CK-12 F OUNDATION Calculus Teacher’s Edition - Common Errors Narasimhan CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the “FlexBook,” CK-12 intends to pioneer the generation and distribution of high-quality educationalcontentthatwillservebothascoretextaswellasprovideanadaptiveenvironmentforlearning, powered through the FlexBook Platform™. Copyright © 2011 CK-12 Foundation, www.ck12.org Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution/Non-Commercial/Share Alike 3.0 Un- ported (CC-by-NC-SA) License (http://creativecommons.org/licenses/by-nc-sa/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. 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Printed: March 23, 2011 Author Ramesh Narasimhan i www.ck12.org Contents 1 Calculus TE - Common Errors 1 1.1 Functions, Limits, and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.3 Applications of Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 1.4 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.5 Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 1.6 Transcendental Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 1.7 Integration Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 1.8 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 www.ck12.org ii Chapter 1 Calculus TE - Common Errors 1.1 Functions, Limits, and Continuity This Calculus Common Errors FlexBook is one of seven Teacher’s Edition FlexBooks that accompany the CK-12 Foundation’s Calculus Student Edition. To receive information regarding upcoming FlexBooks or to receive the available Assessment and Solution Key FlexBooks for this program please write to us at [email protected]. Lesson 1: Equations and Graphs Tobeginthestudyofcalculus, itishelpfultoreviewsomeimportantpropertiesofequationsandfunctions, and how to graph different kinds of functions on an x −y coordinate system. A solid understanding of analytic geometry is essential to developing the techniques of differentiation and integration presented in this textbook. The ability to identify analytic solutions to the points where graphs intersect the x and y axes (e.g. the intercepts), as well as finding the exact points where two graphs or curves cross each other, will be necessary to evaluate limits, derivatives and integrals. An important technique students will need throughout the study of calculus is evaluating functions by substituting in a value for a function’s argument. In simple cases, the argument is given as a number. Given a function f(x) = x2, to find f(4) we substitute the 4 in for x, and get f(4) = 42 = 16. Students must soon become comfortable with substituting entire algebraic expressions in for the argument ofafunctionandevaluatingtheoutput. Forinstance,whencalculatingthederivativeofafunction,students will need to evaluate expressions like f(x+a) for a variety of functions. For instance, if f(x) = x2+2x+3, to calculate f(x+a) we must substitute x+a for the value of x in the original function. This gives us: f(x) = (x+a)2+2(x+a)+3 f(x+a) = (x+a)(x+a)+2x+2a+3 f(x+a) = x2+2ax+a2+2x+3 Because polynomials are usually grouped into like terms, and the letter “a” in this case is a constant (i.e. not a variable), we would rewrite this expression as: f(x+a) = x2+(2a+a2)x+a2+3 1 www.ck12.org In this process, we have done nothing more than apply the rules of algebra to our function, but the process of evaluating functions with algebraic expressions as arguments will be unfamiliar to many students. Many will try to use some sort of shortcut to avoid expanding terms as necessary. An important distinction should be drawn between the terms “function” and “equation”, and how the graphical representation of a function can help us to solve an equation. For example, the table on page 1 displays the output values for f(x) = x2 when evaluated for different values of the input variable x. This enables us to graph the function on the x−y plane for any value of x, as seen on the top of page 2. Alternatively, when we consider an equation with x2 in it, for instance x2 = 4, we are asking for the specific point on the graph of f(x) = x2 that equals 4. Instead of the expression containing the dependent variable y(or f(x)), wearesubstitutingaspecificnumericalvalueforthedependentvariable, anddeterminingwhat value or values of the independent variable x that satisfy this condition. Whereas a function represents the general rule to calculate the value of f(x) for any input value, an equation asks for the value or values of the input that yields a specific output value for y. It therefore usually only has a finite set of answers. On a graph, an equation corresponds to particular points on the curve we have drawn, whereas a function refers to the entire curve. In the case of x2 = 4, there are two points on the parabola where f(x) or y equals 4, so there exist two answers to this equation: x = +2, and x = −2. Similarly, when calculating the y−intercept of a function, such as y = 2x+3, we are asking for the y−value when x = 0. So we would substitute the value 0 for x, and arrive at the equation 2(0)+3 = y which tells us the value of y when x equals 0, called the y-intercept. In this case, the y−intercept is 3. Thegraphonpage3illustratestherelationshipsbetweengraphsandequations,bysettingthevaluesoftwo functions equal to each other. If one function is represented by f(x) = 2x+3, and the other is represented by g(x) = x2+2x−1, to find the points where the graphs of these two curves intersect entails finding the place where f(x) = g(x) for a given value of x. To determine these values, we write the equation 2x+3 = x2+2x−1 and solve for the values of x where this equation is true. This example requires using techniques for solving quadraticequations,asshownonPage3. Againitturnsoutthattherearetwoanswersfor x,corresponding to the two points of intersection for the graphs of the functions f(x) and g(x). Although much of the notation introduced in this and subsequent lessons is very formal, it is important to stress that functions are important because they enable us to model a number of real world phenomena. In the exercises for this chapter, the relationship between the independent variable x, and the example of modeling costs using both linear and nonlinear functions, is emphasized. By using functions to model real world phenomena, we find that properties of functions like slopes and intercepts correspond to actual real world phenomena, like break even points, fixed and variable costs, as well as velocity and acceleration. Lesson 2: Relations and Functions In Lesson 2, the more formal definition of a function is introduced, as are the topics of Domain and Range which provide useful information for analyzing and graphing functions. The classic definition of a function when displayed graphically is that it is a curve on the x−y plane that must satisfy the “vertical line test”, e.g. if you draw a vertical line through the function, it touches the graph in at most one point. This test ensures that for any value of x, there is at most only one value that our function evaluates to for this input. This is sometimes referred to as being “onto” or “surjective”. www.ck12.org 2 The example given in this Lesson of a common graphical representation which is NOT a function is the graph of a circle. There are clearly places where if we were to draw a vertical line on the coordinate plane, it would cross the circle twice. Although the rationale behind this isn’t explained in this Lesson, it may be helpful for students to be shown why an equation like x2 +y2 = 4 will not be a function, whereas an equation like x2+y = 4 does turn out to be a function. The answer becomes clear if we were to isolate the variable y: x2+y2 = 4 y2 = 4−x2 √ y = ± (4−x2) By isolating y, we see that for a given input value of x, there can be two values of y due to the plus/minus in the square root. Because there are two output values for only one input value, this is NOT a function, and thus does not pass the vertical line test. When treating functions in the context of the x−y plane, it often appears that the variety of curves that are functions is very limited, since there are a number of interesting curves which don’t satisfy the vertical line test. These include the circle graphed in the text, and the spiral graphed below. Can the techniques we develop to analyze functions be applied to these non-functions? Although it is outside the scope of this textbook, most students will have been introduced to the concept of a “parametrized curve” in their pre-calculus course. Parametrizing a curve enables us to consider curves that are not functions, like the circle or spiral, and represent them AS functions so that we can analyze them using function-based techniques. This entails creating a new variable, or parameter, and re-writing our expressions for the x− and y−coordinates of our curve using this parameter. For instance, if we were to create a new variable named t, referred to as our parameter, we could describe the circle in this Lesson using the equations: x = cos(t);y = sin(t);0 ≤ t ≤ 2(cid:25) In this case, both of our “parameterized” equations ARE functions: cos(t) and sin(t). By using techniques 3 www.ck12.org like parameterization, we can transform curves that are not functions into representations which ARE functions. This dramatically increases the class of curves and graphs which we can analyze. The bulk of this lesson is devoted to reviewing the topic of a function’s Domain and Range, which define the values of x and y over which a given function extends. Determining the domain of a function is usually much easier for students than finding its range, since there are only a finite number of situations where we cannot evaluate a function at a given x−value. The two most common are dividing by zero and taking the square root of a negative number. In looking at a function to determine its domain, most often we are simply looking for cases where a particular value of x will lead us into one of these conditions of undefinedness, and exclude those values. Take, for instance, the example of the rational function f(x) = 1 given on page ###. In determining the x values of x for which this function exists, clearly we must exclude the value x = 0 since we are not allowed to divide by zero. Since there are no other opportunities for our equation to be undefined through either dividing by zero or taking the square root of a negative number, this is the only point excluded in our domain. We can therefore define the domain as: D = {x|x , 0} Similarly, if we were to look at the rational function: 1 f(x) = (x−2)(x+3) the denominator in this expression will be equal to zero when the product (x−2)(x+3) = 0. This happens when x = 2 or x = −3. In this case, our domain includes all values of x except for x = 2 and x = −3. The determination of a function’s range is much more complicated, since it often requires a great deal of intuition into the behavior of algebraic expressions to understand which values a complicated function can and cannot take. For instance, terms in polynomials which raise x to an even power will always be positive, and the sine or cosine of a variable will always range between −1 and 1. An excellent process to help students identify the range of a function, particularly one that has many terms, is to look at the range of the individual terms, and combine them through logical reasoning to determine the range of the entire functions. For example, consider the following function: f(x) = x2+cos(x) Can we determine what values f(x) will take by just looking at this expression? Looking at the first term, we know that the range of x2 is always greater than or equal 0, since x2 can never be negative. Moving to the next term, cos(x), which we know cosine is always between +1 and −1. Combining these facts, we see that f(x) can never get less than −1, but can grow positively as large as we want due to the term, x2. We can therefore say that the Range of this function is at the very least f(x) > −1, since f(x) can only get as small as −1. Though it turns out that the range is actually more restricted than this, this type of reasoning provides students with an example of bounding a range to a particular interval. To many students, understanding the domain and range often becomes formulaic, with little or no moti- vation as to why these terms are important. In keeping with the importance of understanding both the practical applications of functions, and being able to graph functions and identify functions from their graphs, there are two ways to stress the usefulness of determining the domain and range. First, in many situations in physics, engineering and the natural sciences we derive equations for quantities like cost, weight or distance utilizing functions and algebraic expressions. Understanding the properties of an an- swerweattainforsuchquantities, likeitsdomainandrange, enablesustocheckthevalidityofoursolution through physical intuition. www.ck12.org 4 For instance, when using a function to calculate a quantity that must be strictly positive, like height or weight, if we are using a function whose range contains negative values, we should be wary. In some instances, this is a sign that we have improperly modeled the physical situation at hand. In others, it is a sign that the function we have computed is only valid on certain intervals, for example those values of x which make the function positive. In order to remain consistent with the reality of the physical situation when our range extends to values that seem impossible, we often “restrict the domain”, meaning that we exclude the values of the input variable which lead to the impossible values of the range. This Lesson contains a showcase of the graphs of many important types of functions that we will encounter throughout this textbook. A student should be able to identify these graphs quite easily in the first few weeks of class. They should be able to determine the intercepts and locations of important features of a function, such as the focus of a parabola, the center of a circle, and the domain of the logarithmic function. Most, if not all, of these topics should be review, but a strong understanding of these fundamentals will be important to developing the more complicated topics in this book. Inanticipatingthenextfewlessonsonlimitsandderivatives, itmightbehelpfultohavestudentsrecognize that the function y = |x| is unique amongst the functions showcased. All of the other functions except |x| are smooth, meaning that they have no sharp corners or breaks in them. The absolute value of x has the sharp corner at x = 0, which is an example of a function having a point where the slope approaching from one side isn’t equal to the slope when approaching from the other side. If one were to graph the slope of the function f(x) = |x|, soon to be referred to as its derivative, we would find that the value x = 0 would be excluded, making the derivative of |x| a discontinuous function. The final topic brought up in this chapter is function transformation. This is an important technique for interpreting functions that arise in modeling physical situations to understand the behavior of sys- tems without graphing them. Transformations allow us to take a prototypical function, like one of the 8 showcased in the textbook, and alter their shape to get many different versions of them on the x−y plane. For instance, consider the parabola given by the formula f(x) = x2. What if we wanted to move the graph of our parabola to the right by 3 units? As explained in this lesson, a rightward shift of 3 would be enacted by subtracting 3 from our variable, so instead of f(x) = x2, we would get f(x) = (x−3)2. The graphs of these functions are shown below. An important example of where the shift transformation arises in a physical contexts the solution to the Wave Equation in two dimensions. In that case, if we were to start a wave on the middle of a string that had a particular shape f(x), we would get two copies of that wave, each half in amplitude, that move in opposite directions. This can be written as 5 www.ck12.org 1 1 v(x) = f(x+ct)+ f(x−ct) 2 2 This expression tells us that we have two copies of our original function f(x), divided in half in amplitude, with one copy shifted to the left by the product of c, the wave speed, and t, the time elapsed, and the other copy shifted similarly to the right. As time gets bigger, this shift grows larger, representing the wave moving away from its original position, and traveling along the string. Transformations can create tremendous confusion for students because they appear in some ways the opposite of what one would expect. Take, for instance the shift of the function f(x) to f(x−c). Many students will think that because we are subtracting c, this corresponds to a shift in the negative direction. However, as we see above, by subtracting a constant c, we actually shift the function in the positive direction. ( ) Similarly, if we consider the transformation f(x) to f x , we might expect our original graph to be com- 2 pressed by a factor of 2, since we are dividing by 2; conversely, if we consider the transformation of f(x) to f(2x), we might expect our graph to be expanded by a factor of 2. Ineachofthese3casesoffunctiontransformation, theoppositetowhatseemsimmediatelyapparentturns out to be true. If we transform our f(un)ction f(x) to f(x−c), we are shifting our function to the right by the value c. Transforming f(x) to f x expands our original function by a factor of 2, and transforming 2 f(x) to f(2x) compresses our original function by a factor of 2. These caveats should be emphasized at this stage to ensure that a student is able to easily identify how to graph common functions which have been transformed through these standard operations (called a shift, dilation and contraction, respectively). The rationale for these operations can be deduced algebraically. Lesson 3: Modeling Data with Functions In this lesson, students use their graphing calculators to find curves which best approximate a set of data points on a scatterplot. This technique is often referred to as “regression” or “curve-fitting”. Unlike traditional treatments of regression in statistics classes, which often focus exclusively on the topic of linear regression, Lesson 3 shows students that different sets of data are often best fit by a variety of different functions,dependingonthevisualcharacterofthescatterplot. Thoughalinearapproximationissometimes thebestapproximation(andmostoften,thesimplest),illustratingthatwecanalsomodeldatausinghigher order polynomials, trigonometric functions and transcendental functions may be new to many students. The handling of real world data, even sets as small as provided in this Lesson, is usually handled by a computer or calculator since the calculations involved in determining the curve of best fit can be quite cumbersome. In this Lesson, calculating curves to fit the data is performed through both a graphing calculator as well as using Excel, and both are skills that a student should become comfortable with. It is important, however, to ensure that students understand the underlying reasoning their calculator is using to calculate curves of best fit since the criteria we can use to measure “best fit” can be interpreted very differently. In the examples given, the lines of best fit are calculated by minimizing the least square error between the curve and the data points. This meaning that if we were to add up the distance squared between the curve selected by our curve fitting technique, and all of the data points, the curve that is selected will provide the smallest value for the sum of the squared error. ∑M error = (f(x)−y)2 i i=1 www.ck12.org 6