MOOCULUS. massive open online calculus C A L C U L U S T H I S D O C U M E N T W A S T Y P E S E T O N A P R I L 3,

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1 MOOCULUS massive open online calculus C A L C U L U S T H I S D O C U M E N T W A S T Y P E S E T O N A P R I L 3,

2 2 Copyright c 204 Jim Fowler an Bart Snapp This work is license uner the Creative Commons Attribution-NonCommercial-ShareAlike License. To view a copy of this license, visithttp://creativecommons.org/licenses/by-nc-sa/3.0/ or sen a letter to Creative Commons, 543 Howar Street, 5th Floor, San Francisco, California, 9405, USA. If you istribute this work or a erivative, inclue the history of the ocument. The source coe is available at: This tet is base on Davi Guichar s open-source calculus tet which in turn is a moification an epansion of notes written by Neal Koblitz at the University of Washington. Davi Guichar s tet is available athttp:// calculus/ uner a Creative Commons license. The book inclues some eercises an eamples from Elementary Calculus: An Approach Using Infinitesimals, by H. Jerome Keisler, available athttp:// uner a Creative Commons license. This book is typeset in the Kerkis font, Kerkis c Department of Mathematics, University of the Aegean. We will be gla to receive corrections an suggestions for improvement atfowler@math.osu.eu orsnapp@math.osu.eu.

3 Contents 0 Functions 8 Limits 9 2 Infinity an Continuity 36 3 Basics of Derivatives 47 4 Curve Sketching 64 5 The Prouct Rule an Quotient Rule 82 6 The Chain Rule 90

4 4 7 The Derivatives of Trigonometric Functions an their Inverses 07 8 Applications of Differentiation 2 9 Optimization 46 0 Linear Approimation 6 Antierivatives 78 2 Integrals 97 3 The Funamental Theorem of Calculus Techniques of Integration 29 5 Applications of Integration 235 Answers to Eercises 245 Ine 257

5 List of Main Theorems.3. Limit Laws Squeeze Theorem Intermeiate Value Theorem Differentiability Implies Continuity The Constant Rule The Power Rule The Sum Rule The Derivative of e Fermat s Theorem First Derivative Test Test for Concavity Secon Derivative Test The Prouct Rule The Quotient Rule Chain Rule The Derivative of the Natural Logrithm Inverse Function Theorem The Derivatives of Trigonometric Functions The Derivatives of Inverse Trigonometric Functions L Hôpital s Rule Etreme Value Theorem Rolle s Theorem Mean Value Theorem

6 6.. Basic Antierivatives The Sum Rule for Antierivatives Properties of Definite Integrals Funamental Theorem of Calculus Version I Funamental Theorem of Calculus Version II Integral Substitution Formula Integration by Parts Formula

7 How to Rea Mathematics Reaing mathematics is not the same as reaing a novel. To rea mathematics you nee: (a) A pen. (b) Plenty of blank paper. (c) A willingness to write things own. As you rea mathematics, you must work alongsie the tet itself. You must write own each epression, sketch each graph, an think about what you are oing. You shoul work eamples an fill in the etails. This is not an easy task; it is in fact har work. However, mathematics is not a passive eneavor. You, the reaer, must become a oer of mathematics.

8 0 Functions 0. For Each Input, Eactly One Output Life is comple. Part of this compleity stems from the fact that there are many relations between seemingly inepenent events. Arme with mathematics we seek to unerstan the worl, an hence we nee tools for talking about these relations. A function is a relation between sets of objects that can be thought of as a mathematical machine. This means for each input, there is eactly one output. Let s say this eplicitly. Something as simple as a ictionary coul be thought of as a relation, as it connects wors to efinitions. However, a ictionary is not a function, as there are wors with multiple efinitions. On the other han, if each wor only ha a single efinition, then it woul be a function. Definition A function is a relation between sets, where for each input, there is eactly one output. Moreover, whenever we talk about functions, we shoul try to eplicitly state what type of things the inputs are an what type of things the outputs are. In calculus, functions often efine a relation from (a subset of) the real numbers to (a subset of) the real numbers. Eample 0.. Consier the function f that maps from the real numbers to the real numbers by taking a number an mapping it to its cube: While the name of the function is technically f, we will abuse notation an call the function f () to remin the reaer that it is a function.

9 calculus 9 an so on. This function can be escribe by the formula f ()= 3 or by the plot shown in Figure. Warning A function is a relation (such that for each input, there is eactly one output) between sets an shoul not be confuse with either its formula or its plot. 5 y A formula merely escribes the mapping using algebra. A plot merely escribes the mapping using pictures. 2 2 Eample 0..2 Consier the greatest integer function, enote by f ()=. 5 This is the function that maps any real number to the greatest integer less than or equal to. See Figure 2 for a plot of this function. Some might be confuse because here we have multiple inputs that give the same output. However, this is not a problem. To be a function, we merely nee to check that for each input, there is eactly one output, an this is satisfie. Just to remin you, a function maps from one set to another. We call the set a function is mapping from the omain or source an we call the set a function is mapping to the range or target. In our previous eamples the omain an range have both been the real numbers, enote byr. In our net eamples we show that this is not always the case. Eample 0..3 Consier the function that maps non-negative real numbers to their positive square root. This function is enote by f ()=. Note, since this is a function, an its range consists of the non-negative real numbers, we have that 2 =. Figure : A plot of f ()= 3. Here we can see that for each input (a value on the -ais), there is eactly one output (a value on the y-ais). y Figure 2: A plot of f ()=. Here we can see that for each input (a value on the -ais), there is eactly one output (a value on the y-ais).

10 0 See Figure 3 for a plot of f ()=. Finally, we will consier a function whose omain is all real numbers ecept for a single point. y Eample 0..4 Consier the function efine by f ()= This function may seem innocent enough; however, it is unefine at = 2. See Figure 4 for a plot of this function Figure 3: A plot of f ()=. Here we can see that for each input (a non-negative value on the -ais), there is eactly one output (a positive value on the y-ais). y Figure 4: A plot of f ()= Here we 2 can see that for each input (any value on the -ais ecept for = 2), there is eactly one output (a value on the y-ais).

11 calculus Eercises for Section 0. y 4 () In Figure 5 we see a plot of y=f (). What is f (4)? 3 (2) In Figure 6 we see a plot of y=f (). What is f ( 2)? 2 (3) Consier the following points: (5, 8), (3, 6), ( 6, 9), (, 4), ( 0, 7) Coul these points all be on the graph of a function y=f ()? (4) Consier the following points: (7, 4), (0, 3), ( 2, 2), (, 8), (0, 4) Coul these points all be on the graph of a function y=f ()? Figure 5: A plot of y=f (). (5) Consier the following points that lie on the graph of y=f (): If f ()= 9 fin the value of. ( 5, 8), (5, ), ( 4, 0), (2, 9), (4, 0) (6) A stuent thinks the set of points oes not efine a function: ( 7, 4), (0, 4), (0, 4), (3, 4) They argue that the output 4 has four ifferent inputs. Are they correct? (7) Consier the following points: y (, 5), ( 3, 4), (, 3), (5, 3), (8, 5) 2 Name a value of so that these points o not efine a function. (8) Let f ()= Evaluate f (0). (9) Let f ()= Evaluate f (). (0) Let f ()= Evaluate f ( ). () Let f ()= Evaluate f (w). (2) Let f ()= Evaluate f (+ h). (3) Let f ()= Evaluate f (+ h) f (). 3 4 Figure 6: A plot of y=f ().

12 2 (4) Let f ()=+. What is f (f (f (f ())))? (5) Let f ()=+. What is f (f (f (f (+ h))))? (6) If f (8)=8an g()=3 f (), what point must satisfy y=g()? (7) If f (7)=6an g()=f (8 ), what point must satisfy y=g()? (8) If f ( ) = 7 an f () = g( 6 ), what point must satisfy y = g()?

13 calculus Inverses of Functions If a function maps every input to eactly one output, an inverse of that function maps every output to eactly one input. While this might soun somewhat esoteric, let s see if we can groun this in some real-life contets. Eample 0.2. Suppose that you are filling a swimming pool using a garen hose though because it raine last night, the pool starts with some water in it. The volume of water in gallons after t hours of filling the pool is given by: v(t)=700t+ 200 What oes the inverse of this function tell you? What is the inverse of this function? Solution While v(t) tells you how many gallons of water are in the pool after a perio of time, the inverse of v(t) tells you how much time must be spent to obtain a given volume. To compute the inverse function, first set v=v(t) an write v=700t Here we abuse notation slightly, allowing v an t to simultaneously be names of variables an functions. This is stanar practice in calculus classes. Now solve for t: t= v/700 2/7 This is a function that maps volumes to times, an t(v)=v/700 2/7. Now let s consier a ifferent eample. Eample Suppose you are staning on a brige that is 60 meters above sea-level. You toss a ball up into the air with an initial velocity of 30 meters per secon. If t is the time (in secons) after we toss the ball, then the height at time t is approimately h(t)= 5t t+ 60. What oes the inverse of this function tell you? What is the inverse of this function? Solution While h(t) tells you how the height the ball is above sea-level at an instant of time, the inverse of h(t) tells you what time it is when the ball is at a given height. There is only one problem: There is no function that is the inverse

14 4 of h(t). Consier Figure 7, we can see that for some heights namely 60 meters, there are two times. While there is no inverse function for h(t), we can fin one if we restrict the omain of h(t). Take it as given that the maimum of h(t) is at 05 meters an t= 3 secons, later on in this course you ll know how to fin points like this with ease. In this case, we may fin an inverse of h(t) on the interval [3, ). Write h= 5t t+ 60 0= 5t t+ (60 h) an solve for t using the quaratic formula t= 30± ( 5)(60 h) 2( 5) = 30± (60 h) 0 = (60 h) = (60 h) = 3 2.2h Now we must think about what it means to restrict the omain of h(t) to values of t in [3, ). Since h(t) has its maimum value of 05 when t= 3, the largest h coul be is 05. This means that 2.2h 0 an so 2.2h is a real number. We know something else too, t > 3. This means that the that we see above must be a +. So the inverse of h(t) on the interval [3, ) is t(h)=3+ 2.2h. A similar argument will show that the inverse of h(t) on the interval (, 3] is t(h)=3 2.2h h Figure 7: A plot of h(t)= 5t t+ 60. Here we can see that for each input (a value on the t-ais), there is eactly one output (a value on the h-ais). However, for each value on the h ais, sometimes there are two values on the t-ais. Hence there is no function that is the inverse of h(t). h t We see two ifferent cases with our eamples above. To clearly escribe the ifference we nee a efinition. Definition A function is one-to-one if for every value in the range, there is eactly one value in the omain Figure 8: A plot of h(t)= 5t t+ 60. While this plot passes the vertical line test, an hence represents h as a function of t, it oes not pass the horizontal line test, so the function is not one-to-one. t

15 calculus 5 You may recall that a plot gives y as a function of if every vertical line crosses the plot at most once, this is commonly known as the vertical line test. A function is one-to-one if every horizontal line crosses the plot at most once, which is commonly known as the horizontal line test, see Figure 8. We can only fin an inverse to a function when it is one-to-one, otherwise we must restrict the omain as we i in Eample Let s look at several eamples. y 2 f () Eample Consier the function f ()= 3. Does f () have an inverse? If so what is it? If not, attempt to restrict the omain of f () an fin an inverse on the restricte omain. Solution In this case f () is one-to-one an f ()= 3. See Figure 9. Eample Consier the function f ()= 2. Does f () have an inverse? If so what is it? If not, attempt to restrict the omain of f () an fin an inverse on the restricte omain. Solution In this case f () is not one-to-one. However, it is one-to-one on the interval [0, ). Hence we can fin an inverse of f ()= 2 on this interval, an it is our familiar function. See Figure f () 2 Figure 9: A plot of f ()= 3 an f ()= 3. Note f () is the image of f () after being flippe over the line y=. f () y 2 f () A Wor on Notation Given a function f (), we have a way of writing an inverse of f (), assuming it eists On the other han, f ()=the inverse of f (), if it eists. f () = f (). 2 Figure 0: A plot of f ()= 2 an f ()=. While f ()= 2 is not one-to-one onr, it is one-toone on [0, ).

16 6 Warning It is not usually the case that f ()=f (). This confusing notation is often eacerbate by the fact that sin 2 ()=(sin()) 2 but sin () (sin()). In the case of trigonometric functions, this confusion can be avoie by using the notation arcsin an so on for other trigonometric functions.

17 calculus 7 Eercises for Section 0.2 () The length in centimeters of Rapunzel s hair after t months is given by l(t)= 8t Give the inverse of l(t). What oes the inverse of l(t) represent? (2) The value of someone s savings account in ollars is given by m(t)=900t+ 300 where t is time in months. Give the inverse of m(t). What oes the inverse of m(t) represent? (3) At grauation the stuents all grabbe their caps an threw them into the air. The height of their caps can be escribe by h(t)= 5t 2 + 0t+ 2 where h(t) is the height in meters an t is in secons after letting go. Given that this h(t) attains a maimum at (, 7), give two ifferent inverses on two ifferent restricte omains. What o these inverses represent? (4) The number n of bacteria in refrigerate foo can be moele by n(t)=7t 2 20t+ 700 where t is the temperature of the foo in egrees Celsius. Give two ifferent inverses on two ifferent restricte omains. What o these inverses represent? (5) The height in meters of a person off the groun as they rie a Ferris Wheel can be moele by h(t)=8 sin( π t 7 )+20 where t is time elapse in secons. If h is restricte to the omain [3.5, 0.5], fin an interpret the meaning of h (20). (6) The value v of a car in ollars after t years of ownership can be moele by v(t)= t. Fin v (4000) an eplain in wors what it represents.

18 8 (7) The louness (in ecibels) is given by the equation ( ) I (I)=0 log 0 I 0 where I is the given intensity an I 0 is the threshol soun (the quietest etectable intensity). Determine (85) in terms of the threshol soun. (8) What is the ifference in meaning between f () an f ()? (9) Sort the following epressions into two equivalent groups: sin 2, sin() 2, (sin ) 2, sin( 2 ), sin 2, (sin )(sin ) (0) Sort the following epressions into two equivalent groups: () Is arcsin(), (sin ), sin (), 2 = 3 3? Eplain your reasoning. sin()

19 y 3 2 Limits The Basic Ieas of Limits Consier the function: f ()= While f () is unefine at = 2, we can still plot f () at other values, see Figure.. Eamining Table., we see that as approaches 2, f () approaches. We write this: As 2, f () or lim f ()=. 2 Intuitively, lim f ()=L when the value of f () can be mae arbitrarily close to L a by making sufficiently close, but not equal to, a. This leas us to the formal efinition of a limit. Definition The limit of f () as goes to a is L, 2 3 Figure.: A plot of f ()= f () unefine f () 2 unefine Table.: Values of f ()= lim a f ()=L, if for every ε > 0 there is a δ > 0 so that whenever 0 < a < δ, we have f () L < ε. If no such value of L can be foun, then we say that lim a f () oes not eist. Equivalently, lim a f ()=L, if for every ε > 0 there is a δ > 0 so that whenever a an a δ < < a+δ, we have L ε < f () < L+ ε. In Figure.2, we see a geometric interpretation of this efinition. Limits nee not eist, let s eamine two cases of this.

20 20 L+ ε y Figure.2: A geometric interpretation of the (ε, δ)- criterion for limits. If 0 < a < δ, then we have that a δ < < a+ δ. In our iagram, we see that for all such we are sure to have L ε < f () < L+ε, an hence f () L < ε. L L ε a δ a a+ δ Eample.. Let f ()=. Eplain why the limit oes not eist. lim f () 2 Solution The function is the function that returns the greatest integer less than or equal to. Since f () is efine for all real numbers, one might be tempte to think that the limit above is simply f (2)=2. However, this is not the case. If < 2, then f ()=. Hence if ε=.5, we can always fin a value for (just to the left of 2) such that 0 < 2 < δ, where ε < f () 2. On the other han, lim 2 f (), as in this case if ε=.5, we can always fin a value for (just to the right of 2) such that 0 < 2 < δ, where ε < f (). We ve illustrate this in Figure.3. Moreover, no matter what value one chooses y Figure.3: A plot of f ()=. Note, no matter which δ > 0 is chosen, we can only at best boun f () in the interval [, 2]. With the eample of f ()=, we see that taking limits is truly ifferent from evaluating functions.

21 calculus 2 for lim 2 f (), we will always have a similar issue. Limits may not eist even if the formula for the function looks innocent. ( ) Eample..2 Let f () = sin. Eplain why the limit lim f () 0 y oes not eist. Solution In this case f () oscillates willy as approaches 0, see Figure.4. In fact, one can show that for any given δ, There is a value for in the interval 0 δ < < 0+δ such that f () is any value in the interval [, ]. Hence the limit oes not eist. Sometimes the limit of a function eists from one sie or the other (or both) even though the limit oes not eist. Since it is useful to be able to talk about this situation, we introuce the concept of a one-sie limit: Definition We say that the limit of f () as goes to a from the left is L, ( ) Figure.4: A plot of f ()=sin. lim a f ()=L if for every ε > 0 there is a δ > 0 so that whenever < a an a δ < we have f () L < ε. We say that the limit of f () as goes to a from the right is L, lim a+ f ()=L Limits from the left, or from the right, are collectively calle one-sie limits. if for every ε > 0 there is a δ > 0 so that whenever > a an < a+ δ we have f () L < ε.

22 22 Eample..3 Let f () =. Discuss lim 2 f (), lim f (), an lim f () Solution From the plot of f (), see Figure.3, we see that lim f ()=, an lim f ()= Since these limits are ifferent, lim 2 f () oes not eist.

23 calculus 23 Eercises for Section. y 0 () Evaluate the epressions by referencing the plot in Figure.5. (a) lim 4 f () (b) lim 3 f () (c) lim f () 0 () lim f () 0 (e) lim f () 0+ (f) f ( 2) (g) (h) lim f () 2 lim f () 2 (2) Use a table an a calculator to estimate lim 0 (3) Use a table an a calculator to estimate lim 0 (4) Use a table an a calculator to estimate lim 0 sin(). (i) lim 0 f (+ ) (j) f (0) (k) sin(2). sin ( 3 tan(3) (5) Use a table an a calculator to estimate lim 0 tan(5). (6) Use a table an a calculator to estimate lim 0 ). 2. (7) Use a table an a calculator to estimate lim 0(+) /. (8) Sketch a plot of f ()= ( π (9) Let f ()=sin an eplain why lim oes not eist. 0 ). Construct three tables of the following form (l) lim f ( 4) lim f ( 2) Figure.5: A plot of f (), a piecewise efine function f () where =, 3, 7. What o you notice? How o you reconcile the entries in your tables with the value of lim f ()? 0

24 24 (0) In the theory of special relativity, a moving clock ticks slower than a stationary observer s clock. If the stationary observer recors that t s secons have passe, then the clock moving at velocity v has recore that t v = t s v 2 /c 2 secons have passe, where c is the spee of light. What happens as v c from below?

25 calculus 25.2 Limits by the Definition Now we are going to get our hans irty, an really use the efinition of a limit. Eample.2. Show that lim 2 2 = 4. Recall, lim a f ()=L, if for every ε > 0 there is a δ > 0 so that whenever 0 < a < δ, we have f () L < ε. Solution We want to show that for any given ε > 0, we can fin a δ > 0 such that 2 4 < ε y whenever 0 < 2 < δ. Start by factoring the left-han sie of the inequality above < ε. Since we are going to assume that 0 < 2 < δ, we will focus on the factor + 2. Since is assume to be close to 2, suppose that [, 3]. In this case =5, 4+ε 4 4 ε an so we want 5 2 < ε 2 < ε 5 Recall, we assume ( that [, 3], which is equivalent to 2. Hence we ε ) must set δ= min 5,. 2 δ 2 2+δ Figure.6: The (ε, δ)-criterion for lim 2 = 4. Here ( 2 ε ) δ= min 5,. When ealing with limits of polynomials, the general strategy is always the same. Let p() be a polynomial. If showing lim a p()=l, one must first factor out a from p() L. Net boun [a, a+ ] an estimate the largest possible value of p() L a

26 26 ( ε ) for [a, a+ ], call this estimation M. Finally, one must set δ= min M,. As you work with limits, you fin that you nee to o the same proceures again an again. The net theorems will epeite this process. Theorem.2.2 (Limit Prouct Law) Suppose lim f ()=L an lim g()=m. a a Then lim f ()g()=lm. a Proof Given any ε we nee to fin a δ such that 0 < a < δ implies f ()g() LM < ε. Here we use an algebraic trick, a 0= f ()M+ f ()M: f ()g() LM = f ()g() f ()M+ f ()M LM = f ()(g() M)+(f () L)M f ()(g() M) + (f () L)M = f () g() M + f () L M. Since lim a f ()=L, there is a value δ so that 0 < a < δ implies f () L < ε/(2m). This means that 0 < a < δ implies f () L M < ε/2. We will use this same trick again of aing 0 in the proof of Theorem 5... This is all straightforwar ecept perhaps for the. This follows from the Triangle Inequality. The Triangle Inequality states: If a an b are any real numbers then a+ b a + b. f ()g() LM f () g() M + f () L M. } {{ } ε If we can make f () g() M < ε/2, then we ll be one. We can make g() M smaller than any fie number by making close enough to a. Unfortunately, ε/(2f ()) is not a fie number since is a variable. Here we nee another trick. We can fin a δ 2 so that a < δ 2 implies that f () L <, meaning that L < f () < L+. This means that f () < N, < 2

27 calculus 27 where N is either L or L+, epening on whether L is negative or positive. The important point is that N oesn t epen on. Finally, we know that there is a δ 3 so that 0 < a < δ 3 implies g() M < ε/(2n). Now we re reay to put everything together. Let δ be the smallest of δ, δ 2, an δ 3. Then a < δ implies that so f ()g() LM f () g() M + f () L M. }{{}} {{ }} {{ } <N ε ε < 2N < 2 f ()g() LM f () g() M + f () L M < N ε 2N + ε 2M M = ε 2 + ε 2 = ε. This is just what we neee, so by the efinition of a limit, lim a f ()g()=lm. Another useful way to put functions together is composition. If f () an g() are functions, we can form two functions by composition: f (g()) an g(f ()). For eample, if f ()= an g()= 2 + 5, then f (g())= an g(f ())= ( ) 2 + 5=+ 5. This brings us to our net theorem. Theorem.2.3 (Limit Composition Law) Suppose that lim g()=man a lim f ()=f (M). Then M lim f (g())=f (M). a This is sometimes written as lim f (g())= lim f (g()). a g() M Warning You may be tempte to think that the conition on f () in Theorem.2.3 is unneee, an that it will always be the case that if lim a g()=m an lim M f ()=L then lim a f (g())=l.

28 28 However, consier 3 if = 2, f ()= 4 if 2. an g()=2. Now the conitions of Theorem.2.3 are not satisfie, an lim f (g())=3 but lim f ()=4. 2 Many of the most familiar functions o satisfy the conitions of Theorem.2.3. For eample: Theorem.2.4 (Limit Root Law) Suppose that n is a positive integer. Then lim a provie that a is positive if n is even. n = n a, This theorem is not too ifficult to prove from the efinition of limit.

29 calculus 29 Eercises for Section.2 () For each of the following limits, lim f ()=L, use a graphing evice to fin δ such that a 0 < a < δ implies that f () L < ε where ε=.. (a) lim(3+ )=7 2 (b) lim( 2 + 2)=3 (c) lim sin()=0 π () lim tan()=0 0 (e) lim 3+ =2 (f) 4= 3 lim 2 The net set of eercises are for avance stuents an can be skippe on first reaing. ( ) (2) Use the efinition of limits to eplain why lim sin = 0. Hint: Use the fact that 0 sin(a) for any real number a. (3) Use the efinition of limits to eplain why lim 5)=3. 4(2 (4) Use the efinition of limits to eplain why lim )=. 3( 4 (5) Use the efinition of limits to eplain why lim π= π (6) Use the efinition of limits to eplain why lim = 4. (7) Use the efinition of limits to eplain why lim 3 = (8) Use the efinition of limits to eplain why lim ( )=3. 9 (9) Use the efinition of limits to eplain why lim = (0) Use the efinition of limits to eplain why lim 2 = 2.

30 30.3 Limit Laws In this section, we present a hanful of tools to compute many limits without eplicitly working with the efinition of limit. Each of these coul be prove irectly as we i in the previous section. Theorem.3. (Limit Laws) Suppose that lim f ()=L, lim g()=m, k is a a some constant, an n is a positive integer. Constant Law lim a kf ()=klim a f ()=kl. Sum Law lim a (f ()+g())= lim a f ()+ lim a g()=l+ M. Prouct Law lim a (f ()g())= lim a f () lim a g()=lm. f () Quotient Law lim a g() = lim a f () Power Law lim a f () n = lim a g() = L, if M 0. M ( ) n= lim f () L n. a Root Law lim n f ()= n lim f ()= n L provie if n is even, then f () 0 a a near a. Composition Law If lim g()=m an lim f ()=f (M), then lim f (g())= a M a f (M). Roughly speaking, these rules say that to compute the limit of an algebraic epression, it is enough to compute the limits of the innermost bits an then combine these limits. This often means that it is possible to simply plug in a value for the variable, since lim a = a Eample.3.2 Compute lim. 2

31 calculus 3 Solution Using limit laws, lim = lim lim ( 2) = lim 2 lim 3+ lim 5 lim lim 2 = (lim ) 2 3 lim + 5 lim 2 = = 3+5 = 3. It is worth commenting on the trivial limit lim 5. From one point of view this might seem meaningless, as the number 5 can t approach any value, since it is simply a fie number. But 5 can, an shoul, be interprete here as the function that has value 5 everywhere, f ()=5, with graph a horizontal line. From this point of view it makes sense to ask what happens to the height of the function as approaches. We re primarily intereste in limits that aren t so easy, namely limits in which a enominator approaches zero. The basic iea is to ivie out by the offening factor. This is often easier sai than one here we give two eamples of algebraic tricks that work on many of these limits Eample.3.3 Compute lim. Solution We can t simply plug in = because that makes the enominator zero. However, when taking limits we assume : Limits allow us to eamine functions where they are not efine ( )(+ 3) lim = lim = lim(+ 3)=4

32 32 Eample.3.4 Compute lim Solution Using limit laws, lim = lim = lim = lim = lim (+ )( + 5+2) + (+ )( + 5+2) = Here we are rationalizing the numerator by multiplying by the conjugate. We ll conclue with one more theorem that will allow us to compute more ifficult limits. Theorem.3.5 (Squeeze Theorem) Suppose that g() f () h() for all close to a but not necessarily equal to a. If then lim a f ()=L. lim g()=l=lim h(), a a Eample.3.6 Compute lim 0 sin(). For a nice iscussion of this limit, see: Richman, Fre. A circular argument. College Math. J. 24 (993), no. 2, The limit in this eample will be use in Theorem 7.., an we will give another erivation of this limit in Eample 8..2.

33 calculus 33 Solution To compute this limit, use the( Squeeze Theorem, Theorem.3.5. First π note that we only nee to eamine 2, π ) an for the present time, we ll 2 assume that is positive consier the iagrams below: v v v sin() tan() cos() u u Triangle A Sector Triangle B From our iagrams above we see that u Area of Triangle A Area of Sector Area of Triangle B an computing these areas we fin cos() sin() ( ) π tan() 2 2π 2. Multiplying through by 2, an recalling that tan()= sin() we obtain cos() cos() sin() sin() cos(). Diviing through by sin() an taking the reciprocals, we fin cos() sin() cos().

34 34 Note, cos( )=cos() an sin( ) = sin(), so these inequalities hol for all ( π 2, π ). Aitionally, we know 2 lim cos()== lim 0 0 cos(), sin() an so we conclue by the Squeeze Theorem, Theorem.3.5, lim =. 0

35 calculus 35 Eercises for Section.3 Compute the limits. If a limit oes not eist, eplain why () lim (2) lim (3) lim (4) lim (5) lim (6) lim 0+ (7) lim (8) lim (9) lim 0 2 (0) lim 2 2 () lim 0+ (2) lim 0+ 3 a 3 (3) lim a a (4) lim 2( 2 + 4) 3 5 if, (5) lim 7 if =.

36 2 Infinity an Continuity y Infinite Limits Consier the function f ()= (+ ) 2 While the lim f () oes not eist, see Figure 2., something can still be sai Definition If f () grows arbitrarily large as approaches a, we write lim f ()= a an say that the limit of f () approaches infinity as goes to a. write If f () grows arbitrarily large as approaches a an f () is negative, we lim f ()= a an say that the limit of f () approaches negative infinity as goes to a Figure 2.: A plot of f ()= (+ ) On the other han, if we consier the function f ()= ( ) While we have lim f () ±, we o have one-sie limits, lim f ()= an + lim f ()=, see Figure 2.2.

37 calculus 37 Definition If lim a f ()=±, lim f ()=±, or lim f ()=±, a+ a then the line = a is a vertical asymptote of f (). 40 y Eample 2.. Fin the vertical asymptotes of 20 f ()= Solution Start by factoring both the numerator an the enominator: = ( 2)( 7) ( 2)( 3) Using limits, we must investigate when 2 an 3. Write ( 2)( 7) lim 2 ( 2)( 3) = lim ( 7) 2 ( 3) = 5 = Figure 2.2: A plot of f ()=. Now write ( 2)( 7) lim 3 ( 2)( 3) = lim ( 7) 3 ( 3) 4 = lim 3 3. Since lim 3 approaches 0 from the right an the numerator is negative, 3+ lim f ()=. Since lim 3 approaches 0 from the left an the numerator 3+ 3 is negative, lim f ()=. Hence we have a vertical asymptote at = 3, see 3 Figure y Figure 2.3: A plot of f ()=

38 38 Eercises for Section 2. Compute the limits. If a limit oes not eist, eplain why. () lim (2) lim (3) lim + 3 (4) lim (5) lim 5 ( 5) 4 (6) lim 2 ( ) 2 (7) lim 0 cos() 5 (8) lim 0+ sin() (9) Fin the vertical asymptotes of (0) Fin the vertical asymptotes of f ()= f ()=

39 calculus Limits at Infinity Consier the function: f ()= 6 9 As approaches infinity, it seems like f () approaches a specific value. This is a limit at infinity. y 20 0 Definition If f () becomes arbitrarily close to a specific value L by making sufficiently large, we write lim f ()=L an we say, the limit at infinity of f () is L. If f () becomes arbitrarily close to a specific value L by making sufficiently large an negative, we write lim f ()=L an we say, the limit at negative infinity of f () is L Figure 2.4: A plot of f ()= 6 9. Eample 2.2. Compute 6 9 lim. Solution Write 6 9 lim = lim = lim 6 = lim = / / 6 9 Sometimes one must be careful, consier this eample.

40 40 Eample Compute + lim 2 Solution In this case we multiply the numerator an enominator by /, which is a positive number as since, is a negative number. + lim = lim + / 2 2 / / = lim 2 / 2 =. y 5 Here is a somewhat ifferent eample of a limit at infinity. 4.5 Eample Compute sin(7) lim Solution We can boun our function Since /+ 4 sin(7) + 4 /+ 4. lim /+ 4=4= lim /+ 4 sin(7) we conclue by the Squeeze Theorem, Theorem.3.5, lim + 4= Figure 2.5: A plot of f ()= sin(7) + 4. Definition If lim f ()=L or lim f ()=L, then the line y=l is a horizontal asymptote of f ().

41 calculus 4 Eample Give the horizontal asymptotes of f ()= 6 9 Solution From our previous work, we see that lim f ()=6, an upon further inspection, we see that lim f ()=6. Hence the horizontal asymptote of f () is the line y=6. It is a common misconception that a function cannot cross an asymptote. As the net eample shows, a function can cross an asymptote, an in this case this occurs an infinite number of times! Eample Give a horizontal asymptote of f ()= sin(7) + 4. Solution Again from previous work, we see that lim f ()=4. Hence y=4is a horizontal asymptote of f (). We conclue with an infinite limit at infinity. y 4 3 Eample Compute lim ln() 2 Solution The function ln() grows very slowly, an seems like it may have a horizontal asymptote, see Figure 2.6. However, if we consier the efinition of the natural log ln()=y e y = Since we nee to raise e to higher an higher values to obtain larger numbers, we see that ln() is unboune, an hence lim ln()=. Figure 2.6: A plot of f ()=ln().

42 42 Eercises for Section 2.2 Compute the limits. () lim (2) lim (3) lim (4) lim 2 (5) lim (6) lim ( 4 ) (7) lim + π cos() (8) lim ln() sin ( 7) (9) lim ( (0) lim ) (sin(/2))2 3 () Suppose a population of feral cats on a certain college campus t years from now is approimate by 000 p(t)= 5+2e. 0.t Approimately how many feral cats are on campus 0 years from now? 50 years from now? 00 years from now? 000 years from now? What o you notice about the preiction is this realistic? (2) The amplitue of an oscillating spring is given by a(t)= sin(t). t What happens to the amplitue of the oscillation over a long perio of time?

43 calculus Continuity Informally, a function is continuous if you can raw it without lifting your pencil. We nee a formal efinition. Definition A function f is continuous at a point a if lim a f ()=f (a). y Eample 2.3. Fin the iscontinuities (the values for where a function is not continuous) for the function given in Figure 2.7. Solution From Figure 2.7 we see that lim 4 f () oes not eist as lim f ()= an lim f () Hence lim f () f (4), an so f () is not continuous at = 4. 4 We also see that lim f () 3while f (6)=2. Hence lim f () f (6), an so 6 6 f () is not continuous at = 6. Builing from the efinition of continuous at a point, we can now efine what it means for a function to be continuous on an interval Figure 2.7: A plot of a function with iscontinuities at = 4 an = 6. y Definition A function f is continuous on an interval if it is continuous at every point in the interval. In particular, we shoul note that if a function is not efine on an interval, then it cannot be continuous on that interval. Eample Consier the function ( 5 ) sin if 0, f ()= 0 if = 0, see Figure 2.8. Is this function continuous? Figure 2.8: A plot of ( 5 ) sin if 0, f ()= 0 if = 0.

44 44 Solution Consiering f (), the only issue is when = 0. We must show that lim f ()=0. Note 0 5 f () 5. Since lim 5 =0= lim 5, 0 0 we see by the Squeeze Theorem, Theorem.3.5, that lim 0 f ()=0. Hence f () is continuous. Here we see how the informal efinition of continuity being that you can raw it without lifting your pencil iffers from the formal efinition. We close with a useful theorem about continuous functions: The Intermeiate Value Theorem is most frequently use when = 0. Theorem (Intermeiate Value Theorem) If f () is a continuous function for all in the close interval [a, b] an is between f (a) an f (b), then there is a number c in [a, b] such that f (c)=. In Figure 2.9, we see a geometric interpretation of this theorem. For a nice proof of this theorem, see: Walk, Stephen M. The intermeiate value theorem is NOT obvious an I am going to prove it to you. College Math. J. 42 (20), no. 4, y Eample Eplain why the function f ()= has a root between 0 an. Solution By Theorem.3., lim a f ()=f (a), for all real values of a, an hence f is continuous. Since f (0) = 2 an f () = 3, an 0 is between 2 an 3, by the Intermeiate Value Theorem, Theorem 2.3.3, there is a c [0, ] such that f (c)=0. f (b) f (c)= This eample also points the way to a simple metho for approimating roots. Eample Approimate a root of f ()= to one ecimal place. f (a) a c b Solution If we compute f (0.), f (0.2), an so on, we fin that f (0.6) < 0 an f (0.7) > 0, so by the Intermeiate Value Theorem, f has a root between 0.6 an 0.7. Repeating the process with f (0.6), f (0.62), an so on, we fin Figure 2.9: A geometric interpretation of the Intermeiate Value Theorem. The function f () is continuous on the interval [a, b]. Since is in the interval [f (a), f (b)], there eists a value c in [a, b] such that f (c)=.

45 calculus 45 that f (0.6) < 0 an f (0.62) > 0, so by the Intermeiate Value Theorem, Theorem 2.3.3, f () has a root between 0.6 an 0.62, an the root is 0.6 roune to one ecimal place.

46 46 Eercises for Section 2.3 () Consier the function f ()= 4 Is f () continuous at the point = 4? Is f () a continuous function onr? (2) Consier the function f ()= + 3 Is f () continuous at the point = 3? Is f () a continuous function onr? (3) Consier the function 2 3 if <, f ()= 0 if. Is f () continuous at the point =? Is f () a continuous function onr? (4) Consier the function if 5, f ()= 5 0 if = 5. Is f () continuous at the point = 5? Is f () a continuous function onr? (5) Consier the function if 5, f ()= if = 5. Is f () continuous at the point = 5? Is f () a continuous function onr? (6) Determine the interval(s) on which the function f ()= is continuous. (7) Determine the interval(s) on which the function f ()= is continuous. (8) Determine the interval(s) on which the function f ()= is continuous. 2 9 (9) Approimate a root of f ()= to two ecimal places. (0) Approimate a root of f ()= to two ecimal places.

47 3 Basics of Derivatives 3. Slopes of Tangent Lines via Limits Suppose that f () is a function. It is often useful to know how sensitive the value of f () is to small changes in. To give you a feeling why this is true, consier the following: If p(t) represents the position of an object with respect to time, the rate of change gives the velocity of the object. If v(t) represents the velocity of an object with respect to time, the rate of change gives the acceleration of the object. The rate of change of a function can help us approimate a complicate function with a simple function. The rate of change of a function can be use to help us solve equations that we woul not be able to solve via other methos. The rate of change of a function is the slope of the tangent line. For now, consier the following informal efinition of a tangent line: Given a function f (), if one can zoom in on f () sufficiently so that f () seems to be a straight line, then that line is the tangent line to f () at the point etermine by. We illustrate this informal efinition with Figure 3.. The erivative of a function f () at, is the slope of the tangent line at. To fin the slope of this line, we consier secant lines, lines that locally intersect the curve

48 48 y at two points. The slope of any secant line that passes through the points (, f ()) an (+ h, f (+ h)) is given by Figure 3.: Given a function f (), if one can zoom in on f () sufficiently so that f () seems to be a straight line, then that line is the tangent line to f () at the point etermine by. y = f (+ h) f () (+ h) f (+ h) f () =, h y see Figure 3.2. This leas to the limit efinition of the erivative: Definition of the Derivative The erivative of f () is the function f ()= lim h 0 f (+ h) f (). h If this limit oes not eist for a given value of, then f () is not ifferentiable at. f (+ h) f () Figure 3.2: Tangent lines can be foun as the limit of secant lines. The slope of the tangent line is given f (+ h) f () by lim. h 0 h + h

49 calculus 49 Definition There are several ifferent notations for the erivative, we ll mainly use f ()=f (). If one is working with a function of a variable other than, say t we write t f (t)=f (t). However, if y=f (), y, ẏ, an D f () are also use. Now we will give a number of eamples, starting with a basic eample. Eample 3.. Compute (3 + ). Solution Using the efinition of the erivative, (+ h) 3 + ( 3 + ) f ()= lim h 0 h h+ 3h 2 + h = lim h 0 h 3 2 h+ 3h 2 + h 3 = lim h 0 h = lim( h+ h 2 ) h 0 = 3 2. f () y 4 2 f () See Figure 3.3. Net we will consier the erivative a function that is not continuous onr Eample 3..2 Compute t t. 2 4 Figure 3.3: A plot of f ()= 3 + an f ()=3 2.

50 50 Solution Using the efinition of the erivative, t t = lim h 0 = lim h 0 = lim h 0 t+h t h t t(t+h) t+h t(t+h) t (t+h) t(t+h) h h t t h = lim h 0 t(t+ h)h = lim h 0 h t(t+ h)h = lim h 0 t(t+ h) = t. 2 This function is ifferentiable at all real numbers ecept for t= 0, see Figure 3.4. As you may have guesse, there is some connection to continuity an ifferentiability. y 4 2 f (t) t f (t) 2 Theorem 3..3 (Differentiability Implies Continuity) If f () is a ifferentiable function at = a, then f () is continuous at = a. 4 Figure 3.4: A plot of f (t)= t an f (t)= t 2. Proof We want to show that f () is continuous at = a, hence we must show that lim f ()=f (a). a

51 calculus 5 Consier lim a (f () f (a))= lim a ( ) f () f (a) ( a) a f (a+ h) f (a) = lim h h 0 h ( )( ) = lim h f (a+ h) f (a) lim h 0 h 0 h = 0 f (a)=0. Multiply an ivie by ( a). Set = a+ h. Limit Law. Since lim (f () f (a))=0 a we see that lim a f ()=f (a), an so f () is continuous. y 3 This theorem is often written as its contrapositive: If f () is not continuous at = a, then f () is not ifferentiable at = a. 2 f (t) Let s see a function that is continuous whose erivative oes not eist everywhere. Eample 3..4 Compute Solution Using the efinition of the erivative, = lim h 0 + h. h If is positive we may assume that is larger than h, as we are taking the limit as h goes to 0, + h + h lim = lim h 0 h h 0 h h = lim h 0 h =. f (t) 2 Figure 3.5: A plot of f () = an f if > 0, ()= if < 0. If is negative we may assume that is larger than h, as we are taking the

52 52 limit as h goes to 0, + h h+ lim = lim h 0 h h 0 h h = lim h 0 h =. However we still have one case left, when = 0. In this situation, we must consier the one-sie limits: In the first case, + h lim h 0+ h an lim h 0 + h. h + h 0+h 0 lim = lim h 0+ h h 0+ h h = lim h 0+ h =. On the other han + h 0+h 0 lim = lim h 0 h h 0 h h = lim h 0 h =. Hence we see that the erivative is f if > 0, ()= if < 0. Note this function is unefine at 0, see Figure 3.5. Thus from Theorem 3..3, we see that all ifferentiable functions on R are continuous on R. Nevertheless as the previous eample shows, there are continuous functions onrthat are not ifferentiable onr.

53 calculus 53 Eercises for Section 3. These eercises are conceptual in nature an require one to think about what the erivative means. () If the line y=7 4 is tangent to f () at = 2, fin f (2) an f (2). (2) Here are plots of four functions. y 4 y 4 y 4 y p() q() r() s() Two of these functions are the erivatives of the other two, ientify which functions are the erivatives of the others. (3) If f (3)=6an f (3.)=6.4, estimate f (3). (4) If f ( 2)=4an f ( 2+h)=(h+ 2) 2, compute f ( 2). (5) If f ()= 3 an f ()=2, approimate f (.2). y (6) Consier the plot of f () in Figure 3.6. (a) On which subinterval(s) of [0, 6] is f () continuous? (b) On which subinterval(s) of [0, 6] is f () ifferentiable? (c) Sketch a plot of f () Figure 3.6: A plot of f ().

54 54 These eercises are computational in nature. (7) Let f ()= 2 4. Use the efinition of the erivative to compute f ( 3) an fin the equation of the tangent line to the curve at = 3. (8) Let f ()= + 2. Use the efinition of the erivative to compute f () an fin the equation of the tangent line to the curve at =. (9) Let f ()= 3. Use the efinition of the erivative to compute f (5) an fin the equation of the tangent line to the curve at = 5. (0) Let f ()=. Use the efinition of the erivative to compute f (4) an fin the equation of the tangent line to the curve at = 4.

55 calculus Basic Derivative Rules It is teious to compute a limit every time we nee to know the erivative of a function. Fortunately, we can evelop a small collection of eamples an rules that allow us to compute the erivative of almost any function we are likely to encounter. We will start simply an buil-up to more complicate eamples. The Constant Rule The simplest function is a constant function. Recall that erivatives measure the rate of change of a function at a given point. Hence, the erivative of a constant function is zero. For eample: The constant function plots a horizontal line so the slope of the tangent line is 0. If p(t) represents the position of an object with respect to time an p(t) is constant, then the object is not moving, so its velocity is zero. Hence t p(t)=0. If v(t) represents the velocity of an object with respect to time an v(t) is constant, then the object s acceleration is zero. Hence t v(t)=0. The eamples above lea us to our net theorem. To gain intuition, you shoul compute the erivative of f ()=6using the limit efinition of the erivative. Theorem 3.2. (The Constant Rule) Given a constant c, c= 0. Proof From the limit efinition of the erivative, write c c c= lim h 0 h 0 = lim h 0 h = lim 0=0. h 0

56 56 The Power Rule Now let s eamine erivatives of powers of a single variable. Here we have a nice rule. Theorem (The Power Rule) For any real number n, n = n n. To gain intuition, you shoul compute the erivative of f ()= 3 using the limit efinition of the erivative. Proof At this point we will only prove this theorem for n being a positive integer. Later in Section 6.3, we will give the complete proof. From the limit efinition of the erivative, write Start by epaning the term (+ h) n n = lim h 0 n + ( ) n (+ h) n n n = lim h 0 h n h+ ( n 2 ) n 2 h ( n n. ) h n + h n n h ( ) n Note, by the Binomial Theorem, we write for the coefficients. Canceling the ( ) ( k ) n terms n an n n, an noting = = n, write n n n h+ ( ) n n 2 h ( ) n h n + h n n 2 n = lim h 0 h = lim h 0 n n + ( ) ( n n 2 h+ + 2 ) n n h n 2 + h n. Recall, the Binomial Theorem states that if n is a nonnegative integer, then ( ) ( ) n (a+b) n = a n b 0 + a n b n + + a b n +a 0 b n n where ( ) n n! = k k!(n k)!. Since every term but the first has a factor of h, we see (+ h) n n n = lim h 0 h = n n. Now we will show you several eamples. We begin with something basic.

57 calculus 57 Eample Compute 3. Solution Applying the power rule, we write 3 = 3 2. Sometimes, it is not as obvious that one shoul apply the power rule. Eample Compute 4. Solution Applying the power rule, we write 4= 4 = 4 5. The power rule also applies to raicals once we rewrite them as eponents. Eample Compute 5. Solution Applying the power rule, we write 5 = /5 = 4/5 5. The Sum Rule We want to be able to take erivatives of functions one piece at a time. The sum rule allows us to o this. The sum rule says that we can a the rates of change of two functions to obtain the rate of change of the sum of both functions. For eample, viewing the erivative as the velocity of an object, the sum rule states that the velocity of the person walking on a moving bus is the sum of the velocity of the bus an the walking person.

58 58 y Theorem (The Sum Rule) If f () an g() are ifferentiable an c is a constant, then ( ) (a) f ()+g() = f ()+g (), ( ) (b) f () g() = f () g (), ( ) (c) c f () = c f (). Proof We will only prove part (a) above, the rest are similar. Write f ()+g() f () g() } {{ } h f (a)h+ g (a)h g (a)h f (a)h ( ) f (+ h)+g(+ h) (f ()+g()) f ()+g() = lim h 0 h f (+ h)+g(+ h) f () g() = lim h 0 h f (+ h) f ()+g(+ h) g() = lim h 0 h ( ) f (+ h) f () g(+ h) g() = lim + h 0 h h f (+ h) f () g(+ h) g() = lim + lim h 0 h h 0 h = f ()+g (). Figure 3.7: A geometric interpretation of the sum rule. Since every point on f ()+g() is the sum of the corresponing points on f () an g(), increasing a by a small amount h, increases f (a)+g(a) by the sum of f (a)h an g (a)h. Hence, y f (a)h+ g (a)h = f (a)+g (a). h a Eample Compute ( 5 + ). Solution Write ( 5 + ) = 5 + =

59 calculus 59 Eample Compute ( ). 7 Solution Write ( ) = 3 7 /3 2 /2 + 7 = 4/3 / The Derivative of e We on t know anything about erivatives that allows us to compute the erivatives of eponential functions without getting our hans irty. Let s o a little work with the efinition of the erivative: a +h a a = lim h 0 h a a h a = lim h 0 h = lim a ah h 0 h = a a h lim h 0 h = a (constant) } {{ } lim h 0 a h h There are two interesting things to note here: We are left with a limit that involves h but not, which means that whatever lim(a h )/h is, we know that it is a number, h 0 that is, a constant. This means that a has a remarkable property: Its erivative is a constant times itself. Unfortunately it is beyon the scope of this tet to compute the limit lim h 0 a h. h However, we can look at some eamples. Consier (2 h )/h an (3 h )/h: