Calculus And Analytic Geometry 1 - Final Exam Summary PDF

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University of Wisconsin Madison MATH 221 Fall 2017

Exam Guide

Review

Glimpses: ● For the general limit-finding problems: The very first thing you should do is to plug in the number. If you get the answer, then there is no need to apply any other techniques.

a If ou get a ue oe ∞, that gies ou .  If ou get a ozeo ue oe , that gies ∞.

(c) If ou get ∞+∞ o a ue ties ∞, that is just ∞.

If afte pluggig i, ou get  , ∞ , ∞ − ∞, the ou a eed to use soe tehiues below:

● Cancellation tricks: factoring, conjugate, etc The fist tik is ofte used he ou see . You at to ael soe tes so that ou a appl the liit.  Here are some formulas to help with factoring: a2 −2 =a−a+ a3 − 3 = a − a2 + a + 2) a3 +3 =a+a2 −a+2).

● Dividing by power of x The second trick is often used when you see lim→∞, ad he the futio is a fatio. You ould at to diide the topueato ad otto deoiato  the highest power of x. Two things to keep in mind:

1. You hae to diide the top ad otto  eatl the sae thig. Fo eaple: dot divide the top by x, and the bottom by x2. 2. If there is square root, or square, you have to take them into consideration when finding the highest power. Fo eaple, if the futio is + / √ 2 + , the the highest poe is  = , ad ou want to divide the top and bottom by x.

● Comparison/sandwich theorem The third trick is often used when you have something like sin, cos in your function, or something that can be bounded. Basically, you want to ”sandwich” your function between 2 things: · · · < f (x) < · · · and hopefully the two ends have same limit. Some useful bounds: . −≤si≤ . −≤os≤

● Continuity The condition for a function to be continuous at a point a is . li →a− eists . li →a+ eists 3. f(a) exists All the aoe ae the sae li →a− = li→a+ = fa.

● Asymptotes Keep i id the  tpes of asptotes ee leaed 1. If limx→∞ f = L = ∞ eithe +∞ o −∞, the e sa that f has hoizotal asymptotes at y = L . If li→a f = ∞, e sa that f has etial asptotes at  = a. . If li→∞ f /  = L, e a do poloial diisio to fid the slat asptotes.

Derivative •

The derivative of any constant function is zero Let k(x)= c be a constant function. Then we have k′x = limh→0  kx+h − kx  / h = limh→0 c−c / h = limh→0 0 = 0.

● The derivative of x n for n = 1,2,3,.... fa = li→a f−fa / -a = lim xn −an / -a Simplify the fraction (xn − an /  − a. Fo  = 2, we have x2 − a2 / -a = +a. Hence, d/ d n) = nxn-1.

●

The Differentiation Rules (source: Math Note Packet) We could go on and compute more derivatives from the definition. But each time, we would have to compute a new limit, and hope that there is some trick that allows us to find that limit. This is fortunately not necessary. It turns out that if we know a few basic deiaties suh as d/d = − the e a fid deiaties of aitail complicated functions by breaking them into smaller pieces. In this section we look at rules telling us how to differentiate any function written as either the sum, difference, product or quotient of two other functions.

•

Proof of Sum Rule Suppose that h = f + g fo all  hee f ad g ae diffeetiale. The ha = lim x–> a h(x) - ha / -a = lim x–> a ( f + g - fa+ ga / -a = lim x–> a((f(x) - fa / -a + li x–> a ((g(x) - ga / x-a) = fa + ga

● Proof of produce rule Let h(x) = f(x)g(x). To nd the derivative we must express the change of h in terms of the changes of f and g: h − ha = fg − fafa = f g −f ga + f ga −f aga = f  f  − ga + f  − ga a No diide   − a ad let  → a: li →a h − ha /  - a = li →a fg − ga  /  - a + f − faga /  - a = fag′a + f′aga

● Proof of Quotient rule We a eak the poof ito to pats. Fist e do the speial ase hee h = /g, and then we use the Product Rule to differentiate  = f / g = f· /g . So let h = /g. We a epess the hage i h i terms of the change in g: h−ha =  g−ga  / gga Diidig   − a, e get h − ha / -a = g − ga / g ga .  - a) Using product rule and limits, limx–>a g(x) = g(a). limx–>a h−ha /  - a = g′a / ga2 Theefoe  = fg - fg / g2 .

● REMEMBER THESE: 2 sinx cosx = sin 2x Cos2 = os  +  /  Sin2 x = (cos2x -  / 

● PROBLEM 1: Let f(x) = cos2x - cos x2 Fid : f ad {f} : f = - os si +  si2 = - si + si2 {f} = - os + si2 + 2cosx2

● PROBLEM 2: Let f(x) = x5 Fid f (x) and f ‘(6)(x) : f  = **** =  f ‘(6) =  = 

From the above problem we can establish a general formula: Let f(x) = xn Then f ‘(n)(x) = n*(n-1)*(n-2).....2*1 And f ‘(n+1) (x) = 0

● PROBLEM 3: Let f(x) = x5 +  +  x3 - 2 Fid f ‘(5)(x) and f ‘(6)(x): f ‘(5)(x) = 5! F ‘(6) = 

● PROBLEM 4: Fid (sin x) 7): = ((sin x) (*)) = si  = cos x

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Formulas: o f) = lim h––> f(x + h - f(x / h o f +- g = f +- g o fg = fg + fg o f/g = fg - fg / g2 o fg = fg * g o Cos2 x = (cos2x -  /  o Sin2 x = (cos2x - /  o F[n] ‘(x) = n! o F[+]  =  o f +- g)(n) = f (n)  +- g(n) (x)

● Dealing with equations of the form F1(x, y) = F2(x, y). If the ipliit defiitio of the futio is ot of the fo F ,  =  ut athe of the fo F,  = F,  the ou oe all tes to the left had side, ad poeed to deal with a function y = f (x) that satisfies y2 +  =  =y2 +  - xy =  ad set F, = 2 +  - xy.

● Derivatives of Inverse Trigonometric Functions. Recall that by definition the function y = arc sinx satisfies x = siny , and - π/ ≤  ≤ π/ . = d  / d = d si / d = = os d/d = d/d =  / os  .

As we know, sin2+os 2y=1 =⇒ cosy=± sqrt(1−sin2=± st−2).

Hence, = d a si / d =  / st-x2) Siilal, d a ta / d =  / +2.

GOOD LUCK!

Functions and Derivatives Function 1. DERIVATIVE (Rate of change, slope, velocity) 2. INTEGRAL (length, Area, Volume) •

Derivation is approximation

Example: 32 = 9 But √ = .…(Approximate value)

Example: Use the tangent line of y=f(x) at x=x0

•

Key point: Find the slope of the line at x,y o Slope : y-y0 / x-x0 = f(x) -f(x0 ) / x-x0 o Definition : Slope of the tangent line of y=f(x) at x=x0 o lim f(x)-f(x0) / x-x0 = Δx= x-x0

● Average velocity: We all know what average velocity is. Instead of trying to find the velocity exactly at time t, we find a formula for the average velocity during some (short) tie iteal egiig at tie t. We’ll ite ∆t fo the legth of the tie interval. At time t we are s (t iles fo ou stat. A li le late, at tie t + ∆t e ae s (t + ∆t iles from our start. Theefoe, duig the tie iteal fo t to t + ∆t, e hae oed s (t + ∆t − st iles, Miles per hour: s (t+∆t −s t / ∆t and therefore our average velocity in that time interval was shorter we make the time iteal the salle e hoose ∆t the lose this ue should e to the instantaneous velocity at time t.

So we have the following formula (definition, really) for the velocity at time t v (t= li ∆t→ st+∆t−st / ∆t ● Acceleration as the rate at which velocity changes. As you are driving in your car your velocity may change over time. Suppose v (t) is your velocity at time t (measured in miles per hour). You could try to figure out how fast your velocity is changing by measuring it at one moment in time (you get v (t)), then measuring it a little later (you get v (t + ∆t {Aeleratio at tie t}=a= li∆t→ v (t+∆t − t / ∆t.

Limits ● Defiitio of liit st attept.

o If f is some function then li f = L →a is read “the limit of f(x) as x approaches a is L.” It means that if you choose values of x that are close but not equal to a, then f (x) will be close to the value L; moreover, f (x) gets closer and closer to L as x gets closer and closer to a. o The following alternative notation is sometimes used f(→L as →a; (Read “f (x) approaches L as x approaches a” or “f (x) goes to L as x goes to a”.) o Note that in the definition we require x to approach a without ever becoming equal to a. It’s important that x never actually equals a because our main motivation for looking at limits was the definition of the derivative. In Chapter II, equation (9) we defined the derivative of a function as a limit in which some ue ∆ goes to zeo: f′ = li f + ∆ − f. ∆→ ∆

o The quantity whose limit we want to take here is not even defied he ∆ = . Therefore any definition of limit we come up with had been not depend on what happes at ∆ = ; likeise, the liit li→a f  should ot deped o hat f is.

● Basi properties of liits o

��� [ (�) + (�) ] = ��� (�) + ��� (�) : The limit of the sum of two functions is the sum of their limits.

o

��� [ (�) − (�) ] = ��� (�) − ��� (�) : The limit of the difference of two functions is the difference of their limits.

o ��� [ (� ) ∗ (�) ] = ��� (� ) ∗ ��� (�) : The limit of the product of two functions is the product of their limits. o ��� [ (� ) / (� ) ] = ��� (� ) / ��� (�) ; if li g is ot eual to zeo. The limit of the quotient of two functions is the quotient of their limits if the limit in the denominator is not equal to 0. o ��� ��ℎ ���� [ (� ) ] = ��ℎ ���� [ ��� (� ) ]. If n is even, lim f(x) has to be positive. The limit of the nth root of a function is the nth root of the limit of the function, if the nth root of the limit is a real number.

● The foral, authoritatie, defiitio of liit Our attempted definitions of the limit uses undefined and non-mathematical phrases like “closer and closer”. In the end we don’t really know what those statements really mean, although they are suggestive. Fortunately, there is a good definition, one that is unambiguous and can be used to settle any dispute about the question of whether lim � → � (�) equals some number L or not. Here is the definition.

● Defiitio of ���� → � �(�) = �. If f(x) is a function defined for all x in some interval which contains a, except possil at  = a, the e sa that L is the liit of f as  → a, if fo ee  > , e a d a  >  depedig o  suh that fo all x in the domain of f it is true that 0 < |� − � | < � ������� |(� ) − � | < � .

● Wh the asolute alues? The uatit|−| is the distance between the points x and y on the number lie, ad oe a easue ho lose  is to   alulatig |−|. The ieualit | − | <  sas that the distae etee  ad  is less tha , o that  ad  ae lose tha .

What are  ad ? The uatit  is ho lose ou ould like f to e to its liit L; the uatit  is ho lose ou hae to hoose  to a to ahiee this. To poe that li→a f  = L ou ust assue that soeoe has gie ou a uko  > , ad the fid a positie  fo hih { < | − a| <  iplies |f(x − L| < . } holds. The  ou fid ill deped o .

● Prole for pratie 1. ��� �– > ∞ �2 + 1/ 1010 � + 5 = ��� �– > ∞1 + 1/�2 / 1010 /� − 5/�2 = 1/0 = ∞

2. Show that ��� � → 5(2� + 1) = 11 . We have f(x) = 2x+1, a = 5 and L = 11, and the question we must answer is “how close should  e to  if at to e sue that f =  +  diffes less tha  fo L = ?  To figue this out e t to get a idea of ho ig |f − L| is: |(�) − �| = (2� + 1) − 11 = |2� − 10| = 2 · |� − 5| = 2 · |� − �|

So, if | − a| <  the e hae |f − L| < , i.e. If |−a|< ½  then |f(x − L|  e ae gie ou  ill also e positie, ad if | − | <  the e a guaatee | ( +  − | < . at shos that li →  +  = .

Limits at infinity •

Instead of x approach some finite number, one can let x become “larger and larger” and ask what happens to f(x). If there is a number L such that f (x) gets arbitrarily close to L if one chooses x sufficiently large, then we write

��� � → ∞ (�) = �

(“The limit for x going to infinity is L.”) We have an analogous definition for what happens to f(x) as x becomes very large and negative: we write

��� � → −∞ (�) = �

(“The limit for x going to negative infinity is L.”)

Variatios i Liits •

Not all liits ae fo  → a. Thee ae soe othe aiatios.

Left ad right liits •

When “x approach a” x is to be larger or smaller than a, as long as x “is close to a”. If we explicitly want to study the behavior of f (x) as x approaches a through values larger than a, then we write li ↘a f or li →a+ f or li →a+ f or li→a,>a f

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All four notations are commonly used. Similarly, to designate the value which f(x) approaches as x approaches a through values below we write li↗a f or li→a− f or li→a− f or li→a,  oe a fid a  >  such that a< < a+ = ⇒ |f − L|  oe a fid a  >  such that � −  < � < � =⇒ |�(�) − �| <  Holds for all x in the domain of f. The following theorem tells you how to use one-sided limits to decide if a function f (x) has a limit at x = a.

Theore •

The two-sided limit limx–>a f(x) exists if and only if the two one-sided limits exist and have the same value. li ↘a f, ad li ↗a fX

Proles Questio: limx–>∞ (o.5 3 +  /  5 - 5 +5) Ase: {Diide the ueato ad deoiato  5} = .5/ 5 + / 5 /  - 5/4 + 5/ 5 = / =

Questio: limx–>∞ ( x/ +  / / + ) Ase: {Diide the ueato ad deoiato  /} = x -⅙ + / /  + // {Pluggig i ifiit} = ∞/ =∞

Theorems {Sandwich & Substitution} ● The sadih theore Consider that f ≤ g ≤ h

For all the value of x and both the limits

lim x–>a f ad li x–>a h

Exist and are equal to the same value L. Then

lim x–>a g(x)

also exists and is equal to L.

o The theorem is useful when we want to know the limit of g, and when we can sadih it between two functions f and h whose limits are easier to compute. The Sandwich theorem looks like the first theorem of this section, but there is an important difference: in the Sandwich theorem we don’t have to assume that the limit of g exists. The ieualities f ≤ g ≤ h oied ith the iustae that f and h have the same limit are enough to guarantee that the limit of g exists.

● Sustitutio of liits Given two functions f and g, we can consider their composition h(x) = f (g(x)). To compute the limit Li →a f g

We let u = g(x), so that we want to know

Li →a f u hee u = g.

If we can find the limits

L=li →a g ad li u→L f(u)=M

Then it seems reasonable that as x approaches a, u = g(x) will approach L, and f (g(x)) approaches M. This is in fact a theorem:

Theore: If li→a g = L, ad if the futio f is otiuous at u = L, the

li→a f g = li u→L fu = fL

Another way to write this is

li→a fg = f li→a g .

What this statement means is that we can freely move continuous functions outside of limits.

● Proles Questios: limx–>5 st -  -  / -5

Ase: = limx–> (sqrt(x - 1) - 2) / (x-5) = {Multipl  Cojugate} = (x-5) / ((x-5) * sqrt(x-1) + 2) = /  {pluggig i }

Questio: Poe that li u–>∞ si u/ u =   sustitutio/sadih theoe.

Ase: = - ≤ si u ≤  = - |si u /u | ≤ si u/u ≤ |si u| =0 Questio: lim x–> ∞ (x1/2 - (x2 + x)1/2) / 1 Ase: = ojugate ad sole  pluggig

Questio: limx–>∞ (x - (x2+ x)1/2)

Ase: = Multipl  Cojugate = Divide the numerator and denominator by x = -1/1+1 = -½

Proof and Problems ( lim x––> 0 (Sin x / x = 1)) ● PROOF OF lix–> si  /  o To compute specific formulas for the derivatives of sin x and cos x, we need to udestad the ehaio of si /  ea  = . The defiitio of si θ as the coordinate of a point on the unit circle to poe that liθ→ siθ / θ = . The aiale e’e iteested i is a agle, ot a hoizotal positio, so e ill disuss si θ / θ athe tha si / .

o Coside the adius of the ile , θ is the legth of the highlighted a. This is tue he the agle θ is desied i adias. Also, sie the adius of the ile is , si θ = |opposite| euals the |hpoteuse| legth of the edge idiated (the hypoteuse has legth . I othe ods, si θ/θ is the atio of edge legth to a legth. Whe θ = π/ ad, si θ =  ad si θ / θ = /π ∼= 2/3. When θ = π/ ad, si θ) = √2/2 and sin (θ) / θ = 2 = 9/10. What will happen to the value √2/π ∼ of sin (θ) / θ as the value of θ gets closer and closer to 0 adias? It ill shik as θ shiks, the legth si θ of the seget gets lose ad lose to the legth θ of the ued a. We olude that as θ → ,

si θ / θ ––>  In other words limx––>  sin x / x = 1.

● Proles Questio: lim x––>  x / tan x

Ase: = x / (sin x /cos x) = (x / sinx) * cos x = cos x = 1 (cos 0 = 1)

Questio: lim x––>  sin 5x / x

Ase: = (sin 5x / 5x) / (x / 5x) = 5 x / x =5

Some basic Trigonometric Formulas to solve with ● ● ● ● ●

Sin2 ϴ + cos 2 ϴ = 1 tan2 ϴ + 1 = sec 2 ϴ 1 + cot2 ϴ = cosec 2 ϴ sin (2ϴ) = 2 sinϴ cosϴ Cos (2ϴ) = 1 − 2 sin2 ϴ

. Li x––> (Si  / Si 4x = si / {diidig the ueato ad deoiato  } =  /  {diidig ueato ad deoiato  } =¾

. Li x––> (-os  / 2 = 1-cos2x / x2(1+cos x) =Sin2x / x2(1+cos x) = 1/ 1+ cos x = 1/ (1 + 1) =1/2

Introduction to Continuity ● Defiitio of Cotiuit A function g is continuous at a if lim x → a g(x) = g(a) A function is continuous if it is continuous at every a in its domain. Note that when we say that “a function is continuous on some interval” it has to be defined on that interval. For example, the function f(x) = 1/x2 is continuous on the iteal  <  <  ut is ot otiuous o the iteal − <  < 1 because it isn’t defined at x = 0. Continuous functions are “nice”, and most functions we will want to talk about will be continuous. For example, all polynomials are continuous functions. The ideas in the proof are the same as in the following example.

Eaple A polynomial that is continuous. Let us show that P(x) = x2 + 3x is continuous at x = 2. To sho that e hae to poe that li →2 P(x) = P (2) Solutio We can do this two ways: using the defiitio ith  ad  the had a. It is uh easier to the limit properties (P1)...(P6) from §6: for example, li → 2 +  = li → 2 + li →  = li → ) · (li → ) + (li → ) · (li →  =2·2+3·2 And this is what we wanted.

Asymptotes •

Asymptotes can be vaguely defined by saying that a line is an asymptote of the graph of a function y = f(x) if the graph “approaches the line as x approaches infinity.” For a more precise definition we distinguish between three cases, depending on the line.

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Vertial Asptotes If y = f (x) is a function for which either li↘a f=∞, ...