ExtremeValuesandShapeofCurves

ExtremeValuesandShapeofCurves Philippe B. Laval Kennesaw State University March 23, 2005 Abstract This handout is a summary of the material dealing with finding extreme values and determining the shape of curves. It corresponds to the material covered in 4.2, 4.3 and 4.4 of Stewart s Calculus. In this handout, the derivative is used as a tool to find information about functions in order to have a better understanding of the function. 1 Introduction Now that we have mastered how to compute the derivative, we turn toward some applications of the derivative. We concentrate on how the derivative can be used to study functions. In particular, we will see how the derivative can be used to find where a function is increasing, decreasing, if and where a function has maxima, minima. We first look at the general ideas involved so the reader will have an understanding of what we are trying to do and why it works. Then, we will look at the mathematics behind these ideas. Finally, we implement these ideas to answer the following questions: 1. Given a function f, where is f increasing, decreasing? 2. Given a function f, where is f concave down, up? 3. Given a function f, doesf have maxima, minima? If yes, where are they? 4. Given a function f, what is the shape of its graph? 2 Main Ideas Used In this section, we use the reader s intuitive definition of the various concepts presented. Formal definitions will be given when a formal mathematical explanation is given. 1

2.1 Increasing - Decreasing This is an issue we have already visited, when we first studied the derivative and its various meanings. The sign of f (x) tells us if f is increasing or decreasing. More specifically, if f (x) > 0 on an interval I, thenf is increasing on that interval. If f (x) < 0 on an interval I, thenf is decreasing on that interval. 2.2 Concave up - Concave Down This is an issue we have also already visited, when we first studied the derivative and its various meanings. The sign of f (x) tells us if f is concave up or down. More specifically, if f (x) > 0 on an interval I, then f is concave up on that interval. If f (x) < 0 on an interval I, then f is concave down on that interval. 2.3 Maximum - Minimum Given a function f, we are trying to find the maxima and minima of this function, if any. This is easy when we are looking at the graph of the function. It is more difficult if we just have the formula. The techniques we will develop will allow us to find the values of x where y = f (x) is maximum or minimum. The actual maximum or minimum will then be found by plugging in the value. For example, if we find that f (x) is maximum when x = c, then the maximum will be f (c). To better understand the idea behind the techniques we will develop, letusfirstlookatsomepictures. Figure 1: Local maximum, f (x) =0 We would like to find a set of conditions so that, depending on which condition is satisfied, we will know that the function has a maximum, a minimum or neither. A study of figures 1, 2, 3 and 4, suggests that a function f has a maximum or a minimum at c only if f (c) =0or f (c) is undefined. This might lead us to believe that a function f hasamaximumoraminimumif and only if f (x) =0or f (x) is undefined. However, figure 5 corresponds to a case where f (c) =0,yetf does not have a maximum or a minimum at x = c. 2

Figure 2: Local minimum, f (x) =0 Figure 3: Local maximum, f (x) undefined Figure 4: Local minimum, f (x) undefined 3

Figure 5: f (x) =0, no extreme value Figure 6: f (x) undefined, no extreme value 4

Similarly, figure 6 corresponds to a case where f (c) is not defined, yet f does nothaveamaximumoraminimumatx = c. So, it seems that the criterion f (c) =0or f (c) undefined is not a good one to find maxima or minima. But if we look closer, we see that there is hope. In the case of a maximum. If we compare figures 1 and 3 on one hand with figure 6 or 5, we can see that there is a difference between the cases where a maximum exists and the cases when one does not. When there is a maximum, the function changes direction. It is first increasing, then decreasing. We can make a similar observation in the case of a minimum. When there is no maximum or minimum, the function is always increasing, or always decreasing. We have an easy way to tell if a function is increasing or decreasing, the sign of f (x). It then appears that a correct approach to find the maxima and minima of a function would be: 1. Find values of x where f (x) =0or f (x) is undefined. These will be candidates for possible maximum or minimum. 2. Test these points further to determine which ones correspond to a maximum, a minimum or neither. We now look at the ideas mentioned above in more detail. In particular, we will mention some important theorems which make these ideas work. 3 Maximum and minimum values (4.2) 3.1 Definitions Let us begin by reviewing some elementary concepts about functions, which will be needed for this handout. Given a function y = f (x), the domain of f is the set of values of x also called the set of inputs. The set of values of f (x) or y is called the range or the set of outputs. When we talk about the maximum or the minimum of a function, we are talking about y values, in other words function values. The techniques we develop here, will help us find the values of x at which f has a maximum or a minimum. If c is such a value, then, the actual minimum or maximum will be f (c). Definition 1 (Open interval) An interval of the form (a, b) is called an open interval. Definition 2 (Closed interval) An interval of the form [a, b] is called a closed interval. Definition 3 (Half-closed interval) An interval of the form (a, b] or [a, b) is called a half-closed (or half-open) interval. 5

Definition 4 In the above intervals, the numbers a and b are called endpoints or boundary points. The points of an interval which are not on the boundary are called interior points. Remark 5 As we will see below, whether the domain of f is a closed interval or not, makes a difference when trying to find its maxima or minima. Remark 6 [a, b] contains its boundary points. (a, b) does not contain its boundary points. The interior of [a, b] is precisely (a, b). Definition 7 (Absolute maximum) A function f has an absolute or global maximum at c if f(c) f(x) for every x in the domain of f. The number f(c) is called the absolute maximum. Definition 8 (Absolute minimum) A function f has an absolute or global minimum at c if f(c) f(x) for every x in the domain of f. The number f(c) is called the absolute minimum. Definition 9 (Local maximum) Afunctionf has a local or relative maximum at c if there is an open interval I containing c such that f(c) f(x) for every x in I. Definition 10 (Local minimum) Afunctionf has a local or relative minimum at c if there is an open interval I containing c such that f(c) f(x) for every x in I. Remark 11 If we are studying a function f on a closed interval [a, b], thena local extremum never happens at one of the end points. It always happens inside the interval. An extremum (local or global) is either a max- Definition 12 (Extremum) imum or a minimum. Remark 13 As noted above, it is important to make the distinction between the extremum of a function (a y value) and the value of x where the extremum happens. Definition 14 (Bounded function) Let f be a function. 1. f is said to be bounded from above on an interval I if there exists a constant M such that f (x) M for every x in I. M is called an upper bound. 2. f is said to be bounded from below on an interval I if there exists a constant m such that f (x) m for every x in I. m is called a lower bound. 3. f is said to be bounded if it is both bounded from above and below that is if there exists two constants m and M such that m f (x) M for every x in I. 6

Figure 7 illustrates the various definitions above. We can see that f has local minima at x 2 and x 4. The local minima are f (x 2 ) and f (x 4 ). f has local maxima at x 1 and x 3. The local maxima are f (x 1 ) and f (x 3 ). However, f does not have a global maximum or minimum. We finish this section with a Figure 7: Extreme values theorem about the existence of extreme values. It is an important theorem in mathematics. We will not prove it. Theorem 15 (Extreme value theorem) closed interval [a, b], then: If f is a continuous function on a 1. f is bounded on [a, b] 2. f attains its absolute maximum and its absolute minimum on [a, b]. Remark 16 This theorem gives us conditions under which a function is guaranteed to have extreme values. It does not tell us how to find them. Continuity of the function, and the fact that the interval of study is closed are essential to this theorem. If we remove one of these conditions, the function may not have a global maximum or a global minimum as show figures 8 and 9. We now turn to finding local extreme values. 3.2 Local Extreme Values As we noticed in the preliminary discussion, it appears that good candidates for extreme values are points where either f (c) =0or f (c) is undefined. Because of their importance, such points are given a name. Definition 17 (Critical number) A critical number of a function f is a number c in the domain of f which satisfies one of the conditions below: 7

Figure 8: This function is not continuous on [a, b], it does not have a global maximum Figure 9: A continuous function on an open interval may not have a maximum or a minimum 8

1. f (c) =0 2. f (c) is undefined. If c is a critical number, then f (c) is a critical value, thepoint(c, f (c)) is a critical point. Remark 18 On the graph of a function, a critical number is easy to locate. It is a number where the function has either a horizontal tangent (f (c) =0), a vertical tangent (f (c) is undefined) or a corner point (f (c) is undefined). Remark 19 Note that a critical number is a number in the domain of the function. Example 20 Find the critical numbers of f (x) =3 x 2. To find them, we compute f (x) and find where it is either 0 or undefined. f (x) = 2x Clearly, f (x) is always defined. It is 0 when x =0. 0 is the only critical number. Example 21 Find the critical number of f (x) =2x 5 3 +5x 2 3. We proceed as in the previous problem. f (x) = 10 3 x 2 10 3 + 3 x 1 3 = 10 3 x 1 3 (x +1) = 10 3 x +1 We see that f (x) =0when x = 1 and f (x) is undefined when x =0. 0 and 1 are the critical numbers. The next theorem confirms the facts we had noted in the preliminary discussion. x 1 3 Theorem 22 (Fermat s theorem) f (c) exists, then f (c) =0. If f has a local extremum at c, andif Remark 23 Fermat s theorem is very important. It tells us where to look for in order to find local extreme values. Fermat s theorem can be rephrased as follows: If f has a local extremum at c, thenc is a critical number of f. In other words, when looking for local extreme values, one only has to look for critical numbers. Remark 24 However, Fermat s theorem does not imply that every critical point corresponds to a local extreme value. For example, y = x 3 has 0 as critical number. Yet, it has neither a Max nor a Min. at 0. Also, figures 5 and 6 illustrate that fact. 9

Therefore, to find local extreme values, one must perform the following: Proposition 25 Let f be a given function. To find the local extreme values of f, do the following: 1. Find the critical numbers of f. This will give all the possible candidates where local extremes can happen. Not all these values might be good candidates. 2. Test each critical number to see which ones are local maxima, local minima and neither. 5. The way the critical numbers are tested will be discussed in sections 4 and 3.3 Global Extreme Values The existence of global extreme values is more complicated in general. As noticed on figure 7, it is possible for a function to have several local extreme values, yet no global extreme values. For example, if we study the function on (, ), the function might go to ± as x ±. The If we study the functiononanopeninterval(a, b), the function might approach a higher value than the local maxima or a lower value than the local minima as x approaches the endpoints of this interval. However, the extreme value theorem tells us that if we are studying a continuous function on a closed interval, we are guaranteed the existence of extreme values. The theorem does not tell us where this absolute maximum and minimum are. They can be either one of the critical numbers, or they could be at the endpoints. Example 26 Find the global extreme values of f (x) =3x 4 16x 3 +18x 2 in the interval [ 1, 4] Because we are studying the function on a closed interval, we know it will have both a global maximum and a global minimum. Since these global extremes can either occur at the endpoints or at one of the local extremes, we find all these points, then compare them. First, we find the critical numbers and values. For this, we find where f (x) is either 0 or undefined. f (x) =12x 3 48x 2 +36x =12x ( x 2 4x +3 ) =12x (x 1) (x 3) So, we see that f (x) is always defined. Also, f (x) =0when x =0, x =1, x =3. The corresponding critical values are f (0) = 0, f (1) = 5 and f (3) = 27. Next, we evaluate the function at the endpoints. f ( 1) = 37, f (4) = 32. 10

In conclusion, the global maximum happens when x = 1. Itisf ( 1) = 37. The global minimum happens at 3. Itisf (3) = 27. When we are studying a function on an interval which is not closed, global extremes may not exist as the function may go to infinity as x ±. In this case, one approach is to sketch a graph of the function to see if it has global extremes or not. Another approach is to find the limit of f as x approaches the endpoints of the interval of study. We will revisit this in section 4. Summary: Finding global extreme values There are two cases to consider. 1. To find the global extrema of a continuous function f on a closed interval [a, b] This is the only situation under which we know there will be global extrema. Find all the critical numbers, call them c 1, c 2,..., c n. Compute f (a), f (b), f (c 1 ), f (c 2 ),..., f (c n ) The largest of the values found in the steps above will be the global maximum, the smallest will be the global minimum. 2. To find the global extrema of a continuous function on an open interval (a, b) (notethatwecouldhavea = or b = ) Find all the critical numbers, call them c 1, c 2,..., c n. Then, compute f (c 1 ), f (c 2 ),..., f (c n ). Let M be the largest such value and m be the smallest. Find lim x a+ f (x) and lim x b f (x) If M is larger than both limits, then M is the global maximum. Otherwise, f does not have a global maximum. If m is smaller than both limits, then m is the global minimum. Otherwise, f does not have a global minimum. 3.4 Things to know: Given a function, be able to find its critical numbers and its global extrema. Be able to do problems such as # 1, 3, 5, 7, 23, 25, 27, 35, 37, 39 on pages 276-278 Additional Problems 1. Draw the picture of a function which is not continuous on a closed interval but which has a global maximum and a global minimum. 2. Draw the picture of a continuous function on an open interval which has both a global maximum and a global minimum. 11

3. Draw the picture of a continuous function on an open interval which has a global maximum but no global minimum. 4. Explain how to find the global extrema of a continuous function f on a half-open interval, that is an interval of the form (a, b] or [a, b). 12

4 Using the First Derivative to Test Critical Numbers (4.3) 4.1 Theory: The first derivative is a very important tool when studying a function. It is important to know what kind of information it can provide as well as what it can t. Most of the results in this section are based on a very important theorem, called the Mean Value Theorem. Theorem 27 (Mean Value Theorem) If f is a differentiable function on the interval (a, b) and continuous on [a, b] then there exists a number c in (a, b) such that f f (b) f (a) (c) = b a or equivalently f (b) f (a) =f (c)(b a) This is a very important theorem in mathematics. Not only is it useful here, f (b) f (a) it is also used in many other parts of mathematics. The quantity b a corresponds to the slope of the secant line through the points (a, f (a)) and (b, f (b)). Geometrically, the theorem simply says that there exists a number c between a and b at which the slope of the tangent is the same as the slope of the secant line through the points (a, f (a)) and (b, f (b)). Looking at figure, this seems reasonable. Figure 10: Illustration of the Mean Value Theorem Definition 28 (Increasing/decreasing) A function f is called increasing on an interval I if f(x 1 ) <f(x 2 ) whenever x 1 <x 2 in I. Similarly, a function 13

f is called decreasing on an interval I if f(x 1 ) >f(x 2 ) whenever x 1 <x 2 in I. A function that is either increasing or decreasing is called monotone. Theorem 29 (Test for increasing/decreasing) on [a, b] and differentiable on (a, b). Suppose f is continuous 1. If f (x) > 0 for every x in an interval, then f is increasing on that interval. 2. If f (x) < 0 for every x in an interval, then f is decreasing on that interval. Though this theorem is intuitively easy to understand, proving it is a little bit more challenging. It requires the use of the Mean Value Theorem. The next theorem summarizes the discussion in section 2 on page 1. Theorem 30 (First derivative test) Suppose c is a critical number of a function f and that f is continuous at c. We have: 1. If f changes sign from + to at c, thenf has a local maximum at c. 2. If f changes sign from to + at c, thenf has a local minimum at c. 3. If f does not change sign, then f has neither a local maximum nor a local minimum. Figure 11 illustrates this theorem. Remark 31 The first derivative test has several advantages over other tests we will see later: It can be used to test every critical number. It always gives an answer i.e. we know for sure that a critical number is either an extreme value or is not. Remark 32 To find where a function is increasing/decreasing, or to test critical numbers involves roughly the same type of work since both are determined by knowing the sign of the first derivative. You should follow the following steps Compute the first derivative Find the critical numbers The critical numbers determine intervals on the real line. The sign of the first derivative is constant in each interval. Find the sign. Conclude depending on the sign of the first derivative. We illustrate the above procedure with several examples. 14

Figure 11: Illustration of the First Derivative Test 15

Example 33 Find where f (x) =x 3 3x 2 +1 is increasing, decreasing. For this first example, we explain every detail of the procedure involved. We know that finding where a function is increasing or decreasing amounts to studying the sign of its derivative. This is done by finding the critical numbers of the function. These critical numbers will determine intervals in which the sign of f will be constant. Therefore, to find the sign of f in each interval, it is enough to find the sign of f at a point of each interval. In other words, it is enough to evaluate f (d) where d is a point inside each interval determined by the critical numbers. The sign of the answer will be the sign of f in that interval. This sounds complicated, but it is easy. We will set it up as a table. Finding the critical numbers. For this, we find f (x) and look for points where f (x) =0or f (x) undefined. f (x) = ( x 3 3x 2 +1 ) =3x 2 6x =3x (x 2) Since f (x) is a polynomial, it is always defined. Also, f (x) =0when x =0or x =2. Therefore, 0 and 2 are the two critical numbers. Studying the sign of f. We do it using a table like the one below: Interval <x<0 0 <x<2 2 <x< Test point d 1 1 3 f (d) 9 3 9 Sign of f + + Behavior of f Thus, we can see that f is increasing on (, 0] [2, ). Itisdecreasing on [0, 2]. We can verify this looking at the graph of f below. 16

Remark 34 Let us make several remarks regarding this table. 1. To come up with the correct intervals, simply order the critical numbers found. The intervals will then be the regions between these numbers. If the intervals you come up with overlap, then you made a mistake. 2. The ultimate goal is to find the sign of f in a given interval. If you can find it without plugging in a point, then there is no need to have a test point. In the case above, to find the sign of f (x) =3x (x 2) in the interval (, 0], it is enough to notice that in this interval, both 3x and x 2 are negative and that the product of two negative numbers is positive. Example 35 Find the local extrema of f (x) =x 3 6x 2 +9x +1. From Fermat s theorem, we know that these can only happen at a critical number. From the first derivative test, we know that to test the critical numbers, we need to know the sign of f (x) around the critical points. Therefore, we need to find the critical numbers of f and study the sign of f.thisisexactlywhatwe did in the previous example. First, we find f (x). f (x) = ( x 3 6x 2 +9x +1 ) =3x 2 12x +9 =3 ( x 2 4x +3 ) =3(x 1) (x 3) f (x) is always defined. f (x) =0when x =1or x =3. We now study the sign of f using a table as in the previous example. 17

Interval <x<1 1 <x<3 3 <x< Test point d 0 2 4 f (d) 9 3 6 Sign of f + + Behavior of f Thus, we see that f has a local maximum at x =1. The local maximum is f (1) = 5. f has a local minimum at x =3. The local minimum is f (3) = 1. We can check this by looking at the graph of f below.x 3 6x 2 +9x +1 Remark 36 It should be clear to the reader by looking at the two previous examples that the work to find extreme values or to find where a function is increasing or decreasing is exactly the same. Example 37 Given f (x) =x 4 4x 3, answer the questions below: 1. Find where f is increasing, decreasing. 2. Find the local extreme values of f. 3. Does f have a global minimum? Does it have a global maximum? 4. Finally, find the global extreme values of f on the interval [ 1, 5]. From the two examples above, we should now have a good idea how to proceed, at least for the first two questions. In fact, the work needed to answer the first question will also provide the answer for the second question. 18

Answer to questions 1 & 2 We begin by finding the critical numbers. f (x) = ( x 4 4x 3) =4x 3 12x 2 =4x 2 (x 3) f (x) is always defined. f (x) =0when x =0or x =3. Next, we study the sign of f using a table. Interval <x<0 0 <x<3 3 <x< Test point d 1 1 4 f (d) 16 8 64 Sign of f + + Behavior of f Thus we see that f in decreasing on (, 3) andincreasingon(3, ). The critical number 0 does not correspond to an extreme value because f does not change sign. f has a local minimum at x =3. The local minimum is f (3) = 27. Answer to question 3 If we had the graph of f, we would know the answer. There is another way we can derive it. We can look at the endbehavioroff. lim f (x) = lim ( x 4 4x 3) =. Similarly, ( x x lim x 4 4x 3) =. Therefore, f cannot have a global maximum. x It will have a global minimum, which will be the smallest of its local minima. Since there is only one local minimum, it is also the global minimum. Answer to question 4 In this case, it is easy. We compare the value of f at the endpoints of the given interval with the value of f at the critical numbers. This is summarized in the table below: Point Left endpoint: 1 f ( 1) = 5 Right endpoint: 5 f (5) = 125 Value of f at the point First critical number: 0 f (0) = 0 Second critical number: 1 f (3) = 27 We see that on [ 1, 5], f has a global maximum at x =5. The global maximum is f (5) = 125. f has a global minimum at x =3. The global minimum is f (3) = 27. Remark 38 Thefirstexamplesmightsuggestthatthesignoff is always alternating. If this were true, it would mean that it is enough to find it in the first interval, thus saving quite a bit of time. However, this is not true, as the third example illustrates. So, never assume that the sign of f alternates. You must finditineachinterval. Summary: what the first derivative does for us 19

The first derivative tells us where a function is increasing/decreasing. The first derivative allows us to test the critical numbers of a function f to determine if f has extreme values. 4.2 Things to know: Know how to find the number c in the mean value theorem. Given a function, be able to find where it is increasing, decreasing. Given a function, be able to find its extreme values. Be able to sketch the graph of a function using the information provided by the first derivative. Be able to do problems such as # 3, 7,11, 17, 19, 25, 27, 29, 45, 46 on pages 288-290. In addition, be able to do problems such as: 1. True or False: If c is a critical number of a function f, then f must have an extremum at c. 2. True or False: If a function f has a maximum at x = c, thenc is a critical number. 3. True or False: If a function f has a maximum at x = c, then f (c) =0. 4. True or False: If a function f has a minimum at x = c, thenthe minimum is f(c). 5. Draw the graph of a function y = f(x) such that f (c) =0but the function does not have an extremum at x = c. 6. Draw the graph of a function y = f(x) such that f (c) =0and the function has a maximum at x = c. 7. Draw the graph of a function y = f(x) such that f (c) =0and the function has a minimum at x = c. 8. Draw the graph of a function y = f(x) such that f (c) does not exist and the function does not have an extremum at x = c. 9. Draw the graph of a function y = f(x) such that f (c) does not exist and the function has a maximum at x = c. 10. Draw the graph of a function y = f(x) such that f (c) does not exist and the function has a minimum at x = c. 20

5 Using the Second Derivative to test Critical Numbers (4.3) 5.1 Theory: The sign of the first derivative tells us whether a function is increasing or decreasing. However, this notion is not precise enough. Figure 12 shows us three increasing functions which look very different. Yet, if we only used the first derivative to study these functions, they would appear to us as being identical. The first derivative would be positive for the three functions. Yet, as we look at these three functions, we can see obvious differences. The function on the left is increasing faster and faster. In other words its slope (its first derivative) is increasing. The function in the middle is increasing at a constant rate. In other words, its slope (its first derivative) is constant. The function on the right is increasing slower and slower. In other words, its slope (its first derivative) is decreasing. This is called concavity. Figure 12: 3 increasing functions We have the following definition: Definition 39 (Concavity) Let f be a differentiable function on an interval I. The graph of f is said to be concave up on I if and only if f is increasing on I. Itissaidtobeconcave down if and only if f is decreasing on I. Graphically, it is easy to see if a function is concave up or down in most cases. A function is concave up if its graph always lies above its tangents; it is concave down if it lies below. This is sometimes also used as the definition for concavity. Definition 40 (Concavity) If the graph of y = f(x) lies above (below) all of its tangents in an open interval I, then it is called concave up (down) on I. Figure13 shows the graph of a function which is concave up. Figure 14 shows the graph of a function which is concave down. However, many functions are 21

not either concave up or down. Some are concave up on some intervals, and concave down on the remaining intervals. Figure 15 shows the graph of such a function. At the point c, the concavity changes. Such a point is called an inflection point. At such a point, the tangent would go through the graph. Figure 13: Concave up function Figure 14: Concave down function Definition 41 (Inflection point) point if the concavity changes at P ApointP on a curve is called an inflection The next theorem tells us how one studies the concavity of a function. Theorem 42 (Test for concavity) Suppose f exists on I 1. If f (x) > 0 for all x in I then the graph of f is concave up on I. 2. If f (x) < 0 for all x in I then the graph of f is concave down on I. The next theorem tells us how to find inflection points. 22

Figure 15: Function with a point of inflection at c Theorem 43 If f has a point of inflection at c, then either f (c) =0or f (c) is undefined. Remark 44 However, not every point where f (c) =0or f (c) is undefined correspond to an inflection point. The condition in the theorem gives us candidates for inflection points. We still need to verify that the concavity changes at that point. This is done by studying the sign of f. Example 45 Study the concavity of f (x) = x5 20 x4 12. Since concavity is determined by the sign of f, we compute f and study its sign. We study the sign of f in a way similar to the way we study the sign of f, using a table. Therefore ( x f 5 (x) = 20 x4 12 ) = 5x4 20 4x3 12 = x4 4 x3 3 ( x f 4 (x) = 4 x3 3 = x 3 x 2 = x 2 (x 1) f is always defined. f (x) =0when x =0or x =1. We use the table below to study its sign ) 23

Interval <x<0 0 <x<1 1 <x< Test point p 1.5 2 f (p) 2 0.125 4 Sign of f + Concavity down down up Therefore, we see that f isconcaveupon(, 1) and concave up on (1, ). There were two candidates for inflection points. But the table shows us that concavity only changes at 1. Therefore, f only has an inflection point at 1. The inflection point is (1,f(1)). Figure 16 is the graph of f. Wecanuseittoverify that our findings are correct. Figure 16: Graph of f (x) = x5 20 x4 12 Remark 46 From the example above, we can see that finding inflection points or where a function is concave up/down amounts to the same amount of work since in both cases we must determine the sign of the second derivative. You should follow the following steps: 1. Compute the second derivative. 2. Find where the second derivative is either 0 or undefined. 3. The points found above determine intervals on the real line in which the sign of the second derivative is constant. Determine the sign in each interval. 4. Conclude depending on the sign of the second derivative. Concavity can also be used to test critical numbers to see if they correspond to a local extremum. The next theorem tells us how. Theorem 47 (Second derivative test) exists. We have: Suppose that f (c) =0and f (c) 24

1. If f (c) < 0 then f has a maximum at c. 2. If f (c) > 0 then f has a minimum at c. 3. If f (c) =0you must use another test. Remark 48 The second derivative test has one advantage over the first derivative test; it is easier to use. But it also has draw backs. The first is that it does not work all the time. You can only test critical numbers for which the first derivative is 0. In some cases, the test provides no information, you have to use another test. The second is that it requires the computation of the second derivative. Remark 49 Concavity and inflection points are to the second derivative what being increasing and local extremum are to the first derivative. Weillustratethesecondderivativetestwithanexample. Example 50 Find the local extreme values of f (x) =x 3 6x 2 +9x +1. Since local extreme values can only happen at critical numbers, we first find the critical numbers. For this, we find f (x) and look for points where f (x) =0 or f (x) undefined. f (x) = ( x 3 3x 2 +1 ) =3x 2 6x =3x (x 2) Since f (x) is a polynomial, it is always defined. Also, f (x) =0when x =0 or x =2.Therefore,0 and 2 are the two critical numbers. We now need to test the two critical numbers using the second derivative test. We need the second derivative. f (x) = ( 3x 2 6x ) =6x 6 Note that the conditions of the second derivative test are satisfied. Both make the first derivative equal to zero. The second derivative is defined at them. Since f (0) = 6 < 0, the second derivative test tells us that f has a local maximum at x =0. Since f (2) = 6 > 0, the second derivative test tells us that f has alocalminimumatx =2. Compare the work we did here and the results we obtained with example 33. Which test gave the faster result? Summary: What the second derivative does for us: 1. It allows us to find where a function is concave up, down. 2. It allows us to find where the concavity changes. 3. It allows us to test critical numbers to determine if they are extreme values. 25

5.2 Graphing Using all the information the first and second derivative provide, one can do a precise sketch of the graph of a function. Additional information such as finding the horizontal and vertical asymptotes can help make the graph even more accurate. To graph a function, you should do the following: Find the domain. Find the intercepts, label them on the graph. Find the horizontal and vertical asymptotes, draw them on the graph. Find where the function is increasing, decreasing. Find the extreme values, label them on the graph. Find where the function is concave up, down. Find the inflection points, label them on the graph. Sketch the graph. We illustrate this with a few examples. Example 51 Given f (x) =3x 5 20x 3 1. Find where f is increasing and decreasing. 2. Find the local and global extrema. 3. Find the inflection points and where f is concave up, down. 4. Sketch the graph. Answer of question 1. Find where f is increasing and decreasing. Let us first notice that the domain of this function is (, ), therefore, we will be studying the function on (, ). This is determined by building the table which gives the sign of f.forthis,weneedtofindf,then the critical numbers, then build the table. f is always defined. It is zero when f (x) =15x 4 60x 2 =15x 2 ( x 2 4 ) 15x ( 2 x 2 4 ) =0 15x 2 =0or x 2 4=0 26

The first equation gives us 15x 2 =0 x =0 The second equation gives us x 2 4=0 x = ±2 Therefore, f has three critical values when x =0, ±2. The critical points are ( 2, 64), (0, 0), (2, 64). The table which gives the sign of f and determines where f is increasing and decreasing is: x <x< 2 2 <x<0 0 <x<2 2 <x< test point 3 1 1 3 f at test point 675 45 45 675 sign of f + + f Answer of question 2. Find the local and global extrema. Local Extrema. From the table above, we see that f has a local maximum when x = 2. The local maximum is f ( 2) = 64. f has a local minimum at x =2. The local minimum is f (2) = 64. Global Extrema. Since we are studying the function ( on (, ), we need to study the behavior of f as x ±. lim 3x 5 20x 3) = ( x and lim 3x 5 20x 3) =. Therefore, f has neither a global x maximum, nor a global minimum. Answer of question 3. Find the inflection points and where f is concave up, down. For this, we need to study the sign of the second derivative. We build a table similar to the one we built to study the sign of the first derivative, but we do it using the second derivative. First, we will find the second derivative, then find the points where the second derivative is not defined. Then, build our table. Second derivative. f (x) = ( 3x 5 20x 3) =15x 4 60x 2 f (x) =60x 3 120x =60x ( x 2 2 ) 27

Candidates for inflection points. Remember that inflection points are to the second derivative what critical values are to the first derivative. So, we look for points where the second derivative is either 0 or undefined. f is a polynomial, so it is always defined. f (x) =0whenever 60x =0or x 2 2=0.The first gives The second gives 60x =0 x =0 x 2 2=0 x 2 =2 x = ± 2 So, there are three candidates. f couldhaveaninflectionpointat x =0,x = ± 2. To be an inflection point, the sign of the second derivative has to change. Our table will help us determine that. Build a table x <x< 2 2 <x<0 0 <x< 2 2 <x< test point 2 1 1 2 f at test point 240 60 60 240 sign of f + + f Conclusion: The second derivative changes sign at x =0, x = ± 2. Therefore, f has an inflection point at the three values of x. Answer of question 4. Sketch the graph. All the above information gives us a graph which looks like: 28

Example 52 Consider the function f (x) = 3x2 x 2 1 1. Find the domain of this function? 2. Find the vertical and horizontal asymptotes as well as the behavior of f near its vertical asymptotes. 3. Find where f is increasing and decreasing. 4. Find the local and global extrema. 5. Find the inflection points and where f is concave up, down. 6. Sketch the graph. Answer to 1. f is defined when its denominator is not zero, that is when x ±1. Therefore, the domain of f is R {±1}. Answer to 2. We do each type of asymptote separately. Vertical asymptotes. You will recall that the line x = a is a vertical asymptote for a function f if f (x) approaches or as x a, or as x a or as x a +. In the case of a fraction, this is the case when the denominator of the fraction is approaching zero. For 29

f (x) = 3x2 x 2, this will happen when x = 1 and x =1. Therefore, 1 the vertical lines x = 1 and x =1are the vertical asymptotes of our function. For the behavior of f near its asymptotes, we compute the following limits 3x 2 lim x 1 x 2 1 = To find this limits, we used the fact that a fraction whose denominator approaches 0 and whose numerator approaches a non-zero number, will approach ±. Whether it is or is determined by the sign of the fraction. If x 1 then x< 1 then x 2 > 1 therefore x 2 1 > 0. 3x 2 is also positive. So, when x 1 3x 2, x 2 1 > 0 and thus way. and lim x 1 3x 2 x 2 1 =.The remaining limits are found the same lim x 1+ lim x 1 lim x 1+ 3x 2 x 2 1 = 3x 2 x 2 1 = 3x 2 x 2 1 = These limits will help us when we sketch the graph. Horizontal asymptotes. You will recall that a function y = f (x) has a horizontal asymptote y = a if lim f (x) =a or lim f (x) = x x a. In other words, to find the horizontal asymptotes of a function f, we need to compute the limits lim f (x) and lim f (x). x x Similarly, lim x 3x 2 x 2 1 = lim 3x 2 x x 2 =3 lim x 3x 2 x 2 1 =3 Therefore, the line x =3is the horizontal asymptote for f (x) = 3x 2 x 2 1. 30

Answer to 3 & 4. We need to compute f and find the critical numbers. ( ) 3x f 2 (x) = x 2 1 ( ) 3x 2 ( x 2 1 ) ( 3x 2)( x 2 1 ) = (x 2 1) 2 = 6x ( x 2 1 ) ( 3x 2) (2x) (x 2 1) 2 = 6x3 6x 6x 3 (x 2 1) 2 = 6x (x 2 1) 2 We see that f (x) =0when x =0.Itisundefinedwhenx = 1but these points are not in the domain of f, therefore they are not critical numbers. Therefore, 0 is the only critical number of f. We now need to study the sign of f. We do it using a table. On this table, we will not only put the critical numbers but also the numbers where f is not defined.. x <x< 1 1 <x<0 0 <x<1 1 <x< Sign of f + + Behavior of f Therefore, we see that f is increasing on (, 0) and decreasing on (0, ). We also see that f has a local maximum at 0. The local maximum is f (0) = 0. Answer to 5. This is determined by the second derivative. We need to find where f (x) = 0 or is undefined. ( ) f 6x (x) = (x 2 1) 2 = 6 ( x 2 1 ) 2 ( 6x)2 ( x 2 1 ) (2x) (x 2 1) 4 = 6 ( x 2 1 ) 2 +24x 2 ( x 2 1 ) (x 2 1) 4 = 6 ( x 2 1 ) +24x 2 (x 2 1) 3 = 18x2 +6 (x 2 1) 3 We see that f is never zero. It is undefined at points which are not in the domain of f. Therefore, f has no inflection points. We study the sign 31

of f usingatablesimilartotheoneabove. x <x< 1 1 <x<1 1 <x< Sign of f + + Concavity of f up down up Answer to 6. The graph of f is shown below. 5.3 Things to know: Be able to find where a function is concave up, down. Be able to find the inflection points of a function. Be able to test critical numbers with the second derivative test. Be able to sketch the graph of a function using the information provided by the second derivative. Be able to do problems such as # 3, 7,11, 17, 19, 25, 27, 29, 45, 46 on pages 288-290. In addition, be able to do problems such as: 1. True or False: If f (c) > 0 and f (c) =0then f has a maximum at c. 2. True or False: If f (c) > 0 then f has an extremum at c. 3. True or False: If f (c) =0then f has an inflection point at c. 32

4. Draw the graph of a function which is concave up on (, 3] and concave down elsewhere. 33