Derivative, maxima and minima in a graphical context

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Derivative, maxima and minima in a graphical context a

Antonio Rivera-Figueroa & Juan Carlos Ponce-Campuzano

a

a

Mathematics Education Department, Centre for Research and Advanced Studies, Av. Instituto Politécnico Nacional 2508, Col. San Pedro Zacatenco, Mexico City, 07360 Mexico Version of record first published: 22 Jun 2012.

To cite this article: Antonio Rivera-Figueroa & Juan Carlos Ponce-Campuzano (2012): Derivative, maxima and minima in a graphical context, International Journal of Mathematical Education in Science and Technology, 44:2, 284-299 To link to this article: http://dx.doi.org/10.1080/0020739X.2012.690896

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284

Classroom Notes

Derivative, maxima and minima in a graphical context Antonio Rivera-Figueroa* and Juan Carlos Ponce-Campuzano Mathematics Education Department, Centre for Research and Advanced Studies, Av. Instituto Politećnico Nacional 2508, Col. San Pedro Zacatenco, Mexico City, 07360 Mexico

Downloaded by [UQ Library] at 18:49 10 March 2013

(Received 24 September 2011) A deeper learning of the properties and applications of the derivative for the study of functions may be achieved when teachers present lessons within a highly graphic context, linking the geometric illustrations to formal proofs. Each concept is better understood and more easily retained when it is presented and explained visually using graphs. In this article, we explore the conditions of necessity or sufficiency of the criteria for determining the maxima and minima of a function. The implications for the teaching of derivatives and functions in undergraduate courses are discussed in light of our analysis of textbooks. Keywords: undergraduate courses; calculus; derivative; critical point, maxima; minima

1. Introduction The concept of the derivative has a geometric origin. While the theory of the derivative can be constructed with few geometric illustrations or even without them, geometric figures are a source of ideas for formal proofs. The formal treatment of the derivative and their geometric interpretations are inseparable in the development of the theory and in the process of understanding. In undergraduate calculus courses, the learning of differential calculus presents a myriad of difficulties for students. The literature confirms what many teaching calculus experience, students can acquire the skills for performing calculations while maintaining shallow understanding and, in some cases, fundamental misconceptions (see, e.g. [1–6]). In the middle of the last century, Prenowitz [7] attributed the difficulties of the learning of calculus to the concepts, techniques and applications often involved. For Prenowitz, novice calculus students experience difficulties when they try to understand and apply the concepts of calculus for solving problems. More recent research attributes the origin of these difficulties in teaching itself, that is, the performance of the teacher [8]. A deeper learning of the properties and applications of the derivative to the study of functions may be achieved when teachers present lessons within a highly graphic context, linking the geometric illustrations to formal proofs. In fact, when students are first learning the derivative and its application to functions, they often lack the mathematical maturity to grasp the formal proofs of the results, for example the mean value theorem. For this reason, the teaching of calculus, at the undergraduate

*Corresponding author. Email: [email protected] http://dx.doi.org/10.1080/0020739X.2012.690896

International Journal of Mathematical Education in Science and Technology 285


Figure 1. The relationship between the sign of the derivative and the monotonicity of a function.

level, should rely heavily on graphics specially designed to communicate the ideas that we wish to teach. An example of a theorem, whose proof is based on the mean value theorem, states that if the derivative of a function f is positive (negative) on an interval, then the function f is increasing (decreasing). The relationship between the sign of the derivative and the monotonicity of a function (increasing or decreasing) are interpreted, in textbooks, with images like the ones showed in Figure 1 (see, e.g. [9–14]). The converse of the above theorem is false, a counterexample is the function f ðxÞ ¼ x3 . This function is not only increasing but strictly increasing, that is, the next implication is true x1 , x2 2 R,

x1 4 x2 ) f ðx1 Þ 4 f ðx2 Þ:

However, it is false that f 0 ðxÞ 4 0 for every real number x, since f 0 ð0Þ ¼ 0. A converse version of this theorem is: if a function is increasing and differentiable on an interval, then its derivative is non-negative. One difference between the two theorems is that the latter can be proved directly using the definition of the derivative, while the former requires the mean value theorem. The next statements incorporate both of them (the text above the arrow indicates on what the proof is based): f 0 ðxÞ 4 0 on I

Mean Value Theorem

)

f increasing on I

f increasing and differentiable on I

definition of derivative 0

)

f ðxÞ 0 on I:

Of course, the following implications are valid: f 0 ðxÞ 0 on I ) f increasing on I f 0 ðxÞ 4 0 on I ) f strictily increasing on I: This logical–conceptual difference concerning the nature of the demonstration cannot be perceived in a geometric instruction. Notwithstanding, the graphics are highly recommended for the understanding of various concepts and results about the derivative. We must be aware of their means and limitations and we must be careful when we do interpret the graphs of functions.


286

Classroom Notes

Interpreting graphically a result concerning the derivative can lead to mistaken knowledge. In general, the graphs we can draw with pencil or even visualize are far from showing the generality of the geometric and analytic situations. In practice, the particular graphic representations on which we rely to interpret and assimilate the results of the derivative may limit our understanding and thus lead us to formulate invalid results. Misconceptions held by students can be attributed to misconceptions taught by teachers or to those teachers with accurate knowledge but who are not reflecting on how students are learning. For example, it is common that for finding the maxima and minima values of a function, the students proceed to find the critical points of the function, that is, the points where the derivative vanishes. Usually, when a student applies the second derivative criterion and finds that it vanishes at the critical point, he or she then concludes that there is an inflection point. They do not consider the possibility that even if the second derivative is equal to zero at the critical point, the function could have a maximum or a minimum. This conclusion falls within a misconception that at a critical point one of the three things happens: (i) there is a minimum, (ii) there is a maximum or (iii) there is an inflection point. However, inflection points have nothing to do with the fact that there are, or are not, critical points. Why do students associate inflection points with critical points? This is because usually in the textbooks it is common to find the function f ðxÞ ¼ x3 as an example to show that the condition f 0 ðx0 Þ ¼ 0 is not sufficient to assure that this function has either a maximum or a minimum. In this case the derivative is given by f 0 ðxÞ ¼ 3x2 so that it vanish at x ¼ 0, but it has neither a maximum nor a minimum at this point. On the other hand, it is common to find too, the observation that at this point the function has an inflection point. This may lead students to believe that the inflection points should be looked for at the critical points, that is, at points where the derivative vanishes. Are there functions such that at some critical point have neither a maximum nor a minimum nor an inflection point? The answer is yes, for example, the function ( x4 sin x1 if x 6¼ 0 HðxÞ ¼ 0 if x ¼ 0 has a critical point at x ¼ 0. H is twice differentiable at every real number. Moreover, H0 ð0Þ ¼ 0 and H00 ð0Þ ¼ 0. Nevertheless, H has neither a maximum nor a minimum nor an inflection point at the critical point x ¼ 0. The first derivative criterion provides sufficient conditions to determine the maximum and minimum of a function. The commonly used figures to illustrate this criterion are often simple and eloquent ([9–14]), to the extent that appears, perhaps, to not necessarily be a formal proof (Figure 2). Certainly, the figure helps us to understand and learn the criterion, let us say that its statement totally responds to our intuition. However, the first derivative criterion has a complexity that we cannot perceive in this figure. The figures may communicate the false idea that the theorem is trivial, easy to articulate and to implement. In fact, in practice there may be serious difficulties in applying the first derivative criterion, because this requires determining the sign of a function (the derivative function) in an interval that should be chosen appropriately, when such an interval exists.



Figure 2. Sign of the first derivative around a maximum.

A subtle fact about the first derivative criterion, which certainly stands against our intuition, is that the condition of the sign of the derivative is sufficient but not necessary. For instance, a differentiable function can have a maximum value at a point x; however, it may happen that the derivative is not positive, for the points belonging to any neighbourhood on the left side of the point x. Without any doubt, our intuition and a graph like the above (Figure 2) can lead us to believe that the condition is necessary and sufficient. In the next section, we extend this discussion and provide a counterexample. The second derivative criterion has an advantage over the first derivative criterion as it avoids the problem of determining the sign of a function in an interval. With the second criterion, it is sufficient to determine the sign of the second derivative at a single point. In this sense, it is easier to apply the second derivative criterion than the first derivative criterion, but the second derivative does not always exists, which is either a drawback or a disadvantage. In such cases, the first derivative criterion could be the only option. In this article, we have raised points that are not often considered in textbooks. The first being the concept of the derivative and its geometric interpretation. The second being the conditions of necessity or sufficiency of the criteria for determining the maxima and minima of a function.

2. The process of differentiation We begin our discussion with the definition of the derivative with the geometric interpretation being discussed in the next section. We note that the derivative is a number associated to a function at a point in its domain when the conditions of existence are given: f 0 ðaÞ ¼ lim

x!a

f ðxÞ f ðaÞ : xa

The differentiation is not a process of obtaining a formula (the derivative function) from another source (the function).


288

Classroom Notes

If we conceive the process of differentiation in this way, it can lead us to perform incorrect calculations for obtaining the derivatives in particular situations. Clearly, in a learning environment, this is an obstacle for understanding the concept. The differentiation is a process of obtaining the derivative of a function at each point. The derivative function is obtained as a result of this process rather than as a result of applying a set of algorithmic rules to a formula. Let us call this latter process the formal application of rules of derivation. The derivative function has a domain which is determined previously by finding the points where the function is differentiable. The domain is determined a priori not a posteriori. The domain cannot be determined by getting a function (the so-called derivative function) which was obtained by the formal application of differentiation rules and by determining subsequently the points where the function is defined. This is a misconception usually adopted by students that can be attributed to instances where teaching occurs as described above. pffiffiffiffiffiffiffiffiffiffiffiffiffi For example, let the function f ðxÞ ¼ 1 x2 be defined on 1 x 1, that is, in the closed interval ½1, 1. If we apply formally the differentiation rules, we obtain x f 0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi : 1 x2 It is easy to see that its domain is the open interval 1 5 x 5 1. Outside this interval f 0 ðxÞ is not defined because for jxj 1, we have 1 x2 0. From the formula for f 0 ðxÞ it is often concluded that the function f ðxÞ is differentiable in 1 5 x 5 1 but not on the extreme points of the interval, that is, 1 and 1. The conclusion is correct but the argument is not valid. As another example, consider the function 1 1 : hðxÞ ¼ log 1 þ x2 2 If we apply formally and without consideration1 the differentiation rules to the function hðxÞ, we get h0 ðxÞ ¼

4x : ðx2 þ 1Þðx2 1Þ

The domain of h0 ðxÞ consists of all x not equal to 1 or 1, but this set is not the domain of the function hðxÞ. For instance, the function h0 ðxÞ is defined at x ¼ 2, but the function hðxÞ is not defined at this point, and therefore it makes no sense to compute the derivative of h at x ¼ 2. In other words, the derivative of h does not exist at x ¼ 2. In this case, h0 ðxÞ is not the derivative function of hðxÞ. The above example is a sophisticated form of the trivial case g(x) ¼ log x. This function is defined for all positive real numbers, but it is not defined neither at zero nor for negative numbers. However g’(x) ¼ 1/x is defined for all x not equal to 0, that is, g0 (x) is defined for negative numbers even when the derivative of g(x) does not exist at those points (see Figure 3). In addition, a function h, whose derivative is h0 (x) ¼ 1/ x, is given by h(x) ¼ log|x| which is derivable at every x6¼0. Another example is the function f(x) ¼ log(1/(1+x2) 1). The application of differentiation formal rules leads us to f0 (x) ¼ 2/(x(x2 þ 1)). This function is defined for all x not



Figure 3. (a) Graph g(x) ¼ log x; (b) graph g0 (x) ¼ x1.

equal to 0, however f (x) is not defined at any real number. We might say that the domain of f (x) is the empty set. Consider the absolute value function in relation to the above-mentioned function but from a different perspective. The function f ðxÞ ¼ jxj is a typical example used in the teaching of elementary calculus in order to illustrate the non-differentiability of functions at a point. In this case, f ðxÞ is differentiable for all x 6¼ 0 and its derivative is given by

1 if x 4 0 : f 0 ðxÞ ¼ 1 if x 5 0 Students usually attribute the non-differentiability at a point as a result of the graph having a peak at that point. Hence, the non-differentiability is interpreted as the existence of peaks on the graphs. Now consider the next function

2 if x 0 x : gðxÞ ¼ xjxj ¼ 2 x if x 5 0 In this case, although the students have had the experience of finding the derivative of the absolute value function, some of them do not know how to proceed. Even without making calculations they assume that the function is not differentiable at x 6¼ 0 due to the presence of jxj. Some students proceed by finding the formal derivative, that is

2x if x 0 0 : g ðxÞ ¼ 2x if x 5 0 Fortunately, or maybe we should say unfortunately, they get the right answer despite its lack of analysis and procedural performance characterized by unreflective application of formulas. However, it is common that students do not realize that the point x ¼ 0 deserves special treatment. Figure 4 illustrates the graphs of gðxÞ and g0 ðxÞ. The above examples are particular cases of piecewise-defined functions. The practice of finding the derivative of a function without reflection is revealed when we ask students to find the derivative of such functions. As another example, consider the next two functions

290

Classroom Notes


Figure 4. (a) Graph g(x) ¼ x|x|; (b) graph g0 (x).

Figure 5. Two different piecewise-functions: (a) graph f (x); (b) graph g(x).

83 2 x 1 > > > : ðx 1Þ3 þ 52

if 1 x 1 if 1 x 2 if 2 x 3

;

8 x3 1 if 1 x 1 > > > : 2x þ 2 if 2 x 3

If we ask students to find the derivative of every function, the common answer is 8 8 3x if 1 x 1 3x2 if 1 x 1 > > > > < < f 0 ðxÞ ¼ 3 if 1 x 2 ; g0 ðxÞ ¼ 2 x þ 12 if 1 x 2 : > > > > : : 3ðx 1Þ2 if 2 x 3 2 if 2 x 3 Students usually operate with differential formal rules in every expression. However, f 0 ðxÞ is certainly the derivative function of f ðxÞ but g0 ðxÞ is not the derivative function of gðxÞ. We can understand these situations by looking at the graphs of the functions f ðxÞ and gðxÞ (Figure 5). When we analyze the differentiation of a function of this kind, we should consider the extreme points of the corresponding intervals to the different formulas involved in the definition. At those extreme points, the left and right derivatives as well as the continuity of the function must be considered.

International Journal of Mathematical Education in Science and Technology 291 3. The derivative as slope of a tangent line The derivative at a point in the domain of a function is interpreted as the slope of the tangent line at the corresponding point on the graph. This interpretation is based on the interpretation of the difference quotient f ðxÞ f ðaÞ xa


related to the slopes of secant lines to the graph. According to the above meaning of the derivative, we can point out two observations: (a) We give this meaning to the derivative based on a process involving a limit of geometric objects, such as the secant lines. (b) We give the meaning of the derivative as a slope of a tangent line, assuming implicitly that we have the concept of tangent line to a curve. This assumption is trivialized for curves, which are usually used as an illustrative purpose (Figure 5). The curves are usually simple and in most cases it is not necessary to give any explanation of the meaning of tangent line. It is enough to give an intuitive idea and the visual image for showing that the straight line ‘touches’ the curve at a point. Then it is accepted, without any doubt, that the derivative corresponds to the slope of a tangent line. Hence it is natural that we establish the tangent line, to the graph of the function f at the point ða, f ðaÞÞ, given by the equation y ¼ f 0 ðaÞðx aÞ þ f ðaÞ. In a first course of calculus, in general, teachers and students are not aware that the derivative defines the tangent line, which it is defined through the previous equation. For this reason students are surprised and reluctant to accept the relationship of tangency for curves of more complex forms, this concept of tangent line is much simpler. For example, the function ( 1 2 x sin x1 þ x if x 6¼ 0 hðxÞ ¼ 10 0 if x ¼ 0

Figure 6. Graph of hðxÞ with the tangent line at the point ð0, 0Þ.

292

Classroom Notes

is differentiable at x ¼ 0, so the graph has defined the tangent line at the point ð0, 0Þ and that is precisely the line whose equation is y ¼ x. However, the geometric situation is not easily accepted by the student when he looks at the graph and its tangent line (Figure 6). In this example, the tangent to the curve at ð0, 0Þ does not correspond to the intuitive idea that we use to give meaning to the derivative, in a first calculus course, it is difficult to imagine the meaning that this definition may have.

4. First criterion for maxima and minima


Introductory calculus courses deal with the derivative and its relationship with the maxima and minima of a differentiable function and which is simply established as follows: If a differentiable function f has a maximum or a minimum value at a point a, then f 0 ðaÞ ¼ 0.

This is a first criterion for determining the maxima and minima of a function. In other words, this theorem says that a necessary condition for a differentiable function, having a maximum or a minimum at a point, is that its derivative at that point is zero. To justify this criterion, textbooks usually refer to two figures is shown in Figure 7. The formal proof is very simple and it does not require the mean value theorem as required by other criteria for maximum and minimum, but in a first calculus course may be sufficiently taught via the graphical interpretation. A point where the derivative is zero, the tangent to the graph is horizontal; intuitively this occurs at points where there is a maximum or a minimum value. The above statement may be appropriate for the teaching of calculus in a first undergraduate course; however, it is incomplete and inaccurate because, at first glance, some conditions are not explicit. In that statement, if the domain of the function is an interval of the form ½, , then the point a, where it reaches a maximum or minimum, can be neither nor , i.e. 5 a 5 . If a is one of the endpoints then is not necessarily f 0 ðaÞ ¼ 0. If a is a point on the domain ½, that satisfies 5 a 5 , we say that a is an interior point of the domain f. If the domain of a function consists of all real numbers, then every point is an interior point. In general, a point a is an interior point of the domain of a function, if it belongs to some interval c 5 x 5 d that is contained in the domain. The condition that the point a is an interior point of the domain is important for what we have called the first criterion for maxima and minima. As an example, let the function be pffiffiffiffiffiffiffiffiffiffiffiffiffi f ðxÞ ¼ ðx3 x2 2x þ 1Þ 1 1 x2 :

Figure 7. Maximum and minimum.



Figure 8. (a) Graph of f ðxÞ and (b) graph of f ðxÞ with other scale.

The domain of this function is the closed interval ½1, 1 and f is differentiable at every x such that 1 5 x 5 1. The formula for f ðxÞ does not allows us to define it out of the interval ½1, 1. Now, its derivative is given by pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 ð3x2 2x 2Þ þ 4x4 3x3 7x2 þ 3x þ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffi f 0 ðxÞ ¼ : 1 x2 This derivative can also be written as pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 1 x2 2 3x2 2x þ 4x4 3x3 7x2 þ 3x 1 x pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi f 0 ðxÞ ¼ þ : 1 x2 1 x2 Clearly, f 0 ðxÞ vanishes at x ¼ 0. So x ¼ 0 is a critical point of f ðxÞ. In fact, this function has a minimum at this point with a value f ð0Þ ¼ 0. But we can verify without difficulty that f ð1Þ ¼ 1. So f ð0Þ ¼ 0 is not the minimum of f ðxÞ. The minimum value of f ðxÞ is precisely f ð1Þ ¼ 1. In Figure 8, we observe a change of scale in the y-axis, which allows us to observe that the function actually has a minimum value (local or relative) at x ¼ 0. We can also see from the graph of the function that f reaches a maximum value at some point in the interval 0 5 x 5 1, but the maximum value is not the absolute maximum of f ðxÞ. The function reaches its absolute maximum at x ¼ 1. In short, the function f ðxÞ reaches its maximum value at x ¼ 1 and its minimum value at x ¼ 1, but at these points the derivative is not zero and, in fact, f is not even a differentiable function at these points.

4.1. Criterion of the first derivative (CFD) Since the condition f 0 ðaÞ ¼ 0 is not sufficient for a function f ðxÞ to have a maximum or minimum at the point a, there is the necessity to complement the first criterion and thereby ensure that at a critical point, which meets extra conditions, there is a maximum or minimum. A complementary proposition is called the CFD which we establish as follows: If the derivative of a function f ðxÞ is positive on the left side of a critical point and negative on the right side, then f ðxÞ has a maximum at that critical point. If, on the

294

Classroom Notes

Figure 9. Graph of f ðxÞ.


other side, the derivative is negative on the left side and positive on the right side then f ðxÞ has a minimum at the critical point.

In practice, it is difficult to apply this criterion because this requires finding a suitable neighbourhood of the critical point where the derivative has an algebraic sign on the right side of the critical point and the opposite algebraic sign on the left side. Determining the algebraic sign of a function can be a difficult problem. For example, consider the function f ðxÞ ¼ 2 cos x þ x sin x: The derivative of this function is f 0 ðxÞ ¼ x cos x sin x: It is easy to verify that x ¼ 0 is a critical point, that is f 0 ð0Þ ¼ 0. However, to seek the algebraic sign of the derivative for x 4 0 and for x 5 0 near to x ¼ 0 can be a difficult task. The CFD is applied when, in theory, it is possible to find a neighbourhood of the critical point where the derivative has a sign on the left side of the critical point and opposite sign on the right side. So this criterion may fail for two reasons: (i) the derivative has the same algebraic sign to the left and right sides of the critical point and (ii) there does not exist a neighbourhood of the critical point such that the derivative has a single algebraic sign to the left or to the right sides. These two cases correspond to the negation of the premise in the statement of the criterion. In the case (i), if the derivative has the same sign throughout a neighbourhood of the critical point, with the exception of the critical point where it takes the value zero, then the function is increasing or decreasing in that neighbourhood, and therefore has neither a maximum nor minimum at the critical point. On the other hand, in the case (ii) the CFD is not applicable. An example for the latter case is the function, defined as follows: 8 > x2 : 2 if x ¼ 0

International Journal of Mathematical Education in Science and Technology 295 The graph of this function looks, at first sight, as shown in Figure 9. Looking carefully at the scale on the y-axis, the function has a maximum at x ¼ 0. It is easy to verify that the function is differentiable at x ¼ 0 and that f 0 ð0Þ ¼ 0. Indeed, 2 x 2 1 1þsin x f ðxÞ f ð0Þ ¼ lim lim x!0 x!0 x x x ¼ lim x!0 1 þ sin2 1 x ¼ 0:


The value of the limit follows from the inequality

x

0

jxj:

1 þ sin2 1 x Then f is differentiable at x ¼ 0 and f 0 ð0Þ ¼ 0. Since we have shown that x ¼ 0 is a critical point of the function, maybe we can prove that the function has a maximum value at this point by using the CFD. We should point out that on the graph the oscillations of the sine function cannot be perceived. If the scale is changed slightly, we will observe the expected oscillations (Figure 10). The infinite oscillations around zero are not an impediment for the function having a maximum at x ¼ 0. Actually, for every x 6¼ 0, we have x2 2 2 ¼ f ð0Þ: 1 þ sin2 x1 This means that f ð0Þ ¼ 2 is an absolute maximum of f, that is, f ðxÞ f ð0Þ for every real number. However the derivative takes, on the right side, positive and negative values at points as near to the critical point as desired. In fact, for every natural number n, we have ! 1 2 1 4 16 0 0 5 0 and f 4 0: f ¼ ¼ n þ 2 ð2n þ 1Þ 9 3ð4n þ 3Þ n þ 3 4 Therefore, it is not possible to apply the CFD.

Figure 10. Zoom-in of the graph of f near to (0, 2).

296

Classroom Notes

4.2. Criterion of the second derivative (CSD) A simple statement of this criterion is the following: Let f be a twice differentiable function on an interval and x0 a critical point. Suppose that f 00 ðx0 Þ 6¼ 0. (a) If f 00 ðx0 Þ 4 0, then f has a minimum at x0 . (b) If f 00 ðx0 Þ 5 0, then has f a maximum at x0 . Some authors used to prove the CSD based on the concept of concavity which must be defined previously. However, the criterion can be demonstrated in a simple manner without resorting to the concavity. Suppose, for example, f 0 ðx0 Þ ¼ 0 and f 00 ðx0 Þ 4 0. Since f 0 ðxÞ f 0 ðx0 Þ f 0 ðxÞ ¼ lim , x!x0 x!x0 x x0 x x0


f 00 ðx0 Þ ¼ lim we have

f 0 ðxÞ 40 x x0 for all points x sufficiently close x0 , both to the right and the left sides of x0 . If x lies on the left side, x x0 5 0, then we have f 0 ðxÞ 5 0. If x lies on the right side, x x0 4 0, then we have f 0 ðxÞ 4 0. Using the CFD, it follows that f ðxÞ has a minimum at x0 . Figure 11 contains the relevant information that allows us to recall the CSD. In Figure 11, we see explicitly the sufficient conditions, i.e. f 00 ðx0 Þ 5 0 and 00 f ðx0 Þ 4 0 for the function f having a maximum and a minimum value, respectively; however, these conditions are not necessary. A simple example that illustrates this fact is f ðxÞ ¼ x4 . In this case, f ðxÞ has a critical point at x ¼ 0 and also a minimum at this point too, however f 00 ð0Þ ¼ 0, so that, it is not necessary that the condition f 00 ð0Þ 4 0. This example shows us that the CSD fails because f 00 is zero at the critical point. However, the CFD is applicable. Another example is the function gðxÞ ¼ x3 . In this case g0 ðxÞ ¼ 3x2 for all real number x and the function has a critical point at x ¼ 0. Since g00 ð0Þ ¼ 0, the CSD fails, as well as the CFD, because the first derivative is positive at every point x 6¼ 0. Note that the CFD does not indicate what happens in the case where the derivative has the same algebraic sign to the right and left side of the critical point. We complement this criterion by adding that when the derivative has the same algebraic sign in a neighbourhood of the critical point (except at the critical point where the derivative is zero), the function has neither a minimum nor a maximum at that point.

Figure 11. Graphical representation for the CSD.



With this extended version, we can confirm that the above function has neither a maximum nor a minimum. The second derivative must exist and it must be different to zero at the critical point in order to apply the CSD. Thus, the criterion fails when either the second derivative does not exist or it is equal to zero. An interesting case is when there is no second derivative at the critical point. For example, the function

2 if x 0 x f ðxÞ ¼ xjxj ¼ x2 if x 5 0 has a derivative at all points, including x ¼ 0 and its derivative is given by f 0 ðxÞ ¼ 2jxj 4 0 for all x 2 R, however there is no second derivative, so the CSD cannot applied. This is a case similar to the function gðxÞ ¼ x3 in the sense that the CFD cannot be applied. However, the extended version of this criterion can be applied here. Another example similar to the above one is the function ( 3 3 if x 0 x2 GðxÞ ¼ jxj2 ¼ : 3 ðxÞ2 if x 5 0 This function has a critical point at x ¼ 0, it has a second derivative at every point except at the critical point, therefore the CSD cannot be applied but the function has a minimum at the critical point (Figure 12). In this case the CFD can be easily used because the first derivative is given by ( pffiffiffi 3 if x 0 0 2 x : G ðxÞ ¼ pffiffiffiffiffiffiffi 3 2 x if x 5 0 Note that the graph of GðxÞ might suggest that there is a second derivative pffiffiffiffiffi ffi at x ¼ 0, but G00 ðxÞ ¼ ð3=4Þð1= jxjÞ for all x 6¼ 0 and we can see that G00 ð0Þ does not exist. We also have that limx!0 G00 ðxÞ ¼ þ1, so we can write G00 ð0Þ ¼ þ1:

Figure 12. Graph of GðxÞ ¼ jxj3=2 .

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Classroom Notes

Finally, consider the functions ( x2 sin2 x1 if x 6¼ 0 GðxÞ ¼ 0 if x ¼ 0

( HðxÞ ¼

x4 sin2 x1

if x 6¼ 0

0

if x ¼ 0

:

On one hand, the function GðxÞ has a minimum at x ¼ 0, but G00 ð0Þ does not exist and therefore we cannot apply the CSD. On the other hand, HðxÞ is two times differentiable at the critical point x ¼ 0 and Hð0Þ is a minimum, but none of the criteria can be applied either.


5. Final remark In undergraduate calculus courses, a highly graphical context is recommended to further deeper understanding of various concepts and results about the derivative. However, as teachers, we should be aware that this context has limitations. With the examples discussed in this article, we try to emphasize this fact, which is often neglected in either textbooks or in the classroom.

Acknowledgements The authors would like to thank the reviewers for their valuable comments and suggestions to improve the quality of this article. Also, particular thanks to Kelly E. Matthews, Lecturer in Higher Education at the University of Queensland (Australia), for her insightful and thoughtful comments.

Note 1. Consider: To think about carefully. Merriam Webster Dictionary 2011.

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Computing bisectors in a dynamic geometry environment Francisco Botana*


Department of Applied Mathematics I, University of Vigo, Pontevedra 36005, Spain (Received 1 December 2011) In this note, an approach combining dynamic geometry and automated deduction techniques is used to study the bisectors between points and curves. Usual teacher constructions for bisectors are discussed, showing that inherent limitations in dynamic geometry software impede their thorough study. We show that the interactive sketching of bisectors and an automatic treatment of the algebraic problem involved can give a reasonable knowledge about them. Since some cases are currently out of computational scope, despite the simplicity of the bisector problem, we sketch an alternative method for dealing with them. Keywords: bisector; geometric loci; dynamic geometry

1. Introduction Most of Dynamic Geometry (DG) environments incorporate primitive predicates for computing the perpendicular bisector of two points or a segment, and the parabola defined by a point and its directrix. These cases are particular instances of general bisectors. The bisector of two geometric elements is the locus of a point that moves while remaining equidistant respect to those elements. Since its definition is so simple, it is not surprising that DG users, both teachers and students, frequently deal with bisector constructions. However, most of these constructions are limited to exhibit a graphical trace of the sought locus and there is no algebraic knowledge about them. The aim of this note is to perform a systematic discussion of the construction of bisectors between point and curves in a DG environment. Furthermore, some considerations about automatically obtaining their algebraic characterization are given. We also discuss the differences between answers coming from the DG environment and the symbolic ones. Section 2 deals with bisectors between point and general lines. We study and propose dynamic constructions for bisectors, and a symbolic method for computing the equations of objects containing the bisector is recalled. Section 3 describes the computation of curve/curve bisectors. Since the approach taken here is based on envelopes, we show that a closed-form answer is returned just for low-degree curves. *Email: [email protected] http://dx.doi.org/10.1080/0020739X.2012.690900