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Calculus 1, Week 2 Notes

by: Monica Chang

Calculus 1, Week 2 Notes 21-111

Monica Chang

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Finishing up basics: - long division - intervals and inequalities - unions and intersections - solving inequalities - absolute value - trigonometry
Calculus 1
Deborah Brandon
Class Notes
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This 5 page Class Notes was uploaded by Monica Chang on Thursday September 15, 2016. The Class Notes belongs to 21-111 at Carnegie Mellon University taught by Deborah Brandon in Fall 2016. Since its upload, it has received 12 views. For similar materials see Calculus 1 in Mathematical Sciences at Carnegie Mellon University.


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Date Created: 09/15/16
Week 2 Long Division:  Example 1: 3 2 4 x −4x +x−1Remainder6 x+1 √x +0x −3x +0 x+5 4 3 −(4 x +4 x ) −4x −3x 2 3 2 −(−4x −4x ) x +0 x 2 −(x +x) −x+5 −(−x−1) 6 4 2 So 4x −3x +5 =4x −4x +x−1+ 6 x+1 x+1  Example 2: Factor x −2 x −5x+6 o Use trial and error to find a root x so that x −2 x −5x+6=0 o x=3 is a root, so x-3 is a factor x −2x −5x+6 2 o use long division to find that =x +x−2 x−3 o x −2 x −5x+6= x−( )x +x−2 ) ¿(x−3 )x−1 (x+2) Intervals & Inequalities:  Simple examples: Interval Notation Number Line Inequality (a, b] a < x ≤ b (-∞, b) x ≤ b  Inequalities in graphing example: Unions & Intersections:  Given two sets A, B o Intersection A ∩ B = all objects in both A and B o Union A ∪ B = all objects in A or B  Example: I f A[ −∞,∧B= (0,∞) A∩B= [,1) A∪B=(−∞,∞) Solving Inequalities:  Ex1. 2−5x>7 −5x>5 x<−1 Note: you have to flip the sign x∈(−∞,−1)  Ex2. 4<3−x≤6 4<3−x AND 3−x≤6 1<−x AND −x≤3 x<−1 x ≥−3 AND x∈ −∞,−1 ) AND ¿ x∈ −∞,−1 ) ∩¿ x∈¿  Ex3. x −1≥0 (x−1 )x+1)≥0 (x−1≥0∧x+1≥0 )∨(x−1≤0∧x+1≤0 ) (x≥1∧x≥−1 ∨(x≤1∧x≤−1) −∞ ,−1] x∈ [, ∞∨¿ x∈ −∞ ,−1 ]¿ Absolute value: zif z≥0  |z= −zif z≤0 ¿thecasetha||≤a (this is an “AND” situation) o if a>0then−a≤z≤a  Logically: if we want the distance from z to 0 to be less than or equal to a when a is greater than 0, z has to be between –a and a, inclusive  Mathematical derivation (only doing this for this case as an example): (z>0∧z≤a )∨(z<0∧z≥−a) z∈ 0,a]∨¿ z∈[−a,a] o if a<0then?nosolution  Logically: absolute value cannot be negative o if a=0thenonlysolutionisz=0  Logically: if the distance between z and 0 is less than or equal to a and a is 0, z can only be 0 because absolute value cannot be negative ¿thecasetha||≥a (this is an “OR” situation) o if a>0the(−∞ ,−z]∪ z, )  Logically: if we want the distance from z to 0 to be greater than or equal to a when a is greater than 0, z has to include everything except the numbers between –a and a. o if a<0then−∞≤z≤∞  Logically: if we want the distance from z to 0 to be greater than or equal to a when a is less than 0, z can be anything because distance is always a positive value anyway o if a=0then−∞≤ z≤∞  Logically: if we want the distance from z to 0 to be greater than or equal to a when a is 0, z can be anything because distance is always a positive value anyway  Ex1. |2x+2|≤8 −8≤2 x+2<8 −10≤2x≤6 −5≤ x≤3 −¿ x∈¿ 5, 3]  Ex2. |3x<−8 ?nosolution(becauseabsolutevaluecannot beanegativenumber)  Ex3. |4 x−|>5 4 x−3>5∨4x−3<−5 4 x>8∨4 x<−2 −1 x>2∨x< 2 x∈(−∞ ,− )∪ (,∞ ) 2  Ex4. |2x−4|>−8 2 x−4>−8∨2 x−4<4 2x>−4∨2x<8 x>−2∨x<4 x∈(−∞,∞) Trigonometry:  Sinθ= opp hyp adj cosθ= hyp tanθ= opp = Sinθ adj Cosθ  Pythagorean Theorem: 2 2 2 a +2 =c 2 Sin θ+Cos θ=1  There are 2π radians in a circle (360°)


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