Math Analysis Semester 2 Study Guide Pdf Version

Math Analysis Semester 2 Study Guide Roots/ Polynomials: If f(x) is a polynomial of degree n where n>0, then f has at least one zero in the complex # system. Know Rational Root Theorem: (factors of last co-efficient/factors of 1st co-efficient) Know Polynomial Division: Ex.

𝑓 𝑥 = 𝑥 3 + 11𝑥 2 + 39𝑥 + 29

Use Rational Root Theorem… to get: ±29, ±1 as factors -1 1 11 39 29 -1 -10 -29 1 10 29 0 (𝑥 2 + 10𝑥 + 29)(𝑥 + 1) Then use the quadratic formula

Rational Functions and Asymptotes: Vertical Asymptotes: where denominator = 0 (only after factoring and canceling) Holes (deleted pts): are what cancels out Horizontal Asymptotes: the # you approach as |x|

∞ (only look @ the dominant term then cancel)

Steps: 1.factor numerator + denominator 2. Write restrictions (what makes denominator = 0) 3. Cancel (the holes) 4. Denominator = 0 after canceling are the vertical asymptotes. Ex. 𝑓 𝑥 =

𝑥 2 +2𝑥 2𝑥 2 −𝑥

=

𝑥 𝑥+2 𝑥 2𝑥 −1

1

𝑥 ≠ 0, 2 𝑓 𝑥 =

𝐻𝑜𝑟𝑖𝑥𝑜𝑛𝑡𝑎𝑙 𝐴𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒: 𝑦 =

𝑥 +2 2𝑥 −1

1

ℎ𝑜𝑙𝑒: 𝑥 = 0 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐴𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒: 𝑥 = 2

1 2

***BE ABLE TO GRAPH BY HAND AND STATE THE DOMAIN***

© 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan

Slant Asymptotes: only occur when power in numerator is ONE larger than power in denominator To Find: use Long Division 𝑓 𝑥 =

𝑥 3 +𝑥 2 −2𝑥 −2 𝑥 2+ 3

Y= 𝑥 + 1 +

−5𝑋−4 𝑥 2 +3

𝑥 2 + 3 𝑥 3 + 𝑥 2 − 2𝑥 − 1 𝑥3 3𝑥 2 𝑥 − 5𝑥 − 1 𝑥2 3 −5𝑥 − 4 Slant:

Sequences A sequence is a function whose domain is the set of positive integers Ex. An=3n-2 sequence is 1,4,7,10… Factorials review: n! = 1·2·3...n 0!=1 5! = 5 × 4 = 20 3! Summation notation: 𝑛

𝑎𝑖 = 𝑎1 + a2 … + 𝑎𝑛 𝑖=1

Example: 7

2𝑖 − 1 = 5 + 7 + 9 + 11 + 13 = 45 𝑖=3

Arithmetic Sequences A1,A2,A3…,An A2 – A1 =d A3 – A2 =d you get the idea d is the common difference 1, 4, 7, 10 d=3 an=a1+(n-1)d Sum of finite arithmetic sequence: 𝑆𝑛 =

𝑛 2

𝑎 1 +𝑎 𝑛

(𝑎1 + 𝑎𝑛 ) = 𝑛(

2

)

Geometric Sequences: A2/ A1 = A3/ A2=r An=a1rn-1 Sum of a Finite Geometric Sequence 𝑆𝑛 =

1 − 𝑟𝑛 𝑎1 − 𝑎𝑟 𝑛 = 1−𝑟 1−𝑟

© 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan

Sum of an infinite geometric sequence 𝑆∞ =

𝑎1 1−𝑟

Where |r|<1

Study Guide of Examples for Sequences and Functions: Linear: lim 3x+1 = 3(2)+1 = 7 x2 7 − 𝜖 < 3𝑥 + 1 < 7 + 𝜖 6 − 𝜖 < 3𝑥 < 6 + 𝜖 𝜖 𝜖 2− < 𝑥 < 2+ 3 3 𝜖 𝛿=

Proof: Choose  = E/3 since all steps are reversible whenever x E <2-, 2+> and x  2 then 3x+1 E <7-E, 7+E> and the limit is 7

3

Quadratics: Lim 𝑥 2 +5 = (2)2 +5 = 9 x2

Proof:

|( x2 +5)-(9)|<E | x2 – 4|<E |x+2| |x-2|<E let ≤ 1 |x-2| <1 -1<x-2<1 3<x+2<5 |x+2|<5 |x+2| |x-2|<E |x-2|(5)<E |x-2|<E/5 <E/5 Rationals:

Let ≤1 then |x-2|<1 and |x+2|<5. If we also let |x-2|<E/5 then the product:

Lim (x+4)/(x-2) = (3+4)/(3-2) = 7

Let ≤ ½ then |x-3|<1/2 and |1/(x-2)|<2. If we also let |x3|<E/12 then the product:

x3 |( x+4)/(x-2) – 7|<E |[x+4-7(x-2)]/(x-2)|<E |(x+4-7x+14)/(x-2)|<E |(18-6x)/(x-2)|<E |-6| |x-3| |1/(x-2)| <E |x-3| |1/(x-2)|<E/6 let ≤ ½

|x+2| |x-2|<E |x2 – 4|<E |( x2 +5)-(9)|<E so choose =1 or E/5 whichever is smaller than whenever x E <2-, 2+> and x2 then x2+5 E <9-E, 9+E> and the limit is 9

Proof:

|x-3| |1/(x-2)|<E/6 |-6| |x-3| |1/(x-2)| <E |(18-6x)/(x-2)|<E |(x+4-7x+14)/(x-2)|<E |[x+4-7(x-2)]/(x-2)|<E |( x+4)/(x-2) – 7|<E

so choose =1/2 or E/12 whichever is smaller then whenever x E <3-, 3+> and x3 then (x+4)/(x-2) E <7-E, 7+E> and © 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan the limit is 7

|x-3|<1/2 -1/2 < x-3 < 1/2 ½ < x-2 < 3/2 2 > 1/(x-2) > 2/3 |1/(x-2)|<2 |x-3| |1/(x-2)|<E/6 |x-3| (2) <E/6 |x-3|<E/12 X∞ Lim (x+1)/(x-1) = 1 x∞ Lim (x+1)/(x-1) = 1 x+∞ ***notice the positive sign 1-E< (x+1)/(x-1) < 1+E (x-1)(1-E)<x+1 1-xE-1+E<x+1 -2/E + 1<x true x+1< (1+E)(x-1) x+1<x-1+Ex-E 2/E + E/E < E/E –x 2/E +1 <x Lim (x+1)/(x-1) = 1 x-∞ ***notice the negative sign 1-E< (x+1)/(x-1) < 1+E (1-E)(x-1)>x+1 ** here x is a negative so you flip the sign** x-1-Ex+E>x+1 -1-Ex+E>1 -Ex>2-E x<-2/E +1 x+1>(1+E)(x-1) x+1>x-1+Ex-E 1>1+Ex-E 2+E>Ex x<2/E +1 true since x-∞ so x is a negative Proof: Choose |x0| = |2/E +1| **choose the value farther from zero if there are 2 different answers** then for all |x|>|x0| (x+1)/(x-1) E <1-E, 1+E> and the limit is 1. © 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan

5.3 Properties of Logs y  log a x if x  ay remember e is just the natural log, and can be used like any other number





so log e x  ln x

log a x  change of base

log b x log x ln x   log b a log a ln a

log a(uv)  log a u  log a v

log a 

u  log a u  log a v v

log a un  n log a u



ln(uv)  ln(u)  ln(v)



u ln( )  ln(u)  ln(v) v



ln un  nln u



and the log of 1 is 0



***remember to use these rules to solve logarithmic equations

3.1 Exponential Functions

A  Pe rt

P is the initial investment and r is the interest rate (write % as decimal)

this formula is used for continuous compounding which means that the investment is compounded an infinite amount of times per year (t). A is the ending amount



For n compoundings per year:

© 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan

r A  P(1 ) nt t is the number of years and n is the number of times per year the amount P is n compounded. R is the rate of interest (write 3% as .03)



Annuities: some payment over a length of time PV= Present value

© 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan