Integrated Calculus/Pre-Calculus

Furman University

Department of Mathematics Greenville, SC, 29613 Integrated Calculus/Pre-Calculus

Abstract Mark R. Woodard Contents JJ II J I Home Page Go Back Close Quit Abstract

This work presents the traditional material of calculus I with some of the material from a traditional precalculus course interwoven throughout the discussion. Pre- calculus topics are discussed at or soon before the time they are needed, in order to facilitate the learning of the calculus material. Miniature animated demonstra- tions and interactive quizzes will be available to help the reader deepen his or her understanding of the material under discussion. Integrated This project is funded in part by the Mellon Foundation through the Mellon Furman- Calculus/Pre-Calculus Woﬀord Program. Many of the illustrations were designed by Furman undergraduate Mark R. Woodard student Brian Wagner using the PSTricks package. Thanks go to my wife Suzan, and my two daughters, Hannah and Darby for patience while this project was being produced.

Title Page Contents JJ II J I Go Back Close Quit Page 2 of 191 Contents

1 Functions and their Properties 11 1.1 Functions & Cartesian Coordinates ...... 13 1.1.1 Functions and Functional Notation ...... 13 1.2 Cartesian Coordinates and the Graphical Representa- tion of Functions ...... 20 1.3 Circles, Distances, Completing the Square ...... 22 1.3.1 Distance Formula and Circles ...... 22 Integrated 1.3.2 Completing the Square ...... 23 Calculus/Pre-Calculus 1.4 Lines ...... 24 Mark R. Woodard 1.4.1 General Equation and Slope-Intercept Form .... 24 1.4.2 More On Slope ...... 24 1.4.3 Parallel and Perpendicular Lines ...... 25 Title Page 1.5 Catalogue of Familiar Functions ...... 27 1.5.1 Polynomials ...... 27 Contents 1.5.2 Rational Functions ...... 28 JJ II 1.5.3 Algebraic Functions ...... 28 1.5.4 The Absolute Value and Other Piecewise De- J I fined Functions ...... 29 Go Back 1.5.5 The Greatest Integer Function ...... 30 Close 1.6 New Functions From Old ...... 31 1.6.1 New Functions Via Composition ...... 31 Quit 1.7 New Functions via Algebra ...... 34 Page 3 of 191 1.8 Shifting and Scaling ...... 35 2 Limits 38 2.1 Secant Lines, Tangent Lines, and Velocities ...... 39 2.1.1 The Tangent Line ...... 39 2.1.2 Instantaneous Velocity ...... 41 2.2 Simplifying Difference Quotients ...... 43 2.3 Limits (Intuitive Approach)...... 45 2.4 Intervals via Absolute Values and Inequalities ...... 50 2.5 If – Then Statements ...... 52 2.6 Rigorous Definition of Limit ...... 54 2.7 Continuity ...... 56 2.7.1 Definitions Related to Continuity ...... 56 Integrated 2.7.2 Basic Theorems ...... 57 Calculus/Pre-Calculus 2.7.3 The Intermediate Value Theorem ...... 59 Mark R. Woodard

3 The Derivative 61 3.1 Definition of Derivative ...... 63 3.1.1 Slopes and Velocities Revisited ...... 63 Title Page 3.1.2 The Derivative ...... 64 Contents 3.1.3 The Derivative as a Function ...... 64 3.2 Rules for Taking Derivatives ...... 65 JJ II 3.3 Differentiability ...... 70 J I 3.4 Applications of Derivatives as Rates of Change ...... 72 3.4.1 Overview ...... 72 Go Back 3.4.2 Some Specific Applications ...... 73 Close Rectilinear Motion ...... 73 Biology Applications ...... 74 Quit Chemistry Applications ...... 75 Page 4 of 191 Economics ...... 75 3.5 The Chain Rule ...... 76 3.6 Implicit Differentiation ...... 78 3.7 Higher Derivatives ...... 81 3.7.1 The Second Derivative ...... 81 3.7.2 Third and Higher Derivatives ...... 81 3.8 Related Rates ...... 83 3.9 Linearization ...... 85 3.9.1 Mathematicizing the main idea ...... 85

4 An Interlude – Trigonometric Functions 87 4.1 Definition of Trigonometric Functions ...... 88 Integrated 4.1.1 Radian Measure ...... 90 Calculus/Pre-Calculus 4.2 Trigonometric Functions at Special Values ...... 91 Mark R. Woodard 4.3 Properties of Trigonometric Functions...... 93 4.4 Graphs of Trigonometric Functions ...... 94 4.4.1 Graph of sin(x) and cos(x) ...... 94 Title Page 4.4.2 Graph of tan(x) and cot(x) ...... 94 4.4.3 Graph of sec(x) and csc(x) ...... 95 Contents 4.5 Trigonometric Identities ...... 96 4.6 Derivatives of Trigonometric Functions ...... 97 JJ II 4.6.1 Two Important Trigonometric Limits ...... 97 J I 4.6.2 Derivatives of other Trigonometric Functions ... 99 Go Back 5 Applications of the Derivative 102 Close 5.1 Definition of Maximum and Minimum Values ...... 104 5.2 Overview ...... 110 Quit 5.3 Rolle’s Theorem ...... 111 Page 5 of 191 5.4 The Mean Value Theorem ...... 113 5.5 An Application of the Mean Value Theorem ...... 115 5.6 Monotonicity ...... 117 5.7 Concavity ...... 121 5.8 Vertical Asymptotes ...... 124 5.9 Limits at Infinity; Horizontal Asymptotes ...... 126 5.9.1 A Simplification Technique ...... 127 5.10 Graphing ...... 129 5.11 Optimization ...... 133 5.12 Newton’s Method ...... 137 5.13 Antiderivatives ...... 140 5.14 Rectilinear Motion ...... 142 Integrated 5.15 Antiderivatives Mini-Quiz ...... 143 Calculus/Pre-Calculus Mark R. Woodard 6 Areas and Integrals 145 6.1 Summation Notation...... 147 6.2 Computing Some Sums ...... 149 6.3 Areas of Planar Regions ...... 154 Title Page 6.3.1 Approximating with Rectangles ...... 155 Contents 6.4 The Definite Integral ...... 159 6.4.1 Definition and Notation ...... 159 JJ II 6.4.2 Properties of R b f(x) dx...... 162 a J I 6.4.3 Relation of the Definite Integral to Area ...... 162 6.4.4 Another Interpretation of the Definite Integral .. 164 Go Back 6.5 The Fundamental Theorem ...... 166 Close 6.5.1 The Fundamental Theorem...... 166 6.5.2 The Fundamental Theorem – Part II ...... 168 Quit 6.6 The Indefinite Integral and Substitution ...... 170 Page 6 of 191 6.6.1 The Indefinite Integral ...... 170 6.6.2 Substitution in Indefinite Integrals ...... 173 6.6.3 Substitution in Definite Integrals ...... 176 6.7 Areas Between Curves ...... 179 6.7.1 Vertically Oriented Areas ...... 179 6.7.2 Horizontally Oriented Areas ...... 181

7 Appendix 184 7.1 3 – Minute Reviews ...... 185 7.1.1 The Real Numbers ...... 185 7.1.2 Factoring ...... 186 7.1.3 Solving Equations ...... 186 Integrated 7.1.4 Solving Systems of Equations ...... 187 Calculus/Pre-Calculus 7.1.5 Absolute Value ...... 188 Mark R. Woodard 7.1.6 Quadratics ...... 189 7.1.7 Exponents ...... 190 7.1.8 Inequalities ...... 190 7.1.9 Setting Up Word Problems ...... 190 Title Page Solutions to Exercises ...... 192 Contents Solutions to Quizzes ...... 309 JJ II J I Go Back Close Quit Page 7 of 191 Preface Every teacher of calculus has at one time or another remarked about a particular student or group of students “I think they understand the calculus concepts, but their algebra skills are so weak, they can’t finish the problems.” There are many reasons that explain why this problem is so pre- velant: some students never learned the precalculus material, some learned it but forgot most of it, others are just a little rusty and need some extra practice. For these reasons, some colleges and universities are starting so called “integrated” precalculus/calculus courses. This text is designed to Integrated be used in exactly those kinds of courses, although it could also be used in Calculus/Pre-Calculus a traditional calculus course, with the precalculus material omitted by the Mark R. Woodard instructor, but available nonetheless to the motivated and interested student who needs or desires some extra amplifications of, reminders about, or practice with the precalculus material. An attempt is made to have the precalculus material appear “just in time”, that is, directly before it is Title Page needed for the first time. Contents In addition, this document is unusual in that the online version allows for reading in an interactive style: hypertext links are available, solutions JJ II are offered to many examples and exercises via a hyperlink, and then the J I reader can then return to his or her place in the text via another jump. Miniature quizzes will also be available, which provide immediate feedback Go Back that will hopefully erase misunderstandings and misconceptions before they Close fester and grow. In order to get the reader acclimated to these features, Quit there are some examples provided below. Page 8 of 191 Notes to the reader The electronic version of this document has a number of interactive features, including hyperlinked jumps to solutions of many exercises, as well as interactive quizzes. Examples are below. Answers will be clear to anyone who has carefully read the dedication and acknowledgements. An example of an exercise with a jump to the solution. (You can jump back by via the blue hyperlink on the solutions page.) Exercise 0.0.1. How many children does the author of the document have? Integrated Calculus/Pre-Calculus Here is an example of a short quiz. You will be taken to the “solution” Mark R. Woodard after finding the right answer. You must first click on the colored label “Quiz” to initialize the quiz before it will work.

Quiz Who made most of the illustrations in this document? Title Page (a) Bertolt Brecht (b) Brian Wagner Contents (c) Babe Didrickson (d) Bebe Rebozo JJ II J I Go Back Close Quit Page 9 of 191 Here is an example of a “graded” quiz. You must initialize the quiz by ﬁrst clicking on “Begin Quiz,” and you must end the quiz by clicking on “End Quiz.” To correct your quiz and see the answers, click on the button that says “Correct”. Click to Initialize Quiz Answer the following questions about the author’s aﬃliations. 1. Which university employs the author of this document? Fordham Ferrum Ferris State Furman 2. In what state is the author’s university located? Integrated South Carolina North Carolina Georgia Calculus/Pre-Calculus Mark R. Woodard Click to End Quiz

Title Page

It is hoped that such interactivity will make the reading of the text both Contents more enjoyable, and more eﬀective for the student. Now it’s time to learn JJ II some mathematics! J I Go Back Close Quit Page 10 of 191 CHAPTER 1 Functions and their Properties

Contents

1.1 Functions & Cartesian Coordinates ...... 13 Integrated 1.1.1 Functions and Functional Notation ...... 13 Calculus/Pre-Calculus 1.2 Cartesian Coordinates and the Graphical Rep- Mark R. Woodard resentation of Functions ...... 20 1.3 Circles, Distances, Completing the Square .... 22 1.3.1 Distance Formula and Circles ...... 22 Title Page 1.3.2 Completing the Square ...... 23 Contents 1.4 Lines ...... 24 1.4.1 General Equation and Slope-Intercept Form .... 24 JJ II 1.4.2 More On Slope ...... 24 J I 1.4.3 Parallel and Perpendicular Lines ...... 25 Go Back 1.5 Catalogue of Familiar Functions ...... 27 Close 1.5.1 Polynomials ...... 27 Quit 1.5.2 Rational Functions ...... 28 Page 11 of 191 1.5.3 Algebraic Functions ...... 28 1.5.4 The Absolute Value and Other Piecewise De- ﬁned Functions ...... 29 1.5.5 The Greatest Integer Function ...... 30 1.6 New Functions From Old ...... 31 1.6.1 New Functions Via Composition ...... 31 1.7 New Functions via Algebra ...... 34 1.8 Shifting and Scaling ...... 35

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 12 of 191 1.1. Functions & Cartesian Coordinates 1.1.1. Functions and Functional Notation

The key object of study in a calculus course is a function. Since this is the case, it is important to know what a function is. Many students erroneously think of a function as a formula – this oversimplification and half-truth leads to problems later on in the course, so it is best to get the following definition down pat at the outset. Integrated Definition of Function Calculus/Pre-Calculus Definition 1.1.1 A function is a rule of correspondence Mark R. Woodard between items in one set (called the domain of the function) and items in another set (called the range of the function) so that every element in the domain is paired Title Page with exactly one element in the range. Contents

Note that the definition of function also includes the idea of the domain JJ II of a function and the range of a function. Note also that the definition J I doesn’t say anything about formulas. In fact, a useful way to think about Go Back a function is as a machine. Domain elements are input to the machine, and then the function-machine turns them into (perhaps) something else, Close and these outputs are the range elements. A simple example is illustrated Quit in figure 1.1. An animated version is available here Page 13 of 191 It is important to note that every function comes with three items:

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Every function consists of: • A set called the domain. Title Page • A set called the range. Contents • A rule of correspondence between elements in those JJ II sets J I Go Back In this text, our domain sets will almost always be subsets of the set of Close real numbers (often denoted <), as will our range sets. (There is a review of the set of real numbers in section 7.1.1. Thus, we will think of functions Quit as rules which transform certain numbers into other numbers. There are Page 14 of 191 number of diﬀerent ways to denote or describe a function; we will discuss

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Figure 1.2: An arrow diagram for a function. a few, but will settle on a few to use for the rest of this document. Since we are trying to indicate a relationship between quantities, we can use arrows to show which domain element corresponds to which range element (see figure 1.2.) This is effective when the domain is small, but most of Integrated our domains will have an infinite number of elements, so this approach is Calculus/Pre-Calculus not practical. Another method would be to just write ordered pairs, with Mark R. Woodard domain elements listed first and the corresponding range elements listed second. Thus, the function described by figure 1.2 could instead be denoted {(1, −1), (4, −2), (9, −3)}. Again, this approach is not practical when the domain is infinite. Prob- Title Page ably the best approach for our purposes is to use functional notation. In Contents this notation, the rule of correspondence is given by a formulaic expression, and the domain is either specified directly or is implied. For example, JJ II J I f(x) = x2 + 2, 0 ≤ x ≤ 5 (1.1) Go Back represents the function whose domain consists of the real numbers between Close 0 and 5, and whose rule is such that an input number is paired with the number which is 5 more than the square of the input. Thus, for example, Quit 2 the number 2 would be paired with 2 + 5 = 9. We indicate this by writing Page 15 of 191 f(2) = 9, which you should think of as meaning that the number 2 goes into the function machine f, and 9 comes out. This notation has many advantages, but for now let’s highlight the disadvantages. The range is not mentioned at all, the domain is easy to ignore, and it becomes the habit of many to think of the function as a formula rather than a rule of correspondence between elements in two sets. Resist this habit!. Example 1.1.1. If f(x) = 3x2 − 4, what is f(x + 1)? Solution: Since f is the rule which says that we should square the input, multiply by 3 and then subtract 4, f(x + 1) should be be the result of doing those operations to the quantity (x + 1), thus, the output should be 3(x + 1)2 − 4, so Integrated Calculus/Pre-Calculus f(x + 1) = 3(x + 1)2 − 4 Mark R. Woodard

Note that in the previous example, no domain was specified at all! This will often be the case, even though every function must have a domain. Title Page Thus, there must be a domain implied even when it is not specified. Our Contents rule of thumb for domains will be the following: JJ II Rule of Thumb about Domains Unless otherwise J I specified, the domain of a function written in functional Go Back notation will be taken to be the largest subset of the real Close numbers for which the function is defined. Quit x Page 16 of 191 Example 1.1.2. What is the domain of f(x) = ? x2 − 9

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Figure 1.3: An arrow diagram which does represent a function.

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Integrated Figure 1.4: An arrow diagram which does not represent a function. Calculus/Pre-Calculus Mark R. Woodard Solution: Since there is no domain speciﬁed, the rule of thumb applies.

There is one aspect of the definition of function which we haven’t dis- Title Page cussed yet, and that is the part which says that every domain element “is Contents paired with exactly one element from the range.” Note that this doesn’t say that every range element is paired with exactly one domain element, so JJ II an arrangement like that in figure 1.3 is a function, while that in figure 1.4 J I isn’t. Go Back

Example 1.1.3. If the quantities u and w are related by the relationship Close u2 = w, is u a function of w? Is w a function of u? Quit Solution: It is true that w is a function of u, because for every value of Page 17 of 191 u, there is only one possible value for u2. On the other hand, u is not a function of w, because for a given value of w (like the value 9 for example) there could be two diﬀerent values of u (namely 3 and −3) for which the relationship is met.

In the application of functional notation, we often use the variable x to represent quantities which come from the domain set (this is more formally called the independent variable) and we often use the variable y to represent quantities which come from the range set (more formally called the dependent variable), but there is no reason why these letters necessarily must represent those quantities. For example, it is perfectly OK to write Integrated Calculus/Pre-Calculus x = y2 + 2y Mark R. Woodard and think of x as a function of y, thus making y the independent variable and x the dependent variable for this example. Quiz: Functions and Functional Notation. Title Page Click to Initialize Quiz Answer the following questions about functions and functional notation. When necessary, do you work on a scrap piece of Contents paper, but record your answers on this document. Don’t forget to initialize JJ II the quiz by clicking in the appropriate place before starting, and then click on the appropriate label to ﬁnd you score. J I 1 Go Back 1. If f(z) = z , what is f(z + h)? 1 h 1 Close z + h z z+h 1 Quit 2. If f(y) = y , what is f(z + h) − f(z), in simpliﬁed form? If you need a hint, there is one available here. Page 18 of 191 −h h 1 (z)(z+h) (z)(z+h) h w2+9 3. What is the domain, written in interval notation, for h(w) = w2−2w+1 ? (−∞, 1] ∩ (1, ∞)(−∞, 1) ∪ (1, ∞)(−∞, −1) ∪ (−1, 1) ∪ (1, ∞)

Click to End Quiz

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 19 of 191 1.2. Cartesian Coordinates and the Graph- ical Representation of Functions

The Cartesian Coordinate System is a system by which one can associate numerical quantities with geometric ones, namely, a point in a plane is associated with a pair of real numbers. Our method of associating is as follows: we draw two perpendicular number lines, and call the point where they intersect (0, 0) or, the origin. If we wanted to see which point is associated with (2, 4) for example, we would start at the origin and move Integrated 2 units to the right,and then 4 units up. The ﬁrst coordinate is related Calculus/Pre-Calculus to horizontal movement and the second is related to vertical. The positive Mark R. Woodard horizontal direction is to the right, and the positive vertical direction is up. Note that the most common labels for the axes are x for the horizontal axis and y for the vertical, although these are somewhat arbitrary and on Title Page occasion we will have need for other labels. Contents As we have seen, a function can be thought of as an association of JJ II inputs with outputs, so a function can be thought of as a set of ordered pairs. By coloring the points which correspond to the ordered pairs which J I make up the function, we get what we refer to as the graph of the function. Go Back Note that we can see both the domain and the range in this graphical Close representation: for each point on the graph, if we project straight down (or up) to the horizontal axis, the set of points on that copy of the real Quit number line indicate the domain of the function. Likewise, if we project Page 20 of 191 to the vertical axis, we can see the range of the function. To see the domain of a function from its graph: Project onto the x axis. To see the range of a function from its graph: Project onto the y axis.

We can also graphically see whether or not a given curve really represents y as a function of x via the vertical line test. This test says that if a vertical line can be drawn which intersects the curve in two distinct points (say (x1, y1) and (x1, y2) with y1 6= y2), then the graph does not represent Integrated a function of x, because there are two distinct y values associated with the Calculus/Pre-Calculus same x value x1. Mark R. Woodard

Vertical Line Test If a vertical line can be drawn which intersects the Title Page curve in two distinct points (say (x1, y1) and (x1, y2) with y1 6= y2), then the graph does not represent a Contents function of x. JJ II J I Go Back Close Quit Page 21 of 191 1.3. Circles, Distances, Completing the Square 1.3.1. Distance Formula and Circles

Given two points (x1, y1) and (x2, y2) in the plane, there will be occasions when you are interested in knowing the distance between them. By plotting the two points along with the the point (x2, y1), you can see a right triangle whose hypotenuse has the length we are looking for. The two sides of the triangle have lengths x2 − x1 and y2 − y1. Thus, the pythagorean theorem gives the required distance, and we have Integrated Calculus/Pre-Calculus The distance between the points (x1, y1) and (x2, y2) in Mark R. Woodard p 2 2 the plane is given by d = (x2 − x1) + (y2 − y1)

A circle is the set of points a given distance, the radius, from a given Title Page point, the center. Using this fact and the above distance formula, we see that every point (x, y) on a circle of radius r centered at (h, k) satisﬁes Contents p(x − h)2 + (y − k)2 = r. Squaring both sides yields the standard equa- JJ II tion for a circle, namely: 2 Equation of a Circle with Center at (x0, y0) and radius r:(x − x0) + J I 2 2 (y − y0) = r Go Back Close Equation of a circle with center at (h, k) and radius r: Quit (x − h)2 + (y − k)2 = r2 Page 22 of 191 1.3.2. Completing the Square A helpful algebraic technique when dealing with circles is the technique of completing the square. This technique helps one transform an equation of a circle which is disguised so as to not look like one into the standard form, so one can tell the center and radius. In this technique, an expression of 2 k 2 the form x + kx + ... = c is changed by adding 2 to both sides, yielding 2 k 2 k 2 k 2 k 2 x + kx + 2 + ... = c + 2 . This enables us to write (x + 2 ) + ... = c + 2 , thus changing the left hand side into a perfect square. Exercise 1.3.1. The equation x2 + 4x + y2 + 6y = 10 is actually the Integrated equation of a circle. Complete the square for both the x expression and Calculus/Pre-Calculus the y expression in order to put this equation in standard form. Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 23 of 191 1.4. Lines 1.4.1. General Equation and Slope-Intercept Form Recall that the general equation of a line is an equation of the form Ax + By = C, where A, B and C are real numbers. If we solve for y, (so that A C the expression will give y as a function of x) we have y = − B x + A , which A we usually write as y = mx + b, where the expression − B = m can be C seen to be the slope of the line, and the expression A = b can be seen to be the y - intercept. In the case where A is zero, we have that m is zero as well, and the line has the form y = b which represents a horizontal line. Integrated If B = 0 then we can’s solve for y as above, and we see that our equation Calculus/Pre-Calculus C Mark R. Woodard has the form x = A , which reprsents a vertical line.

General Equation of a Line: Ax + By = C, where A, B and C are real numbers. Title Page Slope-Intercept Form of a Line: Contents y = mx+b, where m is the slope and b is the y-intercept. JJ II 1.4.2. More On Slope J I The slope of a line mentioned above plays a key role in the study of calculus. Go Back The slope of a line is the numerical value which represents the “steepness” Close of the line. Given two points (x1, y1) and (x2, y2) on the line, the slope is Quit deﬁned to be y − y m = 2 1 . Page 24 of 191 x2 − x1 Note that the numerator of this fraction is the vertical “rise” as we move from the point (x1, y1) to (x2, y2), while the denominator is the horizontal “run”. Thus you will hear people describe the slope is being the “rise over run”. From this deﬁnition of slope, it is easy to get point-slope form of the equation of a line. Given a point (x0, y0) on a line of slope m, we know that any other point (x, y) on the line must satisfy y − y 0 = m, x − x0 since the “rise over run” from the point (x0, y0) to (x, y) must be m. Mul- Integrated tiplying both sides by the appropriate expression to clear denominators Calculus/Pre-Calculus leads to the Mark R. Woodard

Point-Slope Form of the Equation of a Line: y − y0 = m(x − x ) 0 Title Page Contents Exercise 1.4.1. A line has slope 5, and goes through the point (−2, −4). Find the equation of this line, and indicate its y - intercept. JJ II Exercise 1.4.2. Find the equation of the line that contains the points J I (−2, 5) and (2, −3). Go Back Close 1.4.3. Parallel and Perpendicular Lines Quit There is a nice relationship between slopes of lines and whether or not they Page 25 of 191 are perpendicular or parallel. Lines which are not vertical are parallel if and only if they have the same slope. Lines which are not vertical or horizontal are perpendicular if and only if their slopes multiply together to be −1.

Exercise 1.4.3. Are the lines 3x + y = 7 and −x − 3y = 9 parallel, perpendicular, or neither? Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 26 of 191 1.5. Catalogue of Familiar Functions 1.5.1. Polynomials We are interested in devloping an informal “catalogue” of the functions we will be using, so we can refer to them by name and understand some of the properties that they might posess. Our ﬁrst category will be the collection of polynomials.

Deﬁnition of Polynomial A polynomial is a function which can be written in the Integrated form Calculus/Pre-Calculus n n−1 Mark R. Woodard anx + an−1x + ... + a1x + a0

where the ai’s are real numbers, an 6= 0, and n is a non- negative integer. Title Page Contents In the above definition, the ai’s are called coefficients and the number n is called the degree of the polynomial. Here are some familiar types of polynomials: JJ II J I 1. The constant function f(x) = k is a degree zero polynomial. Go Back 2. The line f(x) = mx + b is a first degree polynomial. Close 3. A quadratic function f(x) = ax2+bx+c is a second degree polynomial. Quit Quadratics have graphs which are polynomals which open up if the co- Page 27 of 191 efficient on the 2nd degree term is positive, and open down if it is negative. The vertex of a parabola is the lowest point on the parabola if it opens down or is the highest point on the parabola if it opens up. One can find the x value of the vertex by completing the square. If f(x) = ax2 + bx + c, 2 b we can write f(x) − c = a(x + a ), and then completing the square inside the parentheses yields b2 b b2 f(x) − c + = a(x2 + + ), 4a a 4a2 b2 b 2 so f(x) − c + 4a = a(x + 2x ) . From this it is not too hard to see that the vertex will be at x = − b . 2a Integrated Polynomials share many nice properties. For example, every polynomial Calculus/Pre-Calculus has domain equal to , the set of all real numbers. R Mark R. Woodard 1.5.2. Rational Functions

A rational function is one which can be written as the ratio of two polyno- Title Page 3x2+2x 1 3 mials. For example, 5x−3 is a rational function. Also f(x) = x + x2 is a 3+x Contents rational function, because by simplifying it can be written as f(x) = x2 . Rational functions have domain equal to the set of all real numbers except JJ II for those numbers which cause the denominator to be zero. J I 1.5.3. Algebraic Functions Go Back Close An algebraic function is one which can be built out of the standard operations of addition, subtraction, multiplication, division and roots. In par- Quit ticular, everything mentioned in this section so far is algebraic. Some ex- Page 28 of 191 amples of non-algebraic functions are the trigonometric functions (f(x) = sin(x), etc.), the exponential functions (f(x) = 2x, etc.), and the logarith- mic functions (f(x) = log x). We will postpone our study of these types of functions until later.

1.5.4. The Absolute Value and Other Piecewise De- ﬁned Functions The absolute value function, denoted |x| is deﬁned by

x if x ≥ 0 Integrated |x| = −x if x < 0 Calculus/Pre-Calculus Mark R. Woodard This function can be thought of as giving the distance that x is from 0 on the number line. In general, |x − c| always gives the distance between x and c on the number line. Title Page The absolute value function is an example of a piecewise defined func- Contents tion. Here is another example 16 − x2 if x < 2 JJ II f(x) = 2x + 20 if x ≥ 2 J I Note that the domain of this function consists of all real numbers, but Go Back that the function is defined differently over the different “pieces” of the Close domain. To properly evaluate this function at x = 3 for example, one must first ask which piece of the domain 3 falls into. A little thought Quit reveals that 3 is in the category x ≥ 2, so we apply the 2x + 20 rule with Page 29 of 191 input 3 and obtain that f(3) = 26. 1.5.5. The Greatest Integer Function Another interesting and unusual function is the greatest integer function, denoted bxc. This function is not given by a formula, but rather a verbal rule of correspondance. The rule is that the output is always the greatest integer that is less than or equal to the input. Thus b2.5c = 2,b−2.5c = −3, bπc = 3, and so on. This function is sometimes referred to as a step function, since its graph resembles stairs going up from left to right.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 30 of 191 1.6. New Functions From Old

There are a number of ways to get new functions by combining old ones in various ways. These include composing two or more functions to get a new function, or combining familiar functions with addition, subtraction, multiplication or division. We will investigate all of these, and see how the graphs of familiar functions are changed by simple translations and scalings. Integrated Calculus/Pre-Calculus Mark R. Woodard

1.6.1. New Functions Via Composition Title Page Contents Recall that a function is a rule of correspondance which has inputs and JJ II outputs. If we put the output of a particular function into a second function as an input, the resulting output (when thought of as a function of the J I original input) is the composition of the two functions. Notationally, we Go Back write (f ◦ g)(x) = f(g(x)). This notations suggests that if we were to Close evaluate f ◦ g at a number like 2, we would compute f(g(2)), that is, compute g(2) first and then plug the resulting number into f as input. It Quit is important to note that f(g(x)) would not necessarily be equal to g(f(x), Page 31 of 191 so we see that in general g ◦ f 6= f ◦ g. The composition of f and g is defined by f ◦ g(x) = f(g(x)). In this definition we will call g the “inner” function and f the “outer” function. This is not necessarily equal to the composition of g and f, which is defined by g ◦ f(x) = g(f(x)). In this definition, f is the “inner” function and g is the “outer” function.

Exercise 1.6.1. For the functions f(x) = 3x + 2 and g(x) = x2 − 5, Integrated compute (f ◦ g)(x) and (g ◦ f)(x). To see that these are different, compute Calculus/Pre-Calculus f ◦ g(2) and g ◦ f(2). Is this enough information to deduce that the two Mark R. Woodard are different? Exercise 1.6.2. Consider the function h(x) = (4x+1)2 −9. Can you find functions f and g so that h(x) = f(g(x))? Title Page Contents Click to Initialize Quiz Answer the following questions about compositions. JJ II 1. If f(x) = 1 , g(x) = x2, and h(x) = 2x + 1, what is f ◦ g ◦ h(x)? x J I 1 2 2x+1 x2 2x + 1 2x+1 x2 2x+1 x2 Go Back 2. If f(g(h(x))) = 2x + 1 + 1 2, find possible representations for f, 2x+1 Close g, and h. f(x) = x2, g(x) = x2, h(x) = x2, Quit 1 1 1 g(x) = x + x , f(x) = x + x , g(x) = x + x , Page 32 of 191 h(x) = 2x + 1 h(x) = 2x + 1 h(x) = 2x + 1 Click to End Quiz

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 33 of 191 1.7. New Functions via Algebra One can also get new functions from old via the good old-fashioned operations of addition, subtraction, multiplication and division. For example, if f and g are functions, we can form • (f + g)(x) = f(x) + g(x) • (f − g)(x) = f(x) − g(x) • (f · g)(x) = f(x) · g(x) Integrated f f(x) Calculus/Pre-Calculus • g (x) = g(x) . Mark R. Woodard Note that if f and g in the above all have domain equal to the set of all real numbers, then each all of f + g, f − g and f · g do too, but that the f domain of g could be only a subset of the set of all real numbers. In fact, f Title Page any number a for which g(a) = 0 is definitely not in the domain of g . Contents JJ II J I Go Back Close Quit Page 34 of 191 1.8. Shifting and Scaling Suppose you are given a function y = f(x) for which you are familiar with the graph, and a positive constant c. (You might want to think of f(x) = x2 and c = 3 as a good example during this discussion.) How are the graphs of the following related to the graph of f(x)? • f(x + c) • f(x − c) • f(x) + c Integrated Calculus/Pre-Calculus • f(x) − c Mark R. Woodard • cf(x) • −cf(x) Title Page • f(cx) Contents • f(−cx) JJ II A little experimentation and thought should convince you of the following: J I • f(x + c) has the same shaped graph as f(x), but the whole thing is Go Back shifted c units to the left. (To help you remember this, note that Close plugging 0 into f(x) is the same as plugging −c into f(x + c). Quit • f(x − c) has the same shaped graph as f(x), but the whole thing is Page 35 of 191 shifted c units to the right. • f(x) + c has a graph like f(x)’s but shifted c units up. • f(x) − c has a graph like f(x)’s but shifted c units down. • cf(x) is similar to f(x) but the y axis is scaled (if c > 1) to stretch the graph by a factor of c or (if 0 < c < 1) to compress the graph. • −cf(x) is similar to cf(x) but the whole thing is reflected about the x axis. Thus for example when c = 1, the graphs of f(x) and −f(x) can be gotten from one another by reflecting about the x axis.

• f(cx) is similar to f(x) except that the x axis is scaled to stretch the Integrated graph horizontally (if c > 1) or compress the graph horizontally (if Calculus/Pre-Calculus c < 1.) Mark R. Woodard • f(−cx) is similar to f(cx), except that the whole thing is reﬂected about the y axis. Thus for example, f(−x) can be gotten from f(x) by reﬂecting about the y axis. Title Page Note that if a function has the property that it is symmetric about the Contents y axis, then f(x) = f(−x) for that function. Such functions are called even functions, because some examples of these are f(x) = x2, f(x) = x4, JJ II 6 f(x) = x , etc.. Functions for which f(x) = −f(x) are said to be symmetric J I about the origin, and are called odd functions. Some examples of odd functions include f(x) = x, f(x) = x3, etc.. Go Back Exercise 1.8.1. Is the sum of an odd function and another odd function Close always odd? Quit √ Exercise 1.8.2. Draw a simple graph of f(x) = x. Now graph all of Page 36 of 191 the following for c = 3. • f(x + c) • f(x − c) • f(x) + c • f(x) − c • cf(x) • −cf(x) • f(cx) Integrated Calculus/Pre-Calculus • f(−cx) Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 37 of 191 CHAPTER 2 Limits

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2.1 Secant Lines, Tangent Lines, and Velocities ... 39 Integrated 2.1.1 The Tangent Line ...... 39 Calculus/Pre-Calculus 2.1.2 Instantaneous Velocity ...... 41 Mark R. Woodard 2.2 Simplifying Difference Quotients ...... 43 2.3 Limits (Intuitive Approach)...... 45 2.4 Intervals via Absolute Values and Inequalities . 50 Title Page 2.5 If – Then Statements ...... 52 Contents 2.6 Rigorous Definition of Limit ...... 54 2.7 Continuity...... 56 JJ II 2.7.1 Definitions Related to Continuity ...... 56 J I 2.7.2 Basic Theorems ...... 57 Go Back 2.7.3 The Intermediate Value Theorem ...... 59 Close Quit Page 38 of 191 2.1. Secant Lines, Tangent Lines, and Ve- locities One of the main problems that motivated the genesis of calculus was this: What does one mean by the slope of the tangent line to a curve at a given point, and how can we calculate it? Another important motivating question was: What does one mean by instantaneous velocity, and how can we calculute it? We will see in the following discussion that these two apparently unre- lated questions are in fact very closely related. Integrated Calculus/Pre-Calculus Mark R. Woodard 2.1.1. The Tangent Line We start by admitting that we haven’t defined a tangent line but have an intuitive idea of what we are talking about, using the idea of the tangent Title Page line to a circle as a guide. We note that as soon as we know the slope of the tangent line, we can write down its equation. It is Contents

y − b = m(x − a) JJ II J I where the point of tangency is (a, b) and m is the slope of the tangent line. It seems reasonable to approximate the slope of the tangent lines by Go Back using secant lines. More speciﬁcally, we consider the line through the point Close (a, f(a)) and (a + h, f(a + h)). The slope is clearly Quit f(a + h) − f(a) Page 39 of 191 h and we note that this approximation gets better as h gets smaller. Thus we deﬁne the tangent line to be the line whose slope is the result of “taking the limit as h → 0” in the above. We will discuss the technicalities of of this limiting procedure later, but for now will be content with “shrinking h → 0” informally.

The slope of the tangent line is obtained by ﬁrst computing the slope of a secant line between the points (a, f(a)) and (a + h, f(a + h)). This slope is given by Integrated f(a + h) − f(a) Calculus/Pre-Calculus . h Mark R. Woodard This approximation gets better as h → 0, so we will shrink h to zero to obtain the slope of the tangent line. Title Page Contents Exercise 2.1.1. JJ II If f(x) = x3 − 3x, what is the slope of the tangent line at x = 4? What J I is the equation of the tangent line at that point? Go Back Close Exercise 2.1.2. Quit √ Page 40 of 191 If f(x) = x + 3 − 4, what is the slope of the tangent line at x = 6? 2.1.2. Instantaneous Velocity

Suppose that you have a really accurate odometer and a very accurate stop watch but no speedometer, and you want to answer the question: how fast am I moving right now? (Assume you are traveling in a straight line.) Everyone is familiar with the idea that average velocity (i.e. “rate”) is elapsed distance divided by elapsed time. (Think of that familiar formula d = r · t.) Now it seems like a good approximation for instantaneous Integrated velocity would be average velocity over an extremely small time interval. Calculus/Pre-Calculus And the approximation would get better as the time interval gets smaller. Mark R. Woodard So if f(t) is our odometer reading at time t, and if we are interested in our instantaneous velocity at time t = a, we could consider the average velocity between times t = a and t = a + h. Our elapsed distance would be f(a + h) − f(a) and the elapsed time would be h, and thus the average Title Page velocity would be Contents JJ II f(a + h) − f(a) J I . h Go Back Close Quit And the approximation gets better as h → 0 so: HEY WAIT A MINUTE Page 41 of 191 I’M GETTING DEJA VU! The instantaneous velocity at time t = a is obtained by ﬁrst computing the average velocity between the points (a, f(a)) and (a + h, f(a + h)). This average velocity is given by

f(a + h) − f(a) . h This approximation gets better as h → 0, so we will shrink h to zero to obtain the instantaneous velocity. Integrated Calculus/Pre-Calculus Mark R. Woodard Note that this idea of ‘taking the limit as h → 0” needs to be studied in more detail, which we will do in the next few sections.

Exercise 2.1.3. Title Page Contents Suppose an object is moving in a straight line so that it’s position at 3 time t is given by f(t) = t+1 . What is the object’s instantaneous velocity JJ II at time t = 3? J I Exercise 2.1.4. Go Back Close A ball dropped from the bell tower falls so that its position (in feet) Quit above the ground is given by s(t) = −16t2 + 64. How fast is the ball moving after 1 second? Page 42 of 191 2.2. Simplifying Diﬀerence Quotients

f(x+h)−f(x) By a diﬀerence quotient we mean a quantity of the form h or f(x)−f(c) x−c . These kinds of quantities arise when studying slopes of secant or tangent lines, and average and instantaneous velocities. In this section we will focus on simplifying such quantities algebraically. The four main techniques for simplifying diﬀerence quotients are:

1. The “multiply out then cancel and simplify” technique. 2. The “factor then cancel” technique. Integrated Calculus/Pre-Calculus 3. The “multiply the numerator and denominator by the conjugate of Mark R. Woodard either the numerator or denominator and then simplify” technique. 4. The “find a common denominator, then combine and simplify” technique. Title Page Contents Some related comments: the first and second items can be thought of as JJ II two different sides of the same coin. “Multiplying out” and ”factoring” are inverse operations of each other. The appropriate one to do should be J I apparent from the context of the situation. The third item involves using Go Back the conjugate of an expression of the form s + t or s − t, especially where Close either or both of s and t involve square roots of expressions. The√ conjugate Quit √of s+t is s−t, and vice versa. For example, the conjugate of x + h−5 is x + h+5. The fourth item is especially useful when either the numerator Page 43 of 191 or denominator of the difference quotient is a difference of quotients itself. 1 1 x+h − x For example, an expression like x would require the technique in item four.

Examples of each technique follow.

f(x+h)−f(x) 2 Exercise 2.2.1. Find and simplify h for f(x) = 3x + 2x + 3. f(x)−f(3) 3 2 Exercise 2.2.2. Find and simplify x−3 for f(x) = x − x − 5x. f(x+h)−f(x) √ Exercise 2.2.3. Find and simplify h for f(x) = x − 2 + 3. f(x)−f(4) Integrated Find and simplify for f(x) = √1 . Exercise 2.2.4. x−4 x Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 44 of 191 2.3. Limits (Intuitive Approach) We would like to understand this idea of “shrinking h → 0” more formally. To this end, we will deﬁne the important notion of a limit. We ﬁrst give the notation, and then discuss the idea informally and intuitively. A more formal approach will follow in a later section. We will write lim f(x) = L x→c and will read it as “the limit as x approaches c of f(x) is L.” The (informal) meaning of this is as follows: Integrated Calculus/Pre-Calculus Mark R. Woodard When we write

lim f(x) = L x→c Title Page we mean that when x is very close to (but not equal to) Contents c, the corresponding value of f(x) is very close to L. JJ II Example: Since when x is close to 5, f(x) = 2x − 1 is close to 9, it J I would be correct to write limx→5 2x − 1 = 9. In a previous problem, we Go Back √ 1 showed that the slope of the tangent line to y = x + 3 − 4 at x = 6 is 6 . f(6+h)−f(6) 1 Close Thus it would be correct to write limh→0 h = 6 . √ Quit x−3 Exercise 2.3.1. Evaluate limx→9 x−9 . Page 45 of 191 Exercise 2.3.2. Using the diagram below, identify limx→0 f(x) and limx→2 f(x).

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£ ¤ Contents JJ II J I Go Back Close In the second exercise above, it seems that it might be useful to think Quit about when x is close to 2, but on speciﬁcally one side or the other of 2. Page 47 of 191 To this end, we deﬁne the one-sided limits. When we write

lim f(x) = L x→c+ we mean that when x is very close to (but a little bigger than) c, the corresponding value of f(x) is very close to L. When we write

lim f(x) = L Integrated x→c− Calculus/Pre-Calculus Mark R. Woodard we mean that when x is very close to (but a little smaller than) c, the corresponding value of f(x) is very close to L. Title Page Thus in that exercise, even though limx→2 f(x) didn’t exist, we could 1 Contents correctly say that limx→2+ f(x) = 2 and limx→2− f(x) = 2 The following fact should be evident: JJ II

In order for a limit to exist, the two corresponding J I one-sided limits must both exist and be equal. If either Go Back of the two one-sided limits fail to exist, the total limit Close will also fail to exist. Quit

Note that in the following diagram, limx→0 fails to exist, because there Page 48 of 191 is no particular number that x is getting close to as x → 0. We will write limx→0 f(x) = ∞, but that notation will simply mean that the limit doesn’t

exist, because as x → 0, the function increases without bound.

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£ JJ II ¤ J I Go Back Close Does lim x+3 exist? Exercise 2.3.4. x→3 x−3 Quit Page 49 of 191 2.4. Intervals via Absolute Values and In- equalities Before we embark on an attempt to understand the rigorous deﬁnition of limits, we must ﬁrst make sure we are comfortable with the relationship between certain inequalities involving absolute values and the corresponding intervals they describe. The key fact which is necessary to understand what follows is that

|a − b| represents the distance from a to b. Integrated Calculus/Pre-Calculus Because of this fact, we can think (for example) of |x−2| as the distance Mark R. Woodard from x to 2, and the set of points with |x − 2| < 1 as the set of points x so that the distance from x to 2 is less than one. Thus this set represents the interval (1, 3). In general, we have |x − a| < δ is the set of points x so Title Page that the distance from x to a is less than δ, or in other words, the interval Contents (a − δ, a + δ). If instead we were considering the set 0 < |x − a| < δ, we would be JJ II considering the union of intervals (a − δ, a) ∪ (a, a + δ), which doesn’t include a, since we are thinking about points whose distance from a is J I strictly bigger than zero but less than δ. Go Back Close Exercise 2.4.1. Describe the set of points 0 < |x − 3| < .04. Quit Exercise 2.4.2. Consider the set of points (2.4, 2.6), and write it in the Page 50 of 191 form |x − a| < δ for some choice of a and δ. Exercise 2.4.3. It turns out that for f(x) = 2x−1, it is true that |f(x)− 3| < .1 whenever |x − 2| < .05. Can you show that this is true by starting with the inequality |f(x) − 3| < .1 and working backwards to see that |x − 2| < .05? Exercise 2.4.4. Building on the last problem, suppose you were told that whenever |f(x) − 3| < that there was then a corresponding number δ so that |f(x) − 3| < whenever |x − 2| < δ. Mimic the solution to the last problem in order to ﬁnd the value of δ in terms of . Here again Integrated f(x) = 2x − 1. Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 51 of 191 2.5. If – Then Statements An example of an if..then statement is “If I do not understand this review section, then I will read it again.” An if..then statement can be phrased in a variety of ways without changing its meaning. A few examples of how our ﬁrst example can be rephrased are as follows: “My not understanding this review section implies that I will read it again.” “My not understanding this review section is a necessary condition for Integrated my reading it again.” Calculus/Pre-Calculus “I do not understand this review section only if I will read it again.” Mark R. Woodard “I will read this review section again if I don’t understand it.” An if..then statement consists of two parts, the “if” part called the hypothesis and the “then” part called the conclusion. Each of these parts Title Page have a truth value. The truth values of the hypothesis and the conclusion Contents determine the truth value of the if..then statement. The if..then statement is false only when the hypothesis true but the conclusion is false. JJ II Truth tables are a convenient way to determine the truth value of a J I logical statement according to Go Back p q p → q Close T T T T F F Quit F T T Page 52 of 191 F F T When we proof a theorem in class, we are simply showing that the if..then statement is true. To do this we will assume the “if” part is true and attempt to show the “then” part is also true. Since an if..then statement is false only when the hypothesis is true and the conclusion is false, we only need to show that the conclusion is always true when the hypothesis is true. Exercise 2.5.1. Make a truth table for the q → p and compare it with p → q. Are they the same or diﬀerent?

Integrated Exercise 2.5.2. Compare the statement p → q andq ˜ → p˜. Herep ˜ stands Calculus/Pre-Calculus for the negation of p. It has the opposite truth value of that of p. Mark R. Woodard

Exercise 2.5.3. How do the statements “p whenever q” and “p only if q” compare with “if p then q.” Also where does “p if and only if q” ﬁt in? Title Page Contents JJ II J I Go Back Close Quit Page 53 of 191 2.6. Rigorous Deﬁnition of Limit

We want to make rigorous the idea of a limit. We want to say that limx→2 2x − 1 = 3 has a specific meaning, more specific than “when x is close to 2, f(x) is close to 3.” The main idea is the following: no matter how close f(x) is required to be to 3, we can assure that it is that close by Integrated requiring x to be within a certain distance of 2. Suppose we are required to Calculus/Pre-Calculus make f(x) be within .01 of the value 3. We would be happy if there were Mark R. Woodard a number δ so that when x is within δ units of 2 (i.e. |x − 2| < δ), that it would then follow that f(x) is within .01 unit of 3. Or more generally, for any real number > 0, if we are required to have f(x) be within units of 3 (i.e. |f(x) = 3| < ), we would be satisfied if there was a real number δ Title Page (which would of course depend on ), so that as long as |x−2| < δ, it would Contents then follow that |f(x) − 3| < . Now recall that for limits, we don’t really require f(x) to be near the limit when x is equal to 2, we only require it JJ II when x is close to be not equal to 2. Thus we only really want to require J I that |f(x) − 3| < whenever 0 < |x − 2| < δ. If in fact the inequality involving f(x) is satisfied when x = 2 so be it, but it isn’t a requirement Go Back for the limit to exist. Close Quit Page 54 of 191 To formalize all of this: To say that

lim f(x) = L x→c means that for every > 0 there exists δ > 0 so that if 0 < | x − c | < δ, then | f(x) − L | < .

Exercise 2.6.1. Prove rigorously that limx→2 2x − 1 = 3. Integrated Calculus/Pre-Calculus Exercise 2.6.2. Prove rigorously that limx→1 4x + 2 = 6. Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 55 of 191 2.7. Continuity 2.7.1. Deﬁnitions Related to Continuity Graphs of functions that have no holes, gaps or jumps in them seem to have especially nice properties. We would like to deﬁne a concept that would encapsulte these desirable properties. We say that such a function is continuous at a point a if the value of f(a) is what we think it should be, based on what the function near the point is. Thus we say f(x) is continuous when the limit as we approach the point is equal to the value of the function at the point. Integrated Calculus/Pre-Calculus Mark R. Woodard We say that a function f(x) is continuous at a point x = a if the following 3 things are true:

1. f(a) exists (i.e., the function is deﬁned at a, or to Title Page put it another way, a is in the domain of f(x).) Contents 2. limx→a f(x) exists (i.e., the limit of f(x) exists at the point in question.) JJ II 3. The two previously mentioned quantities are equal. J I That is Go Back lim f(x) = f(a). x→a Close Quit Informally, we are calling a function continuous at a point if there are Page 56 of 191 no holes, gaps, or jumps at that point. If a function fails to be continuous at a point, it might be because the function isn’t deﬁned at that point, or because the the limit doesn’t exist at that point, or because these two things do exist but aren’t equal.

If a function is deﬁned at a given point a, and if limx→a− f(x) = f(a), then we say the function is continuous from the left or left continuous. Likewise, if limx→a+ f(x) = f(a) then we say the function is continuous from the right or right continuous. Of course, in order to be continuous at a point a, the function must be both right continuous and left continuous at a that point. Integrated If a function is continuous at every point in an interval (a, b) we say it Calculus/Pre-Calculus is continuous on the interval (a, b). If it is also right continuous at a and Mark R. Woodard left continuous at b, we say it is continuous on [a, b]. You should be able to guess what it means to be continuous on other intervals, like (−∞, 3) and [4, ∞). Title Page Contents 2.7.2. Basic Theorems JJ II Using the limit theorems, it isn’t hard to show that if f and g are continuous J I at a point, then f + g, f − g, and fg are continuous at that point. The Go Back same is true for f as long as g(a) 6= 0. Also for compositions, If g(x) is g Close continuous at a and f(x) is continuous at g(a), then f ◦ g is continuous at a. These facts follows immediately from the various limit theorems. From Quit these facts it is easy to build up a nice catalogue of continuous functions. Page 57 of 191 In particular, note the following:

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¤ JJ II J I Go Back Figure 2.1: This function is continuous from the left at x = 2, but is not continuous there. Close Quit Page 58 of 191 Polynomials are continuous everywhere and rational functions are continuous wherever they are deﬁned. (These follow directly from the plug-in theorem.) Also, the square root function is continuous on its domain, as are other root functions. The absolute value function is continuous everywhere, but the greatest integer functions is discontinuous at each integer.

Integrated 2.7.3. The Intermediate Value Theorem Calculus/Pre-Calculus Mark R. Woodard One example of a consequence of continuity is the IVT. It says that if a function f(x) is continuous on an interval [a, b], and if M is an intermediate value between f(a) and f(b), that there then exists a number c between a and b so that f(c) = M. Thus if you are going 30 mph at one moment Title Page and 40 mph a little later, you must have been going 35 mph at some Contents point in between. This wouldn’t follow necessarily if your speed weren’t a continuous function of time (which it is). JJ II J I q x+1 2 Go Back Exercise 2.7.1. Where is the function h(x) = x2−4 + 1 continuous? Close Exercise 2.7.2. A removable discontinuity is a discontinuity where limx→c f(x) exists, but f(c) doesn’t exist. It is called removable because one could de- Quit ˜ ﬁne a new function f which is equal to f everywhere except at c, and at this Page 59 of 191 point has value equal to limx→c f(x). This new function is now continuous x2−16 at c, we have “removed” the discontinuity. The function f(x) = x−4 has a removable disconinuity at x = 4. What should f˜(4) be deﬁned to be in order to remove the discontinuity?

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 60 of 191 CHAPTER 3 The Derivative

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3.1 Definition of Derivative ...... 63 Integrated 3.1.1 Slopes and Velocities Revisited ...... 63 Calculus/Pre-Calculus 3.1.2 The Derivative ...... 64 Mark R. Woodard 3.1.3 The Derivative as a Function ...... 64 3.2 Rules for Taking Derivatives ...... 65 3.3 Differentiability ...... 70 Title Page 3.4 Applications of Derivatives as Rates of Change 72 Contents 3.4.1 Overview ...... 72 3.4.2 Some Specific Applications ...... 73 JJ II 3.5 The Chain Rule ...... 76 J I 3.6 Implicit Differentiation ...... 78 Go Back 3.7 Higher Derivatives ...... 81 Close 3.7.1 The Second Derivative ...... 81 Quit 3.7.2 Third and Higher Derivatives ...... 81 Page 61 of 191 3.8 Related Rates ...... 83 3.9 Linearization ...... 85 3.9.1 Mathematicizing the main idea ...... 85

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Title Page Contents JJ II J I Go Back Close Quit Page 62 of 191 3.1. Deﬁnition of Derivative 3.1.1. Slopes and Velocities Revisited Recall that the expression f(a + h) − f(a) h represents the slope of a secant line between a and a+h, and that this slope approximates the slope of the tangent line at x = a, and this approximation gets better as h → 0. Thus it make sense to deﬁne the slope of the tangent Integrated line as Calculus/Pre-Calculus f(a + h) − f(a) Mark R. Woodard lim . h→0 h By making the substition h = x − a, we can rewrite a + h as x. Also, as h → 0 we have x → a. Thus an alternative but equivalent expression for Title Page the slope of the tangent line is Contents f(x) − f(a) lim . x→a x − a JJ II As discussed before, these quantities also represent the instantaneous J I rate of change of the function f with respect to x. Go Back Close Exercise√ 3.1.1. Find the equation of the tangent line at x = 9 for f(x) = x + x. Quit 2 Exercise 3.1.2. Find the instantaneous rate of change of y = 3x −5x+3 Page 63 of 191 at x = 3. 3.1.2. The Derivative f(a+h)−f(a) The quantity limh→0 h is so important, it deserves its own name. We will call it the derivative of f(x) at x = a. We will use the symbol f 0(a) to reprsent this quantity. Thus we have

The derivative of f(x) at x = a is denoted f 0(a) and is given by

f(a + h) − f(a) f 0(a) = lim Integrated h→0 h Calculus/Pre-Calculus or Mark R. Woodard f(x) − f(a) f 0(a) = lim . x→a x − a Title Page √ Contents Exercise 3.1.3. For f(x) = x + 2, find f 0(2). JJ II 3.1.3. The Derivative as a Function J I There will be times when we want to think of the slope of the tangent line Go Back to the function f(x) at a general point x. So it would be reasonable to Close 0 f(x+h)−f(x) 0 f(z)−f(x) define f (x) = limh→0 , or f (x) = limz→x . h z−x Quit Exercise 3.1.4. For f(x) = 1 , find f 0(x). x Page 64 of 191 3.2. Rules for Taking Derivatives It turns out that there are a number of theorems which make the process of taking derivatives much easier. We will develop these now. The first rule is that the derivative of a constant function f(x) = c f(x+h)−f(x) is always zero. This is because for such a function, limh→0 h = c−c 0 limh→0 0 = limh→0 h = 0. The second rule involves the functions of the form f(x) = mx, whose graphs are straight lines through the origin. The derivative of such func- Integrated tions turns out to be the constant m, because lim f(x+h)−f(x) = lim mx+mh−(mx) = h→0 h h→0 h Calculus/Pre-Calculus lim mh = lim m = m. h→0 h h→0 Mark R. Woodard The third rule shows us how to differentiate a power function of the form f(x) = xk. Before investigating this rule, we need to review the binomial k theorem, which tells us how to expand an expression like (x+h) . Actually, Title Page the only thing we need to know here is that Contents k k k−1 k−2 2 k−3 3 k−1 k (x + h) = c0x + c1x h + c2x h + c3x h + ... + ck−1xh + ckh JJ II where the ci’s are constants. You may recall that there is a nice formula J I for the constants ci which involves either Pascal’s triangle and factorials, Go Back but for our purposes, we only need to know the first two constants c0 and Close c1. It turns out that c0 = 1 and c1 = k. Thus we have Quit Page 65 of 191 k k k−1 k−2 2 k−3 3 k−1 k (x + h) = x + kx h + c2x h + c3x h + ... + ck−1xh + ckh . 0 (x+h)k−xk Thus we can write f (x) = limh→0 h as xk + kxk−1h + some stuff with a factor of at least h2 − xk f 0(x) = lim h→0 h kxk−1h + some stuff with a factor of at least h2 = lim h→0 h = lim kxk−1 + some stuff with a factor of at least h h→0 = kxk−1 Integrated We will refer to this theorem as the power rule. Calculus/Pre-Calculus Mark R. Woodard Next we investigate the derivative of a funtion f(x) which has the form cg(x) for some constant c and some function g(x). We note that cg(x+h)−cg(x) g(x+h)−g(x) 0 0 limh→0 h = c limh→0 h , so that f (x) = cg (x). We will refer to this theorem as the constant multiple rule. Summarizing our work Title Page so far we have: Contents

The derivative of a function f(x) = cxk is JJ II f 0(x) = ckxk−1. This also holds in the case where k = 0 J I and k = 1 Go Back (where we interpret x0 as 1.) Close How about a function f(x) which is the sum of two functions m(x) and Quit 0 n(x)? So the question is if f(x) = m(x) + n(x), what is f (x)? The answer Page 66 of 191 0 0 m(x+h)+n(x+h)−(m(x)+n(x)) (not surprisingly) turns out to be m (x)+n (x), since limh→0 h = m(x+h)−m(x) n(x+h)−n(x) m(x+h)−m(x) limh→0 h + h which of course can be written as limh→0 h + n(x+h)−n(x) 0 0 limh→0 h which we recognize as m (x) + n (x). We will refer to this theorem as the sum rule. Note that using the rules so far, we can easily take the derivative of any polynomial. For example, to diﬀerentiate f(x) = 4x5 +2x2 −4x+2, we can (by the sum rule) think of the four terms 4x5, 2x2, −4x and 2 separately. They have derivatives 20x4, 4x, −4 and 0 respectively, so (putting these back together) we have f 0(x) = 20x4 + 4x − 4. Integrated Exercise 3.2.1. Find f 0(x) for f(x) = −3x6 + 2x3 − 12x − 123. Calculus/Pre-Calculus Mark R. Woodard

Once we understand the procedure in the last example, we should believe that Title Page Every polynomial is differentiable everywhere. Contents JJ II The two rules which remain are a little less obviously true, and also a little harder to remember. These rules involve functions which are the J I product or quotient of two other functions. To get us started, consider the Go Back function f(x) = (2x + 3)(3x − 1). By multiplying this out, we have that Close f(x) = 6x2 + 7x − 3. Now using the above techniques, we see that f 0(x) = 12x + 7. Note that this result is not equal to what would be obtained by Quit just multiplying the derivatives of 2x+3 and 3x−1 together. Thus whatever Page 67 of 191 is going to become our product rule will have to be a little more complicated. It turns out that if f(x) = m(x)n(x), then the right expression for f 0(x) is m(x)n0(x) + n(x)m0(x), which probably can be remembered best as “the first function times the derivative of the second plus the second times the derivative of the first” . In our example with m(x) = 2x + 3 and n(x) = 3x−1, this expression would be (2x+3)(3)+(3x−1)(2) = 6x+9+6x−2 = 12x − 7. The proof of the product rule (for the truly dedicated reader) can be found elsewhere. Given that the product rule was a little messy and unexpected, you should now expect that the quotient rule will be a doozy. To help us toward the goal of understanding the quotient rule, consider the expression Integrated f(x) = p(x) = 2x+3 . We could find f 0(x) by resorting to the definition. Calculus/Pre-Calculus q(x) 3x−1 Mark R. Woodard Exercise 3.2.2. Use the definition of derivative to compute f 0(x) for 2x+3 f(x) = 3x−1 . q(x)p0(x)−p(x)q0(x) Title Page Now note that the if we were to compute (q(x))2 we would have (3x−1)(2)−(2x+3)(3) 6x−2−(6x+9) −11 0 Contents (3x−1)2 = (3x−1)2 = (3x−1)2 , which is f (x) as deduced in the previous exercise. This is (of course) not a proof, but hopefully gives you JJ II some evidence towards believing in the following quotient rule: J I

p(x) 0 q(x)p0(x)−p(x)q0(x) Go Back If f(x) = q(x) , then f (x) = (q(x))2 . Close Note also that from this theorem follows the fact that all rational func- Quit tions are diﬀerentiable at all points in their domain. Page 68 of 191 We now summarize all of the aforementioned derivative rules. Derivative Rules. 1. If f(x) = c then f 0(x) = 0. (The constant rule.) 2. If f(x) = x, then f 0(x) = 1. 3. If f(x) = xk then f 0(x) = kxkn−1. (The power rule.) 4. If f(x) = cg(x), then f 0(x) = cg0(x). (The constant multiplier rule.)

5. If f(x) = m(x) + n(x), then f 0(x) = m0(x) + n0(x). Integrated (The sum rule.) Calculus/Pre-Calculus Mark R. Woodard 6. If f(x) = m(x)n(x) then f 0(x) = m(x)n0(x) + mh(x)n0(x). (The product rule).

p(x) 0 q(x)p0(x)−p(x)q0(x) 7. If f(x) = q(x) , then f (x) = (q(x))2 . (The Title Page quotient rule.) Contents JJ II J I Go Back Close Quit Page 69 of 191 3.3. Diﬀerentiability Recall that the expression

f(x + h) − f(x) lim h→0 h represents the derivative of the function f(x). Also recall that limits sometimes have the disturbing property of not existing. Thus there could be points in the domain of f(x) where f 0(x) doesn’t exist. At such a point, we say that f is not differentiable. Also, if a is in the domain of f(x), Integrated but f 0(a) doesn’t exist, we say that (a, f(a)) is a singular point of f(x). If Calculus/Pre-Calculus f(x) is differentiable at every point in an interval (a, b), we say that f(x) Mark R. Woodard is differentiable on the interval (a, b). Here is an example of a singular point: consider the function f(x) = |x| |0+h|−|0| at a = 0. To see that this is a singular point, we will see that limh→0 h Title Page doesn’t exist. Of course, this expression is the same as lim |h| , and since h→0 h Contents the absolute value function is defined differently on the two sides of the |h| |h| number zero, it would behoove us to investigate limh→0− h and limh→0+ h . JJ II |h| −h Now recall that |h| = −h if h < 0, so lim − = lim − = h→0 h h→0 h J I limh→0− −1 = −1 (since the limit of a constant is that constant.) On the other hand (sorry, that was a bad pun) we know that when h > 0, Go Back |h| h |h| = h, so we have limh→0+ h = limh→0+ h = limh→0+ 1 = 1. Thus the Close two one-sided limits are different, so the overall limit doesn’t exist. Thus Quit (0, 0) is a singular value for f(x) = |x|. √ Page 70 of 191 Exercise 3.3.1. Show that (0, 0) is a singular value for f(x) = 3 x2. There is an important theorem which relates the differentiability of f(x) at x = a to the continuity of f(x) at x = a. The theorem states that if you know that f(x) is differentiable at x = a, then you know that it is continuous there as well. The reason is as follows: suppose you know 0 f(x)−f(a) that f(x) is differentiable at x = a so f (a) = limx→a x−a exists. In particular you know that f(a) exists, or else this expression wouldn’t make sense. Now to show that f(x) is continuous at x = a, we only need to show that limx→a f(x) = f(a), or (equivalently) show that limx→a f(x) − f(a) = 0. But we can perform the following somewhat devious trick of multiplying x−a by one in a clever way: we can rewrite f(x)−f(a) as (f(x)−f(a))· x−a Now Integrated x−a f(x)−f(a) Calculus/Pre-Calculus taking the limit as x → a we get limx→a(f(x)−f(a))· x−a = limx→a x−a · 0 Mark R. Woodard limx→a(x − a) = f (a) · 0 = 0. Thus putting these ideas together, we have limx→a f(x)−f(a) = 0, so limx→a f(x) = f(a), so f(x) must be continuous at x = a. Note that for the if – then statment above, the converse doesn’t hold. Title Page This is because it isn’t necessarily true that if a function is continuous at a point that it must also be differentiable there. In fact, we’ve already Contents seen two√ such examples. Both the absolute value function and the function JJ II f(x) = 3 x2 are continuous at (0, 0), but are not differentiable there. To summarize: J I Go Back If f(x) is differentiable at x = a then it is continuous at Close x = a. However, it is possible for a function to be continuous at x = a without being differentiable there. Quit Page 71 of 191 3.4. Applications of Derivatives as Rates of Change

3.4.1. Overview

Recall that one of the motivations toward the deﬁnition of the derivative Integrated was in the investigating of the concept of instantaneous velocity as the Calculus/Pre-Calculus limit of average velocity as the time interval shrank to zero. In general, Mark R. Woodard no matter what quantities are represented by y = f(x), we can think of the derivative as a “rate” in the sense that as the quantity x changes, so does the quantity y, and if we write h = (x + h) − x) as ∆x and write ∆y f(x+h)−f(x) Title Page ∆y = f(x + h) − f(x), we can think of ∆x = h as an average rate of change of y with respect to x, since we have the ratio of a change Contents in y to a change in x over a small interval. The limit of this quantity as ∆y JJ II ∆x → 0 (which of course is the derivative) can be written lim∆x→0 ∆x , can be thought of as the limit of the average rate of change y with respect to x J I as the x interval shrinks, so we can think of this as an instantaneous rate Go Back of change of y with respect to x. To make our derivative look more like the limit of a change in y over a change in x, we will sometimes write f 0(x) as Close dy dx . Quit Page 72 of 191 Thus ∆y dy f 0(x) = lim = ∆x→0 ∆x dx is the derivative of f(x) with respect to x, and can be thought of as the instantaneous rate of change of y = f(x) with respect to x.

Integrated 3.4.2. Some Speciﬁc Applications Calculus/Pre-Calculus Rectilinear Motion Suppose an object is moving in a straight line (ei- Mark R. Woodard ther vertically or horizontally) and its position at time t is given by s(t). he theory developed so far indicates that the derivative s0(t) should give us the objects velocity at time t, since the instantaneous rate of change of Title Page position with respect to time is velocity. Thus, we will write v(t) = s0(t). Also, since the rate of change of velocity with respect to time is commonly Contents 0 called acceleration, we can write v (t) = a(t). Now since F = ma where F JJ II is force and m is the mass of an object, we can think of the acceleration at time t as determining the force acting on our object. If velocity were zero J I at a given time, we would know that the object was stopped at that time. Go Back Thus, nonzero velocity indicates movement; we will consider positive veloc- Close ity to indicate movement to the right (if our line is oriented horizontally) or upward movement (if our line is oriented vertically.) Likewise, positive Quit or negative acceleration will be thought of as acting in those directions. Page 73 of 191 When an object’s velocity and acceleration both have the same sign, then the object will speed up. Note that this might occur when they are both positive or both negative (as in the case of a ball thrown downward from a high tower, both its velocity and acceleration are negative, so the ball speeds up.) When velocity and acceleration have opposite signs, the object will be slowing down.

If an object’s position at time t is given by s(t), then 0 ds v(t) = s (t) = dt represents the objects velocity at time t. Also, v0(t) = a(t) represents the object’s acceleration at time t. Integrated Calculus/Pre-Calculus Mark R. Woodard Exercise 3.4.1. An object is moving along a straight line so that its position at time t is given by s(t) = 2t3 − 24t + 2. Analyze the movement of this object. Title Page Biology Applications Many biologists are interested in the growth of Contents populations, of humans, animals, cells, etc.. If p(t) is the size of a given population at time t, then p0(t) is the population growth rate at time t. JJ II J I Exercise 3.4.2. A population√ of bacterial cells in a petri dish is growing so that at time t, there are t(t2 + 3t) cells present. Go Back 1. What is the change in the size of population between time t = 4 and Close t = 9? Quit 2. What is the instantaneous growth rate of the population at time t = 4? Page 74 of 191 Chemistry Applications Chemists are often interested in rates of chem- ical reactions, as well as the relationship between various physical quantities of various types of matter. For example, one form of Boyle’s law states that the absolute temperature T , pressure P and volume V of an ideal gas of ﬁxed mass m are related by PV = mRT where R is constant. Exercise 3.4.3. In the above equation for Boyle’s law, suppose you were able to hold temperature T constant. What would happen to pressure P as volume V increased? What would be the instantaneous rate of change of pressure with respect to volume? Integrated Calculus/Pre-Calculus Economics Suppose C(x) represents the cost of producing x units of Mark R. Woodard a certain good. The quantity C(n + 1) − C(n) would represent the cost of producing the (n + 1)st unit after already producing n units. This item is sometimes referred to as the marginal cost of producing the (n + C(n+1)−C(n) Title Page 1)st item. It can be approximated by the derivative, since 1 ≈ C(n+h)−C(n) 0 limh→0 h . Thus economists often refer to C (x) as the marginal Contents cost function. JJ II Exercise 3.4.4. Suppose that the cost of producing x units of a certain 2x+.004x2+.008x3 J I good is 2000 . Find the actual cost of producing the 101st unit, and compare with C0(100). Go Back Close Quit Page 75 of 191 3.5. The Chain Rule

We need one more shortcut rule for taking derivatives. The question is: If the function f(x) is really the result of composing functions m(x) and n(x), what is the derivative of f in terms of information about m and n? For example, how can we take the derivative of (2x+3)2 without expanding the expression ﬁrst. This would be useful to know, because we wouldn’t want to have to expand something like (2x + 3)100 in order to diﬀerentiate it! Going back to the example of f(x) = (2x + 3)2, if we do expand it we get f(x) = 4x2 +12x+9, so f 0(x) = 8x+12. Note that this is not the same Integrated thing as 2(2x + 3)1, so that can’t be the right thing. But using the fact Calculus/Pre-Calculus ∆m ∆m ∆n Mark R. Woodard that a derivative is almost a quotient, how about if we write ∆x = ∆n · ∆x . (Here we are pretending that ∆n 6= 0, so that this makes sense.) Now by shrinking appropriate quantities to zero, one might hope that we would have Title Page dm dm dn f 0(x) = = · . Contents dx dn dx JJ II Let’s see if this works for the case f(x) = (2x + 3)2. We can write J I f(x) = m(n(x)) if m(x) = x2 and n(x) = 2x + 3. Then dm would be the dn Go Back derivative of the squaring function at (2x + 3), so we would get 2(2x + 3)1 dn 0 Close for this portion of the result. Now dx would be n (x) which would be the 0 1 constant 2. Thus the total derivative f (x) would be 2(2x + 3) · 2 which Quit would be 8x + 12, which is the result desired as we noticed above. Page 76 of 191 To summarize: df dm dn If f(x) = m(n(x)), then dx = dn · dx , or in other notation, f 0(x) = m0(n(x)) · n0(x).

√ Exercise 3.5.1. Find f 0(x) for f(x) = 12x5 − 3x2 + 4x + 3. p √ Exercise 3.5.2. Find the derivative of x + x. Exercise 3.5.3. By using the chain rule twice in sucession, ﬁnd a formula for the derivative of a function of the form f(x) = r(s(t(x))). (This could Integrated be called the triple chain rule.) Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 77 of 191 3.6. Implicit Diﬀerentiation

Consider the function f(x)2. The chain rule tells us that the derivative of this (with respect to x) is 2f(x)1f 0(x). Suppose we were to call the function f(x) by the alternative name y, and write y2 for the expression f(x)2. Then 0 dy the derivative would have to be 2yy or 2y dx , again by the chain rule. If you were thinking that the derivative of y2 should be simply 2y, you were probably thinking of the derivative of the expression y2 with respect to y, which would indeed be simply 2y. Thus we see that when differentiating, we must always be aware of which quantity represents the variable we Integrated are differentiating with respect to, and which quantity or quantities are Calculus/Pre-Calculus actually functions of that variable. Mark R. Woodard This comes into play when considering an equation like the circle x2 + y2 = 1. Of course, we could solve this for y in terms of x (and consider the top half and the bottom half of the circle separately), but it may be Title Page more convenient to deal with this equation in its standard form. Now even though the circle doesn’t represent y as a function of x (since it fails the Contents vertical line test), it is still the case that there is a tangent line at every JJ II point on the circle and all but two of those points have non-vertical tangent dy J I lines. Thus, we should be able to compute the slope of the tangent line dx at all points except the two with vertical tangents. When dealing with a Go Back function which is not in the usual y = f(x) form with y written explicitely Close in terms of the variable x, we say we have an implicit function. The process of finding the derivative of a function (without resorting to solving for y Quit explicitely in terms of x) is called implicit differentiation. This is carried Page 78 of 191 out by simply differentiating both sides of the equation with respect to x, and recalling (as above) that we are to think of y as a function of x. Thus, for example, in the case of the implicit equation x2 + y2 = 1, if we differentiate both sides of the equation with respect to x, we would obtain dy 2 2 2x + 2y dx = 0, since the derivative of x is 2x, the derivative of y with dy respect to x is 2y dx (by the chain rule) and the derivative of the constant dy dy −x 1 is 0. This equation can now be solved for dx to obtain dx = y . Thus the√ slope√ of the tangent line at a point like (0, 1) is 0, while at a point like 2 2 ( 2 , 2 ), the slope is −1. We can even see that the points on the circle where the tangent line has undefined slope are the points where y = 0, namely (1, 0) and (−1, 0). Integrated Calculus/Pre-Calculus 3 Mark R. Woodard Exercise 3.6.1. Consider the curve y = x , but for practice with implicit dy differentiation, write it as xy = 3. Find dx by differentiating both sides with respect to x. Be sure to use the product rule on the left hand side dy of the equation. When finished, check your answer by finding dx for the Title Page explicit function y = 3 . x Contents

When differentiating a reasonably complex implicit function, the result JJ II dy often involves several terms, some which have factors of dx and some of J I which don’t, and these factors could be on either side of the equal sign. Go Back dy To solve such an equation for dx , first identify which terms have factors of dy dy Close dx , and which ones don’t. Gather all the terms which do have a dx factor on one side of the equal sign, and put all other terms on the other side. Quit dy Then factor out the dx factor, and finish solving by dividing both sides of Page 79 of 191 the equation by the appropriate factor. For example, consider the function x3y2 +4x+5y2 −2 = xy. Differentiating both sides with respect to x yields dy dy dy x3 · 2y + y2 · 3x2 + 4 + 10y − 0 = x + y(1). dx dx dx Analyzing this equation, we see that there are 4 non-zero terms on the left hand side, and 2 terms on the right hand side. Of these terms, there are dy two terms on the left hand side which have a dx factor, and one term on dy the right hand side which has one. Gathering all the terms with a dx factor on the left (and the others on the right) we obtain Integrated dy dy dy Calculus/Pre-Calculus 2x3y + 10y − x = y − 3x2y2 − 4. dx dx dx Mark R. Woodard

dy Factoring out the dx factor yields

dy Title Page 2x3y + 10y − x = y − 3x2y2 − 4, dx Contents so dy y − 3x2y2 − 4 JJ II = . dx 2x3y + 10y − x J I √ dy Exercise 3.6.2. Suppose xy = x + y + 1. Find dx . Go Back Close Quit Page 80 of 191 3.7. Higher Derivatives 3.7.1. The Second Derivative Since the derivative of a function is another function, it makes sense to think about differentiating the derivative to get another function, which we will call the second derivative of the original function. We will write f 00(x) d2y or dx2 for the second derivative of y = f(x) with respect to x. Because of our previous dealing with the derivative, we should expect that there is both a geometric significance to the second derivative (related to the Integrated graph of the function) and a physical intepretation (related to the rate of Calculus/Pre-Calculus change of the function.) We now explore the latter, leaving the discussion Mark R. Woodard of the significance of the second derivative in relation to geometry for a later chapter. Suppose y = f(t) represents the position of an object at time t. We have already seen that f 0(t) represents the instantaneous rate of change Title Page of f (position) with respect to t (time.) Thus f 0(t) represents velocity. Contents Likewise, f 00(t) should represent the instantaneous rate of change of f 0 (velocity) with respect to t (time.) The rate of change of velcocity with JJ II respect to time is what we generally call acceleration. Thus we can think J I of f 00(t) = a(t) as the acceleration of the object at time t. Go Back Close 3.7.2. Third and Higher Derivatives Quit 00 000 Of course, we can differentiate f (x) to obtain f (x) and so on. Since it Page 81 of 191 is impractical to write f 0000000000000000(x) for the sixteenth derivative of f with respect to x, we instead write f (16)(x). Of course, this could lead to con- fusion with the expression (f(x))16, but hopefully we can use the context to “differentiate” between the two. (Sorry, bad pun.) In Leibniz notation, dny we write dxn to stand for the nth derivative of y with respect to x. Exercise 3.7.1. Explain why the nth derivative of a degree k polynomial is always the constant zero, if n > k. 1 Exercise 3.7.2. Find the first few derivatives of f(x) = x2 . See if you dkf can find a pattern, so that you can write down a general formula for dxk . Hint: the number (1) · (2) · (3) · ... · (n) is sometimes written as n!. This Integrated is read “n factorial.” For example, 8! = 8 · 7 · 6 · ... · 1 = 40320. Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 82 of 191 3.8. Related Rates One application of implicit differentiation and the chain rule is in the solution to problems like the following: Suppose two cars leave an intersection at the same time, with one car headed north and the other car headed east, with the northbound car travelling at 50 mph and the eastbound car travelling at 30 mph. How far are they separating after 30 minutes? We can solve such a problem by trying to find a relationship between the rate at which the two cars are separating and the rates they are going. To this end, we could let x represent the distance the northbound car has gone after t Integrated hours, and y the distance the eastbound car has gone after t hours. Note Calculus/Pre-Calculus that these are both functions of time t. Also, we can related the distance Mark R. Woodard between the cars by the pythagorean theorem: x2 + y2 = z2, where z is the distance between the cars at time t. Differentiating implicitely with respect x dx +y dy to t yields 2x dx +2y dy = 2z dz , and solving for dz yields dz = dt dt . Now dt dt dt dt dt z Title Page dx = 50 and dy = 30 and when t = .5, x = 25 and y = 15 and thus (using dt dt √ √ the pythagorean theorem) z = 252 + 152 = 850. Thus at the moment Contents in question, dz = 25·50+15√ ·30 . dt 850 JJ II Exercise 3.8.1. A right circular cylinder with radius 2 feet and length J I 10 feet is being filled with water at a rate of 3 cubic feet per minute. At what rate is the height of the water in the cylinder increasing? Go Back Close Exercise 3.8.2. A swimming pool which is 3 feet deep at the shallow end and 8 feet deep at the deep end is being filled with water at the rate of 10 Quit cubic feet per minute. The pool is 25 feet long and 10 feet wide, and the Page 83 of 191 shallow end descends along a straight line to the deep end. How fast is the water level rising when the water level (measure in the deep end) is 3 feet? Here are some suggestions which you might find useful when trying to solve related rates problems.

Steps for Solving Related Rates Problems 1. Draw a picture if appropriate. 2. Label things in the picture which are functions of time with a letter, and label things that are con- Integrated stants with the appropriate number. Calculus/Pre-Calculus 3. Write down the rates which you know and the rate Mark R. Woodard you want to know. This helps keep you focused on the task at hand. 4. Write down an equation which relates the variables Title Page which is valid for all time t. Contents 5. Diﬀerentiate both sides of the equation with respect JJ II to t, using the chain rule. J I 6. After diﬀerentiating, substitute the values of the Go Back quantities that are known at the desired time. Close 7. Solve for the desired rate. Quit Page 84 of 191 3.9. Linearization Near the point of tangency, a given curve and its tangent line look quite similar. This can be seen using the “zoom” feature on a graphing calculator or computer on which a curve and a tangent line to the curve have been graphed together. If we zoom in on the point of tangency, the curve and the tangent line are hard to distinguish from each other. Thus we could say that the tangent line is a good approximation to the curve near the point of tangency. In fact, it can be proven that the tangent line is the best linear approximation to the curve which goes through the given point. Integrated Calculus/Pre-Calculus 3.9.1. Mathematicizing the main idea Mark R. Woodard Recall that the tangent line to y = f(x) at the point x = a is given by

0 y − f(a) = f (a)(x − a). Title Page Now instead of calling this line “y”, let’s call it L(x). Thus we have Contents

L(x) = f(a) + f 0(a)(x − a). JJ II J I And the above main idea says that when x is close to a, Go Back 0 f(x) ≈ f(a) + f (a)(x − a). Close The quantity (x − a) in the above discussion is sometimes called ∆x Quit or dx, since it represents a small change in x when x is near a. The Page 85 of 191 quantity (f(x) − f(a)) in the above discussion is sometimes called ∆y. It represents the corresponding change in y for the function f(x). The quantity f 0(a)(x − a) = f 0(a)dx is sometimes called dy. This represents the corresponding change in y when looking at the tangent line rather than the curve. Thus for a given point a near x, dx is determined, and this produces ∆y (which is a diﬀerence between y values on the curve,) and dy (which is a diﬀerence between y values on the tangent line.) One way of re-expressing the main idea can then be rewritten as

∆y ≈ dy Integrated because this is equivalent to Calculus/Pre-Calculus Mark R. Woodard f(x) − f(a) ≈ f 0(a)dx.

Exercise 3.9.1. Show how to use the linear approximation√ of the square root function near the point (36, 6) to approximate 36.05 without the use of a calculator. Title Page Contents Exercise 3.9.2. Draw a sketch of y = x2, and draw the tangent line at x = 1. Using a value of dx = .05, ﬁnd dx, dy and ∆y, and draw them at JJ II appropriate places on the sketch. J I Go Back Close Quit Page 86 of 191 CHAPTER 4 An Interlude – Trigonometric Functions

Contents Integrated Calculus/Pre-Calculus 4.1 Definition of Trigonometric Functions ...... 88 Mark R. Woodard 4.1.1 Radian Measure ...... 90 4.2 Trigonometric Functions at Special Values .... 91 4.3 Properties of Trigonometric Functions ...... 93 Title Page 4.4 Graphs of Trigonometric Functions ...... 94 Contents 4.4.1 Graph of sin(x) and cos(x) ...... 94 4.4.2 Graph of tan(x) and cot(x) ...... 94 JJ II 4.4.3 Graph of sec(x) and csc(x) ...... 95 J I 4.5 Trigonometric Identities ...... 96 Go Back 4.6 Derivatives of Trigonometric Functions ...... 97 Close 4.6.1 Two Important Trigonometric Limits ...... 97 Quit 4.6.2 Derivatives of other Trigonometric Functions ... 99 Page 87 of 191 4.1. Definition of Trigonometric Functions You may at some point in your mathematical career have defined the trigonometric functions using ratios of sides of a right triangle. Under this definition, you will recall that the sine of an angle θ in a right triangle is the ratio of the side opposite θ to the hypotenuse. The other trigonometric ratios are defined in box below.

If a is the length of the side opposite θ in a right triangle, and b is the length of the side adjacent to θ, and c is the Integrated length of the hypotenuse, then Calculus/Pre-Calculus a Mark R. Woodard sin(θ) = c b cos(θ) = c a tan(θ) = b b cot(θ) = a Title Page c sec(θ) = b c Contents csc(θ) = a JJ II Although these definitions are accurate, there is a sense in which they J I are lacking, because the angle θ in a right triangle can only have a measure Go Back between 0◦ and 90◦. We need a definition which will allow the domain of Close the sine function to be the set of all real numbers. Our definition will make use of the unit circle, x2 + y2 = 1. We first associate every real number t Quit with a point on the unit circle. This is done by “wrapping” the real line Page 88 of 191 around the circle so that the number zero on the real line gets associated with the point (0, 1) on the circle. The positive portion of the number line is wrapped counterclockwise around the circle, while the negative portion is wrapped counterclockwise. Another way of describing this association is to say that for a given t, if t > 0 we simply start at the point (0, 1) and move our pencil counterclockwise around the circle until the tip has moved t units. The point we stop at is the point associated with the number t. If t < 0, we do the same thing except we move clockwise. If t = 0, we simply put our pencil on (0, 1) and don’t move. Using this association, we can now define cos(t) and sin(t). Integrated Using the above association of t with a point (x(t), y(t)) Calculus/Pre-Calculus on the unit circle, we define cos(t) to be the function Mark R. Woodard x(t), and sin(t) to be the function y(t), that is, we define cos(t) to be the x coordinate of the point on the unit circle obtained in the above association, and define Title Page sin(t) to be the y coordinate of the point on the unit circle obtained in the above assocation. Contents

Using the above deﬁnition of sin(t) and cos(t), we can deﬁne JJ II sin(t) J I tan(t) = cos(t) cos(t) Go Back cot(t) = sin(t) 1 Close sec(t) = cos(t) 1 csc(t) = sin(t) Quit Exercise 4.1.1. What point on the unit circle corresponds with t = π? Page 89 of 191 What therefore is cos(π) and sin(π). 3π Exercise 4.1.2. What point on the unit circle correspond with t = 2 ? 3π 3π What therefore is cos 2 and sin 2 ?

4.1.1. Radian Measure In calculus, we will use radian measure rather than degree measure. There are numerous reasons for this, most of them based on the fact that radians are more “natural” (even if it may not seem that way to you at ﬁrst.) A radian is deﬁned to be the measure of an angle cut of in the circle of radius one by an arc of length one. Thus, a 90◦ angle corresponds to an angle of Integrated π radian measure 2 , since the distance one fourth of the way around the unit Calculus/Pre-Calculus π ◦ circle is 2 . It is also useful to note that an angle of measure 1 corresponds Mark R. Woodard π with an angle of radian measure 180 , since 90 of these would correspond to a right angle. Also, an angle of radian measure 1 would correspond to 180 ◦ π an angle of measure π , since 2 of these would correspond to a right angle. These facts are enough to help you convert from degrees to radians Title Page and back, when necessary. Contents π Exercise 4.1.3. What is the degree measure of the angle θ = 6 ? JJ II J I Exercise 4.1.4. What is the radian measure of the angle 225◦? Go Back Close Quit Page 90 of 191 4.2. Trigonometric Functions at Special Val- ues There are a few special angles for which you should know the values of the trigonometric functions, without having to resort to a table or a calculator. These are summarized in the following table. θ cos(θ) sin(θ) 0 1 0 √ π 3 1 Integrated 6 √2 √2 π 2 2 Calculus/Pre-Calculus 4 2 √2 π 1 3 Mark R. Woodard 3 2 2 π 2 0 1

Title Page You should also be able to use reference angles along with these values to compute the values of the trigonometric functions at related angles in Contents the second, third and fourth quadrants. For example, the point associated 5π JJ II with t = 6 is directly across the unit circle from the point associated π π with 6 . (In this case we say that we are using 6 as a reference angle.) J I 5π Thus the coordinates of the point associated with 6 has the same y value Go Back and the opposite x value of the point associated with π . Thus cos( 5π ) = √ 6 6 π 3 5π π 1 Close − cos( 6 ) = − 2 , and sin( 6 ) = sin( 6 ) = 2 . 7π Quit Exercise 4.2.1. Evaluate sin( 6 ). Page 91 of 191 −3π Exercise 4.2.2. Evaluate tan( 4 ). Exercise 4.2.3. Solve sin(x) + cos(x) = 0 for x. Exercise 4.2.4. Solve 2 cos(2x) + 1 = 0 for x.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 92 of 191 4.3. Properties of Trigonometric Functions

1. Pythagorean Identity: sin2(x) + cos2(x) = 1. 2. Pythagorean Identity Rephrased: 1 + tan2(x) = sec2(x), 1 + cot2(x) = csc2(x). 3. Range of sin(x) and cos(x): −1 ≤ sin(x) ≤ 1, −1 ≤ cos(x) ≤ 1.

4. cos(x) is Even: cos(−x) = cos(x). Integrated Calculus/Pre-Calculus 5. sin(x) is Odd: sin(−x) = − sin(x). Mark R. Woodard 6. Periodicity: sin(x + 2π) = sin(x), cos(x + 2π) = cos(x).

Title Page Contents JJ II J I Go Back Close Quit Page 93 of 191 4.4. Graphs of Trigonometric Functions 4.4.1. Graph of sin(x) and cos(x) Note that from the deﬁnition of sine and cosine, it is clear that the domain of each of these is the set of all real numbers. Also, from the properties above, we know that the range of both of these is the set of numbers between −1 and 1, and that the functions are periodic. This information, together with a few points plotted as a guide, are enough to graph the two π functions. Note that if we shift the graph of the sine function by 2 units to the left, we get the graph of the cosine function. This is related to the Integrated fact that sin(x − π ) = cos(x). Calculus/Pre-Calculus 2 Mark R. Woodard 4.4.2. Graph of tan(x) and cot(x)

Note that the domain of tan(x) is the set of all real numbers except those Title Page π 3π at which cos(x) = 0. Thus, the points 2 , 2 , and so on aren’t in the domain of tan(x). An easy way to characterize these points is to say that these are Contents π all the points which have the form 2 + kπ, where k is any integer. Thus π JJ II the domain of the tangent function is {x : x ∈ R, x 6= + kπ, k ∈ Z}. 2 J I Exercise 4.4.1. What is the domain of cot(x)? Go Back We can get a good grasp on the graph of tan(x) by plotting a few Close points and doing a careful analysis of the limiting behavior when x is π Quit near 2 and the other points that aren’t in the domain. Note that when π x is a little less than 2 , sin(x) is close to 1, while cos(x) is close to zero Page 94 of 191 (but is positive.) Thus, the ratio of these two things should be big. Thus −π lim π − tan(x) = +∞. A similar analysis of points to the right of helps x→ 2 2 us believe that lim −π + tan(x) = −∞. This information together with x→ 2 −π a few strategically plotted points (like tan( 4 = −1 and tan(0) = 0 and π tan( 4 ) = 1) should help us to graph the tangent function on the interval π π (− 2 , 2 ). Now using the fact that the tangent function is periodic with period π, we can graph the whole tangent function. Exercise 4.4.2. Do a similar analysis to graph the cotangent function.

Integrated 4.4.3. Graph of sec(x) and csc(x) Calculus/Pre-Calculus Mark R. Woodard Like the tangent function, the domain of the secant function is the set of all real numbers except those which make cos(x) equal to zero. Thus the domain of the secant function is the same as the domain of the tangent function. Also, the fact that the cosine function always has values between Title Page 1 −1 and 1 tells us that sec(x) = cos(x) always has values less than −1 Contents or greater than 1. An analysis of the limiting behavior of sec(x) near π −π JJ II x = 2 and 2 and a few strategically plotted points leads to the graph of y = sec(x). J I Exercise 4.4.3. Finish this analysis and draw a graph of y = sec(x). Go Back Then do a similar analysis and graph y = csc(x). Close Quit Page 95 of 191 4.5. Trigonometric Identities

1. Addition Identity for sin(x): sin(x + y) = sin(x) cos(y) + sin(y) cos(x). 2. Addition Identity for cos(x): cos(x + y) = cos(x) cos(y) − sin(x) sin(y).

From these, we can get other identities like the double angle formulas. For example, Integrated Calculus/Pre-Calculus sin(2x) = sin(x + x) = sin(x) cos(x) + sin(x) cos(x) = 2 sin(x). Mark R. Woodard Exercise 4.5.1. Show that cos(2x) = cos2(x)−sin2(x) in a manner similar to the above. 2 1−cos(2x) Title Page We can also show that sin (x) = 2 as follows: since cos(2x) = 2 2 1−cos(2x) 1−(cos2(x)−sin2(x)) 1−cos2(x)+sin2(x) Contents cos (x) − sin (x), we have 2 = 2 = 2 . Now using the fact that 1 − cos2(x) = sin2(x), we have that the above 2 2 JJ II expression is equal to sin (x)+sin (x) = sin2(x). 2 J I 2 1+cos(2x) Exercise 4.5.2. Show that cos (x) = 2 , using methods similar to Go Back the above. Close Quit Page 96 of 191 4.6. Derivatives of Trigonometric Functions If f(x) = sin(x), what is f 0(x)? We have no alternative but to use the deﬁnition of derivative. Thus we have that

sin(x + h) − sin(x) f 0(x) = lim h→0 h sin(x) cos(h) − sin(h) cos(x) − sin(x) = lim h→0 h sin(x) cos(h) − sin(x) − sin(h) cos(x) = lim Integrated h→0 x Calculus/Pre-Calculus sin(x) (cos(h) − 1) − cos(x) sin(h) = lim Mark R. Woodard h→0 h (cos(h) − 1) sin(h) = lim sin(x) − cos(x) h→0 h h Title Page cos(h) − 1 sin(h) = sin(x) lim − cos(x) lim . Contents h→0 h h→0 h

cos(h)−1 sin(h) JJ II If we denote limh→0 h by a and limh→0 h by b, then we have deduced that the derivative of sin(x) is a sin(x) + b cos(x). It remains to J I compute these two limits. Go Back Close 4.6.1. Two Important Trigonometric Limits Quit sin(h) We will need to use a little geometry to help us compute limh→0 h . Page 97 of 191 Consider the picture below:

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Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Recall that the area of a sector of a circle of radius r and angle h is 1 2 1 2 Contents A = 2 hr . Now consider the sector OSQ. It has area 2 h cos (h). On the other hand, triangle OPQ has area 1 cos(h) sin(h), and sector OPR has 2 JJ II area 1 h. From the picture we can tell that even when h is small, 2 J I area OSQ ≤ area OPQ ≤ area OPR Go Back so Close 1 1 1 h cos2(h) ≤ cos(h) sin(h) ≤ h. Quit 2 2 2 Page 98 of 191 If we multiply through by 2 and divide by cos(h) · h, we see that sin(h) 1 cos(h) ≤ ≤ . h cos(h) Now as h → 0, cos(h) → 1, so this inequality together with the squeeze sin(h) theorem tell us that limh→0 h = 1. sin(h) Exercise 4.6.1. Use the now established fact that limh→0 h = 1, to cos(h)−1 show that limh→0 h = 0. Hint: Try multiplying the numerator and the denominator by the conjugate of the numerator. Integrated Calculus/Pre-Calculus 4.6.2. Derivatives of other Trigonometric Functions Mark R. Woodard d d Knowing that dx sin(x) = cos(x) and dx cos(x) = − sin(x), and knowing the quotient rule is enough to enable one to compute the derivatives of the other trigonometric functions, since they are all quotients involving these Title Page two trigonometric functions. For example Contents d tan(x) = d sin(x) dx dx cos(x) JJ II cos(x) cos(x)−sin(x)(− sin(x)) = cos2(x) 1 J I = 2 cos (x) Go Back = sec2(x) Close Quit d d d Exercise 4.6.2. Compute dx sec(x), dx cot(x), and dx csc(x) in a manner Page 99 of 191 similar to the previous example. Of course, it would be a good idea to practice these new trigonometric derivatives in conjuction with out familar rules such as the chain rule, quotient rule, product rule, etc.. Exercise 4.6.3. Compute the derivative with respect to x of f(x) = x2+1 sin 2x−3 . √ √ Exercise 4.6.4. Compute g0(x) for g(x) = ( x2 − 4) · tan( x2 − 4).

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 100 of 191 Click to Initialize Quiz Answer the following questions about trigonometric derivatives.

tan(x) 1. What is the derivative of f(x) = cos(x) ? sec2(x) cos(x) sin2(x)+1 cos(x) sin(2x) cos3(x)

2. What is the derivative of f(x) = tan(cos(x))? sec2(cos(x) tan(− sin(x)) sec2(cos(x))·− sin(x) Integrated 3. If f(x) = (sin(x2 + cos(x2)))2, what is f 0(x)? Calculus/Pre-Calculus 2 sin(x2 + cos(x2)) 2 sin(x2 + cos(x2)) · 2 sin(x2 + cos(x2)) · Mark R. Woodard (cos(x2 + − sin(x2)) (cos(x2 + cos(x2)) · (2x + − sin(x2) · 2x) Title Page 4. Find d pcot(2x + 1). dx Contents 1 −1 1 −1 p 2 2 2 − csc (2x + 1) 2 (cot(2x + 1) · 2 (cot(2x + 1) · 1 − csc2(2x + 1) · 2 JJ II Click to End Quiz J I Go Back Close Quit Page 101 of 191 CHAPTER 5 Applications of the Derivative

Contents

5.1 Definition of Maximum and Minimum Values . 104 Integrated 5.2 Overview ...... 110 Calculus/Pre-Calculus 5.3 Rolle’s Theorem ...... 111 Mark R. Woodard 5.4 The Mean Value Theorem ...... 113 5.5 An Application of the Mean Value Theorem .. 115 Title Page 5.6 Monotonicity ...... 117 Contents 5.7 Concavity ...... 121 5.8 Vertical Asymptotes ...... 124 JJ II 5.9 Limits at Infinity; Horizontal Asymptotes ..... 126 J I 5.9.1 A Simplification Technique ...... 127 Go Back 5.10 Graphing ...... 129 Close 5.11 Optimization ...... 133 Quit 5.12 Newton’s Method ...... 137 Page 102 of 191 5.13 Antiderivatives ...... 140 5.14 Rectilinear Motion ...... 142 5.15 Antiderivatives Mini-Quiz ...... 143

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 103 of 191 5.1. Deﬁnition of Maximum and Minimum Values An important application of calculus is that it can be used to maximize and minimize functions. One might be trying to minimize the cost of producing some item, maximize the area or volume of a geometric region, or minimize the time it takes for a certain biological reaction to occur. In all of these, the theoretical mathematical basis is the same. We begin with the deﬁnition of what we mean by a maximum and minimum value. Integrated Calculus/Pre-Calculus A function f has a maximum value of M at x = c if Mark R. Woodard

f(c) = M, and f(x) ≤ f(c) for all x in the domain of f(x). Title Page Likewise, we say that f has a minimum value of m at x = c if Contents

f(c) = m, and f(x) ≥ f(c) for all x in the domain of f(x). JJ II J I Go Back Exercise 5.1.1. Using what you know about the graph of y = x2, ﬁnd the absolute minimum of y = (x − 3)2 + 2 and the absolute maximum of Close 2 y = −3 − (x + 2) . Quit Sometimes the fact that the maximum is begin considered over the Page 104 of 191 entire domain is emphasized by putting the adjectives global or absolute in front of the words maximum or minimum. Thus one might say “there is a global maximum of 3 at x = 4” or “there is an absolute minimum of 2 at x = 5.” In contrast, you might sometimes want to discuss the fact that a certain function value is larger than those near it, even if it isn’t the largest in the whole domain. We call such points relative or local maximums or minimums.

We say that there is a relative or local maximum value Integrated M at x = c if there exists an interval I containing c so Calculus/Pre-Calculus that Mark R. Woodard f(c) = M, and f(x) ≤ f(c) for all x in I.

Likewise we say that here is a relative or local minimum Title Page value m at x = c if there exists an interval I containing Contents c so that JJ II f(c) = m, and f(x) ≤ f(c) for all x in I. J I Go Back Close Quit Exercise 5.1.2. Consider the function pictured. Where does it have rela- Page 105 of 191 tive extrema? (Note: the word extrema means “maximum or minimum.”)

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Calculus/Pre-Calculus £ ¥ Mark R. Woodard Note that not all functions have maximums or minimums (local or global.) For example, the straight line y = 2x + 1 has neither kind of maximum or minimum over its entire domain. Even if you restrict the Title Page domain to be a finite interval like (2, 5), there is still no maximum or min- Contents imum value. If we restrict the domain to be a closed interval like [2, 5], this example does have a maximum of 11 at x = 5 and a minimum of 5 at JJ II x = 2, but a function like J I  x + 2 if 2 ≤ x ≤ 3  Go Back g(x) = −20x + 80 if 3 < x < 4 Close x + 2 if 4 ≤ x ≤ 5 Quit has no maximum or minimum on [2, 5]. Page 106 of 191 Exercise 5.1.3. Explain why f(x) = 2x+1 has no maximum or minimum on (2, 5). Then explain why the function g(x) defined in the previous paragraph has no maximum or minimum on [2, 5]. Of course, the fact that g(x) above has no maximum or minimum on [2, 5] is related to the fact that it is not continuous. What turns out to be true (but is not obvious and is actually quite difficult to prove) is that if we restrict ourselves to continuous functions defined on closed intervals, we can be sure we have a maximum and a minimum. This is called the extreme value theorem.

Integrated The Extreme Value Theorem Calculus/Pre-Calculus A continuous function deﬁned on a closed interval always Mark R. Woodard has a maximum and a minimum value on that interval.

The examples above show the necessity of the hypotheses: if the interval Title Page is open, the maximum or minimum might not exist. If the function is not continuous, the maximum or minimum might not exist. Contents Of course, the question still remains: how can one find the value of x JJ II which gives a maximum or a minimum? Here are some examples that you are already familiar with: the function f(x) = x2 has a minimum at x = 0. J I Of course, the interesting thing about this function is that the tangent line Go Back at x = 0 is a horizontal line. In other words, the slope of the tangent line Close at this point is zero. Thus we see that extrema can occur at points where the derivative is zero. We will call such points stationary points. Another Quit example is h(x) = |x|, which also has a minimum at x = 0. However, this Page 107 of 191 function does not have a horizontal tangent line at x = 0, in fact, there is no tangent line there at all! Recall that h(x) = |x| is not differentiable at x = 0, so we see that extrema can occur at points where the derivative doesn’t exist. We call such points singular points. A third example is to reconsider the function f(x) = 2x+1 on the interval [2, 5]. We have already seen that the maximum value for this function occurs at the right endpoint, while the minimum occurs at the left endpoint. Thus we see that extrema can occur at endpoints. What is not obvious (but is true nonetheless,) is that this is the complete list of types of points where extrema can occur. This fact is called “Fermat’s Theorem.” Integrated Calculus/Pre-Calculus Fermat’s Theorem: Mark R. Woodard If f(x) has a maximum or a minimum at a point (c, f(c)) where c is not an endpoint of the domain, then either f 0(c) = 0 or f 0(c) doesn’t exist. Title Page Contents We can apply this theorem as follows: given a continuous function f(x) and a closed interval [a, b], we can find all singular and stationary points by JJ II studying f 0(x) and finding where it is equal to zero or where it is undefined. J I Since the maximum and minimum must occur at either a stationary point, a singular point, or an endpoint, by comparing the function values at these Go Back points, we can determine the maximum and minimum for the function. Close

Exercise 5.1.4. Find the absolute maximum and absolute minimum of Quit f(x) = 4x5 − 5x4 + 3 on the interval [−3, 2]. Page 108 of 191 Click to Initialize Quiz Answer the following questions about maximum and minimum values.

1. What are the values of the relative extrema of f(x) = 2x3 + 3x2 − 120x + 30? −5, 4 455, −274 4, 0

2. What is the maximum value of f(x) = |x − 3| on [−1, 5]? There is no 0 4 2 maximum Integrated value. Calculus/Pre-Calculus Mark R. Woodard 2 3. Does the function f(x) = 3x 3 + 2 have a minimum on [0, 4]? If so, what is it? Yes, it is 2. No, there isn’t one. Yes, it is 5. Title Page

4. Does f(x) = −x2 + 4x have a maximum on (1, 5)? How about a Contents minimum? There is a There is neither a There is a minimum JJ II maximum but no maximum nor a but no maximum. J I minimum. minimum. Go Back Click to End Quiz Close Quit Page 109 of 191 5.2. Overview The Mean Value Theorem is an enigma, in that it seems to be a somewhat innocuous statement of fact, but turns out to be a key element in the proof of many important theorems. It is important because it relates the derivative (which is not actually a quotient but is rather the limit of a certain quotient) to a particular quotient. As a warm-up, we ﬁrst investigate a related theorem due to the French mathematician Michel Rolle.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 110 of 191 5.3. Rolle’s Theorem Try to draw a function f(x) with the following characteristics: It is defined on the interval [2, 5], it is continuous and differentiable on its domain, and it satisfies f(2) = 0 = f(5). No matter how you draw such a function, you will find that you there must have been someplace on the domain where there was a horizontal tangent line. In other words, there must have been a point c between 2 and 5 where f 0(c) = 0. This fact is called Rolle’s Theorem: Integrated Rolle’s Theorem Calculus/Pre-Calculus If f(x) is continous on [a, b] and differentiable on (a, b), Mark R. Woodard and if f(a) = 0 = f(b), then there is a point c with a < c < b and f 0(c) = 0.

Title Page The reason Rolle’s Theorem is true is quite simple. Since f(x) is a continuous function defined on a closed interval, the Extreme Value The- Contents orem says that it has a maximum and a minimum on [a, b]. If either the JJ II maximum or minimum or the minimum occur at some point c other than the endpoints, then Fermat’s Theorem says that either f 0(c) = 0 or f 0(c) J I doesn’t exist. But one of our hypotheses is that f(x) is differentiable on Go Back [a, b], so f 0(c) must exist, so f 0(c) = 0. If neither the maximum nor the Close minimum occur at a point other than an endpoint, then f(a) = 0 = f(b) is both the maximum and the minimum. But this means that our function is Quit the constant function f(x) = 0, and this function has lots of values where Page 111 of 191 the derivative is zero! (In fact, the derivative is zero at every point on (a, b).) x2−2x−3 Exercise 5.3.1. Consider the function f(x) = x+8 on the interval [−1, 3]. Does Rolle’s theorem apply? If not, explain why not. If so, find one or more of the promised values of c where f 0(c) = 0. Exercise 5.3.2. The next two exercises will show you that the hypotheses for Rolle’s theorem are necessary. Consider the function defined by g(x) = 2 0 x 3 on the interval [−1, 1]. Show that there is no value of c so that g (c) = 0 on [−1, 1]. Why does this not violate Rolle’s Theorem? Integrated Exercise 5.3.3. Consider the function defined by Calculus/Pre-Calculus ( Mark R. Woodard 0 if x = 0 g(x) = −2x + 2 if 0 < x ≤ 1.

Does Rolle’s Theorem apply to this function on [0, 1]? Title Page Contents JJ II J I Go Back Close Quit Page 112 of 191 5.4. The Mean Value Theorem One way to think of Rolle’s Theorem is this: if a function f(x) meets the hypotheses of the theorem, then there must be a point c between a and b where the tangent line is parallel to the x axis. The Mean Value Theorem is a generalization of Rolle’s Theorem. It doesn’t require that f(a) = 0 = f(b), but instead says that as long as f(x) is continuous on [a, b] and diﬀerentiable on (a, b), there will be a point c between a and b where the tangent line is parallel to the line containing (a, f(a)) and (b, f(b)). Stated algebraically we have Integrated Calculus/Pre-Calculus The Mean Value Theorem Mark R. Woodard If f(x) is continuous on [a, b] and diﬀerentiable on (a, b), then there is a point c so that a < c < b with

f(b) − (a) Title Page f 0(c) = . b − a Contents

Note that you can think of the right hand side of the above equation JJ II as representing the slope of the line between (a, f(a)) and (b, f(b)), so the J I above says that there is a point c where the slope of the tangent line is Go Back equal to the slope of the secant line, so the two lines are parallel. The Mean Value Theorem can be seen to be true using Rolle’s Theorem Close and a clever trick. Given f(x) which meets the hypotheses of the theorem, Quit f(b)−f(a) consider the function g(x) = f(x) − f(a) − b−a · (x − a). One can Page 113 of 191 check that this function is continuous on [a, b] and diﬀerentiable on (a, b), and g(a) = 0 = g(b). (You should check this yourself!) Thus, g(x) meets the hypotheses of Rolle’s theorem. So there exists c with g0(c) = 0. But 0 0 f(b)−f(a) 0 g (c) = f (c) − b−a , so if g (c) = 0, we get the conclusion of the Mean 0 f(b)−f(a) Value Theorem, that f (c) = b−a . This function g(x) seems to be quite mysterious, it can be derived geometrically, however, by taking the picture for the Mean Value Theorem and trying to make it look like the picture for Rolle’s Theorem.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 114 of 191 5.5. An Application of the Mean Value The- orem

We claimed above that the Mean Value Theorem is often used to proof other interesting and important theorems. We will now see an example of one such theorem. By way of introduction, note that there are many functions which have derivative equal to 2x. Some of them are y = x2, y = x2 + 4, y = x2 − 123, and so on. The question is: do all functions whose derivative is equal to 2x have the form y = x2 + C where C is a constant? The answer turns out to be yes, and we will prove it using the Integrated Mean Value Theorem. Calculus/Pre-Calculus Mark R. Woodard

The “+C” Theorem If f(x) and h(x) are diﬀerentiable functions, and if f 0(x) = h0(x), then f(x) = h(x) + C. Title Page Contents

We can see that the “+C” theorem holds by considering g(x) = f(x) − JJ II 0 0 0 h(x). Note that g (x) = f (x) − h (x) = 0. We will show that g must J I be a constant function. Suppose x1 is any point in the domain of g, and Go Back g(x1) = k. Let x2 be any other point in the domain of g with x1 6= x2. Consider the quantity g(x2)−g(x1) . By the Mean Value Theorem, there must Close x2−x1 0 g(x2)−g(x1) 0 be a number c between x1 and x2 with g (c) = . But g (x) = 0 Quit x2−x1 for all x, so in particular g0(c) = 0. Thus we have g(x2)−g(x1) = 0, but then x2−x1 Page 115 of 191 g(x2)−g(x1) = 0, so g(x2) = g(x1) = k. We have shown that for any point x2 in the domain of g, the output is the same. Thus g(x) = f(x) − h(x) is a constant function. So f(x) = h(x) + C for some constant C.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 116 of 191 5.6. Monotonicity We all have an intuitive idea about what it means for a function to “go up” or increase (or to do the opposite: “go down” or decrease.) Let’s formalize this idea with a deﬁnition:

We say that a function is increasing on an interval I if for all x1, x2 ∈ I with x1 < x2 we have f(x1) < f(x2). We say that f(x) is decreasing on an interval I if for all x1, x2 ∈ I with x1 < x2 we have f(x1) > f(x2) Integrated Calculus/Pre-Calculus For example, f(x) = x2 is increasing on the interval [0, ∞) and decreas- Mark R. Woodard ing on the interval (−∞, 0]. A function is said to be monotonic on an interval I if it is strictly increasing or strictly decreasing on I. In attempt- ing to understand the geometry of the graph of a particular function f(x), Title Page it is desirable to ﬁnd the intervals on which the function is monotonic. Of course, it is most interesting to ﬁnd the largest such intervals. There is a Contents clear and almost obvious relationship between the derivative of a function JJ II and the monotonicity. This can be seen by drawing a number of strictly increasing functions, and drawing tangent lines on any of them at any point, J I and thinking about what is always true about the slopes of the tangent Go Back lines. You should be able to informally convince yourself pretty quickly Close that they all have positive slopes. Thus there seems to be a relationship between an increasing function and a positive derivative. Likewise, there Quit seems to be a relationship between a decreasing function and a negative Page 117 of 191 derivative. The following seems plausible: Monotonicity Theorem If f(x) is so that f 0(x) > 0 for all x in the interval I, then f(x) is increasing on I. If f 0(x) < 0 for all x in I, then f(x) is decreasing on I.

The proof of the Monotonicity Theorem uses the Mean Value Theorem. 0 Integrated The idea is that if f (x) > 0 on I, and if x1 < x2 on I, then there is f(x2)−f(x1) 0 0 Calculus/Pre-Calculus a point c between x1 and x2 with = f (c). But if f (c) > 0, x2−x1 Mark R. Woodard we must have f(x2) − f(x1) > 0, since the denominator x2 − x1 > 0. So f(x1) < f(x2), so f(x) is increasing. A similar proof works for f(x) decreasing when f 0(x) < 0 on I. This result can be used to find the intervals of monotonicity for a given function: by finding the largest intervals on Title Page which the derivative of f(x) is positive, we are also finding the largest Contents intervals on which f(x) is increasing. A similar statement can be made replacing the word “increasing” by “decreasing” and the word “positive” JJ II by “negative.” J I Go Back x Example 5.6.1. Find the largest intervals on which f(x) = x2+2 is mono- Close tonic. Quit Page 118 of 191 x Solution: Find the largest intervals of monotonicity for f(x) = x2+2 . First we need to compute and simplify f 0(x). We have

(x2 + 2)(1) − (x)(2x) f 0(x) = Quotient Rule (x2 + 2)2 2 − x2 = Simplifying (x2 + 2)2

√ Integrated 0 Note that the numerator of f (x) is zero when x = ± 2, and the de- Calculus/Pre-Calculus 0 nominator is never zero. Thus, f √(x) is either√ √ strictly positive√ or strictly Mark R. Woodard negative on the intervals (−∞, − 2), (− 2√, 2),√ and ( 2, ∞). Using 0 test values, we see that f (x) is positive on (− 2, √2) and√ negative on the other two intervals. Thus f(x) is increasing on (− 2, 2) and decreasing √ √ Title Page on (−∞, − 2) and on ( 2, ∞). Contents JJ II One upshot of the Monotonicity Theorem is that we can now see how to use the ﬁrst derivative to help us locate relative maximum and relative J I minimums. This is because if f(x) is continuous on the interval (a, b), and Go Back is increasing on an interval (a, c) and decreasing on (c, b) where c is between Close a and b, then there must clearly be at least a relative maximum at x = c. A similar statement can be made about relative minimums if f(x) changes Quit from decreasing to increasing in the middle of some interval on which f is Page 119 of 191 continuous. These ideas are collectively know as The First Derivative Test Suppose that f(x) is continuous on (a, b) and diﬀeren- tiable on (a, b) except perhaps at a point c with a < c < b. Suppose also that either f 0(c) = 0 or f 0(c) doesn’t exist. Then 1. If f 0(x) < 0 on (a, c) and f 0(x) > 0 on (c, b), then there is a relative minimum value at x = c. 2. If f 0(x) > 0 on (a, c) and f 0(x) < 0 on (c, b), then Integrated there is a relative maximum value at x = c. Calculus/Pre-Calculus 3. If the sign of f 0(x) is the same on both (a, c) and Mark R. Woodard (c, b), then there is no relative extremum at x = c.

Title Page Exercise 5.6.1. Use the analysis done previously to ﬁnd the relative ex- x Contents trema for f(x) = x2+2 . JJ II J I Go Back Close Quit Page 120 of 191 5.7. Concavity We’ve had such good luck relating the sign of the derivative of a function to information about the graph of the function that one can’t help but wonder whether or not the sign of the second derivative would yield any nice information about the graph of the function. Consider the following

graph of y = sin(x). ¤

¢ Integrated

£ Calculus/Pre-Calculus

¢ ¢

¡ Mark R. Woodard

¤ ¡ Title Page

Note that between −π and 0, the slopes of the tangent lines are in- Contents creasing as we go from left to right. (The lines start out having negative JJ II slopes, and then the slopes increase to zero, and then the lines continue to get steeper.) Since the derivative of y = sin(x) tells us the slope of the J I tangent line at a given point, we can see that the derivative of the deriva- Go Back tive of y must be positive here. Thus the second derivative is positive. In Close this situation, we say that y is concave up on the interval (−π, 0). On the other hand, between 0 and π we see that the slopes of the tangent lines are Quit decreasing as we move from left to right, so the derivative of the derivative Page 121 of 191 must be negative here, so y00 < 0 on (0, π). Summarizing: We say that f(x) is concave up on an interval I iff f 00(x) > 0 on I. We say that f(x) is concave down on I iff f 00(x) < 0 on I. If c is a point in the domain of f(x), and if f(x) has one type of concavity on (a, c) and the other type on (c, b), then we say that the point (c, f(c)) is a point of inflection of f(x).

Integrated Calculus/Pre-Calculus Mark R. Woodard x Exercise 5.7.1. Continue studying f(x) = x2+2 by analyzing its concavity. Title Page Contents JJ II It turns out that for stationary points, we can use the second derivative J I to help analyze whether a given point is a relative maximum or a relative Go Back minimum. This is because if f(x) has a relative minimum and a horizontal Close tangent line at x = c, then it must be concave up there. (Try drawing an example and see for yourself!) On the other hand if it has a relative Quit maximum at x = c and a horizontal tangent line, it must be concave down Page 122 of 191 there. Thus we have: The Second Derivative Test Suppose that f(x) is so that f 00(x) exists in some interval containing c, and f 0(c) = 0.

1. If f 00(c) < 0 then there is a relative maximum at (c, f(c)). 2. If f 00(c) > 0 then there is a relative minimum at (c, f(c)). 3. If f 00(c) = 0, we make no conclusion. Integrated Calculus/Pre-Calculus Mark R. Woodard Exercise 5.7.2. Use the analysis done previously to ﬁnd the relative ex- x trema for f(x) = x2+2 using the second derivative test. Title Page Contents JJ II J I Go Back Close Quit Page 123 of 191 5.8. Vertical Asymptotes We have already noted in this course that sometimes the limit of a function as x → c doesn’t exist because when x is close to but not equal to c, the function gets very large. In this case we write limx→c f(x) = ∞. Geometrically, this means that there is a vertical asymptote at x = c. For 1 example, the function f(x) = x2 has a vertical asymptote at x = 0, since 1 limx→0 x2 = ∞. For the functions we have studied, vertical asymptotes n(x) tend to occur when we have a rational expression of the form f(x) = d(x) , and there is a value c for which d(c) = 0, but n(c) 6= 0. Then as x → c, Integrated the ratio f(x) would increase without bound. Of course, it could be the Calculus/Pre-Calculus case that both n(c) and d(c) are zero, but the ratio could increase without Mark R. Woodard bound, as the example below shows. sin(x) Exercise 5.8.1. Does f(x) = x2 have a vertical asymptote at x = 0 or not? Title Page Contents JJ II J I Go Back Close Quit Page 124 of 191 Click to Initialize Quiz For each of the following, ﬁnd all of the vertical asymptotes.

5x 1. f(x) = x−3 x = 3 x = 0 x = 0, x = 3 x3 2. f(x) = x2−1 x = 1 x = 0, x = 1 x = ±1 x = ±1, x = 0 3. f(x) = x cot(x), x ∈ [−4, 4]. Integrated There aren’t x = ±π x = ±π, x = 0 Calculus/Pre-Calculus any. Mark R. Woodard Click to End Quiz

Title Page Contents JJ II J I Go Back Close Quit Page 125 of 191 5.9. Limits at Inﬁnity; Horizontal Asymp- totes

We now study a different kind of limit involving infinity, which will be related to horizontal asymptotes. A horizontal asymptote occurs when a curve levels off at a certain y value as x gets very large in either the postive or negative direction. Thus, it seems natural to define limx→∞ f(x). We need a technical definition akin to the epsilon/delta definition of ordinary limits. We want to say that when x is really big, the function f(x) should be close to the constant function g(x) = L. Thus, we want to say that for Integrated any > 0, if x is large enough, we should have |f(x) − L| < . Formally Calculus/Pre-Calculus we have: Mark R. Woodard

Technical Deﬁnition of limx→∞ f(x) = L. Title Page We say that limx→∞ f(x) = L if it is true that for any > 0 there exists M so that if x > M, then |f(x)−L| < . Contents JJ II

1 J I Exercise 5.9.1. Use the above definition to show that limx→∞ x = 0. What does this tell us about horizontal asymptotes? Go Back Close The definition above doesn’t really address the question of “how does one compute limits as x → ∞?” Fortunately, our limit theorems are still Quit applicable, and this with the result of the previous exercise allow us to draw Page 126 of 191 some nice deductions. Also, we will introduce a simplification technique that will proof to be advantageous.

Exercise 5.9.2. Use the limit theorems and the above result to show that 1 limx→∞ xr = 0, for r a rational number, r ≥ 0.

Integrated Calculus/Pre-Calculus Mark R. Woodard

5.9.1. A Simpliﬁcation Technique Title Page Contents JJ II When trying to compute limx→∞ f(x) where f(x) is a rational expression J I like n(x) , it is often useful to simplify the rational expression by dividing d(x) Go Back both the numerator and denominator by the same quantity. This is especially useful if you can ﬁnd a quantity q(x) so that you know the limit of Close n(x) d(x) x2 either q(x) or q(x) . For example, consider limx→∞ 3x2+2x+1 . As the reader Quit x2 3x2+2x+1 will shortly see, we know the limits of both x2 (of course) and x2 , so Page 127 of 191 it would be prudent to divide both the numerator and denominator by x2. Thus we have

2 2 1 x x 2 lim = lim · x divide by one in a clever way x→∞ 2 x→∞ 2 1 3x + 2x + 1 3x + 2x + 1 x2 x2 x2 = lim 2 Rewriting x→∞ 3x +2x+1 x2 1 = lim 2 Simplifying x→∞ 3x 2x 1 x2 + x2 + x2 1 = lim Simplifying Integrated x→∞ 2 1 3 + x + x2 Calculus/Pre-Calculus 1 Mark R. Woodard = Taking the Limit 3 + 0 + 0 1 = Arithmetic! 3 Title Page Contents JJ II J I Go Back Close Quit Page 128 of 191 5.10. Graphing

Integrated Calculus/Pre-Calculus We have now devised techniques which, when put together, enable us to Mark R. Woodard graph almost any function in our catalogue. To do this, we carefully analyze our function for domain, asymptotes, monotonicity, concavity and extrema. We then plot any “important” points such as inflection points, relative Title Page maximum and minimum points, and indicate any asymptotes on our graph, and then draw a sketch, connecting the points we have plotted and being Contents sure to respect the information obtained so far. JJ II J I Go Back Close Quit A checklist to make sure we don’t miss anything in this somewhat long Page 129 of 191 process follows: Steps for Graphing f(x). 1. Note the domain of f(x). Any points or intervals not in the domain should be carefully noted. Any isolated points which are not in the domain should appear on any sign chart created while analyzing this function. 2. Look for vertical and horizontal asymptotes. Recall that a vertical asymptote occurs at x = c if limx→c± f(x) = ±∞. Such things often (but not always) occur in rational expressions where the denominator is zero. Horizontal asymptotes occur when Integrated Calculus/Pre-Calculus limx→±∞ f(x) = L, in which case there is a horizontal asymptote at y = L. Mark R. Woodard 3. Investigate the Monotonicity of f(x). This will involve a study of the sign of f 0(x). Be sure to carefully make a sign chart over the domain which includes any values at which f 0(x) = 0 or Title Page at which f 0(x) doesn’t exist. Contents 4. Locate Relative Extrema of f(x). Since you have already made JJ II a sign chart for f 0(x), you can easily apply the first derivative test to locate the relative extrema of f(x). J I Go Back Close Quit Page 130 of 191 Steps for Graphing f(x), continued. 5. Investigate the Concavity of f(x). This will involve a study of the sign of f 00(x). Be sure to make a sign chart. From the sign chart you should be able to find points of inflection. It would also be prudent to double check your conclusions about relative extrema by applying the second derivative test to any points where f 0(x) = 0. 6. Preparation for Graphing. Integrated (a) Plot a few important points. Plot all extrema and inflec- Calculus/Pre-Calculus tion points. (Be sure to use f(x) and not f 0(x) or f 00(x) to Mark R. Woodard determine the y values of points you are plotting.) (b) Intercepts. Plot the point (0, f(0)), since this is usually easy to compute, and represents the y-intercept. If easy to do so, Title Page set f(x) = 0 and solve, and then plot any of these x-intercepts that you find. Contents JJ II J I Go Back Close Quit Page 131 of 191 Steps for Graphing f(x), continued. 6. Preparation for Graphing. (Continued.) (c) Asymptotes. Using a dashed line, indicate the asymptotes on the graph, and for vertical asymptotes, think about whether the function is approaching +∞ or −∞ as the x value is approached from either the left or the right. (d) Symmetry. Think for a moment about symmetry. If f(−x) = f(x), then the function is symmetric about the x Integrated axis. If f(−x) = −f(x), then the function is symmetric about Calculus/Pre-Calculus the origin. In most cases, of course, the function will have nei- Mark R. Woodard ther of these two kinds of symmetry. 7. Sketch the Graph. Using all the information gathered so far, draw a sketch of the function. Be sure to respect the domain, all Title Page the intervals of monotonicty and concavity, the extrema and so forth. If it does not seem possible to draw a sketch which respects Contents all the information you have, there is likely a mistake in some JJ II calculation. Check over your work to try and find the mistake before proceeding. J I Go Back

x2−2 Close Exercise 5.10.1. Sketch a graph of f(x) = x3 , following the above guidelines. Quit Page 132 of 191 5.11. Optimization

Suppose that you wanted to make a rectangular aquarium out of 4 pieces of glass, and you wanted it to hold exactly 32 cubic feet of water. Also, the aquarium is going to have two square sides (not the bottom), and the aquarium doesn’t need a top. What dimensions should you make it, in order to minimize the amount of glass needed? This kind of question is a constrained optimization problem: a problem in which your goal is to maximize or minimize some quantity subject to one or more given constraints. In this problem, we are constrained by the fact that our aquarium must Integrated have a certain volume. Our objective is to minimize the amount of glass Calculus/Pre-Calculus used. In such problems, we need to give variable names to the quanti- Mark R. Woodard ties involved, and then come up with an equation which represents each constraint, and one which represents the objective to be maximized or minimized. Then typically, we use the constraint equation to reduce the Title Page number of variables in the objective equation, and then use the techniques of this chapter to maximize or minimize the objective. For example, if we Contents let the base of our aquarium have dimensions x × y, then we should let the JJ II third dimension also be x, so that there is an x × x square on two of the sides. Note that x2y = 32, since the volume is x2y. The surface area repre- J I sents the amount of glass we require, and the bottom has dimensions x×y, Go Back and two of the sides have dimensions x × x and two have dimensions y × x. Close A rough sketch of the aquarium will help in making this determination. Thus, our objective is to minimize S = xy + 2x2 + 2xy = 2x2 + 3xy. This Quit 32 function has two variables in it, but our constraint tells us that y = x2 , Page 133 of 191 2 32 2 96 so we can rewrite S as S = 2x + 3x( x2 ) = 2x + x . This function is continuous for x > 0, and we shall shortly see that it only has one critical number for x > 0, and that it is strictly decreasing from 0 to this critical number, and strictly increasing from this critical number on. Thus, this only critical number will yield a minimum, as desired. Calculating f 0(x), we have that f 0(x) = 4x − 96 , which is zero only when x3 = 24, that √ x2 √ 3 0 3 0 is,√ when x = 2 3. Note that f (x) < 0 on (0, 2 3) and√ f (x) > 0 on (2 3 3, ∞). Thus, there is a absolute minimum at x = 2 3 3. At this value of x, we have that y = 32 = 32√ . Thus, the dimensions of the aquarium √ √ x2 2 3 3 3 3 16 should be 2 3 × 2 3 × √3 . 3 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 134 of 191 Steps for Solving Optimization Problems 1. Read the problem carefully. Draw a diagram if appropi- ate. As you read, think about the constraints and the objective. You may need to read the problem thorugh more than once. 2. Write down a constraint equation. In doing so, be sure that you have carefully identified whatever variables you are using. There might be more than one constraint equation for a given problem. Integrated 3. Write down an objective function. This is the function to be Calculus/Pre-Calculus maximized or minimized. Mark R. Woodard 4. Simplify the objective function, using the constraint equation. Do this especially if your objective function has more than one variable. Title Page 5. Maximize or Minimize the objective function. This involves Contents finding critical numbers of the objective function. Be sure to pay attention to the natural domain of the objective function. JJ II J I 6. Be sure that your solution is a maximum or a minimum, as desired. This will likely involve the first derivative test, or Go Back similar analysis of where the function is monotonic. Close 7. Be sure to answer the question that is asked. Sometimes the Quit critical numbers are desired, other times the value of the objective Page 135 of 191 function at these values will be required. Exercise 5.11.1. Farmer Brown wants to build a rectangular pen, and then subdivide it into 4 identical pens by building two parallel walls as shown. He has 200 feet of fence at his disposal. What dimensions should he use in order to maximize the area of the entire enclosed rectangle?

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 136 of 191 5.12. Newton’s Method Newton’s Method is a method for ﬁnding roots of an equation. We study it now simply because it is another interesting and useful application of the derivative. The necessary information to begin the method consists of the function you are trying to ﬁnd the root of, and an initial “guess”, or approximation to the root. The end result of the method (or algorithm) is a sequence of numbers which in many cases will be converging to a root of the equation. The outline of the steps in the process are as follows: Integrated Calculus/Pre-Calculus 1. Call the initial estimate x0. Mark R. Woodard 2. Consider the tangent line to y = f(x) at the point (x0, f(x0)). As long as this tangent line isn’t parallel to the x−axis, it should intersect the x−axis at some point. Call this point of intersection (x1, 0). Title Page 3. Consider the tangent line to y = f(x) at the point (x , f(x )). As long 1 1 Contents as this tangent line isn’t parallel to the x−axis, it should intersect the x−axis at some point. Call this point (x2, 0). JJ II

4. Repeat this process. If the sequence x0, x1, x2,... converges, it likely J I is converging to a root r. Go Back An animation demonstrating this idea and the following derivation of Close the iterative formula is at the following link: Quit Page 137 of 191 Click Here to Visit The Newton Animation To generate a formula for this process, the key fact is that the slope of 0 the tangent line at x0 is given by f (x0), while the slope of the line between 0−f(x0) (x1, 0) and (x0, f(x0)) is given by . Thus we need to set these equal x1−x0 to each other and solve for x1, to ﬁnd out what it is in terms of information about x0. We have 0 − f(x0) 0 = f (x0), x1 − x0 and multiplying both side by the denominator on the right yields

0 −f(x0) = f (x0)(x1 − x0). Integrated Calculus/Pre-Calculus We then have Mark R. Woodard 0 −f(x0) = f (x0)(x1 − x0) 0 0 −f(x0) = f (x0)(x1) − f (x0)(x0) Distributing 0 0 0 f (x0)(x0) − f(x0) = f (x0)(x1) Adding f (x0)(x0) to both sides Title Page 0 f (x0)(x0) − f(x0) Contents 0 = (x1) Dividing f (x0) JJ II In general, our recursive formula is given by J I

0 Go Back f (xn)(xn) − f(xn) 0 = (xn+1) f (xn) Close Quit 2 For example, suppose you wanted to ﬁnd a root of f(x) = x − 4, and Page 138 of 191 decided to start with an initial estimate of x = 3. (Of course, we already know that the roots of this are ±2, but this example is just to illustrate 0 0 the method.) Since f (x) = 2x, we have that f (xn) = 2xn. Thus we have

0 2 f (xn)(xn) − f(xn) (2xn)(xn) − (xn − 4) xn+1 = 0 = . f (xn) 2xn

2 If we simplify this, we see that x = xn+4 . Starting with x = 3, we get n+1 2xn 0 the following table:

i x f(x ) i i Integrated 0 3 5 Calculus/Pre-Calculus 1 2.16667 0.694444 Mark R. Woodard 2 2.00641 0.0256821 3 2.00001 0.0000409602

From this is it pretty clear that the xi’s are converging to 2, as expected. Title Page Exercise 5.12.1. Use Newton’s Method to ﬁnd a root for f(x) = x3 + Contents x − 4. JJ II J I Go Back Close Quit Page 139 of 191 5.13. Antiderivatives

We have spent much effort learning how to carry out the following problem: given y = f(x), find y0 = f 0(x), and interpret its significance. In this section, we consider the inverse problem: given y = f(x), find a function F (x) so that F 0(x) = f(x). Such a function is called an antiderivative of f(x). For example, an antiderivative of y = 2x is F (x) = x2, since the derivative of F (x) = x2 is F 0(x) = 2x. An antiderivative of 2 −1 3 0 −1 2 y = cos (x) sin(x) is F (x) = 3 cos (x), since then F (x) = 3 · 3 cos (x) · − sin(x) = cos2(x) sin(x). Of course, this last example may leave you Integrated scratching your head . . . , you can see that it is correct, but how does Calculus/Pre-Calculus one figure it out based on the function you were given. That will be dis- Mark R. Woodard cussed at length in the sections to come. For now, we will focus on some basic facts about antiderivatives, and will think a little about what they might be good for. Title Page It is worth noting at the outset that if one knows a single antiderivative of a function, he in fact knows all of them, by the “+C” theorem. Thus, Contents 2 since F (x) = x represents a single antiderivative of y = 2x, the collection JJ II x2 +C represents all possible antiderivatives of y = 2x. This is often called the general antiderivative of y = 2x. J I Go Back Exercise 5.13.1. Find the general antiderivative of y = 2x + 8. Close In the preceeding exercise we used the idea that the antiderivative of Quit a sum should be the sum of the antiderivatives. This can be stated more Page 140 of 191 formally as follows: The Sum Rule and Constant Multiplier Rule for Antiderivatives If F (x) is an antiderivative of f(x), and G(x) is an antiderivative of g(x), then F (x) + G(x) + C is the general antiderivative of f(x) + g(x). Also, the if k is a constant, the general antiderivative of kf(x) is kF (x) + C.

Exercise 5.13.2. What is the general antiderivative of y = 5x3 +8x+10? Integrated The previous exercise suggests a rule. Calculus/Pre-Calculus Mark R. Woodard Power Rule for Antiderivatives If f(x) = xn where n is an a rational number other than −1, then the general antiderivative of f(x) is F (x) = Title Page xn+1 . n+1 Contents The above is easily checked by differentiating. Note that the exclusion of −1 is necessary, as otherwise the formula is undefined. JJ II J I Go Back Close Quit Page 141 of 191 5.14. Rectilinear Motion Suppose that a particle is moving in a straight line, and the function a(t) which describes the object’s acceleration at time t is given. This is a reasonable scenario, since the physics formula F = ma tells us that the acceleration is essentially given by the forces acting on the particle. Thus, if one were to understand all of the forces acting on the particle, he would know a(t). The question, is: given a(t), can we recover v(t) and s(t), the objects velocity at time t and position at time t? The answer turns out to be yes, as long as we have a few more pieces of Integrated information. For example, suppose that a(t) = 2t + 1. Since acceleration Calculus/Pre-Calculus is the derivative of velocity, velocity must be the antiderivative of accelera- Mark R. Woodard tion. Thus v(t) = t2 +t+C for some constant C. If we know the velocity at time t = 0 (let’s say that it was 5 feet per second) then we could compute C. This is because v(0) = 5 = 02 +0+C, so C = 5. The fact that v(0) = 5 is called an initial condition. If we also had an initial condition for position, Title Page we could compute it as well. For example, if s(0) = 2, then since s(t) is an Contents t3 t2 antiderivative of v(t), we would have s(t) = 3 + 2 + 5t + C1, where C1 is some constant which is potentially different from C. The initial condition JJ II tells us that s(0) = 2 = 03 + 02 + 5(0) + C , so C = 2. 3 2 1 1 J I Exercise 5.14.1. The acceleration due to gravity near the surface of the Go Back earth is approximately −32 feet per second per second. (We sometimes say “‘feet per second squared.”) If an object is thrown upward from a height Close of 3 feet with an initial velocity of 12 feet per second, find a formula for Quit s(t), the position of the object at time t. How high does the ball go? Page 142 of 191 Exercise 5.14.2. Find s(t) if a(t) = cos t + sin t, s(0) = 4, v(0) = 1. 5.15. Antiderivatives Mini-Quiz

Click to Initialize Quiz Answers the following questions about antiderivatives. 1. For the initial value problem “Find s(t), given a(t), where v(0) = m and s(0) = n,” the following is always true: The constant of We can use m and m and n have integration are m n to ﬁnd the nothing to do with and n. constants of the constants of Integrated integration. integration. Calculus/Pre-Calculus Mark R. Woodard 2. The “+C” Theorem implies which of the following: Every function has The antiderivative The derivative of a an antiderivative. of the zero function polynomial has Title Page is a constant. lower degree than the original Contents function. JJ II

3. If m(x) = k1g(x) − k2f(x), and if G(x) is an antiderivative of g(x), J I and F (x) is an antiderivative of f(x), then the general antiderivative Go Back of m(x) is: Close k1G(x)−k2F (x)+CK1G(x) − G(x) − F (x) + C Quit K2F (x) + C Page 143 of 191 4. If m(x) = f(x)g(x), and F 0(x) = f(x) and G0(x) = g(x), then an antiderivative of m(x) is: M(x) = F (x)G(x) M(x) = None of the above. F (x)g(x)+G(x)f(x) Click to End Quiz

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 144 of 191 CHAPTER 6 Areas and Integrals

Contents

6.1 Summation Notation ...... 147 Integrated 6.2 Computing Some Sums ...... 149 Calculus/Pre-Calculus 6.3 Areas of Planar Regions ...... 154 Mark R. Woodard 6.3.1 Approximating with Rectangles ...... 155 6.4 The Definite Integral ...... 159 Title Page 6.4.1 Definition and Notation ...... 159 R b Contents 6.4.2 Properties of a f(x) dx...... 162 6.4.3 Relation of the Definite Integral to Area ...... 162 JJ II 6.4.4 Another Interpretation of the Definite Integral .. 164 J I 6.5 The Fundamental Theorem ...... 166 Go Back 6.5.1 The Fundamental Theorem...... 166 Close 6.5.2 The Fundamental Theorem – Part II ...... 168 Quit 6.6 The Indefinite Integral and Substitution ...... 170 Page 145 of 191 6.6.1 The Indefinite Integral ...... 170 6.6.2 Substitution in Indefinite Integrals ...... 173 6.6.3 Substitution in Definite Integrals ...... 176 6.7 Areas Between Curves ...... 179 6.7.1 Vertically Oriented Areas ...... 179 6.7.2 Horizontally Oriented Areas ...... 181

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 146 of 191 6.1. Summation Notation We all know how to add. In this section, we will simply talk about adding, but will focus on the case where we have many terms in our sum, each of which can be described by a formula of some sort. For example, consider the sum 1 + 2 + 3 + 4 + ... + 100 which has as its ith term the number i. We will discuss a convenient way to write this sum, and also a way to compute this sum. We will use the greek capital letter Σ to stand for “sum” (they both Integrated have something to do with the letter “s”,) and will use subscripts and Calculus/Pre-Calculus superscripts to indicate where the sum begins and ends. Thus, if we write Mark R. Woodard

25 X i2 i=1 Title Page we are talking about the sum whose ith term is i2, and which starts with Contents 2 2 1 = 1 and ends with 625 = 25 . Thus JJ II 25 X J I i2 = 1 + 4 + 9 + ... + 625. i=1 Go Back In general, we have Close Quit m X ai = an + an+1 + an+2 + ... + am. Page 147 of 191 i=n Exercise 6.1.1. How would you write the sum 4 + 9 + 16 + 25 + ... + 100 in summation notation?

Exercise 6.1.2. Write the following sum out in the form a1 +a2 +...+an: P8 2 i=3 i + i + 1.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 148 of 191 6.2. Computing Some Sums There is an interesting story about the mathematician Gauss(1777-1855) which goes something like this: Gauss was such a child prodigy that he sometimes was diﬃcult to control during class, so one day his teacher decided to keep him busy by having him add up the ﬁrst 100 integers. Gauss went to work and was done almost immediately! His teacher was perplexed and asked Gauss how he could have done it so quickly, and Gauss answered that he noticed that if he wrote the numbers this way:

1 2 3 . . . 50 Integrated 100 99 98 . . . 51 Calculus/Pre-Calculus that the columns added to 101. Since there are 50 columns, the answers Mark R. Woodard must be (101)(50) = 5050. This amazingly insightful (for a child) concept can be generalized to compute the sum of the first n integers as follows. Assume that n is odd. Then write the first n − 1 terms as follows: Title Page n−1 1 2 3 . . . 2 n+1 Contents n − 1 n − 2 n − 3 . . . 2 n−1 and note that the columns add to be n, and there are 2 of them. Thus JJ II the sum of the first n − 1 terms is ( n−1 )(n). Now if we add the last term 2 J I n to this sum, we obtain Go Back n − 1 n − 1 n + 1 (n) + n = (n) + 1 = (n) . 2 2 2 Close Quit Exercise 6.2.1. Explain why in the previous example, the 2nd row ends Page 149 of 191 n+1 Pn n(n+1) with 2 . Then show that the formula i=1 i = 2 holds for all n by showing it is true in the case where n is even. (This case is more like Gauss’ calculation than the previous example is.)

Pn 2 It turns out that there are few more nice formulas like Gauss’ for i=1 i Pn 3 and i=1 i , although these are not nearly as easy to prove. We will need them anyway, so we will list them here, together with the prevoius result Pn about i=1 i.

n X n(n + 1) Integrated i = (6.1) Calculus/Pre-Calculus 2 i=1 Mark R. Woodard n X (n)(n + 1)(2n + 1) i2 = (6.2) 6 i=1 Title Page n 2 X n(n + 1) i3 = (6.3) Contents 2 i=1 JJ II J I P50 2 Suppose that you wanted to compute something like i=1 2i + 4i + 8. Go Back We can use the above formulas together with some basic facts about sums Close to compute this. The facts we need are simply that sums are commutative, that we can factor constants out of sums, and that the sum of n terms of Quit a constant k is just the product of n with k. For completeness, we write Page 150 of 191 this ideas using summation notation. n n n X X X ai + bi = ai + bi (6.4) i=1 i=1 i=1 n n X X kai = k ai (6.5) i=1 i=1

n n terms X z }| { k = k + k + k + ... + k = nk (6.6) i=1 Integrated Calculus/Pre-Calculus P50 2 Thus we can compute i=1 2i + 4i + 8 as follows: Mark R. Woodard 50 50 50 50 X X X X 2i2 + 4i + 8 = 2i2 + 4i + 8 By 6.4 i=1 i=1 i=1 i=1 50 50 50 Title Page X 2 X X = 2 i + 4 i + 8 By 6.5 Contents i=1 i=1 i=1 50 50 JJ II X 2 X = 2 i + 4 i + 400 By 6.6 J I i=1 i=1 (50)(51)(101) (50)(51) Go Back = 2 + 4 + 400 By 6.1 and 6.2 6 2 Close = 91350 By arithmetic. Quit Pn 3 Exercise 6.2.2. Compute i=1 4i +6i−1. Your answer will be a function Page 151 of 191 of n. 2 2i Example 6.2.1. Suppose that f(x) = x +1 and xi = 2+ n is a sequence. n X 2 Compute both: a) f(x ) · ,and b) The limit as n → ∞ of the answer i n i=1 to a).

Solution: Since f(x) = x2 + 1 and x = 2 + 2i , we have that f(x ) = Integrated i n i Calculus/Pre-Calculus 2i 2 4i 2 + n + 1. If we expand this expression we get that f(xi) = 4 + n + Mark R. Woodard 4i 4i2 8i 4i2 n + n2 + 1 which is the same expression as 5 + n + n2 . We are sup- 2 posed to sum the product of this expression with n , so we must compute n X 10 16i 8i2 + + . We see that Title Page n n2 n3 i=1 Contents JJ II

n n n n J I X 10 16i 8i2 10 X 16 X 8 X + + = 1 + i + i2 n n2 n3 n n2 n3 Go Back i=1 i=1 i=1 i=1 10 16 n(n + 1) 8 n(n + 1)(2n + 1) Close = · n + + n n2 2 n3 6 Quit 8(n + 1) (4)(n + 1)(2n + 1) = 10 + + Page 152 of 191 n 3n2 This is in a form which makes it reasonably easy to compute the limit as n → ∞. We have 8(n + 1) (4)(n + 1)(2n + 1) lim 10 + + = n→∞ n 3n2 n + 1 2n2 + 3n + 1 2 10 + 8 lim + 4 lim = 10 + 8(1) + 4 n→∞ n n→∞ 3n2 3 62 = 3 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 153 of 191 6.3. Areas of Planar Regions

In the course of your human experience, you have encountered and learned some basic facts about areas of planar regions. You know, for example, that the area of a rectangle is the length times the width, and if you think for a moment about this fact, it makes perfect sense. For example, in a 2 × 3 rectangle, you can actually see the 6 square units that make up the 1 region. You also know that the area of triangle is 2 of the base times the height, which again makes sense if you draw a rectangle with the same Integrated base and the same height as the triangle, and work to convince yourself Calculus/Pre-Calculus that there is just as much area outside the triangle but inside the rectangle Mark R. Woodard is there is inside the triangle, and thus the area of the triangle is one half the area of the corresponding rectangle. You also “know” that the area of a circle is πr2 where r is the radius, but it is healthy to admit that this is a diﬀerent kind of knowledge. In fact, the only reason you probably Title Page “know” this fact is that someone who you trusted convinced you to believe Contents it on blind faith! If true confessions are in order, you probably will have to admit that you don’t really know anything about the area of anything JJ II with a curved side. In this section, we will investigate such area, and use J I calculus to see how to compute them. Go Back We will begin by considering area of regions which are curved on only Close one “side.” Suppose f(x) is a function with f(x) > 0 on the interval (a, b). Let R be the region under y = f(x), above the x axis, and between the Quit lines x = a and x = b. This region has 3 straight “sides” and one curved Page 154 of 191 “side”. We will attempt to ﬁnd the area of such a region. 6.3.1. Approximating with Rectangles

In order to have a specifice region R to work with, let’s let f(x) = x2+1 and let a = 2 and b = 4. The first good idea we will introduce toward finding the area of R is to approximate the area with something we know the area of. We will use rectangles, since (as mentioned before) we understand the areas of those. Let’s use 4 rectangles to start. If we divide the interval from a to b up into 4 subintervals, the endpoints would be at 2, 2.5, 3, 3.5 and 4. If we were to draw four rectangles whose bases are on these four subintervals, and whose height is determined by the function value using the endpoint on Integrated the right-hand side of the subinterval, we would get a decent approximation Calculus/Pre-Calculus to the area of R. The good news is that we can compute the sum of the Mark R. Woodard areas of these four rectangles, and it would be approximately the area of R. The bad news is that it wouldn’t be exactly right. To compute the area we 1 would need to compute the widths of the rectangles, which is 2 for each, and the height of each which are f(2.5), f(3), f(3.5), and f(4) respectively. Title Page 29 53 These values can be computed to be 4 , 10, 4 , and 17 respectively. If we Contents 1 95 multiply each of these by the width 2 and add we arrive at 4 . This is a very rough approximation, but at least it’s a start. Now, what if we JJ II were to increase the number of rectangles to eight? A quick sketch reveals J I what looks like a better approximation. There is a nice demo of this at Go Back http://math.furman.edu/ dcs/java/quad.html. Close One can’t help but extrapolate along these lines: if eight rectangles yields a better approximation than four, and 16 yields a better approxi- Quit mation than eight, why not use infinitely many rectangles? Of course, this Page 155 of 191 statement doesn’t really make sense as stated, but we can make it precise by requiring n rectangles, and then taking the limit as n → ∞. Toward this end, we want to divide the the interval from a to b up into n distinct subintervals. It will be useful to make them all the same width, b−a and to call this width ∆x. A little thought shows that ∆x = n . It will also be useful to let a = x0 and b = xn, and the other endpoints of the subintervals be called xi. In this case we would have xi = a + i∆x. This is summarized as

b−a ∆x = n is the width of each subinterval. The collection xi = a + i∆x are the endpoints of the subintervals. Integrated Calculus/Pre-Calculus Mark R. Woodard Now each rectangle has width ∆x, and height f(xi) for some endpoint xi. To encompass all of the n rectangles and using the right-hand side of each subinterval, we would have the approximate area given by Title Page f(x1)∆x + f(x2)∆x + f(x3)∆x + ... + f(xn)∆x which is better written as Contents n n X X b − a b − a JJ II f(x )∆x = f a + i . i n n i=1 i=1 J I Note that in this sum the f(x−i) factor of each term represents the height Go Back of the ith rectangle, while ∆x represents the width of the ith rectangle. In Close our current example, we would have Quit n n 2 ! X 2 2 X 2 2 f 2 + i = 2 + i + 1 . Page 156 of 191 n n n n i=1 i=1 This sum is exactly the one we computed in the previous section. If we let An be this approximation using n rectangles, you will recall that we 8(n+1) 4(n+1)(2n+1) computed An = 10 + n + 3n2 . Finally, to compute the area, we need to ﬁnd limn→∞ An, which we 62 2 previously saw was 3 . Thus, the area under y = x + 1, above the x axis 62 and between a = 2 and b = 4 is 3 . To summarize, we note the steps in this somewhat arduous process.

Computing Areas of Simple Regions These are the steps to compute the area under y = f(x), Integrated above the x axis and between x = a and x = b when Calculus/Pre-Calculus f(x) > 0 on (a, b). This is sometimes called the Right- Mark R. Woodard Hand Rule.

b−a 1. Use rectangles of width ∆x = n . Title Page 2. Let x = a + i∆x be the endpoints of the subinter- i Contents vals. Pn JJ II 3. Let An = i=1 f(xi)∆x be the approximation using n rectangles. Simplify and compute An using the J I techniques of the previous section. Go Back

4. Compute limn→∞ An, yielding the area. Close Quit 2 Exercise 6.3.1. Compute the area under f(x) = 4x + 2x + 1 between Page 157 of 191 a = 3 and b = 6. Exercise 6.3.2. Compute the area under f(x) = 3x + 5 between a = 0 and b = 4 two diﬀerent ways: by using the techniques of this section and by simply drawing the region and using the fact that it is made up of straight sides. Note that answer is the same either way.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 158 of 191 6.4. The Deﬁnite Integral

6.4.1. Deﬁnition and Notation

In the previous section we have seen that for a given function f(x) and n X interval [a, b], then quantity lim xi∆x is an important quantity, since n→∞ i=1 Integrated it gives the area under y = f(x) and above the x axis between x = a and Calculus/Pre-Calculus x = b if f(x) > 0 on [a, b]. This “limit of a sum of areas of rectangles” is Mark R. Woodard important enough that it deserves its own special name and notation. We will call this quantity the definite integral of f(x) with respect to x from a to b. Our definition will actually be slightly more general. We don’t actually have to have all subintervals the same with, and we don’t have Title Page to use the right-hand side of each subinterval. As long as we have points Contents a = xo, x1, x2, ldots, xn = b, we can let ∆xi = xi+1 − xi be the width of ∗ JJ II the ith subinterval, and we can let xi be any point on the ith subinterval. n X ∗ J I Then we can consider lim f(xi )∆xi as the limit of a sum of areas n→∞ i=1 Go Back of rectangles with various widths and (perhaps) using various points to Close determine the heights of the rectangles. Of course, the important theorem (whose proof we will omit) is that we get the same result whether the xi’s Quit are left-hand endpoints, right-hand endpoints, midpoints, or whatever, as Page 159 of 191 long as the widths ∆xi shrink to zero as n → ∞. Thus we have: Definition of Definite Integral Let the interval [a, b] be divided up into subintervals by a set of points xi so that

a = xo < x1 < x2 < ··· < xn = b.

Let ∆xi = xi+1 − xi be the width of the ith subinterval. ∗ ∗ Let xi be any number so that xi ≤ xi ≤ xi+1. Then the definite integral of f(x) with respect to x from a to b is given by Integrated n Calculus/Pre-Calculus X ∗ lim f(xi ) ∆xi. n→∞ Mark R. Woodard i=1 This quantity is a number which doesn’t depend on the ∗ choice of endpoints xi or the choice of xi . The definite integral is defined whether or not f(x) > 0 on [a, b], but Title Page only represents the area under f(x) and above the x axis Contents between a and b when f(x) ≥ 0 on (a, b). JJ II J I Exercise 6.4.1. Compute the definite integral from 2 to 5 of f(x) = 5x3 + 3x. Does this represent an area? Go Back Close Since it is somewhat painful to write out the quantity represented by the definite integral, it will be useful to introduce some notation to represent Quit it. The notation will necessarily have to involve the function f(x) and the Page 160 of 191 endpoints a and b. We choose a mathematical symbol which will remind us of an “S” for “sum.” The symbol is actually two symbols which are always used in conjuction: the integral sign “R ” and the differential “dx.” Z b R b Thus we write a f(x) dx or f(x) dx. Thus we have: a

Notation for the Definite Integral We write the definite integral of f(x) with respect to x from x = a to x = b as Z b Integrated f(x) dx. Calculus/Pre-Calculus a Mark R. Woodard The symbol R is called the integral sign, and the a and b are called the lower limit of integration and the upper limit of integration, respectively. The symbol dx is called Title Page the differential, and is always used in conjuction with the integral sign. The x in the symbol dx represents that the Contents definite integral is to be considered with respect to x. JJ II The function f(x), which is placed between the integral sign and the differential, is called the integrand. J I Go Back

Thus, for example, the previous exercise could have been worded: find Close R 5 3 2 5x + 3x dx. Quit R 3 Page 161 of 191 Exercise 6.4.2. Compute −1 5x + 2 dx. R b 6.4.2. Properties of a f(x) dx. Because the definite integral is defined to be a limit of a sum of areas of rectangles, and because limits and sums have nice properties, we might expect for the definite integral to have some nice properties, especially as related to constant multipliers and sums. We have the following:

Properties of the Definite Integral 1. R b kf(x) dx = k R b f(x) dx, where k is a constant. a a Integrated 2. R b f(x) ± g(x) dx = R b f(x) dx ± R b g(x) dx. Calculus/Pre-Calculus a a a Mark R. Woodard R b R c R c 3. a f(x) dx + b f(x) dx = a f(x) dx R b R a 4. a f(x) dx = − b f(x) dx Title Page Contents R 5 2 Thus, for example, to compute 1 3x + 3x + 1 dx we could instead R 5 2 R 5 R 5 JJ II compute 3 1 x dx + 3 1 x dx + 1 1 dx. J I 6.4.3. Relation of the Definite Integral to Area Go Back R b Close We have seen that when f(x) ≥ 0 on [a, b] that a f(x) dx is the area under f(x), above the x-axis and between x = a and x = b. What if f(x) is not Quit greater than or equal to zero on [a, b]? If f(x) ≤ 0 on [a, b], In this case, Page 162 of 191 think about the function g(x) = −f(x). This function would be greater R b than or equal to 0 on [a, b], so a g(x) dx would be the area under g(x). But the graph of g(x) = −f(x) is obtained from the graph of f(x) by flipping around the x-axis. Thus the area under g(x) between a and b is the same as the area above f(x), below the x-axis and between x = a and x = b. If a function h(x) is above the x-axis between 0 and 1 and is below the x-axis R 2 between 1 and 2, then 0 f(x) dx, which by the above properties is equal R 1 R 2 to 0 f(x) dx+ 1 f(x) dx, which yields the area below f(x) between 0 and 1 minus the area above f(x) between 1 and 2. This idea can be generalized to see that in general, the definite integral gives the area above the x-axis minus the area below the x-axis. To get the total area bounded by f(x), Integrated one would need to compute R b |f(x)| dx. Summarizing: Calculus/Pre-Calculus a Mark R. Woodard

If Aup is the area below f(x) and above the x-axis, and Adown is the area above f(x) and below the x-axis, then Title Page Z b Contents f(x) dx = Aup − Adown. a JJ II The total area bounded by f(x) is given by R b |f(x)| dx. a J I Go Back R 1 3 Close Exercise 6.4.3. Compute −1 x dx. Does this represent an area? Ex- plain. Quit R 3 Exercise 6.4.4. In a previous exercise we computed −1 5x+2 dx. Think- Page 163 of 191 ing only in terms of the area(s) represented here, recompute this using a sketch as your guide. (Hint: This works because all sides of the region are straight lines.)

6.4.4. Another Interpretation of the Definite Integral We used area to motivate the definition of the definite integral, but it turns out that there are other applications and interpretations of this entity. Just like the two main interpretations of the derivative (slope of tangent line R b and instantaneous rate of change) there is an interpretation of a f(x) dx in terms of “physics” in addition to the geometric interpretation we have Integrated already discussed. Suppose that an object is moving in a straight line, but Calculus/Pre-Calculus with a velocity that varies with time t. Let v(t) represent the velocity at Mark R. Woodard time t. Our intuition tells us that if we know the velocity at any time t, we should be able to recover the distance the object travels between time a and b. For the sake of simplicity, let’s assume the object is always Title Page moving to the right, so its velocity is always positive on [a, b]. If v(t) were constant, we know that distance = rate · time, but here the rate (velocity) Contents isn’t constant. However, on a very small time interval [ti, ti+1] the velocity ∗ ∗ JJ II is almost constant, and we could approximate it by v(ti ) for some ti on [ti, ti+1]. Thus for that small time interval, the distance is given approx- J I ∗ imately by v(ti ) · ∆ti. (This simply comes from distance = rate · time.) Go Back Adding up all of the distances over the various small time intervals gives Pn ∗ Close i=1 v(ti ) · ∆ti, and this approximation will get better as the time inter- n ∗ vals get shorter, so the exact distance is given by limn→∞ sumi=1v(ti )·∆ti, Quit R b but this is exactly a v(t) dt. Thus in this context, the definite integral of Page 164 of 191 velocity gives distance travelled. If v(t) is sometimes positive and sometimes negative on [a, b], the definite integral gives the difference between the distance travelled to the right and the distance travelled to the left, sometimes called displacement. To get the total distance travelled, one R b would have to compute a |v(t)| dt.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 165 of 191 6.5. The Fundamental Theorem 6.5.1. The Fundamental Theorem

We turn now to a theorem with an ominous sounding name – The Funda- mental Theorem of Calculus. This theorem will connect the theory of the definite integral with differentiation in a fundamental way, by linking the definite integral of a function with information related to its antiderivative. The key idea is the following: consider a small subinterval of the closed interval [a, b], say the interval from xi to xi+1. The portion of the Integrated definite integral that has something to do with this interval is the term of Calculus/Pre-Calculus ∗ the form f(xi )∆xi, where ∆xi = xi+1 − xi. Now, suppose that F (x) is any Mark R. Woodard antiderivative of f(x), and apply the Mean Value Theorem to F (x) on the interval [xi, xi+1]. The MVT says that there is a point ci between xi and xi+1 with Title Page

0 Contents F (xi+1) − F (xi) = F (ci) · ∆xi = f(ci)∆xi.

In other words, there is a point ci so that the area of the rectangle with JJ II base ∆xi and height determined by the function f evaluated at ci is exactly J I the diﬀerence between the antiderivative F evalauated at the right-side Go Back endpoint and F evaluated at the left side endpoint. If we do this on each Close subinterval, then by choosing appropriate values of ci, it must follow that Quit n n X X f(ci)∆xi = F (xi) − F (xi−1), Page 166 of 191 i=1 i=1 but this sum on the right can be expanded to be

F (x1) − F (x0) + F (x2) − F (x1) + ... + F (xn) − F (xn−1), which simpliﬁes to F (xn) − F (x0) = F (b) − F (a). Thus we have

The Fundamental Theorem of Calculus (Part I) R b If f(x) is continuous on [a, b], then a f(x) dx = F (b) − F (a), where F (x) is any antiderivative of f(x).

Note that the fundamental theorem draws a connection between the Integrated definite integral of f and its antiderivative, F . It also provides a much Calculus/Pre-Calculus easier way to calculate definite integrals, since antiderivatives are often Mark R. Woodard easier to find than it is to go through the ardous process of finding the limit of a sum of areas of rectangles. Here is an example: R 3 2 Example 6.5.1. Use the fundamental theorem to calculate 1 x + 4x dx. Title Page 2 x3 2 Solution: Since an antiderivative of f(x) = x + 4x is F (x) = 3 + 2x , we Contents see that 1 1 74 F (3) − F (1) = 9 + 18 − ( + 2) = 225 − = JJ II 3 3 3 J I . A word on notation: since the expression F (b) − F (a) is so important Go Back R b Close in the practice of using the fundamental theorem to compute a f(x) dx, it b will be useful to introduce notation for it. We will use the notation F (x)|a Quit to represente F (b) − F (a). Page 167 of 191 R π Example 6.5.2. Use the above notation to compute 0 sin(x) dx. R π π Solution: 0 sin(x) dx = − cos(x)|0 = − cos(π) − − cos(0) = −(−1) + 1 = 2.

6.5.2. The Fundamental Theorem – Part II Integrated Calculus/Pre-Calculus Mark R. Woodard The first part of the fundamental theorem says that there is a close relationship between definite integrals and antiderivatives. The second part of the fundamental theorem will reiterate this relationship, but in a slightly Title Page different way. We will create a kind of “definite integral function”, and see that it is actually the antiderivative of the integrand. Specifically, consider Contents the function A(x) = R x f(t) dt. Note that in this definition, the indepen- 0 JJ II dent variable for the function A is x. We choose the letter A because the function is sort of an area function – if f(t) is above the t axis, then A(x) J I gives the area under f(t) between t = 0 and t = x. Of course, if F (t) is Go Back an antiderivative (with respect to t) of f(t), then another name for A(x) Close is F (x) − F (0). Then the derivative with respect to x of A(x) is F 0(x) − 0 (since F (0) is a constant), but since F is an antiderivative of f, we have Quit 0 0 that F (x) = f(x). Thus, A (x) = f(x). This is the Fundamental Theorem Page 168 of 191 – Part II. Fundamental Theorem of Calculus – Part II Suppose that A(x) is the function defined by A(x) = R x 0 a f(t) dt. Then A (x) = f(x). That is, the derivative of a function defined so that the independent variable is the upper limit of integration of a definite integral has as its derivative the integrand of the definite integral as a function of that upper limit of integration.

R x p Integrated Example 6.5.3. What is the derivative with respect to x of 2 sin(t) dt? p Calculus/Pre-Calculus Solution: The FTOC – Part II applies. The derivative is simply sin(x). Mark R. Woodard R 4 1 0 Example 6.5.4. If A(x) = x t dt with x > 0, what is A (x)? Solution: We can’t apply the FTOC – Part II to A(x) as written, but Title Page R 4 1 R x 1 we could apply it to −A(x) = − x t dt = 4 t dt. Thus the derivative of 1 1 Contents −A(x) is x , so the derivative of A(x) = − x . 2 JJ II Example 6.5.5. What is d R x sec(t) dt ? dx 1 J I Solution: The chain rule applies, together with the FTOC – Part II. Note R x Go Back that the function is the result of composing the function A(x) = 1 sec(t) dt with the function g(x) = x2. In fact, the original function is then A(g(x)). Close Thus, by the chain rule, the derivative must be A0(g(x))g0(x). By the Quit FTOC – Part II, A0(x) = sec(x), so A0(g(x)) = sec(x2). Also, g0(x) = 2x, 2 so the derivative of the original function must be sec(x )2x. Page 169 of 191 6.6. The Indeﬁnite Integral and Substitu- tion

One thing that arises from the fundamental theorem is the idea that antiderivatives are quite important, since they help us to evaluate definite integrals. Thus, we probably should study them in more detail. In this section, we will introduce a new notation for the general antiderivative, based on the fact that it is closely related to the definite integral. We also Integrated learn a technique for finding antiderivatives, and thus definite integrals Calculus/Pre-Calculus which is essentially the inverse of the chain rule. Mark R. Woodard

Title Page Contents JJ II 6.6.1. The Indefinite Integral J I Go Back Close The FTOC indicates that antiderivatives and integrals are closely related. Because of this, and because of our need for better notation for the an- Quit tiderivative of a function, we introduce the following new notation for the Page 170 of 191 general antiderivative. The Indefinite Integral We will call the general antiderivative of a function f(x) the indefinite integral of f(x), and will denote it by R f(x) dx. Thus Z f(x) dx = F (x) + C,

where F 0(x) = f(x). We choose this notation (which is similar to the notation for the deﬁnite integral) because Integrated the fundamental theorem of calculus tells us that the Calculus/Pre-Calculus general antiderivative is related to the deﬁnite integral. Mark R. Woodard

R b R Note the diﬀerence between a f(x) dx and f(x) dx. The former is Title Page a number, while the latter is a collection of functions. They are related Contents not because their symbols are similar but rather their symbols are similar because they are related. To reiterate: they are related because one JJ II R b can compute a f(x) dx by using any of the functions from the collection R f(x) dx, and evaluating at b and a and subtracting. J I Go Back R 2 Example 6.6.1. Compute sec (x) + x dx. Close R 2 x2 Quit Solution: sec (x) + x dx = tan(x) + 2 + C. Page 171 of 191 Example 6.6.2. Compute R (x2 + 3x + 1)(2x + 8) dx. Solution:

Z Z (x2 + 3x + 1)(2x + 8) dx = 2x3 + 6x2 + 2x + 16x2 + 24x + 8 dx Multiplying Out the Integrand Z = 2x3 + 22x2 + 26x + 8 dx Combining Like Terms 2x4 22x3 26x2 Integrated = + + + 8x + C Integrating Calculus/Pre-Calculus 4 3 2 Mark R. Woodard x4 22x3 = + + 13x2 + 8x + C Simplifying 2 3

Title Page Contents

JJ II J I Go Back Close Quit It might be useful to list everything we know about antiderivatives to Page 172 of 191 this point. We have f(x) R f(x) dx sin x − cos x + C cos x sin x + C sec2 x tan x + C csc2 x − cot x + C sec x tan x sec x + C

csc x cot x − csc x + C Integrated k, k a constant kx + C Calculus/Pre-Calculus n xn+1 Mark R. Woodard x , n 6= −1 n+1 + C kg(x) k R g(x) dx R R g(x) + h(x) g(x) dx + h(x) dx Title Page Contents

6.6.2. Substitution in Indeﬁnite Integrals JJ II J I We now encounter a technique of integration which will help us to integrate Go Back more complex functions, helping us to expand the above list of functions Close we know how to integrate. The technique is called substitution, because in the technique we will replace (or substitute) some quantities for other Quit quantities. The goal is to perform a substitution which will simplify the Page 173 of 191 integrand, hopefully to one on the above list. The motivation for the technique comes from the chain rule. Recall that the chain rules says that

d f(g(x)) = f 0(g(x))g0(x). dx

Thus, it must be true that Z f 0(g(x))g0(x) = f(g(x)) + C

Since these statements are equivalent. For example, the derivative Integrated of (3x2 + 2x)100 is 100(3x2 + 2x)99(6x + 2). Thus it must be true that Calculus/Pre-Calculus R 100(3x2 + 2x)99(6x + 2) dx = (3x3 + 2x)100 + C, but how does one rec- Mark R. Woodard ognize this? Or worse, what if we were given the equivalent R 200(3x + 1)(3x2 + 2x)99 dx? The idea behind the technique of substitution is the following: suppose that we gave the quantity 3x2 + 2x a simpler name. du Title Page What if we called it simply u? Then since dx = 6x + 2, it might be a good idea to call du by the name “(6x + 2) · dx.” Then we could write the Contents integral u du JJ II Z z }| { z }| { 100((3x2 + 2x))99 (6x + 2) dx J I Go Back as Z 100u99 du, Close Quit u100 and this last integral is one we know how to do. It is equal to 100 100 +C = Page 174 of 191 u100 + C. Of course, we should report our answer in terms of x, since the original problem was in terms of x. Thus we could write (3x2 + 2x)100 + C as our answer.

Substitution Technique – Indeﬁnite Integrals Given an integral of the form R f 0(g(x))g0(x) dx, we let u = g(x) and du = g0(x) dx. Then we write the integral as R f 0(u) du, which integrates to be f(u) + C. This should be reported as f(g(x)) + C, which is the correct antiderivative of f 0(g(x))g0(x), as can be checked by the Integrated chain rule. Calculus/Pre-Calculus Mark R. Woodard

Example 6.6.3. Compute R sin(cos(x)) · sin(x) dx. Title Page Solution: Let’s try letting u = cos(x). Then du = − sin(x) dx, so sin(x) dx = −du. If we substitute, we obtain the integral R sin(u) · −du, which can be Contents R written as − sin(u) du. This is equal to −(− cos(u)) + C, so the desired JJ II antiderivative is cos(sin(x)) + C. J I √ Example 6.6.4. Compute R x 5x2 + 8 dx. Go Back

Solution: If we let u = 5x2 + 8, then du = 10x dx. It might be helpful to Close du R 1 √ write this as 10 = x dx. Thus our integral can be rewritten as 10 u du = Quit 1 √ 1 2 3 1 3 R 2 2 2 10 u du. Upon integrating, we obtain 10 3 u + C = 15 (5x + 8) + C. Page 175 of 191 How do I know what to let u equal? This is a common question asked by many beginning calculus students. The best answer is: let u be something so that by substituting u and du, you are able to completely transform the integral into an integral in terms of u which you know how to compute. A more practical answer may be this: try to let u be something whose derivative is a factor of the integrand. This is a necessary (but not suﬃcient) ingredient of a successful substitution! Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents 6.6.3. Substitution in Definite Integrals JJ II J I Go Back Can the technique of substitution be used to evaluate definite integrals? Close Of course the answer is yes – since it helps us to find antiderivatives and that is what is needed to use the FTOC. Care must be taken, however, Quit to be sure write the substitution correctly in the presence of the limits of Page 176 of 191 integration. Substitution Technique – Definite Integrals R b 0 0 Given an integral of the form a f (g(x))g (x) dx, we let u = g(x) and du = g0(x) dx. Then we write the inte- R g(b) 0 g(b) gral as g(a) f (u) du, which integrates to be f(u)|g(a) = f(g(b)) − f(g(a)). This is the numerical value of the original definite integral, so no further modifications are necessary.

Integrated R 4 2 Calculus/Pre-Calculus Example 6.6.5. Compute 0 x sin(x + 4) dx. Mark R. Woodard Solution: It seems reasonable to let u = x2 + 4, since the derivative of this expression is 2x which is a factor of the integrand (if we think of x 2x du as x .) Since du = 2x dx, we can write dx = 2 . Substituting we obtain R 20 1 1 du Title Page the integral 4 2 sin(u) du. The 2 factor comes from the 2 substitution, and the limits of integration come from the fact that u has value 4 when Contents x = 0 and has value 20 when x = 4. This new integral has exactly the same numerical value as the original, so I need only compute it. But we JJ II have J I Z 20 1 1 20 Go Back sin(u) du = − cos(u)|4 4 2 2 Close 1 = − [cos(20) − cos(4)] , Quit 2 Page 177 of 191 so this is the value of the original integral. Assignment:

1. Stewart, section 5.4, 1 - 41 odd. 2. Stewart, section 5.5, 1 - 51 every other odd. You do not need to graph the functions.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 178 of 191 6.7. Areas Between Curves

6.7.1. Vertically Oriented Areas

The theory developed in this chapter so far indicates a close relationship between certain kinds of areas and related deﬁnite integrals. In fact, we Integrated have seen that in general, the deﬁnite integral of f(x) from x = a to x = b Calculus/Pre-Calculus gives us the signed area trapped between the x axis and the curve, and Mark R. Woodard between the lines x = a and x = b. We now try to extend this theory to enable us to compute area of regions like the following. Title Page Let R be the region bounded by the curves g(x) = x2 −5 and f(x) = 1− Contents 2 2 2 2 x . Note that these√ curves cross when x −5 = 1−x , which is√ when√ 2x = 6, or when x = ± 3. Now imagine taking the interval from [− 3, 3], and JJ II Pn ∗ dividing it up into subintervals and then computing limn→∞ i=1(f(xi ) − J I ∗ g(xi )) ∆xi. This should be thought of as a limit of a sum of areas of Go Back rectangles, which are approximating the area between the√ curves√ f(x) and g(x). It is important to note that over the interval [− 3, 3], f(x) is Close greater than g(x), so that f(xi) − g(xi) is a positive number, and we really Quit are getting the area (and not the signed area). Of course, this limit of a √ Page 179 of 191 3 R √ sum of areas of rectangles can be written as − 3(f(x) − g(x)) dx. This can be computed as √ √ Z 3 Z 3 2 2 √ (f(x) − g(x)) dx = √ 1 − x − (x − 5) dx − 3 − 3 √ Z 3 2 = √ 6 − 2x dx Simplifying the Integrand − 3 √ 3 3 x = 6x − 2 By the FTOC 3 √ √− 3 √ √ 3 3 √ −3 3 Integrated = 6 3 − 2 − (6(− 3 − 2 ) Simplifying Calculus/Pre-Calculus √ √3 3 = 12 3 − 4 3 More Simplifying Mark R. Woodard √ = 8 3

The general principal illustrated in the previous example can be sum- Title Page marized as follows. Contents Vertically Oriented Areas Between Curves JJ II If f(x) ≥ g(x) over [a, b], then the area bounded by x = R b J I a, x = b, f(x), andg(x) is a f(x)−g(x) dx. If in addition, f(a) = g(a) and f(b) = g(b), then this represents the Go Back area of a single planar region bounded above by f(x) and Close below by g(x). We call such an area vertically oriented because the approximating rectangles are upright. Quit Page 180 of 191 Example 6.7.1. Let a be the smallest positive number where cos(x) = sin(x), and let b > a be the next smallest positive number where cos(x) = sin(x). Compute the area bounded by the curves f(x) = sin(x) and g(x) = cos(x) between a and b. π Solution: A quick analysis of the unit circle leads us to believe that a is 4 5π and b = 4 . Between a and b, it is clear that sin x > cos x. Thus we need 5π R 4 to compute π sin x − cos x dx, which by the FTOC is equal to 4 √ √ √ √ 5π √ 4 2 2 2 2 − cos x − sin x| π = −(− − ) − (− − ) = 2 2. 4 2 2 2 2 Integrated Calculus/Pre-Calculus Mark R. Woodard 6.7.2. Horizontally Oriented Areas

For certain planar regions, it is easier to think of the approximating rect- Title Page angles as laying on their sides – that is, we will use horizontal rectangles to yield our definite integral. An example of such a planar region could be Contents 2 the region bounded by x = 1 − y and x + y = −1. This region is most JJ II easily thought of as boiunded on the left by the line x + y = −1 and on the right by the “sideways” parabola x = 1 − y2. If instead you look for the J I function on top and the function on the bottom, you find that it switches Go Back part of the way through. Thus, it is best to think “left and right” rather Close than “up and down” for this region. Because we are thinking horizontally, we imagine dividing the y axis up into subintervals rather than the x axis. Quit This will lead to an integral with respect to x rather than one with respect Page 181 of 191 to y. Since the two functions cross when y = −1 and when y = 2, we want to divide the interval from [−1, 2] along the y axis into subintervals. The horizontal approximating rectangles we want to use to find the area are bounded on the right by 1 − y2 and on the left by −1 − y. Thus our Pn 2 approximating sum has the form i=1(1 − yi ) − (−1 − yi)∆yi, and after taking the limit as n → ∞, we end up with the integral

Z 2 (1 − y2) − (−1 − y) dy. −1

In general, we have Integrated Calculus/Pre-Calculus Mark R. Woodard Area of Regions usings Horizontal Approximations The area of a region bounded on the left by x = g(y) and on the right by x = h(y) between y = c and y = d is given by Title Page Z d h(y) − g(y) dy. Contents c JJ II J I Example 6.7.2. Find the area bounded by the curves x = 3 − y2 and Go Back y = x − 1. Close Solution: These curves cross when y+1 = 3−y2, which is when y2 +y−2 = 0. Thus they cross at y = −2 and y = 1. Over this region, the left-hand Quit boundary of the region is formed by the line x = y + 1, and the right-hand Page 182 of 191 boundary is formed by the parabola x = 3 − y2. Thus the area is of the R 1 2 region is given by −2 3 − y − (y + 1) dy. Computing we get

Z 1 Z 1 3 − y2 − (y + 1) dy = 3 − y2 − y − 1 dy −2 −2 Z 1 = 2 − y − y2 dy −2 2 3 1 y y = 2y − − 2 3 −2 1 1 −8 Integrated = 2 − − − −4 − 2 − 2 3 3 Calculus/Pre-Calculus Mark R. Woodard = 4.5

Title Page Assignment: Stewart: Section 6.1, 1 - 29 odd. Contents JJ II J I Go Back Close Quit Page 183 of 191 CHAPTER 7 Appendix

Contents

7.1 3 – Minute Reviews ...... 185 Integrated 7.1.1 The Real Numbers ...... 185 Calculus/Pre-Calculus 7.1.2 Factoring ...... 186 Mark R. Woodard 7.1.3 Solving Equations ...... 186 7.1.4 Solving Systems of Equations ...... 187 7.1.5 Absolute Value ...... 188 Title Page 7.1.6 Quadratics ...... 189 Contents 7.1.7 Exponents ...... 190 JJ II 7.1.8 Inequalities ...... 190 J I 7.1.9 Setting Up Word Problems ...... 190 Go Back Solutions to Exercises ...... 192 Solutions to Quizzes ...... 309 Close Quit Page 184 of 191 7.1. 3 – Minute Reviews 7.1.1. The Real Numbers It is convenient to think of the real numbers as being points on a line. Points on the number line represent real numbers. The real numbers can be further divided into two distinct sets of numbers– rational numbers and irrational numbers. Rational numbers are numbers a that can be written in the form b where a and b are integers and b 6= 0. They include terminating decimals and repeating decimals. Irrational num- Integrated bers are√ those which are not rational. Some examples of irrational numbers include 3 and π. A point to keep in mind is that it is impossible to write Calculus/Pre-Calculus Mark R. Woodard an irrational number as a finite decimal. Irrationals can only be approximated by a finite decimal. When it becomes necessary to approximate an irrational√ number be certain to use√ correct notation. For example, never write 2 = 1.414. Instead, write 2 ≈ 1.414. Title Page The operations of addition, subtraction, and multiplication can be per- Contents formed on pairs of real numbers with the result also being a real number. Division can be performed on pairs of real numbers with the resulting quo- JJ II tient being real as long as the divisor is not 0. Division by zero is undefined J I for the real numbers. The square roots of negative numbers are also not real numbers. More generally, the even indexed roots of negative numbers Go Back are not real numbers. Close √ 4 Some examples that are not real numbers are −5 and x when x = 0. √ Quit 5 Exercise√ 7.1.1. Which of the following are not real numbers? π, −243, Page 185 of 191 4 −16, √−5 2 Exercise 7.1.2. Let a be a rational number and let β and γ be irrational. What can be said about a + β? aβ? β + γ?

7.1.2. Factoring Common Factors: ab + ac = a(b + c) Diﬀerence of Squares: a2 − b2 = (a + b)(a − b) Diﬀerence of Cubes: a3 − b3 = (a − b)(a2 + ab + b2) Sum of Cubes: a3 + b3 = (a + b)(a2 − ab + b2) 2 2 2 Trinomial Squares: a ± 2ab + b = (a ± b) Integrated Factor Theorem: Let P (x) be a function. If r is a real number such Calculus/Pre-Calculus that P (r) = 0, then (x − r) is a factor of P (x). Mark R. Woodard Exercise 7.1.3. Factor 2x3 + 3x2 − 8x − 12 completely. Exercise 7.1.4. Factor x4 − x2 − 6x − 9 completely. Title Page 7.1.3. Solving Equations Contents Exercise 7.1.5. Solve for x: JJ II x − 7 3x + 5 = 2 J I Go Back Exercise 7.1.6. Solve for x: Close

b2y Quit = C xy − ax Page 186 of 191 7.1.4. Solving Systems of Equations For example, solve the system of equations:

x + 2y = 10 (7.1) 3x − 5y = 8 (7.2)

Solve equation (1) for x: x = 10 − 2y Now substitute 10 − 2y in for x in equation (2): 3(10 − 2y) − 5y = 8. Now solve for y: Integrated 30 − 6y − 5y = 8 Calculus/Pre-Calculus −11y = −22 Mark R. Woodard So y = 2 Now substitute 2 in for y in the equation: x = 10 − 2y. x = 10 − 2(2) Title Page So x = 6 Hence the solution to the system is (6, 2). Contents For example, solve the system of equations: JJ II 3x − 2y = 5 (7.3) J I 4x − 3y = 9 (7.4) Go Back Close Multiply equation (3) by −4 and equation (4) by 3. Quit − 12x + 8y = −20 (7.5) Page 187 of 191 12x − 9y = 27 (7.6) Add equations (5) and (6) and solve for y: −y = 7 So y = −7. Now substitute y = −7 into equation (1) and solve for x: 3x − 2(−7) = 5 3x = −9 So x = −3. Hence the solution to the system of equations is (−3, −7)

Exercise 7.1.7. Solve the systems of equations: Integrated 7x + 4y = 8 (7.7) Calculus/Pre-Calculus 6x + 3y = 4 (7.8) Mark R. Woodard

Title Page 7.1.5. Absolute Value Contents x if x ≥ 0 |x| = −x if x < 0 JJ II Properties of Absolute Value: Let x and y be real numbers. J I Property 1: |x||y| = |xy| Go Back Property 2: | − x| = |x| Property 3: |x| + |y| ≥ |x + y| Close Quit The greatest integer function which is represented x is the function Page 188 of 191 that sends each real number x to the largest integer thatJ K is less than x. For example, 5.71 = 5 and −23.9 = −24. J K J K Exercise 7.1.8. Evaluate: a) | − 4.5| b) J| −4.5 |K c) Jπ K J K

Exercise 7.1.9. For each of the following statements, either show the statement is true and provide a example when it is false. Let x and y be real numbers. Integrated a) x y ≤ xy Calculus/Pre-Calculus b) J| xKJ| Ky ≤J xyK Mark R. Woodard J K J K J K

Exercise 7.1.10. Show or provide a counterexample: Title Page |x| |y| ≤ |xy| Contents J KJ K J K JJ II 7.1.6. Quadratics J I Go Back Quadratic Formula: Let ax2 +bx+c = 0 where a,b, and c are real numbers Close and a 6= 0. Then √ −b ± b2 − 4ac Quit x = 2a Page 189 of 191 . Equation of Parabola with Vertex (h, k): y − k = a(x − h)2 If a > 0 then the parabola opens up. If a < 0 then the parabola opens down.

7.1.7. Exponents Let a,b,c,d be real numbers where a > 0 and b > 0. −c 1 Property 1: a = ac Property 2: acad = ac+d Property 3: (ac)d = acd Integrated Property 4: (ab)c = acbc Calculus/Pre-Calculus Note that in general (a + b)c 6= ac + bc. Mark R. Woodard ac c−d Property 5: ad = a a c ac Property 6: ( b ) = bc Title Page 7.1.8. Inequalities Contents 7.1.9. Setting Up Word Problems JJ II J I Go Back Close Quit Page 190 of 191 Hints 1. You should have the diﬀerence between two fractional expressions. To combine them, you would need to ﬁnd a common denominator (which would be the product of the two denominators) and combine them. Return to the problem.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 191 of 191 Solutions to Exercises Exercise 0.0.1. The author of the document is blessed to have two won- derful daughters, Hannah and Darby. Exercise 0.0.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 192 of 191 Exercise 1.3.1. We need to add 4 to both sides, since half of 4 is 2, and 22 = 4. (This will complete the square in the x expression.) We also need to add 9 to both sides, since half of 6 is 3, and 32 = 9. (This will complete the square in the y expression.) Thus we have x2+4x+4+y2+6y+9 = 10+4+9, 2 2 which becomes (x + 2) + (y +√ 3) = 23, which we recognize as a circle centered at (−2, −3) of radius 23. Exercise 1.3.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 193 of 191 Exercise 1.4.1. According to the point-slope form of the equation of a line, our line must have the form y − (−4) = 5(x − (−2). Simplifying this yields the equation y + 4 = 5(x + 2), or y = 5x + 10 − 4, or y = 5x + 6. Thus the y - intercept is 6. Exercise 1.4.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 194 of 191 5−(−3) Exercise 1.4.2. First ﬁnd the slope. m = −2−2 = −2 Now take the slope and one of the points (2, −3) and put them into the point-slope formula. y−(−3) = −2(x−2) By simplifying and solving for y, we get the equation in slope-intercept form: y = −2x−1 Exercise 1.4.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 195 of 191 Exercise 1.4.3. The line 3x + y = 7 can be written y = −3x + 7, so it −1 has slope −3. The line −x − 3y = 9 can be written as y = 3 x − 3, so −1 it has slope 3 . These two slopes are not the same, nor do they multiply together to be −1, so these lines are neither parallel nor perpendicular. Exercise 1.4.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 196 of 191 Exercise 1.6.1. We have f(g(x)) = f(x2−5). Now since f is the “multiply by 3 and add 2” rule, f(x2 − 5) = 3(x2 − 5) + 2. Simplifying this gives 3x2 − 15 + 2 = 3x2 − 13. Thus f(g(x)) = 3x2 − 13. On the other hand, g(f(x)) = g(3x + 2), and since g is the “square and subtract 5” rule, we have g(3x+2) = (3x+2)2−5, and simplifying this gives (9x2+12x+4)−5 = 9x2 + 12x − 1. Thus g(f(x)) = 9x2 + 12x − 1. Now f(g(2)) = −1 and g(f(2)) = 59. This is enough to deduce that f(g(x)) 6= g(f(x)), since for functions to be equal, the outputs must be equal for all their inputs. Exercise 1.6.1 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 197 of 191 Exercise 1.6.2. Note that the ﬁrst things that are “done” by h to an input x is that x is multiplied by 4 and then 1 is added. Later, this number is squared and 9 is subtracted. Thus one possible answer is to let the “inner” function g(x) = 4x + 1 and the “outer” function f(x) = x2 − 9. Then f(g(x)) = f(4x + 1) = (4x + 1)2 − 9. Exercise 1.6.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 198 of 191 Exercise 1.8.1. Yes. If f and g are both odd, then (f + g)(−x) = f(−x) + g(−x) = −f(x) + −g(x) = −(f(x) + g(x)) = −(f + g)(x). Exercise 1.8.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 199 of 191 Exercise 1.8.2.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 200 of 191 Exercise 2.1.1.

f(4+h)−f(4) 3 First compute h . We have f(4+h) = (4+h) −3(4+h), which can be expanded to 64 + 48h + 12h2 + h3 − 12 − 3h = h3 + 12h2 + 45h + 52. Thus f(4 + h) − f(4) h3 + 12h2 + 45h + 52 − 52 (h)(h2 + 12h + 45) = = = h2+12h+45. h h h

Now if we shrink h → 0, we are left with 45. Thus the slope of the Integrated tangent line at x = 4 is 45. The equation of the tangent line would thus Calculus/Pre-Calculus be y − 52 = 45(x − 4). Mark R. Woodard

Exercise 2.1.1 Title Page Contents JJ II J I Go Back Close Quit Page 201 of 191 Exercise 2.1.2.

√ The slope of the secant line is given by f(6+h)−f(6) = 6+h+3−4−(−1) . √ h h 9+h−3 Simplifying this yields h To simplify this expression further, we need the technique√ of multiplying by the conjugate.√ The conjugate of the expression 9 + h − 3 is the expression 9 + h + 3. So we multiply both the numerator and the denomator by this conjugate quantity, yielding √ √ 9 + h − 3 9 + h + 3 (9 + h) − 9 h 1 ·√ = √ = √ = √ . Integrated h 9 + h + 3 h · ( 9 + h + 3) h · ( 9 + h + 3) 9 + h + 3 Calculus/Pre-Calculus Mark R. Woodard 1 Then as h → 0, we obtain a result of 6 . Thus the slope of the tangent line 1 at x = 6 is 6 .

Title Page Exercise 2.1.2 Contents JJ II J I Go Back Close Quit Page 202 of 191 Exercise 2.1.3.

f(3+h)−f(3) 3 3 We need to compute h . We have f(3 + h) = 3+h+1 = 4+h . Thus we have f(3 + h) − f(3) 3 − 3 = 4+h 4 . h h To simplify this last expression, we should ﬁnd a common denominator between 4 + h and 4, which should be 4(4 + h). Thus, the above expression is equal to Integrated 12 (4+h)(3) 12−12−3h − −3h Calculus/Pre-Calculus 4(4+h) 4(4+h) = 4(4+h) = . h h 4(4 + h)h Mark R. Woodard −3 −3 Cancelling the h’s leaves 4(4+h) . Shrinking h → 0 yields a value of 16 .

Title Page Exercise 2.1.3 Contents JJ II J I Go Back Close Quit Page 203 of 191 Exercise 2.1.4.

f(1+h)−f(1) −16(1+h)2+64−48 −16−32h−16h2+64−48 If we compute h we get h which simpiﬁes to h = −32h−16h2 (h)(−32−16h) h = h = −32 − 16h. Shrinking h → 0 yields a value of −32.

Exercise 2.1.4 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 204 of 191 Exercise 2.2.1.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 205 of 191 Exercise 2.2.1 Exercise 2.2.2.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 206 of 191 Exercise 2.2.2 Exercise 2.2.3.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 207 of 191 Exercise 2.2.3 Exercise 2.2.4.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 208 of 191 Exercise 2.2.4 Exercise 2.3.1. First note that it might pay to multiply both the numerator and denominator by the conjugate of the numerator. Thus we would have √ √ x − 3 x + 3 x − 9 1 lim · √ = lim √ = lim √ . x→9 x − 9 x + 3 x→9 (x − 9)( x + 3) x→9 x + 3

Now it isn’t hard to believe that when x is really close to 9, √ 1 is really √ x+3 1 x−3 1 close to , so limx→9 = . Exercise 2.3.1 6 x−9 6 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 209 of 191

Exercise 2.3.2.

Integrated

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Calculus/Pre-Calculus

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£ ¤ JJ II J I It is clear from the diagram that when x is close to 0, f(x) is close to 0 Go Back too, so limx→0 f(x) = 0. However, when x is close to but a little less than 1 Close 2, f(x) is 2 , while when x is close to but a little bigger than 2, f(x) is close to 2. Thus there is not a single value for which it can be said that Quit when x is close to 2, f(x) is close to that value. Thus limx→2 f(x) Page 210 of 191 does not exist. Exercise 2.3.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 211 of 191

Exercise 2.3.3.

Integrated

Calculus/Pre-Calculus

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£ ¤ JJ II J I Note that when x is close to (but not equal to) −1, f(x) is close to 1. Go Back It doesn’t matter what f(x) is deﬁned to be at −1, or even whether or not Close it is deﬁned at −1. It is still true that when x is close to but not equal to Quit −1, f(x) is close to 1. Thus limx→−1 f(x) = −1. Exercise 2.3.3 Page 212 of 191 Exercise 2.3.4.

x+3 x+3 No. In fact, both limx→3+ x−3 and limx→3− x−3 fail to exist, because x+3 x+3 limx→3+ x−3 = ∞ and limx→3− x−3 = −∞. We can see this either from the graph of this function, or by analyzing the quotient algebraically near x = 3. We see that when x is close to but a little bighger than 3, the numerator is close to 6 and the denominator is close to 0 (but is a positive number.) Such numbers are very big, and get even bigger when choosing a value for x closer to 6. On the other hand, when x is close to but a little less than 3, the numerator is still close to 6, and the denominator is still Integrated close to 0, but the denominator is a negative number. Since the ratio of a Calculus/Pre-Calculus positive number and a negative number is always a negative number, we Mark R. Woodard know that the output is a “large” negative number. (“Large” in the sense that its absolute value is large,) and the numbers get even “larger” when x is chosen closer to 3. Thus we see that both one-sided limits fail to exist. Title Page Exercise 2.3.4 Contents JJ II J I Go Back Close Quit Page 213 of 191 Exercise 2.4.1. The set of points 0 < |x − 3| < .04 is the set of points (3 − .04, 3) ∪ (3, 3 + .04), or in other words, (2.96, 3) ∪ (3, 3.04). Exercise 2.4.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 214 of 191 Exercise 2.4.2. Note that the midpoint of the interval (2.4, 2.6) is 2.5, and the interval has “radius” .1. Thus this set can be described as the set of all x so that |x − 2.5| < .1. Exercise 2.4.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 215 of 191 Exercise 2.4.3. Suppose that |f(x) − 3| < .1. Then |2x − 1 − 3| < .1. Then |2x − 4| < .1, so .1 |x − 2| < 2 , so |x − 2| < .05, and these steps can all be reversed. Exercise 2.4.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 216 of 191 Exercise 2.4.4. Suppose that |f(x) − 3| < . Then |2x − 1 − 3| < . Then |2x − 4| < , so |x − 2| < 2 , so the corresponding value of δ should be 2 . Exercise 2.4.4

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 217 of 191 p q p → q q → p T T T T Exercise 2.5.1. T F F T We can see that they aren’t the F T T F F F T T same. Thus a statement p → q might be true but its converse q → p false. An example of this is the statement “ If f(x) is diﬀerentiable on (a, b), then it is continuous on (a, b). This statment is true but its converse “If f(x) is continuous on (a, b) then it is diﬀerentiable on (a, b) is false. Exercise 2.5.1 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 218 of 191 p q p → q q˜ → p˜ T T T T Exercise 2.5.2. T F F F We see that these are the same. F T T T F F T T The statementq ˜ → p˜ is called the contrapositive of p → q. Exercise 2.5.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 219 of 191 Exercise 2.5.3. “p whenever q” means the same thing as “q → p.”, and “p only if q” means the same thing as “p → q.” The statement “p if and only if q” means p → q AND q → p. This statement is often abbreviated “iﬀ” and perhaps written p ↔ q. Exercise 2.5.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 220 of 191 Exercise 2.6.1. We must show that for any > 0, there is a corresponding δ > 0 so that |2x − 1 − 3| < whenever 0 < |x − 2| < δ. Working backwards, Suppose that |f(x) − 3| < . Then |2x − 1 − 3| < . Then |2x − 4| < , so |x − 2| < 2 , so the corresponding value of δ should be 2 . Now to give a nice forwards formal presentation we could say: Integrated Given > 0, let δ = 2 . Calculus/Pre-Calculus Then if 0 < |x − 2| < 2 , Mark R. Woodard Then |2x − 4| < , so |2x − 1 − 3| < so |f(x) − 3| < . Title Page Exercise 2.6.1 Contents JJ II J I Go Back Close Quit Page 221 of 191 Exercise 2.6.2. We must show that for any > 0, there is a corresponding δ > 0 so that |4x + 2 − 6| < whenever 0 < |x − 1| < δ. Working backwards, Suppose that |f(x) − 6| < . Then |4x + 2 − 6| < . Then |4x − 4| < , so |x − 1| < 4 , so the corresponding value of δ should be 4 . Now to give a nice forwards formal presentation we could say: Integrated Given > 0, let δ = 4 . Calculus/Pre-Calculus Then if 0 < |x − 1| < 4 , Mark R. Woodard Then |4x − 4| < , so |4x + 2 − 6| < so |f(x) − 6| < . Title Page Exercise 2.6.2 Contents JJ II J I Go Back Close Quit Page 222 of 191 Exercise 2.7.1. Since the square root function is continous on its domain, and the expression underneath the square root is positive, this doesn’t x+1 present any problems. Also, x2−4 is a rational function, which is continuous wherever it is defined, but note that it isn’t defined at x = 2 or x = −2. Thus, at these points h(x) is definitely not continuous, since it is not defined there. At every other point h(x) is the composition of continuous functions, and is thus continuous. So the only discontinuities are at x = 2, −2. Exercise 2.7.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 223 of 191 ˜ Exercise 2.7.2. We need to have f(4) = limx→4 f(x), and this can be shown to be 8. (Do it!) Thus if we deﬁne f˜(4) = 8, then f and f˜ are the same everywhere except at 4, and f˜ is continuous at x = 4. Exercise 2.7.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 224 of 191 Exercise 3.1.1.

Integrated Calculus/Pre-Calculus First note that the point (9, 12) is on the tangent line, so the equation Mark R. Woodard of the tangent line is

Title Page Contents JJ II y − 12 = m(x − 9), J I Go Back Close Quit Page 225 of 191 f(x)−f(9) where m is the slope of the tangent line. Thus we must compute limx→9 x−9 . We have √ f(x) − f(9) x + x − 12 lim = lim x→9 x − 9 x→9 x − 9 √ x + x − 12 = lim x→9 x − 9 √ √ x + x − 12 x − (x − 12) = lim · √ x→9 x − 9 x − (x − 12) Integrated x − (x − 12)2 = lim √ Calculus/Pre-Calculus x→9 (x − 9)( x − (x − 12)) Mark R. Woodard

x − (x2 − 24x + 144) = lim √ x→9 (x − 9)( x − (x − 12)) Title Page −(x2 + 25x − 144) = lim √ Contents x→9 (x − 9)( x − (x − 12)) JJ II −(x2 − 25x + 144) = lim √ J I x→9 (x − 9)( x − (x − 12)) Go Back −(x − 9)(x − 16) Close = lim √ x→9 (x − 9)( x − (x − 12)) Quit −(x − 16) 7 Page 226 of 191 = lim √ = x→9 ( x − (x − 12)) 6 7 Thus the equation of the tangent line is given by y − 12 = 6 (x − 9). Exercise 3.1.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 227 of 191 Exercise 3.1.2. f(3 + h) − f(3) 3(3 + h)2 − 5(3 + h) + 3 − 15 We must compute lim . This is equal to lim h→0 h h→0 h 3(3 + h)2 − 5(3 + h) + 3 − 15 3(9 + 6h + h2) − 15 − 5h + −12 lim = lim h→0 h h→0 h

(27 + 18h + 3h2) − 5h − 27 = lim h→0 h Integrated 12h + 3h2 Calculus/Pre-Calculus = lim h→0 h Mark R. Woodard

= lim 12 + 3h = 12 h→0 Thus the instantaneous rate of change of y with respect to x is 12. Title Page Exercise 3.1.2 Contents JJ II J I Go Back Close Quit Page 228 of 191 Exercise 3.1.3. √ f(2 + h) − f(2) 4 + h − 2 We must compute lim . This is equal to lim . h→0 h h→0 h We have √ √ √ 4 + h − 2 4 + h − 2 4 + h + 2 lim = lim · √ h→0 h h→0 h 4 + h + 2

4 + h − 4 = lim √ Integrated h→0 h( 4 + h + 2) Calculus/Pre-Calculus h Mark R. Woodard = lim √ h→0 h( 4 + h + 2)

1 1 = lim √ = Title Page h→0 4 + h + 2 4 Contents 0 1 Thus f (2) = 4 . Exercise 3.1.3 JJ II J I Go Back Close Quit Page 229 of 191 Exercise 3.1.4.

f(x + h) − f(x) 1 − 1 We must compute lim . This is equal to lim x+h x . h→0 h h→0 h We have

1 − 1 x−(x+h) lim x+h x = lim (x)(x+h) h→0 h h→0 h

x − (x + h) = lim Integrated h→0 hx(x + h) Calculus/Pre-Calculus −h Mark R. Woodard = lim h→0 hx(x + h)

(−1) = lim Title Page h→0 (x)(x + h) Contents −1 = JJ II x2 0 −1 J I Thus f (x) = x2 . Exercise 3.1.4 Go Back Close Quit Page 230 of 191 Exercise 3.2.1. For f(x) = −3x6 + 2x3 − 12x − 123, we have f 0(x) = −18x5 + 6x − 12. Exercise 3.2.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 231 of 191 Exercise 3.2.2.

2x+2h+3 − 2x+3 lim 3x+3h−1 3x−1 h→0 h 2x+2h+3 · 3x−1 − 2x+3 · 3x+3x−1 = lim 3x+3h−1 3x−1 3x−1 3x+3h−1 h→0 h (2x+2h+3)(3x−1)−(3x+3x−1)(2x+3) = lim (3x−1)(3x+3h−1) h→0 h Integrated (6x2 + 6hx + 9x − 2x − 2h − 3) − (6x2 + 6hx − 2x + 9x h − 3) = lim 9 Calculus/Pre-Calculus h→0 (h)(3x − 1)(3x + 3h − 1) Mark R. Woodard −11h = lim h→0 (h)(3x − 1)(3x + 3h − 1) −11 Title Page = lim h→0 (3x − 1)(3x + 3h − 1) Contents −11 = JJ II (3x − 1)2 Exercise 3.2.2 J I Go Back Close Quit Page 232 of 191 Exercise 3.3.1. √ √ 3 (0+h)2− 3 0 Our job is to show that limh→0 h doesn’t exist. We can write √ 2 3 2 h 3 2 2 3 3 −1 h as h , and using the rules of exponents, we can write h as h = −1 1 1 1 h 3 = 1 = √3 . Now limh→0 √3 doesn’t exist, so (0, 0) is a singular point h 3 h h for this function. Exercise 3.3.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 233 of 191 Exercise 3.4.1. s0(t) = v(t) = 6t2 − 24. Now, this equation is zero when t = 2 or t = −2, so those times represent when the object is stopped. An analysis of the sign of v(t) shows that v(t) is positive on the interval (−∞, −2) and on (2, ∞). Thus the object is moving to the right during those intervals. On the interval (−2, 2), the object is moving to the left, since velocity is negative on that interval. Now a(t) = v0(t) = 12t, which is zero at time t = 0. A quick check shows that the acceleration is positive for t > 0 and negative for t < 0. Thus during the time interval (−∞, −2) the object is moving to the right but slowing down. At t = −2, the object is stopped. From t = −2 to t = 0 the object is moving to the left and Integrated speeding up. At time t = 0 the acceleration becomes positive, so the Calculus/Pre-Calculus object continues to move to the left, but starts to slow down. At t = 2, Mark R. Woodard the object is stopped, and thereafter moves to the right and speeds up. Exercise 3.4.1

Title Page Contents JJ II J I Go Back Close Quit Page 234 of 191 Exercise 3.4.2. √ 2 1. At time t = 4, the population has size p(4) = 4(4 + 3(4))√ = 2(16 + 12) = 56. At time t = 9, the population has size p(9) = 9(92+3(9) = 3(81 + 27) = 324. Thus over that time interval of length 5, the change in p is 324 − 56 = 268. Thus the average rate of change of the 268 population is 5 . 2. The instantaneous rate of√ change of the population would be given 0 0 2 1 −1 0 by p (4). We have p (t) = t(2t + 3) + (t + 3t) t 2 . Thus p (4) = √ 2 √1 28 Integrated 4 · 11 + (28) · = 22 + 4 = 29. 2( 4) Calculus/Pre-Calculus Exercise 3.4.2 Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 235 of 191 mRT Exercise 3.4.3. We can rewrite the equation as P = V . Recall that all of m, R, and T are being treated as constants in this discussion. Thus we dP −2 can ﬁnd dV = −(mRT )V . Exercise 3.4.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 236 of 191 Exercise 3.4.4. C(101) = 4.242606 (Use your calculator.) C(100) = 4.12 0 1 2 Thus the diﬀerence is .122606. Now C (x) = 2000 (2 + .008x + .024x ). So 0 1 242.8 C (100) = 2000 (2 + .8 + 240) = 200 = .1214. Exercise 3.4.4

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 237 of 191 √ 5 2 Exercise 3.5.1. For f(x) = 12x − 3x + 4x + 3, the inside√ function n(x) is 12x5 − 3x2 + 4x + 3, while the outside function is x. Thus the expression m0(n(x) would be √ 1 , while n0(x) = 60x4 − 6x + 4. 2 12x5−3x2+4x+3 4 Thus f 0(x) = √ 60x −6x+4 . Exercise 3.5.1 2 12x5−3x2+4x+3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 238 of 191 1 1 Exercise 3.5.2. Let f(x) = (x + x 2 ) 2 . The inside function is given by 1 1 x + x 2 while the outer function is given by x 2 . Thus the derivative of f is −1 1 1 2 1 2 √ given by 2 x + x · 1 + 2 x . Exercise 3.5.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 239 of 191 Exercise 3.5.3. Let s(t(x)) be called n(x) for the moment. Then f(x) = r(n(x)), so f 0(x) = r0(n(x)) · n0(x). Now n0(x) = s0(t(x)) · t0(x), by the chain rule. Thus (substituting) we get

f 0(x) = r0(s(t(x)) · s0(t(x)) · t0(x).

Exercise 3.5.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 240 of 191 Exercise 3.6.1. If xy = 3, then differentiating the left hand side with respect to x dy (recalling that y is a function of x) we get x dx + y(1) from the product rule. The derivative of the right hand side is zero, since we have a constant on that side of the equation. Thus we have dy x + y = 0, dx so dy −y = . Integrated dx x Calculus/Pre-Calculus 3 Mark R. Woodard As suggested, since this function is easy to write explicitly as y = x , we can dy −3 check our answer by differentiaing this expression. We get dx = x2 . But if −3 −y −3 y = x , then the expressions x and x2 are the same. Exercise 3.6.1 Title Page Contents JJ II J I Go Back Close Quit Page 241 of 191 Exercise 3.6.2. √ Differentiating xy = x + y + 1 with respect to x yields 1 −1 dy dy (xy) 2 x + y = 1 + . 2 dx dx

dy There are two terms on the left hand side, one of which has a factor of dx . dy That is also true for the right hand side. Gathering terms with a dx factor on the left yields Integrated 1 −1 dy dy 1 −1 (xy) 2 · x − = 1 − (xy) 2 y. Calculus/Pre-Calculus 2 dx dx 2 Mark R. Woodard dy Factoring out the dx factor and solving yields

1 −1 dy 1 − (xy) 2 y = 2 . Title Page 1 −1 dx 2 2 (xy) x − 1 Contents Exercise 3.6.2 JJ II J I Go Back Close Quit Page 242 of 191 Exercise 3.7.1. Suppose you have a degree k polynomial like f(x) = k akx + ... + a1x + a0. Each derivative of this will result in a polynomial of lower degree, in fact it will have degree one less than what came before. So the ﬁrst derivative has degree k − 1, the second has degree k − 2 and so on. The kth derivative would thus have degree k − k = 0. But a zeroth degree polynomial is a constant function, and the derivative of a constant function is zero. Thus the k + 1st derivative (and thus all those after that) result in the constant zero. Exercise 3.7.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 243 of 191 0 −2 00 2·3 000 −2·3·4 Exercise 3.7.2. Note that f (x) = x3 , f (x) = x4 , f (x) = x5 . It looks like the numerator for the kth derivative has a (k + 1)! in it, as well as either a plus or minus 1 factor. In fact, the sign looks to be positive when k is even and negative when k is odd, so a factor of −1k would take care of this alternating sign. The denominator is always x to a power, and the power looks to be k + 2 for the kth derivative. Thus it looks like we have the following formula for the kth derivative:

dky (−1)k(k + 1)! k = k+2 . dx x Integrated Exercise 3.7.2 Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 244 of 191 Exercise 3.8.1. Draw a picture of such a cylinder, and label the height h. The volume of the water in the cylinder is the area of the base times dV dh dV the height, so it is V = 4πh. Thus dt = 4π dt , and since dt = 3, we have dh dt = 3 · 4π = 12π. This would be measured in feet per minute. Exercise 3.8.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 245 of 191 Exercise 3.8.2. Note that when the water is h feet deep, the volume of water is the area of a triangular cross section times the width (which is 10 feet). Let x be the length of water in the pool. Then the area of the triangle 1 1 is 2 xh, so the volume of water is given by V = 2 · 10xh = 5xh. What do dx we do about the fact that we don’t know anything about dt ? Using similar x 25 triangles, we can write h = 5 , so x = 5h. Thus we can rewrite the volume 2 dV dh dh dh 3 as V = 5(5h)h = 25h . Thus dt = 50h dt , so 3 = 50h dt , so dt = 50h . dh 1 When h = 3, we have dt = 50 . Exercise 3.8.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 246 of 191 Exercise 3.9.1.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 247 of 191 Exercise 3.9.2.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 248 of 191 Exercise 4.1.1.

Since it is 2π units around the unit circle, half-way around would correspond with t = π. Thus the point on the unit circle which corresponds with t = π would be (−1, 0). Therefore, cos(π) = −1 and sin(π) = 0. Exercise 4.1.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 249 of 191 Exercise 4.1.2.

3π The point on the circle corresponding with t = 2 would be the point 3 3π that is 4 of the way around the circle, since 2 is three fourths of 2π. Thus 3π 3π the point in question is (0, −1), and thus sin 2 = −1 and cos 2 = 0 Exercise 4.1.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 250 of 191 Exercise 4.1.3.

180 π Since one radian corresponds with π degrees, 6 radians would corre- π 180 spond with 6 · π = 30 degrees. Exercise 4.1.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 251 of 191 Exercise 4.1.4.

π Since one degree corresponds with 180 radians, 225 degrees would cor- π 5π respond with 225 · 180 = 4 radians. Exercise 4.1.4

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 252 of 191 Exercise 4.2.1.

7π The point on the unit circle corresponding with t = 6 is directly opposite the point on the unit circle corresponding with t = π . This is because √ 6 7π = π + π . The point associated with π is ( 3 , 1 ), so the point directly 6 6 √6 2 2 3 1 π −1 opposite the circle from this point is (− 2 , − 2 ). Thus sin 6 = 2 . Exercise 4.2.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 253 of 191 Exercise 4.2.2.

−3π The point on the unit circle associated with 4 is directly opposite the point associated with t = π . This is because −3π + π = π . The 4√ √ 4 4 point associated with t = π is ( 2 , 2 ). Thus the point associated with √ √ 4 2 2 −3π 2 2 −3π 4 is (− 2 , − 2 ), and tan 4 = 1, since it is the ratio of these two coordinates. Exercise 4.2.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 254 of 191 Exercise 4.2.3.

If sin(x) + cos(x) = 0, then sin(x) = − cos(x). Of course, we also know that sin2(x) + cos2(x) = 1, so by substituting) we have that (− cos(x))2 + 2 2 cos√ (x) = 1, so 2 cos (x) = 1, from which it is easy to see that√| cos(√ x)| = 2 . Thus we are looking at two points on the unit circle: (− 2 , 2 ) and 2√ √ 2 2 ( 2 , − 2 ). These points are directl opposite the circle from each other, 2 2 √ √ 2 2 and the ﬁrst one is the reﬂection about the y axis of the point ( 2 , 2 ), π 3π which corresponds with x = 4 . Thus we are looking at x = 4 , and the sum of this with π, −π, 2π, −2π, . . .. Thus we can say that the solutions Integrated 3π Calculus/Pre-Calculus all have the form x = + kπ, k ∈ Z. Exercise 4.2.3 4 Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 255 of 191 1 Exercise 4.2.4. Clearly if 2 cos(2x) + 1 = 0, then cos(2x) = − 2 . But an analysis of the unit circle shows that the points where the cosine function 1 2π 4π is − 2 are associated with 3 and 3 . Thus, since we are looking for where 1 2π 4π cos(2x) is − 2 , we must have 2x = 3 and 2x = 3 . Thus we get that π 2π x = 3 and x = 3 . Also, if z is a solution, then z + π is a solution, because cos(2(z + π)) = cos(2z + 2π) = cos(2z). Thus the solutions have the form π 2π 3 + kπ, k ∈ Z and 3 + kπ, k ∈ Z. Exercise 4.2.4

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 256 of 191 Exercise 4.4.1.

cos(x) The cotangent function is deﬁned as cot(x) = sin(x) . This will be un- deﬁned when sin(x) = 0, which occurs at 0, ±π, ±2π, . . .. Every other real number is in the domain of both sin(x) and cos(x), and so is in the domain of the quotient of these two. Thus the domain is {x : x 6= kπ, k ∈ Z}. Exercise 4.4.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 257 of 191 Exercise 4.4.2.

cos(x) The cotangent function is deﬁned by cot(x) = sin(x) . This function is undeﬁned where sin(x) = 0, which is at all multiples of π. If x is a little greater than 0, then cos(x) is near one, while sin(x) is very small (but is positive.) Thus the ratio is a large number. We can see from this analysis that limx→0+ cot(x) = +∞. On the other hand, if x is a litle less than 0, then sin(x) is negative, while cos(x) is still close to 1, so is positive. Thus the ratio of the two, while large in absolute value, is negative. Thus limx→0− cot(x) = −∞. Using this information along with a few key plotted Integrated π π points (like cot( 2 ) = 0 and cot( 4 ) = 1) leads to the graph. It is also useful Calculus/Pre-Calculus to note that cot(x) is periodic with period π. Mark R. Woodard Exercise 4.4.2

Title Page Contents JJ II J I Go Back Close Quit Page 258 of 191 Exercise 4.4.3.

The sketches are coming soon. Exercise 4.4.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 259 of 191 Exercise 4.5.1.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 260 of 191 Exercise 4.5.2.

First note that we can rewrite cos(2x) as follows:

cos(2x) = cos(x + x) = cos(x) cos(x) − sin(x) sin(x) = cos2(x) − (1 − cos2(x)) = 2 cos2(x) − 1 Thus we have that cos(2x) = 2 cos2(x) − 1, and solving this expression for cos2(x) yields the desired identity. Exercise 4.5.2 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 261 of 191 Exercise 4.6.1. cos(h) − 1 cos(h) − 1 cos(h) + 1 lim = lim · h→0 h h→0 h cos(h) + 1 cos2(h) − 1 = lim h→0 h · (cos(h) + 1) − sin2(h) = lim h→0 h · (cos(h) + 1) sin(h) − sin(h) = lim · lim Integrated h→0 h h→0 cos(h) + 1 −0 Calculus/Pre-Calculus = 1 · Mark R. Woodard 1 + 1 = 0

Exercise 4.6.1 Title Page Contents JJ II J I Go Back Close Quit Page 262 of 191 Exercise 4.6.2.

d d 1 dx sec(x) = dx cos(x) cos(x)·0−1·(− sin(x)) = cos2(x) sin(x) = cos2(x) = sec(x) tan(x)

Integrated d d cos(x) dx cot(x) = dx sin(x) Calculus/Pre-Calculus sin(x)(− sin(x)−cos(x) cos(x)) Mark R. Woodard = sin2(x) −1 = sin2(x) = − csc2(x) Title Page Contents d d 1 dx csc(x) = dx sin(x) sin(x)·0−1·cos(x) JJ II = sin2(x) − cos(x) J I = sin2(x) = − cot(x) csc(x) Go Back Close

Exercise 4.6.2 Quit Page 263 of 191 Exercise 4.6.3.

x2+1 0 x2+1 (2x−3)(2x)−(x2+1)(2) If f(x) = sin 2x−3 , then f (x) = cos 2x−3 · (2x−3)2 .

Exercise 4.6.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 264 of 191 Exercise 4.6.4. √ √ √ √ 2 2 0 2 2 2 If g(x) = ( x − 4)·√tan( x − 4), then g (x) = ( x − 4)·sec ( x − 4)· 1 −1 1 −1 2 2 2 2 2 2 (x − 4) · 2x + tan( x − 4) · 2 (x − 4) · 2x.

Exercise 4.6.4

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 265 of 191 Exercise 5.1.1. The graph of y = (x − 3)2 + 2 has the same shape as the graph of y = x2, except that it is shifted 3 units to the right and 2 units up. Thus, the vertex is now at (3, 2), and the graph opens upward. Thus there is an absolute minimum of 2 at x = 3. On the other hand, the graph of y = −3−(x+2)2 is the same as the graph of y = x2 except that it is ﬂipped about the x axis, then shifted 2 units to the left and then 3 units down. Thus the vertex is now at (−2, −3), and the parabola opens downward. Thus the absolute maximum value is −3 at x = −2. Exercise 5.1.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 266 of 191

Exercise 5.1.2.

¤ £ ¢ ¡ ¡ ¢ £ ¤

¥ ¥ ¥ ¥ ¥

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¥ Integrated ¢ ¥ Calculus/Pre-Calculus

Mark R. Woodard

£ ¥ It looks like the pictured function has local minimums at about x = −3.1, x = .8 and x = 4.7. (It is hard to be too precise on such a graph.) It looks like there are local maximums at x = 0 and x = 1.6. Title Page Exercise 5.1.2 Contents JJ II J I Go Back Close Quit Page 267 of 191 Exercise 5.1.3. Note that the function never reaches a value of 11, although it gets very close. Since it doesn’t actually reach the value of 11, this cannot be the maximum value. It is tempting to say something like: “the maximum value is whatever real number is the closest real number to 11 which is less than 11”, but a little thought about the nature of real numbers leads one to realize that that statement is silly. There is no such real number, so this function does not have a maximum. Likewise, the function is always greater than 5, but as x → 2+, we see that we can get as close to a value of 5 as we desire. Thus there is no minimum value of this function. Integrated The function g(x) as similar problems. The function almost gets as high Calculus/Pre-Calculus as 20, and almost gets as low as 0, but never quite makes it to these values. Mark R. Woodard Thus, there is no maximum or minimum. Exercise 5.1.3 Title Page Contents JJ II J I Go Back Close Quit Page 268 of 191 Exercise 5.1.4. First note that the function is continuous (since it is a polynomial), and we are given a closed interval. Thus we know that this function has a maximum and a minimum on this interval. The derivative is f 0(x) = 20x4 − 20x3 = 20x3(x − 1). Setting f 0(x) = 0, we get stationary points at x = 0 and x = 1. Comparing the values of the function at these points we have: x f(x) −3 −1374 0 3 Integrated 1 2 Calculus/Pre-Calculus 2 51 Mark R. Woodard From this chart we can see that there is a maximum value of 51 at x = 2 and a minimum value of −1374 at x = −3. Exercise 5.1.4 Title Page Contents JJ II J I Go Back Close Quit Page 269 of 191 Exercise 5.3.1.

Yes, Rolle’s Theorem does apply because since this is a rational function, it is continuous and differentiable on its domain, which consists of all real numbers except x = −8. Since −8 isn’t part of the interval −1, 3], this function is continuous and differentiable on the given interval. Differenti- ating we have

(x + 8)(2x − 2) − (x2 − 2x − 3)(1) f 0(x) = Quotient Rule Integrated (x + 8)2 Calculus/Pre-Calculus (2x2 + 14x − 16) − (x2 − 2x − 3) = Expanding in the numerator Mark R. Woodard (x + 8)2 x2 + 16x − 13 = 2 Combining like terms (x + 8) Title Page

2 This derivative is zero when its numerator is zero, which is when √x + Contents 16x − 13 = 0, which (by the quadratic formula is when x = −8 ± 77. JJ II Now according to the theorem,√ at least one of these must√ be in the interval [−1, 3], and since −8 + 77√≈ .775, we see that −8 + 77 is the promised J I value of c. Note that −8 − 77 isn’t in the interval [−1, 3], so we disregard Go Back it. Exercise 5.3.1 Close Quit Page 270 of 191 Exercise 5.3.2.

2 −1 2 0 3 √ Note that g (x) = 3 x = 3 3 x . This is never zero since its numerator is never zero. This does not violate Rolle’s Theorem, because g(x) is not differentiable at x = 0, so it is not diﬀerentiable on (−1, 1). Rolle’s Theorem only applies to functions which are continuous on the given closed interval, and diﬀerentiable on the corresponding open interval. Exercise 5.3.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 271 of 191 Exercise 5.3.3.

No, it does not apply. The function is diﬀerentiable on (0, 1), but it is not continuous at 0, so is not continuous on [0, 1]. Note that g0(x) = −2, which is never zero. Exercise 5.3.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 272 of 191 Exercise 5.6.1.

Use the analysis done previously to ﬁnd the relative extrema for f(x) = x . x2+2 √ We already found that there√ were√ stationary points at x = ±√2, and that√f(x) is increasing on (− 2, 2) and decreasing on (−∞, − 2) and on ( 2, ∞). Using the√ ﬁrst derivative test, we know see that√ there is a local maximum at √x = 2 and a local minimum at x = √− 2. The local 2 − 2 maximum value is 4 while the local minimum value is 4 . Exercise 5.6.1 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 273 of 191 Exercise 5.7.1.

x Continue studying f(x) = x2+2 by analyzing its concavity. 0 2−x2 Recall that f (x) = (x2+2)2 . So

(x2 + 2)2(−2x) − (2 − x2)(2)(x2 + 2)(2x) f 00(x) = Quotient Rule (x2 + 2)4 (x2 + 2)((x2 + 2)(−2x) − (4x)(2 − x2)) = Factoring the Numerator (x2 + 2)4 Integrated (x2 + 2)((−2x3 + −4x) − (8x − 4x3)) Calculus/Pre-Calculus = Distributing (x2 + 2)4 Mark R. Woodard (−2x3 + −4x) − (8x − 4x3) = Cancelling Factors (x2 + 2)3 (2x3 − 12x) Title Page = Combining Like Terms (x2 + 2)3 Contents (2x)(x2 − 6) = Factoring the Numerator JJ II (x2 + 2)3 J I Go Back Now an analysis of this second derivative for places where the sign might √ Close change shows that x = 0 and x = ± 6 are possibilites, since that is where the numerator (and thus f 00(x)) is zero. Also note that there are no places Quit where the denominator√ √ is zero.√ The potential√ intervals of concavity are Page 274 of 191 thus (−∞, − 6), (− 6, 0), (0, 6), and ( 6, ∞). Using test values we √ √ see that the second derivative√ is negative√ on (−∞, − 6) and on (0, 6), and is positive√ on (− 6√, 0) and on ( 6, ∞). Thus f√(x) is concave down√ on (−∞, − 6) and on (0, 6), and is concave up on (− 6, 0) and on ( 6, ∞). √ √ √ √ − 6 6 There are points of inﬂection at (0, 0), (− 6, 8 ), and ( 6, 8 ). Exercise 5.7.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 275 of 191 Exercise 5.7.2.

Use the analysis done previously to ﬁnd the relative extrema for f(x) = x . x2+2 √ We already found that there were stationary points at x = ± 2, and √ √ 00 (2x)(x2−6) 00 −2 2(−4) we found f (x) = 2 3 . Since f (− 2) = 3 > 0 there must be (x +2) √ √ 4 √ a relative minimum at x = − 2. Since f 00( 2) = 2 2(−4) < 0, there must √ 43 be a relative maximum at x = 2. Note that this coincides with what we obtained using the ﬁrst derivative test. Exercise 5.7.2 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 276 of 191 Exercise 5.8.1.

sin(x) + We must compute limx→0 x2 . Let’s ﬁrst focus on the limit as x → 0 . sin(x) sin(x) sin(x) x Recall that limx→0 x = 1, so it would be useful to write x2 as x . Then as x → 0+, the numerator is approaching 1, while the denominator is approaching zero (but is positive), thus the ratio is increasing without sin(x) x bound. Thus limx→0+ x = ∞. This is enough to show that there is a vertical asymptote at x = 0. A careful analysis of the limit as x → 0− shows that that limit is −∞. Exercise 5.8.1 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 277 of 191 Exercise 5.9.1.

1 We are trying to show that limx→∞ x = 0. We must show that for any > 0, there is a value of M so that if x > M, |f(x) − 0| < . Working 1 backwards from this last expression, we see that we need x < , or in other 1 1 words, we need x > . So let’s take M = . Then (starting over) if x > M, 1 1 we have x > so x < . Also since x is some positive number, we can 1 1 write x < or x − 0 < . Thus we have shown what we set out to show. 1 Since limx→∞ x = 0, the line y = 0 must be a horizontal asymptote. Exercise 5.9.1 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 278 of 191 1 1 r 1 1 r Exercise 5.9.2. Note that xr = x . Thus limx→∞ xr = limx→∞ x = 0r = 0. Exercise 5.9.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 279 of 191 Exercise 5.10.1.

x2−2 Sketch a graph of f(x) = x3 , following the above guidelines.

1. Note the domain of f(x). Since f(x) is a rational function, its domain is the set of all real numbers except those which make the denominator 0. Thus any value for which x3 = 0 is not in the domain, but only 0 meets this requirement. Thus the domain consists of all real numbers except 0. We will be sure to include 0 on any sign chart we make. Integrated Calculus/Pre-Calculus 2. Look for vertical and horizontal asymptotes. We need to con- Mark R. Woodard x2−2 sider the limits as we approach 0 from each side. Consider limx→0+ x3 . Since the numerator is approaching a number other than 0, and the denominator is approaching zero, the ratio is getting “big”, but is Title Page a negative number since the numerator is near −2 and the denom- x2−2 Contents inator is postive. Thus limx→0+ x3 = −∞. On the other hand, x2−2 limx→0− x3 = ∞. We have a vertical asymptote at x = 0. To look x2−2 JJ II for horizontal asymptotes, consider limx→∞ x3 . A quick check shows that this limit is 0. (Divide the numerator and denominator by x3 to J I see this.) Thus y = 0 (the x-axis) is a horizontal asymptote. Go Back

3. Investigate the Monotonicity of f(x). A little work (do it!) shows Close 2 √ that f 0(x) = 6−x . This is zero at x = ± 6. By making the sign chart, Quit x4 √ √ we see that f(x√) is decreasing on√ (−∞, − 6) and on ( 6, ∞). It is Page 280 of 191 increasing on ( 6, 0) and on (0, 6). 4. Locate Relative Extrema of f(x). Using√ the sign charts, we see that there√ is a local maximum at x = 6 and a local minimum at x = − 6.

5. Investigate the Concavity of f(x). A little√ work (do it!) shows that f 00(x) = 2x2−24 . This is zero at x = ± 12. By making the sign x5 √ 00 chart for f√(x), we see that f(x) is concave√ down on (−∞√, − 12) and on (0, 12). It is concave up on√ (− 12, 0) and on ( 12, ∞). There are inﬂection points√ at x = ± 12. Also, applying the second derivative test to x = ± 6 conﬁrms our conclusions that there a local √ √ Integrated minimum at x = − 6 and a local maximum at x = 6. Calculus/Pre-Calculus 6. Preparation for Graphing. Mark R. Woodard

(a) Plot a few√ important√ points. We will√ want to√ plot all the points (− 12, − 10√ ), ( 12, 10√ ), (− 6, − √4 ), ( 6, √4 ). 12 12 12 12 6 6 6 6 Title Page (b) Intercepts. There is no y intercept since x = 0 is not in the Contents domain√ of f(x). A quick check for x intercepts shows that x = ± 2 are intercepts. JJ II (c) Asymptotes. Since the asmptotes are the axes, these are already drawn. Recall that f(x) is approaching +∞ as x → 0− and f(x) J I is approaching −∞ as x → 0+. Go Back (−x)2−2 x2−2 (d) Symmetry. Since f(−x) = (−x)3 = − x3 = −f(x), this Close function is symmetric about the origin. We will think about this Quit when sketching. Page 281 of 191 7. Sketch the Graph. It should look like this: 1 0.75 0.5 0.25

-4 -2 2 4 -0.25 -0.5 -0.75 -1 Integrated Calculus/Pre-Calculus Exercise 5.10.1 Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 282 of 191 Exercise 5.11.1.

Let the outside of the entire pen structure have dimensions x × y. Then the objective is to maximize the area A = xy, subject to the constraint that there is only 200 feet of fence available. A diagram shows that in building this pen, the farmer will use 4x feet to build the “vertical” walls and 2y feet to build the “horizontal” walls. Thus the total number of feet used is 4x+2y, which must equal 200. Thus 4x+2y = 200 is the constraint equation. Solving this for y yields y = 100 − 2x, and substituting this into 2 the area equation gives an objective function A = x(100−2x) = 100x−2x . Integrated It is quite easy to maximize this downward opening parabola — the vertex Calculus/Pre-Calculus is at the x value where A0(x) = 0, so we need A0(x) = 100 − 4x to be zero, Mark R. Woodard which occurs when x = 25. Note that while the natural domain for this parabola is the set of all real numbers, it is only true that x ≥ 0 and y ≥ 0 if x is between 0 and 50. One can easily check that this is a maximum, in fact, a global maximum, since the area function A(x) is increasing on Title Page [0, 25] and decreasing on [25, 50]. Of course when x = 25, the constraint Contents equation tells us that y = 100 − 2(25) = 50. Thus the outer dimensions of the pen should be 25 × 50. Exercise 5.11.1 JJ II J I Go Back Close Quit Page 283 of 191 0 f (xn)(xn)−f(xn) Exercise 5.12.1. The formula 0 = x becomes f (xn) n+1

2 3 (3xn + 1)(xn) − (xn + xn − 4) xn+1 = 2 3xn + 1 which simpliﬁes to 3 2xn + 4 xn+1 = 2 . 3xn + 1

If we start with an initial guess of x0 = 1, we have the following table: Integrated i xi f(xi) Calculus/Pre-Calculus 0 1 -2 Mark R. Woodard 1 1.5 0.875 2 1.38709677 0.0559229297 3 1.37883895 0.000283202346 Title Page 4 1.3787967 0.00000000738259587 Contents So the root seems to be pretty close to 1.3787. Exercise 5.12.1 JJ II J I Go Back Close Quit Page 284 of 191 Exercise 5.13.1.

A general antiderivative of y = 2x + 8 should involve the sum of the antiderivatives of 2x and 8. We have already seen that an antiderivative of 2x is x2, and a little thought shows that an antiderivative of a constant like “8” is the linear function 8x. Thus one speciﬁc antiderivative of y = 2x+8 would be F (x) = x2 +8x. By the “+C” theorem, the general antiderivative would be x2 + 8x + C. Exercise 5.13.1

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 285 of 191 Exercise 5.13.2.

By the sum and constant multiplier rules, we can simply try to ﬁnd antiderivatives for x3 and x, and then string these together with the proper constant factors and plus signs. A little thought reveals that an antiderivative of x3 should be a 4th degree monomial, and some trial and error yields x4 x2 the correct result of 4 . Similarly, an antiderivative of x is 2 . Thus the 3 x4 x2 general antiderivative of y = 5x + 8x + 10 is 5 · 4 + 8 · 2 + 10x + C = 5 4 2 4 x + 4x + 10x + C. Exercise 5.13.2 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 286 of 191 Exercise 5.14.1.

We are given a(t) = −32, so v(t) = −32t + C1. since the initial velocity is 12, we know that v(0) = 12. Thus v(0) = 12 = −32(0) + C1, so C1 = 12. t2 2 We then have that s(t) = −32 2 + 12t + C2 = −16t + 12t + C2. Also, 2 2 s(0) = 3 = −16(0) + 12(0) + C2, so C2 = 3. Thus s(t) = −16t + 12t + 3. The ball is at its maximum height when its velocity is zero. Thus, the 12 3 maximum occurs when −32t + 12 = 0, or when t = 32 = 8 . The maximum 9 3 21 is −16 64 + 12 8 + 3 = 4 . Exercise 5.14.1 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 287 of 191 Exercise 5.14.2.

We have v(t) = sin t + − cos t + C1. Also, v(0) = 1 = sin(0) + cos(0) + C1 = 1 + C1, so C1 = 0. Thus v(t) = sin t + − cos t. Therefore, s(t) = − cos t + − sin t + C2. Also, s(0) = 4 = − cos(0) + − sin(0) + C2 = −1 + C2, Thus C2 = 5. We therefore have s(t) = − cos t + − sin t + 5 Exercise 5.14.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 288 of 191 Exercise 6.1.1.

We recognize the numbers 4, 9, 16 and 25 as the squares of the integers 2, 3, 4, and 5 repsectively. Thus, we suspect that the terms involve the square of i. We must start with i = 2 and end with i = 10 to get the correct sum. We have

10 X 4 + 9 + 16 + ... + 100 = i2. i=2 Integrated Exercise 6.1.1 Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 289 of 191 Exercise 6.1.2.

P8 2 2 2 2 i=3 i + i + 1 = (3 + 3 + 1) + (4 + 4 + 1) + ... + (8 + 8 + 1) = 13 + 21 + 31 + 43 + 57 + 73. Exercise 6.1.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 290 of 191 Exercise 6.2.1.

n+1 First of all, the reason the 2nd row ends with 2 is that that is the number which is one more than what the ﬁrst row ends with. Speciﬁcally, n − 1 n − 1 2 n + 1 + 1 = + = . 2 2 2 2 Pn n(n+1) To see that the formula i=1 i = 2 holds when n is even, we write n 1 2 3 . . . 2 Integrated n+2 n n − 1 n − 2 . . . 2 Calculus/Pre-Calculus n Mark R. Woodard and note that the columns add to n+1, and that there are 2 such columns. n(n+1) Thus, the sum is 2 . Exercise 6.2.1

Title Page Contents JJ II J I Go Back Close Quit Page 291 of 191 Exercise 6.2.2.

n n n n X X X X 4i3 + 6i − 1 = 4i3 + 6i − 1 By 6.4 i=1 i=1 i=1 i=1 n n n X X X = 4 i3 + 6 i − 1 By 6.5 i=1 i=1 i=1 n n X X = 4 i3 + 6 i − n By 6.6 Integrated Calculus/Pre-Calculus i=1 i=1 Mark R. Woodard (n)(n + 1)(2n + 1) (n)(n + 1) = 4 + 6 − n By 6.1 and 6.2 6 2

Exercise 6.2.2 Title Page Contents JJ II J I Go Back Close Quit Page 292 of 191 Exercise 6.3.1.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 293 of 191 Exercise 6.3.2.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 294 of 191 Exercise 6.4.1.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 295 of 191 Exercise 6.4.2.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 296 of 191 Exercise 6.4.3.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 297 of 191 Exercise 6.4.4.

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 298 of 191 √ √ Exercise 7.1.1. Only 4 −16 is not a real number. 5 −243 = −3 which is an integer. Both π and √−5 are irrational numbers. Exercise 7.1.1 2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 299 of 191 Exercise 7.1.2. a + β is an irrational number. aβ can be either rational or irrational√ depending on√ the values of a and β. For example, if a =√ 1 and β = 2, then aβ = 2 which is irrational. If a = 0 and β = 3, then aβ = 0 which is rational. Similarly, β + γ can be either rational or irrational. For example, if √β = π and γ √= −π, then β + γ =√ 0 which is rational. However, if β = 2 and γ = 2, then β + γ = 2 2 which is irrational. Exercise 7.1.2

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 300 of 191 Exercise 7.1.3. Group the first two terms and the last two terms together. (2x3 + 3x2) + (−8x − 12) Notice that an x2 can be factored out of the first group and a −4 can be factored out of the second group. x2(2x + 3) − 4(2x + 3) Now notice that each group has the common factor (2x + 3). Factor out the (2x + 3). (2x + 3)(x2 − 4) The factor (x2 − 4) is a difference of squares which can be factored into (x + 2)(x − 2). Finally we end with (2x + 3)(x + 2)(x − 2) as the factored form. Exercise 7.1.3

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 301 of 191 Exercise 7.1.4. Group the second, third, and ﬁnal terms together. x4 + (−x2 − 6x − 9) Factor out a −1 from the group. x4 − (x2 + 6x + 9) Notice that x2 + 6x + 9 is a trinomial square which can be factored into (x + 3)2. So now we have x4 − (x + 3)2. Since x4 = (x2)2, x4 − (x + 3)2 is a diﬀerence of squares. So it may be factored into (x2 + x + 3)(x2 − x − 3). Exercise 7.1.4

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 302 of 191 Exercise 7.1.5. Multiply both sides of the equation by 2. 2(3x+5) = x−7 6x + 10 = x − 7 Now subtract x from both sides: 5x + 10 = −7 Now subtract 10 from both sides: 5x = −17 17 x = − 5 Exercise 7.1.5

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 303 of 191 Exercise 7.1.6. Multiply both sides of the equation by xy − ax: b2y = C(xy − ax) Factor out the x on the left side of the equation: b2y = Cx(y − a) Now divide both sides of the equation by C(y − a):

b2y x = C(y − a) Exercise 7.1.6 Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 304 of 191 Exercise 7.1.7. Multiply equation (7) by −3 and equation (8) by 4:

− 21x − 12y = −24 (7.9) 24x + 12y = 16 (7.10)

Add equations (9) and (10) and solve for x: 3x = −8 8 So x = − 3 . 8 Now substitute x = − 3 into equation (8) and solve for y: Integrated 8 6(− 3 ) + 3y = 4 Calculus/Pre-Calculus 3y = 20 Mark R. Woodard 20 So y = 3 . 8 20 Hence the solution to the system is (− 3 , 3 ). Exercise 7.1.7 Title Page Contents JJ II J I Go Back Close Quit Page 305 of 191 Exercise 7.1.8. a) | − 4.5| = 4.5 = 4 b) J| −4.5 |K= |J − 5|K= 5 c) Jπ = 3K Exercise 7.1.8 J K

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 306 of 191 Exercise 7.1.9.

1 1 a) False, let x = − 2 and y = − 2 . 1 1 So − 2 − 2 = (−1)(−1) = 1 J KJ1 K1 1 But (− 2 )(− 2 ) = 4 = 0 < 1 b) False,J let x = 2K.5 andJ Ky = −2.5. So | 2.5 | −2.5 = |2|(−3) = −6 ButJ (2.K5)(J −2.5)K = −6.25 = −7 < −6 J K J K Integrated Exercise 7.1.9 Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 307 of 191 Exercise 7.1.10. True. Proof: By the deﬁnition of absolute value, |x| > 0 and y > 0. Then |x| ≥ |x| and |y| ≥ |y| . Since |x| ≥J 0,K take the inequalityJ K |y| ≥ |y| and multiply the leftsideJ K by |x| and the right side by |Jx| .K This yields the inequality: |x||y| ≥ |x| J|y|K By a property of absolute value, |x||Jy|=KJ|xy|K Now apply the greatest integer function to both sides of Integrated the inequality. The resulting inequality is Calculus/Pre-Calculus |xy| ≥ ( |x| |y| ) . Mark R. Woodard SinceJ K |xJ| J|y|KJis anK K integer, |xy| J≥ KJ( |x|K |y| ) = |x| |y| HenceJ K |xyJ J| ≥KJ |x|K K|y| J. KJ K J K J KJ K Title Page Exercise 7.1.10 Contents JJ II J I Go Back Close Quit Page 308 of 191 Solutions to Quizzes Solution to Quiz: Most of the illustrations were done by Brian Wagner, an undergraduate at Furman University. See the acknowledgments for the credits. End Quiz

Integrated Calculus/Pre-Calculus Mark R. Woodard

Title Page Contents JJ II J I Go Back Close Quit Page 309 of 191