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## math notes etc

by: Brenna Graham

6

0

9

# math notes etc 191g

Brenna Graham
NMSU
GPA 3.9

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notes
COURSE
Calculus I
PROF.
Dr. Andres Contreras
TYPE
Class Notes
PAGES
9
WORDS
CONCEPTS
UCSB
KARMA
25 ?

## Popular in Math

This 9 page Class Notes was uploaded by Brenna Graham on Sunday September 25, 2016. The Class Notes belongs to 191g at New Mexico State University taught by Dr. Andres Contreras in Fall 2016. Since its upload, it has received 6 views. For similar materials see Calculus I in Math at New Mexico State University.

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Date Created: 09/25/16
Contents:   Index  of  theorems   and  rules, List  of practice   problems,   and  Exam   review  link.    Trig:    √ 2 √2 Look   atterns:   at adians  (±   2 ,  ± 2 )  At   ?/  r   ?(± 1,0) or (0,± 1)  Etc  Radians   to   degree     #  πx etc  opposite adjacent sinx = hypotenuse   cosx = hypotenuse   tanx = sinx cotx = cosx or  1  secx = 1   cscx = 1   cosx sinx tan cosx sin sin  x  +  cos  x = 1          Chapter       Change   in   position=   velocity   x n   time  i Only   if   velocity   is   constant  change in position Average   velocityLength of time interval   s(t1 − s(t0   Δt = t  t    =  slope of secant line  1− 0 Instantaneous   is   derivative/    ine nt  l sin?x/?x=?  p(x) q(x) q = 0  One  Sided  Limits:  From   th eft    ight   x→c −(x)    ε < 0    ∣f(x) − L < ε    δ > 0    x ε(c − δ,c)    From   ight   eft l lim f(x)    x→c+ x ε(c,c + δ)    ∣x c∣ < δ   − if and  only   f  ε(c δ,cδ)  x   −   lim = lim < L  If x→c − x→c+   Then    imf(x) = L    x→c Infinite  limits  x→cf(x) =+ ∞  f(xif   increases     ithout  bounds,   as   → c    Vertical Asymptote   at x=c if at  oth x→    +  nd    →    ­ t approaches    /+∞    Properties  of   imits:  1. If  x→cf(x)  +  x→cG(x) exist then limx→cx) + g(x)) exists  and   ame   or  the reverse  2. limk f(x) = klimf(x)  assuming  k is a constant  x→c x→c 3. x→cf(x) × g(x) = lix→cx)  × lix→cx)    f(x) x→c f(x) 4. x→c g(x)= lim g(x)  x / 0    x→c 5. h  s  ositive  a. lim[f(x)]  = (limf(x))    h x→c x→c b. lim f√x) = n limf(x)  assuming    im   s positive  x→c √ x→c  →  6. Identities  limk = k  a. k→c   b. limx = c    x→c p p c. limx  = c  assume   c>0   f q  is  ven  x→c Example:  5 x→2x  − 2x + 1 =?  limx  + lim( 2−x  + 1  x→2 x→2 5 (limx)  + ( 2−limx  + 1  x→2 x→2 32 4 + 1 = 29    − I the equation  iscontinuous    ustplug  c in   It  scontinuous   i  a,b)  nd let c (a,b)  AKA:   x→cf(x) = f(c)    TA  NOTES:  2.1  1. Average   ate of change    (x) on (a,b)  Δf(x) = f(b− f(a)  Δx b−a 2. Instantaneous    ateof change  Δf(x) lim Δx   Δx→o Example:  Δx Δy on [2,6),  =4x­9  Solution   Δx = (4(5)+9)−(8+9) = 29−17 = 12 =    Δy 5 − 3 3 Find  nstant  ate of change  at   =2  (x) = 4x + 9      [2,2.01]  [2, 2.005)  [2, 2.001)  Using: Δf(x) = f(b)−f(a)   Using: Δf(x) = f(b− f(a)  Using: Δx b−a Δx b−a Δf(x) f(b)−f(a) f(2.01)−f(2) =  f(2.005)−f(2) =  Δx = b−a   (4(2.01)+9) (4(2)+9) (4(2.005)+9) (4(2)+9) f(2.001)−f(2) −   −   2.001−2 =  0.041 0.0205 (4(2.001)+9− (4(2)+9)  = 0.01 =  4  = 0.005 = 4  0.001 = 0.004 = 4  0.001   Rate  of   hange    s (via squeeze  theorem)         If x→cf(x) = f(c)   thus  limit exists  if equal   at both  sides  If  (x)  is continuous  at I then   continuous  on  line  L   I =[a,b)  If on  (a,b)  and   x→a+ /−x) = f(a)  f(x)if   is continuous   at point  a.    Example  f(x) = x    2 limx    x→1 Continuous    unctions:  1. y  =  x  1/n  2.     =  sinx,  y     osx,   x 3. y = b   if b > 0, b = 1 /   4. y = log  x bf x > 0, b > 0, b = 1  /   If function     and    g are  ontinuous    t   =c   hen   ollowing    ont.   I  f + g or    − g   II.  kf, k  is   constant   III.       IV.  f/g   g(c) = 0 /   V.  (even   if   isn’t  ontinuous)    F(x) = f(g(x))  is continuous  − 1 If    is continuous  with   range   R   f     s    ith domain   R  He   showed   proof   for continuity   but  I’m  lazy  and   itis not  necessary   for  you  to  learn.  It’s in linear  algebra   at khan   academy.   All  of the  above’s  proofs  are  at:  http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx  lim f(x) = L if f(x) → L as x increases     x→+∞ x→ − f(x) = L if f(x) → L as x decreases     lim f(x) means evaluate both sides separately     x→+ /− lim =+ ∞  Ifx  becomes    arger   ithout    ound  x→+∞ Y=L    s he    orizontal  asymptote  lim x =   + ∞    x→∞ lim x −h = 0    x→∞ ∞ if n is even  − ∞ if odd    Ifp+q    s a   ositive  nteger  lim x p/q = { 50 if odd undefined if q is even     x→ − − Polynomials  Example  5−2x 5x 6+19   9x 9/2 −1/2 lim 2x 2−31/2   x→∞ x +x lim 4x−3   x→∞ √25x+4x 3x +2 1. lx→3 x3−1   Solve   by plugging   in  3(27)+2 83 x→3 27 − = 26   √x−3     2. lim x−9   x→9 Solve  by  multiplying   by conjugate  So    lim √x−3 • x+9 = √x + 1   x→9 x−9 x+9 x 3 Plug    n 3(1)+2                  lim √x−3   x→9 √ x− −2   √x−3 √x+5+2 lim •   x→9 √ x− −2 √x+5+2 √ x−3     lim x−9   this was on top but other will be canceled  > − √ x + 5 + 2 = 4    x→9 Bottom= 1 = 1 /    √ −3 4.   lim sin 5x  x→0 sin7x 1 x→0 cos 7x• sin 5x=1  7x 10 5.   lim −2   x→ − √ 16x +1 7x−10  so  − 7 /     4x+1 To   complete   limits  1) Evaluate   if nd    here   f   is   continuous  2) Algebraic   manipulation   to   produce   1  3) Reduction  t o   known imits  4) Squeeze   theorem  IVT  f(a,b) = R and f(a) < 0 < f(b)  then it crosses x = 0 and f(x) =  0 if continuous   So    ontinuous   function   on   E(a,b)   on   alue   between   ) and f(b)    EX:  x2   f(x) = x +1 Takes   on  alue  ,4  Solution:  x2 x +1     2 = (x + 1)(x − x + 1) continuous on ∣all real∣  A • B = 0   1± 3 1 1 ± √ − 4 / √2√ −      a=0    =1  x2 Is f(x) = x +1 [0,1)  1 f(0) = 0  f(1) =2  0 < 0.4 < 0.5  IVT   tells        etween   0      1   HORIZONTAL     SYMPTOTES   n For   f(x) = axm+...   bx +... n      m      0.      n   =   m,   y   =   a/b.          m  o  orizonta symptote  But  or imits   > m, n − m odd =   − ∞   n > m, n −m even =   − ∞      Various  Equations:  1 −osx x→0 x = 0    sinh lim h − 1   h→0 h lim eh1 = 1  h→0 Derivatives:  Definition   = f(x+h) −(x)   h Line  equation:y − f(a) = f′(a)(x − a) if a is a point   f′(c) = 0 if c =  constant number      f′(x) = 1   f′(x ) = nx n−1   (f + g)′ = f′+ g′    (cf)′ = c(f′)   x x f′(e ) = e    Product   rule  (fg)′(x) = f′(x)g(x) + f(x)g′(x)    Quotient   rule  ( ) ′(x) = g(x)f′(x) − f(x)g′(x)   g g(x) 2 Questions  in  extbook    o   ractice:  2.3 10­21  2.5   ­36  2.6   ­26  2.7  17­32    Link  to  eview   Problems:  https://nmsu.instructure.com/courses/1018142/discussion_topics/3230610

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