### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# CALCULUS I MATH 181

GPA 3.92

### View Full Document

## 46

## 0

## Popular in Course

## Popular in Mathematics (M)

This 63 page Class Notes was uploaded by Hoyt Beer on Sunday October 25, 2015. The Class Notes belongs to MATH 181 at University of Nevada - Las Vegas taught by Bellomo in Fall. Since its upload, it has received 46 views. For similar materials see /class/228626/math-181-university-of-nevada-las-vegas in Mathematics (M) at University of Nevada - Las Vegas.

## Similar to MATH 181 at

## Popular in Mathematics (M)

## Reviews for CALCULUS I

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/25/15

Chapter 2 Section 6 Page 1 of 1 Section 26 Implicit Differentiation More Limitations So far we have seen only functions defined explicitly in terms of one variable For example fx xcosx2 Some functions are not de ned explicitly and the second variable the fin the above case is hidden inside the relation itself That is f2 2x cos x These are called implicit relations It is sill advantageous to find the derivative of such relations without solving for the variable explicitly The Chain Rule in Disguise What you need to remember is that in the example above f is actually a function and should be treated as such Which means you need to use the chain rule when differentiating it To nd the derivative of an implicit relation 1 Differentiate the entire line as you would any other function keeping the sign in place 2 Be sure to use the chain rule 3 Solve for the derivative f39y39 etc Example Find y39 if x2y cosx 3 Realize that x2 y is a product and you need to use the chain rule with y x2 y39 y2x1 sinx 0 x2 y39 2xy sinx 2xy sinx y Z x Example Final y39 if xzy3 cosy 3 x23y2y39 y32x sinyy39 0 y393x2y2 sin y 2xy3 i 2xy3 y 32 x y s1ny Example Final the derivative ofthefunction x2 2xy y2 x 2 at 12 2x2xy39y 2yy39l 0 y392x 2y 2x 2y l 2x 2y 1 2x 2y At we have y39 w 1 21 22 2 C Bellomo revised 16Aug08 Chapter3 SectionZ Page 1 of5 Inverse Iklations 0 To find an inverse relationyou are looking to 39undo39 the process that was done with the original relation 0 For example the inverse of the function fx L2 is the function gx 1 2 x 7 x x 71 3K i l Notice that 3 1 and g1 3 Also notice in the previous example we have the ordered pair 31 for x and the ordered pair 13 for gx In fact being able to interchange the first and second coordinates of each ordered pair in a relation is another way to recognize an inverse relation M y 39 y s in the equation and if possible solve fory 0 Example Find the equation of the inverse of y x2 7 6x 9 x y 76y 9 x y 73 y J 3 0 Q Is the original problem a function Does it pass the vertical line test Is the inverse a function A Yes the original problem is a polynomial of degree 2 which is a function It passes the vertical line test The inverse though is not a function for eachy there are 2 differentx s Inverse Functiom and OnetoOne o The concepts above hold for functions as well You can find an inverse function by switching the order of all ordered pairs or you can switch thex s andy s in the equation and so ve o The notation for inverse functions is a linle bit different because for a function x whose inverse is also a function the notation for the inverse is fquotx 1 f x 7 0 Note this does NOT mean butfinverse C Bellomo revised 17Aug08 Chapter 3 Section 2 Page 2 of 5 0 Notice in the problem given previously f x x2 7 6x 9 is a lnction But the inverse 43 3 is NOT a function Q How can you tell the inverse is not a function A Lookng at the graph it does not pass the vertical line test We can t use the f 1 notation Recall that for yx to be a function each x has only one y yx is onetoone ifit is l a function and 2 eachy maps back to only one x For onetoone eachx has only one y and each y has only one x It has to pass not only the vertical line test but also a horizontal line test 0 Q Which of the following are functions Which are onetoone R a b c A a is not a function so it cannot be 11 b is a function passes the vertical line test but is not ll since it fails the horizontal line test c passes the vertical and horizontal line test so it s ll The following lnctions are always onetoone Linear square roots The following lnctions are not ever onetoone Quadratic absolute value C Bellomo revised 17Aug08 Chapter 3 Section 2 Page 3 of5 0 Example Graph XTT determine ifit is 11 ifsofind the inverse x What is the root of x W39hen5x73 0 orx35 What is the unde ned value for x Wlien2x l 0 orx It passes the vertical and horizontal line tests It is one to one Find the inverse 5 y 7 3 x 2yl 2xyx5y73 y2x757x73 TX 3 71 y 2x75 f NOTE The domain for x is all values except 712 The range is all values except 52 The domain for f 391x 1s all values except 52 and the range is all values except 7 2 0 Q What is the relationship between the domain and ranges for functions and their inverses A They switch the domain for one is the range for the other and viceversa Since an inverse function undoes the original function if we compose the two ie take fofquotx or fquotofxwewillgetx Example Show x43 f 1x 4x75 are inverses ffquotx 7 4 x Log Functions Recall by x ifand onlyif y 10gb x y is the exponent b is the base andx is the argument 0 Special Log Bases Log base e is natural log written In Log base 10 is common log written log C Bellomo revised 17Aug08 Chapter 3 Section 2 Page 4 of5 Log Rules log I 1 log 1 0 log by p logbUWV log Mlogb N 10gb M plogb M logbMN log M logb N blag 17 6b8 Example Express loga 6127 in terms of sums anal differences a la b8 loga azbs logaa4b312 l lo a lb3 2 ga loga a4 loga b3 1 E4loga a3loga b Example simplify Zlogs x log5 y 3log5 2 Since they are all the same base we can use the rules backwards to combine terms 210g5 x logs y 3log5 z logs x2 log5 y 1 log5 z393 2 C 10 5 yZ3 Solving Exponential Eguations Equations with variables in the exponent are called exponential equations If the base is the same on both sides of the equation you can equate the exponents Example Solve 3 i 27 3x24r 373 x2 4x 3 0 x 3 1 More work comes when the base is not the same You will then have to solve the equation using logs C Bellomo revised 17Aug08 Chapter 3 Section 2 Page 5 of 5 0 Example Solve 2 40 ln2x ln40 In 40 ln2 x 0 Example Solve ex 6equot 1 e e 6 0 e 3e 2 0 So either e 3 3x ln3 or e 2 never So x In 3 Solving Log Equations 0 Often it is useful to change to exponential in form 0 Example Solve 10g2 x 3 2393 x x 8 0 Example Solve 10g5 8 7x 3 53 8 7x 7x 1 17 117 Z T 0 Example Solve 10g x 10gx3 1 10 1 gx3 101i x3 01x3 x 09x03 1 x 3 C Bellomo revised 17Aug08 Chapter 2 Section 5 Page 1 of 3 Section 25 The Chain Rule Our Limitations 2xl x2 3 0 So far we have only looked at simple functions products and quotients ie f x 0 Q For our skill set as it stands how would we nd the derivative of f x 2x l2 A We would have to use the product rule 2xl2xl or multiply it out o This doesn t seem so bad but what if we wanted to differentiate f x 2x l12 We would prefer another quicker way In a similar way we don t have the tools to differentiate f x l4xl or f x sinx2 e What is so special about these functions above They are composite functions meaning you have one function inside another function That is f x V4xl can be thought of as the function 4x 1 inside the function J The Chain Rule 0 The chain rule is what you use when differentiating composite functions This is why we memorized all of our previous rules as text instead of symbols Let s review them the derivative For example the derivative of a function to a power is the power times the function to the one less power times the derivative of that function Example Find the derivative of fx 2x 112 First you must realize that you have overall a function 2xl to a power 12 f39x122x l 12xl39 122x l11 2 242x l11 C Bellomo revised 16Aug08 Chapter 2 Section 5 Page 2 of 3 Example Find the derivative of f x l4x1 I 1 12 f x dx4xl 4x1i 14x139 71 l4x12 4 2 L 4xl As another example the derivative of the sine of a function is the cosine of that function times the derivative of that function Example Find the derivative of f x sinx2 e Again realize that overall you have the sine of a function f39x cosx2 e x2 e 39 cosx2 e 2x e 2x e cosx2 e Practice Makes Perfect Realize that you have to practice this to get it down It may not seem natural to you at rst The other thing that is essential for success is that you must be able to understand what type of function you have overall Let s work on this f x is rst a quotient Then you have a function to a power on the bottom x gx cos2xsinx2 2x is rst a product then you have both sine and cosine functions with functions inside each hx x2 1 tane is rst a sum Then you have tangent with a function inside kx xxlx l is rst a function to a power Then you have another function to a power Don t let the Jl confuse you it is just a constant C Bellomo revised 16Aug08 Chapter 2 Section 5 Page 3 of 3 0 Let s differentiate these 4x3 2x 1 xlx3 2x 12 3x f x 2 x3 2x gx cos2x cosx2 2x 2x 2 sinx2 2x sin2x 2 h39x 2x 0 sec2 ex e kx xx 1 2 2 k39x xxl12 12xxl1239 xx1 1x1 1 0 Note I did not clean up any of these problems so you could clearly see how the chain rule works Derivative of General Exgonential o In order to differentiate f x ax we need to know three things 1 air emf 2 gang 2 ie ex dx 3 The chain rule 0 The derivative of a general exponential is d d air gang dx dx e1quot ln ax elna x In a aXlna C Bellomo revised 16Aug08 Chapter 1 Section 6 Page 1 of 3 Section 16 7 Limits at In nig Recall 17x l1x You have already studied rational functions that is f x You should already be able to identify the domain of this function in other words the values of x that areare not allowed for inputs of f Q How do we nd any unde ned points in the domain A Set the denominator equal to zero and solve for x These points if they are not removable discontinuities are called in nite discontinuities and the limit of the function there would tend to plus or minus in nity These types of in nite discontinuities are also called vertical asymptotes l x Example For fx What is the limit as x tends to 2from the right From the left First note that the domain is restricted when x7 2 0 2 x 2 2 7 Also note that the top does not go to 0 at x 2 x 2 is an in nite discontinuity lim f x oo xgt2 1n fx foo Horizontal A mgtotes Think of an asymptote in general being an invisible line that a function tends to if you were to keep drawing You never quite get there For example in the above function the graph as x tends to 2 gets closer and closer to the vertical line at x 2 but never quite reaches it And it will not cross over either The above example function has another asymptote that is horizontal If you were to let x get larger and larger x a too the function would get closer and closer to an invisible horizontal line Q What value does the function tend to as x a too What is the equation of this line A It tends to 0 or the equation y 0 C Bellomo revised 16Aug08 Chapter 1 Section 6 Page 2 of 3 Finding Horizontal Asym grates As we extend the function to the left and right as far as we want we are essentially taking the limit as x tends to plus and minus in nity To nd the horizontal asymptote if it exists take the limit as x tends to plus and minus in nity E 100 1 Find the leading term of the numerator px and denominator qx 2 Write as a quotient and simplify 3 If the reduction is 7 a constant then this is the value of your horizontal asymptote HOW TO nd horizontal asymptotes for rational functions f x positive value 7 constant then the horizontal asymptote is zero x positive value 7 constant x then there is no horizontal asymptote It will tend to plus or minus in nity plug in to see which 2 Example Find lim3x2 x5 Hoe x The leading term of the top is 3x The leading term of the bottom is x 2 3x 2 So we have 3 x This problem has a horizontal asymptote at y 3 2 lim3x 2x53 Hoe x 4 3 2 5 Note that llme x3 Xgt7oo x 4 3 2x35 0 Example Find 11m 4 H ltgt2x x 4 The leading term of the top is x3 The leading term of the bottom is x4 3 x 1 So we have 4 x x This problem has a horizontal asymptote at y 0 3x2x35 111117 Hwa x 4 3x2x35 Note that lim 0 2x2x4 4 C Bellomo revised 16Aug08 Chapter 1 Section 6 Page 3 of 3 2 0 Example Find lim Hoe x The leading term of the top is x3 The leading term of the bottom is 2x2 3 So we have x2 lx 2x 2 This problem has no horizontal asymptote 3x2 x3 5 l llmz llm x Hoe 2x 4 Hoe 2 00 2 0 Example Find lim W Him x The leading term of the top is x2 xx x4 The leading term of the bottom is 7x 4 So we have x x3 x There is no horizontal asymptote 2 hm x x lx 2 H40 3 x 335 oo C Bellomo revised 16Aug08 Chapter3 SectionZ Page 1 of5 Inverse Iklations o Tofindan quot undu quot with the original relation 0 For example the inverse of the function fx L2 is the function gx 1 2 x 7 x x 71 3K l H Notice that 3 1 and g1 3 0 Also notice in the previous example we have the ordered pair 31 for x and the ordered pair 13 for gx In fact being able to interchange the first and second coordinates of each ordered pair in a relation is another way to recognize an o Jul 39 39 y s in the equation and if possible solve fory 0 Example Find the equation of the inverse of y x2 7 6x 9 0 Q Is the original problem a function Does it pass the vertical line test Is the inverse a function A Inverse Functiom and OnetoOne o The concepts above hold for functions as well You can find an inverse function by switching the order of all ordered pairs or you can switch thex s andy s in the equation and so 2 n o The notation for inverse functions is a linle bit different because for a function x whose inverse is also a function the notation for the inverse is fquotx 1 fx39 0 Note this does NOT mean butfinverse C Bellomo revised 17Aug08 Chapter 3 Section 2 Page 2 of 5 0 Notice in the problem given previously f x x2 7 6x 9 is a lnction But the inverse 43 3 is NOT a function Q How can you tell the inverse is not a function A Recall that for yx to be a function each x has only one y yx is if it is l a function and 2 eachy maps back to only one x For onetoone eachx has only one y and each y has only one x It has to pass not only the vertical line test but also a 0 Q Which of the following are functions Which are onetoone A The following lnctions are always onetoone Linear square roots The following lnctions are not ever onetoone Quadratic absolute value C Bellomo revised 17Aug08 Chapter 3 Section 2 Page 3 of 5 5x 3 2x1 0 Example Graph fx determine ifit is 1 1 ifsofmal the inverse 0 Q What is the relationship between the domain and ranges for functions and their inverses 0 Since an inverse function undoes the original function if we compose the two ie take f of 1x or f 1 0 fx we will get x x Example Show f x 5 4 f 1x 4x 5 are inverses Log Functions 0 Recall by x ifand only if y 10gb x y is the exponent b is the base and x is the argument Special Log Bases Log base e is written In Log base 10 is written log C Bellomo revised 17Aug08 Chapter 3 SectionZ Page 4 of5 0 Log Rules log I 1 log 1 0 log by p logbUWV log Mlogb N 10gb M plogb M logbMN log M logb N blag P 6b8 0 Example Express loga 6127 in terms of sums anal differences a 0 Example simplify 210g5 x log5 y 3log5 z Solving Exponential Equations 0 Equations with variables in the exponent are called 0 If the base is the same on both sides of the equation you can equate the exponents 1 E 0 Example Solve 3 o More work comes when the base is not the same You will then have to solve the equation using logs C Bellomo revised 17Aug08 Chapter 3 Section 2 Page 5 of 5 0 Example Solve 2 40 0 Example Solve er 6e I 1 Solving Log Eguatz39ons o Often it is useful to change to exponential in form 0 Example Solve 10g2 x 3 0 Example Solve 10g 8 7x3 0 Example Solve 10gx 10gx3 1 C Bellomo revised 17Aug08 Chapter 2 Section 6 Page 1 of 1 Section 26 Implicit Differentiation More Limitations So far we have seen only functions defined explicitly in terms of one variable For example fx xcosx2 Some functions are not de ned explicitly and the second variable the fin the above case is hidden inside the relation itself That is f2 2x cos x These are called It is sill advantageous to find the derivative of such relations without solving for the variable explicitly The Chain Rule in Disguise What you need to remember is that in the example above f is actually a function and should be treated as such Which means you need to use the chain rule when differentiating it To nd the derivative of an implicit relation 1 Differentiate the entire line as you would any other function keeping the sign in place 2 Be sure to use the chain rule 3 Solve for the derivative f39y39 etc Example Find y39 if x2ycosx 3 Example Final y39 if xzy3 cosy 3 Example Final the derivative ofthefunction x2 2xy y2 x 2 at 12 C Bellomo revised 16Aug08 Chapter 4 Section 7 Page 1 of 2 Section 47 Antiderivatives A Recag of Some Derivatives 0 Let s recall some derivatives we have seen so far at s1n x cos x dx d 1 ln x dx x 184x 4841 dx dicosax2 x 4x l sin2x2 x x o If we wanted to undo the derivative we could say Antiderivative cos x sin x Antiderivative In x x Antiderivative4e lx e Antiderivative 4x l sin2x2 x cos2x2 x o In this section we will not focus much on nding this antiderivative we will just be identifying the relationship between them The Arbitrary Constant 0 Look at the functions below with their derivatives what do you notice isinxl7 cosx dx isinx l cosx dx d l 12lnx dx x d l elnx dx x e 7 46 d 4 4 e 8 4e Z 0 Adding or subtracting a constant to a function only shifts it up or down It does not affect the derivative or shape of that function C Bellomo revised 17Sep08 Chapter 4 Section 7 Page 2 of 2 In that regard any time you take the antiderivative of a function you end up with what is called an arbitrary constant Antiderivative cos x sin x c l Antlderlvatlve In x c x Antiderivative4e e c Antiderivative 4x l sin2x2 x cos2x2 x c The Easy Antiderivative Recall ixquot nxH dx Shifting the value ofn by l we see 611x 1 nlxquot x n1 Dividing by n1 we nd i x xquot dx n1 So the antiderivative of a function to a power is given by n1 x Antlderlvatlvexquot C n1 Example Find the antiderz39vative of f39x 2x2 3x7 fmd Example Find the antiderz39vative of f39x 5x 3 8x2 7x fmd Don t forget the arbitrary constant Q What value of n will NOT work in the formula above Have we seen it already C Bellomo revised 17Sep08 Chapter 2 Section 4 Page 1 of 4 Section 24 The Product and g Quotient Rules Product Rule Let s look at the limit function of the derivative of a product fxh gxh fx39gx h fxgxlhi g hmfxhgxhfxgx fxhgxfxhgx h 2 Km fxhgxh fxhgx fxgxfxhgx hao h zlim fxh gxh gx 1 gx fxhfx 1 hgt0 h h h fxlgggx I g gx1hgquotngfx I fxfoeg39xgxf39x d 0 So 1n general Efxgx fxg39xgx39f39x my llwzquot e 3939Him l plus the o The derivative of a product M W 7 Remember it is essential for future success that you memorize the words and not the symbolic formula QuotientRule o In general you can think of a quotient as a product d l d l l fx fx f39x dx gx dx gx gx d 1 g39x Now we don t actually have the tools yet to nd 2 61x gx gxl but let s take it on faith for now and plug it in and simplify d l g39x l dici x goal x gx2gx x 2 fxg39x gxf39x goof goof z gxf x fxg39x goof C Bellomo revised 16Aug08 Chapter 2 Section 4 Page 2 of 4 d 0 So 1n general we have gxf39xfxg39x dx gx gx2 f x is o The derivative of a quotient gx Some Example Problems 0 Example Differentiate f x x 1x 2 0 Example Differentiate gx2L x 1 x2 0 Example Differentiate hx 1 1 x C Bellomo revised 16Aug08 Chapter 2 Section 4 Example Find the equation of the tangent line to the curve yx x Page 3 of 4 23 x1 atx 1 Derivatives of Other Trig Functions sine and cosine itanx i Slnx dx dx cosx We can nd the derivative of all other trig functions using our knowledge of how they relate back to Derivative of tangent cos xcos x sin x sin x cos2 x Z 39 Z cos xs1n x cos2 x sec2 x Derivative of cosecant d d 1 J cscx dx dx s1nx sin x0 1cos x sin2 x 1 cos x sin x sin x csc x cotx C Bellomo revised 16Aug08 Chapter 2 Section 4 Page 4 of 4 o Derivative of secant d d 1 sec x dx dx cos x cos x0 1 sin x cos2 x 1 sin x cos x cos x sec x tan x quiz 1 11mm n39wgm n o Derivative of cotangent d d cos x cot x dx dx sm x sin x sin x cos xcos x sin2 x sin2 x cos2 x sin2 x csc2 x 0 Example Find the derivative of fx ex sec x 1 0 Example Find the derivative of f x m NOTE this may not be in simple terms but has demonstrated differentiation completely C Bellomo revised 16Aug08 chapterl Sectlon 2 Page 1 of6 Section 1 A Catalog ofEssential Functions Lmear Madels 0 All linear equations have the form rise chan ein hon39zontal o The lettermls the ofthe hne or It can beposrtrve run change In vertical negative or zero It can also be very large or very small 0 Q Whatwould the line look like in each one ofthese cases7 smallpositive largepositive srnallnegatlve largenegative zero o The letters and y are rneanlng they vary or change along the line At least one of them must be present in the equation Together they represent the x y The letter b represents the this is where the line crosses the horizontal axis Q What are the restrictions on this value A o Emme Below are three th erehtgrophs alung wzth A dszemmequanuns Match them Equanuns lyeo5r 2y 3 3 x 73 4y Zx 5 hhlh c Bellomo revtsed l rAugrOS Chapter 1 Section 2 Page 2 of 6 Polynomials 0 Recall a is an expression of the form anxquot aHx 1 azx2 a1x a0 where the a s are real number 0 For nonzero an the expression is said to be of nth degree the highest power is n the is anx and the is an 0 Examples of polynomials that are common Degree Name Form 0 Constant f x c 1 Linear f x mx b 2 Quadratic f x ax2 bx c 3 Cubic fx ax3 bx2 cx al Power Functions 0 A is of the form f x x where a is any constant 0 Q What is the difference between a power function and a polynomial A 0 Q Is a polynomial a power function A o If the value of a is a fraction the power function is also called a 0 Example fx x 2 J o If the value of a is negative it is the 0 Example f x x 1 l T Rational Functions 0 A rational function is a ratio of two polynomials pgx qx o If the rational function has a root it would be when pc 0 so long as qc isn t zero 0 The rational function will be unde ned at c whenever qc 0 C Bellomo revised 16Aug08 Chapter 1 Section 2 Page 3 of 6 T ri gonometric Functions 0 The siX trig functions are de ned in terms of right triangles see gure The two most importan being sine and cosine o osite ad39acent men s1n6 L cos 6 J h use opposite hypotenuse hypotenuse 9 adjacent o All other trig functions can be de ned in terms of sine and cosine so remember the de nitions above and the relationships between them and the others sin 6 cos 6 1 sec6 tan6 cot6 s1n6 cos6 cos6 s1n6 csc6 0 To evaluate trig functions remember some fundamentals 0 First remember the basic shape of sinx and cosx along with when they are 0 and i1 0 From the gure you can see they are periodic both with a period of 271 0 Q When is sine equal to zero 0 Q When is cosine equal to zero C Bellomo revised 16Aug08 Chapter 1 Section 2 Page 4 of 6 Exponential Functions 0 y 2 kb 0 Variables are y and t y is the dependent variable and t is the independent variable 0 bisthe bgt0b l o k is the when t 0 o A commonly used base is e 27182818284 0 Rules for Exponentials bxby 17 W b w i bx Log Functions 0 We want to undo the exponential function by x o This is true if and only if y 10gb x y is the exponent b is the base and x is the argument 0 So the log function is the of the exponential function 0 Q What are our conditions on x and b 0 Example Change to exponential form to solve log10 0 Example Convert e 4000 to log 0 Special Log Bases is natural log written In is common log written log 0 These will be the only two on your calculator So if you need to calculate say log 2 you have to loga M loga b I use the change of base formula long C Bellomo revised 16Aug08 Chapter 1 Section 2 Page 5 of 6 0 Example use the change of base formula to evaluate log4 2 Algebraic anal T l Functions 0 An is constructed with algebraic operations addition subtraction multiplication and division A is anything else 0 Q Classify all the functions we have looked at so far as algebraic or transcendental Log Exponential Trig Rational Power Polynomials A Transformations of Functions 0 NOTE There are ways of transforming functions by shifting them left right up or down Or by stretching them or shrinking them Or by re ecting them BUT once we learn the actual tools for graphing after differentiation it usually more straightforward to graph what you are given So we will not go over transformations in class but I expect you will look over the material and bring me any questions you might have Some functions like trigs exponentials and logs are admittedly easier to do with transformations Composition of Functions 0 Sometimes it is helpful to break things into different parts and recognize how they fit together 0 A composite function works to accomplish this goal because it identifies functions inside functions For example we can think of the function f x x12 as the composite of two functions one function is the inside piece inx x1 and the other is the outside piece outx x2 So we see that fx outinx o The f o gx the composition offand g is defined as f o gx fgx The domain of g is x and the domain of f is gx Example Final g o h12 ifgx x2 2x 6 anal hx 3 C Bellomo revised 16Aug08 Chapter 1 Section 2 Page 6 of 6 0 Example Find the composites and domains if fx g and gx C Bellomo revised 16Aug08 Chapter 1 Section 5 Page 1 of3 Section 15 7 Continuity Recall Limits and Function Values We have already studied all the concepts necessary to understand continuity To recap let s look at the following graphs and answer three questions Q Does a exist Does lim fx exist Does lim fx fa 1 Ha Hg 3 u a l o U f A 1 g exists g fx exists and fx fa 2 g exists fx exists and fx f a 3 g dne f x exists and fx f a 4 g exists fx exists and fx f a 5 g exists fx dne and fx fa For a function to be continuous it has to pass all three tests above if it fails even one of these tests it is NOT continuous Q Which of the above functions are continuous at a A 1 is the only one that is continuous Formal Definition at Continuigg A functionfis continuous at x a if and only if 1fa is de ned 2 lim f x exists 3 lim fx fi1 If a function f is not continuous at x a it is said to be discontinuous at x a C Bellomo revised 8Sep08 Chapter 1 Section 5 Page 2 of 3 Types of Discontinuz39ties o If the limit of the function exists and is nite at x a then continuity fails because either at is not defined or lim f x at f a The discontinuity is classi ed as removable This is because by simply redefining a we can make f continuous at a 0 Q Which of the above pictures has a removable discontinuity How would you x this A Number 2 fix by setting fa lim fx 0 If the limit of the function is in nite then it is called an inf39mite discontinuity 0 Q Which of the above pictures has an in nite discontinuity A 3 and 4 are in nite discontinuities o If the limit from the right and left of a are not the same then the function is said to have a jump discontinuity 0 Q Which of the above pictures has a jump discontinuity A Number 5 Continuous on an Interval o A function can be continuous from the right and continuous from left at a number basically you only look at a right or left hand limit 0 A function can be continuous on an interval if it is continuous for all the points in that interval 0 A function is said to be continuous if it is continuous on its entire domain 0 While the same word ie continuous is used to describe all of these situations it is important to keep in mind what you are referring to Here they are from least strict to most strict 7 continuous at a from the left or right 7 continuous at a 7 continuous on an interval 7 continuous o The following functions are continuous on their domains polynomials rational functions root functions trig functions inverse trig functions exponential functions log functions 0 But wait I thought log functions were not de ned for negative values Q How could say logx satisfy the above statement if it isn t even de ned for x lt 0 A Well it is continuous for every value in its domain so it is said to be continuous 0 Some books and mathematicians differ in how they interpret this some will say a function is not continuous unless it is continuous on the entire real line Others like our book say that as long as the function is continuous on its domain it is continuous C Bellomo revised 8Sep08 Chapter 1 Section 5 Page 3 of 3 Intermediate Value Theorem If f is continuous on the closed interval ab and N is strictly between fa and b then there is a number 0 between a and b such that fc N Example Let s say you want to nd the root of the function x x 7 tanx This isn t so easy because it is not an algebraic equation and can t be solved explicitly We have to estimate the roots There are actually an in nite number of them starting with the trivial root at x 0 So it looks like there is a root somewhere near x 45 We can prove this is true because of the IVT f is continuous on 4 46 f4 28 andf46 7426 N 0 is between 28 and 7426 So by the IVT there is a 0 between 4 and 46 where fc 0 We can continue to choose le and right values to move closer and closer to the actual root C Bellomo revised 8Sep08 Chapter 1 Section 6 Page 1 of 3 Section 16 Limits at lnfinig Recall that is fx px qx You have already studied You should already be able to identify the domain of this function in other words the values of x that areare not allowed for inputs off Q How do we nd any unde ned points in the domain A These points if they are not are called of the function there would tend to plus or minus in nity and the limit These types of in nite discontinuities are also called Example For fx What is the limit as x tends to 2from the right From the left x Horizontal Asym Qtotes Think of an asymptote in general being an invisible line that a function tends to if you were to keep drawing You never quite get there For example in the above function the graph as xtends to 2 gets closer and closer to the vertical line at x 2 but never quite reaches it And it will not cross over either The above example function has another asymptote that is horizontal If you were to let x get larger and larger x gt ice the function would get closer and closer to an invisible horizontal line Q What value does the function tend to as x gt ioo What is the equation of this line C Bellomo revised 16Aug08 Chapter 1 Section 6 Page 2 of 3 Finding Horizontal Asym grates As we extend the function to the left and right as far as we want we are essentially taking the limit as x tends to plus and minus in nity To nd the horizontal asymptote if it exists take the limit as x tends to plus and minus in nity E 100 1 Find the leading term of the numerator px and denominator qx 2 Write as a quotient and simplify 3 If the reduction is 7 a constant then this is the value of your horizontal asymptote HOW TO nd horizontal asymptotes for rational functions f x then the horizontal asymptote is zero positive value 7 constant x positive value 7 constant x then there is no horizontal asymptote It will tend to plus or minus in nity plug in to see which 3x2 x 5 2 Example Find lim Hoe x 3 2 35 Example Find lim Hwa x 4 C Bellomo revised 16Aug08 Chapter 1 Section 6 Page 3 of 3 Z 0 Example Find limw H 2x 4 2 0 Example Find limw Him x C Bellomo revised 16 Aug 08 Chapter4 Section 1 Page 1 of 5 Section 41 7 Max and Min Values Horizontal Tangents I We have looked at graphs and identified horizontal tangents or places where the slope of the tangent line is zero I Q For whichx values does the following function have a horizontal tangent 1 b c d e f g A There is a horizontal tangent at b d and f Note that c is not a horizontal tangent I Take a look at each horizontal tangent again Notice that to the left and to the right the sign of the derivative is different I For example to the left of point b the derivative is positive function is increasing and to the right of point b the derivative is negative function is decreasing I Also notice that when we say to the left of b and to the right of b we mean values that are really close to the value of b Another point say a or 0 would be considered too far away I Also note that the point imelf is a maximum or minimum compared to the values near it This is called a relative max or relative min sometimes called local max or local min I Notice point c It is not a horizontal tangent but to the le of c the function is decreasing and to the right of c the function is increasi It is also considered to be a relative min but the derivative there is undefined recall the derivative is undefined at sharp corners Critical Points I A critical point sometimes called a critical number of a function x is when f x 0 or f x is undefined I Q What are the critical poinm of the graph pictured above A b c d and f I Critical points include both horizontal tangenm derivative is zero and places where the derivative is undefined which could be corners jumps or asyrnptotes This does not mean the point is a maximum or minimum value But if a point is a max or min it will be a critical point C Bellomo revised 14Apr09 Chapter 4 Section 1 of 5 Example Identi the critical points horizontal tangents local maxmin in the picture below a b c d e f Critical points b d e points between e and f Horizontal tangents b d and e through f Relative Max b Relative Min d Notice that e could be considered a max but it for sure considered a critical point Also the points between e and f are critical points because the derivative is zero there but they are not max or min va ues So to recap relative max or min implies critical point but critical point does not imply relative maximum or minimum Formal D ettnitions A functionf has a local maximum at e if fe 2 fx for all x near 0 A functionf has a local minimum at e if fe g fx for all x near 0 Note that the equality allowed in the de nition according to our book is what makes e in our example above a maximum value Some peoplebooks will not allow for equality in the de nitions above and use strict inequalities For purposes of tlus class we will stick with the books de nition A functionfhas an absolute max or global max at e if fe 2 fx for all x in the domain off A functionfhas an absolute min or global min at e if fc g fx for all x in the domain off Again some peoplebooks would use strict inequalities Q What is the difference between a global max and a local max A A global max will be larger than all values in the domain and a local max will only be larger than the values around it Q Is it possible to have more than one global max or min A Yes Because we allow for equality it is possible that we can have more than one global maximum or min C Bellomo revised 14Apr09 How to Fmd and Idem chaptenA Secuml Page afS Example mm cnncalpam 6 bode f a 15 a1oea1 mm 1H5 not aglobal mm even though the llmltls less than all other values beeause et wd and a1oea1 max agam beeause we allow for equahty tits a1oea1 mm e 15 a ennea1 pomt a1oea1 max and aglobal max fxs a 1oea1 mm thee 15 gomg by the de nmon m our book whteh allows h n th graph 15 eut off Fermat s Thenmn says that f has a1oea1 max or mm ate andlf e exests then e 0 anxeaz Palms tan E uanzm or unde ned you have a entteal pomt Kthe emttea1 pomt e eomes from when the denvanve 15 zero e 0 it could be 7A 1oea1 max e lt0 Also f c gt 0 and e lt0 ens avalue immed ately to the le of e e 15 nght 7A 1oea1 mm e gt 0 Also f c lt0 and e gt 0 7A global max on 11 e lt0 fxs eontmuous on the mtemal ab and f z d v1 dab 7A global mm on a e gt 0 J15 eontmuous on the mtemal ab and f Sd v1 dab rAn m eenon pomt quotr 0 Kthe emttea1 pomt eomes from when the denvanve 15 unde ned et couldbe 7 sh quot 7A jump dumnunmty rAn m mte duconnnmty asymptote All of these will be edentt able m the ongmal funenon c Eeunmo nemsenmAnnna Chapter 4 Section 1 Page 4 of 5 2 Example Find the 017 s and identify for f x x2 on 44 x x2 42x x2 42x x2 42 16x x2 42 f 3900 Notice the original function is de ned for all x but the domain is restricted to 44 f 39x is never unde ned f 39x 0 when x 0 This is the only critical point Let s classify this critical point using the second derivative test 16x2 42 64x2 x2 4 f x x 4 l6x2 4 4x2 x2 43 164 3x2 x2 43 f 0 gt 0 so x 0 is a relative min Is it a global min on 44 There is only one critical point so to test for global max we need only test the value we found and the endpoints lt 4gt2 4 f4 424 06 42 4 f4 m 06 0 4 f0m 1 Yes it is a global min Is there a global max Yes Recall that f4 f4 06 This means thatfhas a global max at 4 and 4 Q Why could there not be a value that exceeds 06 on 44 A Because in that range there is only one critical point and no undefmed values This means there is nothing special going on other than the localglobal min we found already In fact the picture of the function is below C Bellomo revised 14Apr09 Chapter 4 Section 1 Page 5 of 5 Example Find and identify any critical points for f x e 652 on 01 f x e 2652 Notice the original function is de ned for all x but the domain is restricted to 01 f39x is never unde ned f39x 0 when e 2872 0 2e72x eix 26 xln2 This is the only critical point Let s classify this critical point using the second derivative test f quotx e 4872 f ln2 lt 0 so x 0 is a relative maX Is ln2 a global maX on 01 We need to test the endpoints and any critical points recall the function is de ned in the domain f0e0 e 0 f1 6 1 e 2 023 fln2 6 8721112 025 Yes it is a global maX Is there a global min f has a global min at 0 since there are no other critical points and f0 is less than the rest In fact the function looks like C Bellorno revised 14Apr09 Chapter 4 Section 5 Page 1 of 4 Section 45 Ogtimization Problems 0 Example Find two numbers whose difference is 100 and whose product is a minimum 0 Example Find the dimensions of a rectangle with area 1000 sq meters whose perimeter is as small as possible C Bellomo revised 17Sep08 Chapter 4 Section 5 Page 2 of 4 0 Example A box with a square base and open top must have a volume of32000 cubic cm Find the dimensions of the box that minimize the amount of material used C Bellomo revised 17Sep08 Chapter 4 Section 5 Page 3 of 4 0 Example A rectangular storage container with an open top is to have a volume of 10m3 The length of its base is twice the width Material for the base costs 1 0 per square meter Material for the sides costs 6 per square meter Final the cost of the materials for the cheapest such container 0 Q Why do I say about A C Bellomo revised 17Sep08 Chapter 4 Section 5 Page 4 of 4 0 Example Find the point on the line 6x y 9 that is closest to the point 31 C Bellomo revised 17Sep08 Chapter 4 Section 7 Page 1 of 2 Section 47 Antiderivatives A Recag of Some Derivatives 0 Let s recall some derivatives we have seen so far at s1n x cos x dx d 1 ln x dx x 184x 4841 dx dicosax2 x 4x l sin2x2 x x o If we wanted to undo the derivative we could say Antiderivative cos x sin x Antiderivative In x x Antiderivative4e lx e Antiderivative 4x l sin2x2 x cos2x2 x o In this section we will not focus much on nding this antiderivative we will just be identifying the relationship between them The Arbitrary Constant 0 Look at the functions below with their derivatives what do you notice isinxl7 cosx dx isinx l cosx dx d l 12lnx dx x d l elnx dx x e 7 46 d 4 4 e 8 4e Z 0 Adding or subtracting a constant to a function only shifts it up or down It does not affect the derivative or shape of that function C Bellomo revised 17Sep08 Chapter 4 Section 7 Page 2 of 2 In that regard any time you take the antiderivative of a function you end up with what is called an arbitrary constant Antiderivative cos x sin x c l Antlderlvatlve In x c x Antiderivative4e e c Antiderivative 4x l sin2x2 x cos2x2 x c The Easy Antiderivative Recall ixquot nx39quot1 dx Shifting the value ofn by l we see 611x 1 nlxquot x n1 Dividing by n1 we nd i x xquot dx n1 So the antiderivative of a function to a power is given by n1 x Antlderlvatlvexquot C n1 Example Find the antiderz39vative of f39x 2x2 3x7 fmd x3 x2 x 2 3 7xc f 3 2 Example Find the antiderz39vative of f39x 5x 3 8x2 7x fmd x72 x3 x2 x 5 8 7 c f 2 3 2 Don t forget the arbitrary constant Q What value of n will NOT work in the formula above Have we seen it already xn1 A Antiderivativexquot n at l n1 l Thls 1s a spec1al case and Antlderlvatlve In x x C Bellomo revised 17Sep08 Chapter 5 Section 5 Page 1 of 3 Section 55 Substitution A Look Back 0 Recall the chain rule where for example ie z e 2 2x 0 Clearly since we know integrals undo derivatives we would have Ie 2 2xalx e 2 c Ifwe let u x2 we would get ill 1 2x or alu 2x dx Substituting u and alu into the integral we would have Ie alu e c e 2 c o This method is the chain rule used backwards 0 You CANNOT evaluate an integral of a product like above without using substitution 0 Formally the de nition of the substitution rule is I f gx g39x alx I f u alu where u gx Substitutions with Constants al al 0 Ifgx cx then settmg u cx y1elds 61 c or alx u x c 0 Example Evaluate Iezxdx Let u 2x Then dx alu2 Iezxalx Ie 1 u e 0 2 ieuc 2 Other 39 39 39 ReauireMorePieces 0 Example Evaluate Jde x2 12 2 alu Let ux l Then alu2xdx xdx l 1 al J 2 2 dezj z u x l u 2 1 2 EJu alu l u391 c 2 l l C 2x2 1 C Bellomo revised 14Apr09 Chapter 5 Section 5 Page 2 of 3 0 Example Evaluate If l2y4 1 dy Let u2y4 1 Then alu8y3aly 3 y3dy alu I l2y4 1y3alyx 1u32 c 832 1 2 4132c 12 y x2 JG Let u1 x Then alu alx and x1 u dx 0 Example Evaluate 2 2 0 du I d1 232quot M J u J39u7122u12u32 d 12 32 52 21 x1 1 xm 1 x5 c Definite Integrals 0 So long as you change back to x these are no different than the above problems 6 0 Example Evaluate J1 xcosx2dx 2 alu Let ux Then alu2xalx3 xalx alu 1 Then cos x2 xalx cosu j j 2 1 s1nu s1nx2c 2 2 J7 J J1 xcosx2 alx sinx2 0 sin n39 sin 0 C Bellomo revised 14Apr09 Chapter 5 Section 5 Page 3 of 3 Average Value 1 b a 0 Example Final the average value of fx e on 12 0 We de ne the average value of f on the interval ab as I fx alx 2 L 282 dx1ielr 2 1 1 2 1 4 1 e e 2 C Bellomo revised 14Apr09 Chapter 3 Section 6 Page 1 of 5 Section 36 Hyperbolic Functions De nitions 0 Recall the trig functions are related to the unit circle x2 y2 l 01 sin pi2 cos piZ l0 sin pi cos pi 1 0 sin 0 cos 0 K 0 1 sin 3pi2 cos 3pi2 o In a similar way hyperbolic functions are related to the hyperbolic function x2 y2 1 cosht 39 t They can be expressed in terms of linear combinations of exponential growth and decay e i sinh x A i e e coshx o For the regular trig functions sine and cosine give rise to tangent cotangent secant and cosecant o In a similar way hyperbolic sine and hyperbolic cosine give rise to tanh x m cschx sechx coth x C sh x osh x s1nh x cosh x s1nh x C Bellomo revised 26Oct09 chapcea Secdon Page 2 of5 Greeks otH rbolic Sine and Coxine C Bellomo revised 26Oct09 Chapter 3 Section 6 Page 3 of 5 Properties of H ggerbolic Functions x 7 7r 7H o sinh x e 28 sinhx e39x 64quot e e39x 2 o cosh x cosh x 0 Q From our de nitions of even and odd how can we classify sinh and cosh A Sinh is an odd function cosh is an even function 0 Q What can we then say about their symmetry A Sinh is symmetric about the origin cosh is symmetric about the y axis 0 cosh2 x sinh2 x 1 0 And the last two properties are sinhx y sinh xcosh y cosh xsinh y coshx y cosh xcosh y sinh xsinh y Derivatives r A 1 H d d sinh x j cosh xcosh x sinh xsinh x 1 2 2 sechzx cosh x cosh x 0 Q Use your understanding of the chain rule and the property that cschx lsinhx to nd cschx A icschx 1 dx dx sinh x 1sinh x Z cosh x cosh x sinh2 x coth xcschx C Bellomo revised 26Oct09 Chapter 3 Section 6 Page 4 of 5 d Q Use your understandlng of the property csch x x 7 to nd d cschx e x A icsch x dx dx ex ex 2 1e equot 2 e equot e ex 2 I eat 84 ex 84 coth x cschx Q Use your understanding of the chain rule and the property that sechx lcoshx to nd disechx x A isechx i 1 dx dx cosh x 1cosh x 2 sinh x sinh x I cosh2 x tanhxsechx Q Use your understanding of the chain rule and the property that cothx coshxsinhx to nd icothx dx A icothx i cfmh x dx dx s1nh x sinh xsinh x cosh xcosh x sinh2 x l sinh2 x csch2x Inverse H Myerboli 0 Functions There are functions that undo the hyperbolic functions y sinh 1 x cgt sinhy x ycosh391xcgt coshyx yZO y tanh391 x cgt tanhy x Because hyperbolic functions are related to exponentials their inverses are related to logs sinh391 x lnx x e R cosh 1 x lnx x 21 lx tanh391 x 1In 2 l x J lltxltl C Bellomo revised 26Oct09 Chapter 3 Section 6 Page 5 of 5 0 Let s try to see why the rst equation works We want to nd the inverse of the function y sinh x As we usually proceed to nd an inverse we swap x and y and solve for y y eey xsinhye 2xey e39y 2xey ezy l ey2 2xey l0 2 iquot4 2 41 l eywxialx21 21 So either eyx Ix2l Butx Vx2llt0forallx This is notvalid or ey xxlx2l 3 ylnxxx2l 0 We can also differentiate any of the inverse trig functions these formulas are in your book by the quadratic formula C Bellomo revised 26Oct09 Chapter 4 Section 4 Page 1 of 4 Section 44 Curve Sketching A Quick Guideline 0 When looking to sketch a function it is a good idea to start with the function type 0 Find the domain 0 Find any asymptotes 0 Find any limits if needed 0 Find the rst and second derivative 0 Find any critical points in ection points 0 Determine when it is increasingdecreasing concave up down 0 See the supplemental to sketching for a more detailed guideline C Bellomo revised 14Apr09 Chapter 4 Section 4 Page 2 of 4 x Example Sketch y W This is a rational expression We expect a restricted domain and verticalhorizontal asymptotes Domain x712 0 when x 1 Since x 1 does not make the top go to zero it is a vertical asymptote Roots x 0 is aroot Note x712 0 7t 0 Horizontal Asymptote x 2 as x gets large 1t behaves llke 2 li So there is a HA at 0 Hm x1 x x Evaluating the rst derivative 7 x 7l2 7 2xx 7 l1 y x 71 7 7l 7 x 7 x 71f x 1 xi D3 y39 0 when x l undefined when x l vertical asymptote Evaluating the second derivative I N x D3 x 13x 12 i y 271 2x4 1P 2 022 x l W I There is an in ection point at x 72 CPmin 1 25 So doing a sign diagram we see the basic shape 1 f 70 g f 0 5 2 l l Evaluatin l 1 70 25 and 2 2 0 22 g y 1 12 39 y 2 12 39 We can also evaluate the limit as x tends to 1 from the right and left to verify C Bellomo revised 14Apr09 Chapter 4 Section 4 asymptote Page 3 of 4 Example Sketch y x 7 5 This is a composite of a root and a rational expression Domain We require What is under the radical to be positive i 0 u xS 0 5 We restrict our domain to x g 0 x gt 5 Asymptotes There will be a vertical asymptote at x 5 and lim y 00 x7 There will be a horizontal asymptote at 1 Also note that lim y 0 xaU First derivative 7 l l x 7 5 7 x y 2 f x x 7 52 x 7 5 7 i 2 quotxx 7 53 There are critical points atx 5 and x 0 but no horizontal tangents Notice that the derivative is always negative Second derivative I 75 712 73 752 732 71 732 x x75 x75 x y 2 2 2 7 75 57 4x 4 1 x3x 7 55 There could be an in ection point at 57 4x 0 x but it is not in our domain Sign diagram and graph f H u i f i i u 7 7 horizontal E asymptote i I 5 LP LU I E vertical I I I a Chapter 4 Section 4 Page 4 of 4 Example Sketch y xln xZ Domain The function is only de ned for x gt 0 Limit Let s look at the limit as x tends to zero 2 lim xln xZ limw xao xao x 2ln x x 1 7x72 m 401 x x li xgt0 lim Xgt0 72thlfZ xgt0 71x First derivative y39 2ln x ln xZ 2lnxln x y390 When 2lnx0 lnx0 or xe zz0l4 xl Second derivative yquot 2lnxlllnx x x max 014 054 l221nx x yquot0 When 2Zlnx0 xe391z037 Sign diagram and rough sketch 01 l 0 min IP 037 137 C Bellorno revised 14Apr09 Chapter 1 Section 4 Page 1 of 4 Section 14 Calculating Limits In this section we will be using methods to nd limits not prove them as in the last section Sometimes it is helpful to break up limits and work with pieces rather than the whole Example Find lim 2 2 H2 x 2 lim 2 2 and lim 2 00 1692 xgt2 x 2 l 2 00 x 2 Most of these rules are intuitive in that you would think of using them anyway because you are familiar with the rules of addition subtraction multiplication and division lim 0 0 Therefore lin12 131fmquot 13334545 1imfxi gx1im fxi1imgx Qo xmg b ggrfxgxggg xiggggm fx mfx 1m xgta xgta g x a gt 0 for 71 even lim gx EelWquot 132 le lim quotlfx nllim fx A very useful theorem is called the squeeze theorem and it says that if f x S gx S hx when x is close to a and lim fx lim hx L then limgx L limgx 0 lim fx gt 0 for 71 even Essentially what is happening is that f and h are tending to L and since g is squeezed in between them it must also tend to L C Bellomo revised 8Sep08 Chapter 1 Section 4 Page 2 of 4 Example Problems 0 Example Evaluate 41h 1 41h 1 1h1 11m H0 h H0 h 1 h 1 lim 1h 1 lim H hJlh1 lim H 1h1 1 7 71 71 0 Example Evaluate 1 1 lim3h 1 3 1Lm3h 33h ho h o h 33h m3 3h 11 H0 3h3h 1 l1m H0 33h l 9 0 Example Evaluate til I 9 2 lim t it t H0 tt 1 Aim H0 t2t I 1 11m 2 H0 t r t lim 0 11 1 C Bellomo revised 8Sep08 chapteri Section4 Page 3 of4 x o Em 2 Evaluate 1irn quot9 Hzlx l asxaz x72x72 74 m x2x72 X So lim Hr m x2 HT 1 4 Butas xgt21x727x72 e4 s272 72 2 So 1irn 1irn Hzrlxezl H27 X x274 Therefore 1irn H2 I 2 Lzmzts u Tn Functzuns sin 7 A famous limit is given by 1irn 1 Ho X o This is pretty clear from the graph ofthe function i smAX J n But it can also be found with the squeeze theorem because nearx 0 cos lt 5 quot X lt1 And both x limcosx1 and 1irn11 Ho Ho c Benorno revised seSepeos Chapter 1 Section 4 Page 4 of4 39 2 0 Example Evaluate 11mM raO x hm 2s1n2x 2 lim s1n2x xgt0 x 2xgt0 2x 21 2 cos x 1 0 Example Evaluate l1m HO s1nx cosx 1 cosx l cosx1 H0 sin x H0 sin x cos x1 cos2 x 1 m H0 s1n xcos x 1 sin2 x l1m H0 s1n xcos x 1 sin x 0 H0 cos x 1 C Bellomo revised 8Sep08

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.