 2. x² - 3x > 10 * Solution: x² - 3x -10 > 0 (x - 5)(x + 2) > 0 (x - 5)(x + 2) = 0 Our zeros are x = 5 and x = -2. (-3 - 5)(-3 + 2) > 0, (0 - 5)(0 + 2) < 0, (6 - 5)(6 + 2) > 0 Solution is x ∈ (-∞, -2) ∪ (5, ∞) = IR - [-2, 5] 3. x² - 9 ≤ 0 * exercise 4. 3 + 7x ≤ 2x - 9 * exercise