Polynomial Functions. Chapter Graphs of Polynomials

Transcription

1 Chapter 3 Polnomial Functions 3.1 Graphs of Polnomials Three of the families of functions studied thus far: constant, linear and quadratic, belong to a much larger group of functions called polnomials. We begin our formal stud of general polnomials with a definition and some eamples. Definition 3.1. A polnomial function is a function of the form f() = a n n + a n 1 n a a 1 + a 0, where a 0, a 1,..., a n are real numbers and n 1 is a natural number. polnomial function is (, ). The domain of a There are several things about Definition 3.1 that ma be off-putting or downright frightening. The best thing to do is look at an eample. Consider f() = Is this a polnomial function? We can re-write the formula for f as f() = ( 3) ( 5). Comparing this with Definition 3.1, we identif n = 5, a 5 = 4, a 4 = 0, a 3 = 0, a 2 = 3, a 1 = 2 and a 0 = 5. In other words, a 5 is the coefficient of 5, a 4 is the coefficient of 4, and so forth; the subscript on the a s merel indicates to which power of the coefficient belongs. The business of restricting n to be a natural number lets us focus on well-behaved algebraic animals. 1 Eample Determine if the following functions are polnomials. Eplain our reasoning. 1. g() = p() = q() = f() = 3 5. h() = 6. z() = 0 1 Enjo this while it lasts. Before we re through with the book, ou ll have been eposed to the most terrible of algebraic beasts. We will tame them all, in time.

2 236 Polnomial Functions Solution. 1. We note directl that the domain of g() = 3 +4 is 0. B definition, a polnomial has all real numbers as its domain. Hence, g can t be a polnomial. 2. Even though p() = 3 +4 simplifies to p() = 2 + 4, which certainl looks like the form given in Definition 3.1, the domain of p, which, as ou ma recall, we determine before we simplif, ecludes 0. Alas, p is not a polnomial function for the same reason g isn t. 3. After what happened with p in the previous part, ou ma be a little sh about simplifing q() = 3 +4 to q() =, which certainl fits Definition 3.1. If we look at the domain of 2 +4 q before we simplified, we see that it is, indeed, all real numbers. A function which can be written in the form of Definition 3.1 whose domain is all real numbers is, in fact, a polnomial. 4. We can rewrite f() = 3 as f() = 1 3. Since 1 3 polnomial. is not a natural number, f is not a 5. The function h() = isn t a polnomial, since it can t be written as a combination of powers of even though it can be written as a piecewise function involving polnomials. As we shall see in this section, graphs of polnomials possess a qualit 2 that the graph of h does not. 6. There s nothing in Definition 3.1 which prevents all the coefficients a n, etc., from being 0. Hence, z() = 0, is an honest-to-goodness polnomial. Definition 3.2. Suppose f is a polnomial function. Given f() = a n n + a n 1 n a a 1 + a 0 with a n 0, we sa The natural number n is called the degree of the polnomial f. The term a n n is called the leading term of the polnomial f. The real number a n is called the leading coefficient of the polnomial f. The real number a 0 is called the constant term of the polnomial f. If f() = a 0, and a 0 0, we sa f has degree 0. If f() = 0, we sa f has no degree. a a Some authors sa f() = 0 has degree for reasons not even we will go into. The reader ma well wonder wh we have chosen to separate off constant functions from the other polnomials in Definition 3.2. Wh not just lump them all together and, instead of forcing n to be a natural number, n = 1, 2,..., allow n to be a whole number, n = 0, 1, 2,.... We could unif all 2 One which reall relies on Calculus to verif.

3 3.1 Graphs of Polnomials 237 of the cases, since, after all, isn t a 0 0 = a 0? The answer is es, as long as 0. The function f() = 3 and g() = 3 0 are different, because their domains are different. The number f(0) = 3 is defined, whereas g(0) = 3(0) 0 is not. 3 Indeed, much of the theor we will develop in this chapter doesn t include the constant functions, so we might as well treat them as outsiders from the start. One good thing that comes from Definition 3.2 is that we can now think of linear functions as degree 1 (or first degree ) polnomial functions and quadratic functions as degree 2 (or second degree ) polnomial functions. Eample Find the degree, leading term, leading coefficient and constant term of the following polnomial functions. 1. f() = g() = h() = p() = (2 1) 3 ( 2)(3 + 2) Solution. 1. There are no surprises with f() = It is written in the form of Definition 3.2, and we see that the degree is 5, the leading term is 4 5, the leading coefficient is 4 and the constant term is The form given in Definition 3.2 has the highest power of first. To that end, we re-write g() = = , and see that the degree of g is 3, the leading term is 3, the leading coefficient is 1 and the constant term is We need to rewrite the formula for h so that it resembles the form given in Definition 3.2: h() = 4 5 = = The degree of h is 1, the leading term is 1 5, the leading coefficient is 1 5 and the constant term is It ma seem that we have some work ahead of us to get p in the form of Definition 3.2. However, it is possible to glean the information requested about p without multipling out the entire epression (2 1) 3 ( 2)(3 + 2). The leading term of p will be the term which has the highest power of. The wa to get this term is to multipl the terms with the highest power of from each factor together - in other words, the leading term of p() is the product of the leading terms of the factors of p(). Hence, the leading term of p is (2) 3 ()(3) = This means that the degree of p is 5 and the leading coefficient is 24. As for the constant term, we can perform a similar trick. The constant term is obtained b multipling the constant terms from each of the factors ( 1) 3 ( 2)(2) = 4. Our net eample shows how polnomials of higher degree arise naturall 4 in even the most basic geometric applications. 3 Technicall, 0 0 is an indeterminant form, which is a special case of being undefined. The authors realize this is beond pedantr, but we wouldn t mention it if we didn t feel it was neccessar. 4 this is a dangerous word...

4 238 Polnomial Functions Eample A bo with no top is to be fashioned from a 10 inch 12 inch piece of cardboard b cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. Let denote the length of the side of the square which is removed from each corner. 12 in height depth width 10 in 1. Find the volume V of the bo as a function of. Include an appropriate applied domain. 2. Use a graphing calculator to graph = V () on the domain ou found in part 1 and approimate the dimensions of the bo with maimum volume to two decimal places. What is the maimum volume? Solution. 1. From Geometr, we know that Volume = width height depth. The ke is to find each of these quantities in terms of. From the figure, we see that the height of the bo is itself. The cardboard piece is initiall 10 inches wide. Removing squares with a side length of inches from each corner leaves 10 2 inches for the width. 5 As for the depth, the cardboard is initiall 12 inches long, so after cutting out inches from each side, we would have 12 2 inches remaining. As a function 6 of, the volume is V () = (10 2)(12 2) = To find a suitable applied domain, we note that to make a bo at all we need > 0. Also the shorter of the two dimensions of the cardboard is 10 inches, and since we are removing 2 inches from this dimension, we also require 10 2 > 0 or < 5. Hence, our applied domain is 0 < < Using a graphing calculator, we see that the graph of = V () has a relative maimum. For 0 < < 5, this is also the absolute maimum. Using the Maimum feature of the calculator, we get 1.81, This ields a height of 1.81 inches, a width of inches, and a depth of inches. The -coordinate is the maimum volume, which is approimatel cubic inches (also written in 3 ). 5 There s no harm in taking an etra step here and making sure this makes sense. If we chopped out a 1 inch square from each side, then the width would be 8 inches, so chopping out inches would leave 10 2 inches. 6 When we write V (), it is in the contet of function notation, not the volume V times the quantit.

5 3.1 Graphs of Polnomials 239 In order to solve Eample 3.1.3, we made good use of the graph of the polnomial = V (), so we ought to turn our attention to graphs of polnomials in general. Below are the graphs of = 2, = 4 and = 6, side-b-side. We have omitted the aes to allow ou to see that as the eponent increases, the bottom becomes flatter and the sides become steeper. If ou take the the time to graph these functions b hand, 7 ou will see wh. = 2 = 4 = 6 All of these functions are even, (Do ou remember how to show this?) and it is eactl because the eponent is even. 8 This smmetr is important, but we want to eplore a different et equall important feature of these functions which we can be seen graphicall their end behavior. The end behavior of a function is a wa to describe what is happening to the function values (the -values) as the -values approach the ends of the -ais. 9 That is, what happens to as becomes small without bound 10 (written ) and, on the flip side, as becomes large without bound 11 (written ). For eample, given f() = 2, as, we imagine substituting = 100, = 1000, etc., into f to get f( 100) = 10000, f( 1000) = , and so on. Thus the function values are becoming larger and larger positive numbers (without bound). To describe this behavior, we write: as, f(). If we stud the behavior of f as, we see that in this case, too, f(). (We told ou that the smmetr was important!) The same can be said for an function of the form f() = n where n is an even natural number. If we generalize just a bit to include vertical scalings and reflections across the -ais, 12 we have 7 Make sure ou choose some -values between 1 and 1. 8 Herein lies one of the possible origins of the term even when applied to functions. 9 Of course, there are no ends to the -ais. 10 We think of as becoming a ver large (in the sense of its absolute value) negative number far to the left of zero. 11 We think of as moving far to the right of zero and becoming a ver large positive number. 12 See Theorems 1.4 and 1.5 in Section 1.7.

6 240 Polnomial Functions End Behavior of functions f() = a n, n even. Suppose f() = a n where a 0 is a real number and n is an even natural number. The end behavior of the graph of = f() matches one of the following: for a > 0, as, f() and as, f() for a < 0, as, f() and as, f() Graphicall: a > 0 a < 0 We now turn our attention to functions of the form f() = n where n 3 is an odd natural number. (We ignore the case when n = 1, since the graph of f() = is a line and doesn t fit the general pattern of higher-degree odd polnomials.) Below we have graphed = 3, = 5, and = 7. The flattening and steepening that we saw with the even powers presents itself here as well, and, it should come as no surprise that all of these functions are odd. 13 The end behavior of these functions is all the same, with f() as and f() as. = 3 = 5 = 7 As with the even degreed functions we studied earlier, we can generalize their end behavior. End Behavior of functions f() = a n, n odd. Suppose f() = a n where a 0 is a real number and n 3 is an odd natural number. The end behavior of the graph of = f() matches one of the following: for a > 0, as, f() and as, f() for a < 0, as, f() and as, f() Graphicall: a > 0 a < 0 13 And are, perhaps, the inspiration for the moniker odd function.

7 3.1 Graphs of Polnomials 241 Despite having different end behavior, all functions of the form f() = a n for natural numbers n share two properties which help distinguish them from other animals in the algebra zoo: the are continuous and smooth. While these concepts are formall defined using Calculus, 14 informall, graphs of continuous functions have no breaks or holes in them, and the graphs of smooth functions have no sharp turns. It turns out that these traits are preserved when functions are added together, so general polnomial functions inherit these qualities. Below we find the graph of a function which is neither smooth nor continuous, and to its right we have a graph of a polnomial, for comparison. The function whose graph appears on the left fails to be continuous where it has a break or hole in the graph; everwhere else, the function is continuous. The function is continuous at the corner and the cusp, but we consider these sharp turns, so these are places where the function fails to be smooth. Apart from these four places, the function is smooth and continuous. Polnomial functions are smooth and continuous everwhere, as ehibited in the graph on the right. corner cusp hole break Pathologies not found on graphs of polnomials The graph of a polnomial The notion of smoothness is what tells us graphicall that, for eample, f() =, whose graph is the characteristic shape, cannot be a polnomial. The notion of continuit is what allowed us to construct the sign diagram for quadratic inequalities as we did in Section 2.4. This last result is formalized in the following theorem. Theorem 3.1. The Intermediate Value Theorem (Zero Version): Suppose f is a continuous function on an interval containing = a and = b with a < b. If f(a) and f(b) have different signs, then f has at least one zero between = a and = b; that is, for at least one real number c such that a < c < b, we have f(c) = 0. The Intermediate Value Theorem is etremel profound; it gets to the heart of what it means to be a real number, and is one of the most oft used and under appreciated theorems in Mathematics. With that being said, most students see the result as common sense since it sas, geometricall, that the graph of a polnomial function cannot be above the -ais at one point and below the -ais at another point without crossing the -ais somewhere in between. The following eample uses the Intermediate Value Theorem to establish a fact that that most students take for granted. Man students, and sadl some instructors, will find it sill. 14 In fact, if ou take Calculus, ou ll find that smooth functions are automaticall continuous, so that saing polnomials are continuous and smooth is redundant.

8 242 Polnomial Functions Eample Use the Intermediate Value Theorem to establish that 2 is a real number. Solution. Consider the polnomial function f() = 2 2. Then f(1) = 1 and f(3) = 7. Since f(1) and f(3) have different signs, the Intermediate Value Theorem guarantees us a real number c between 1 and 3 with f(c) = 0. If c 2 2 = 0 then c = ± 2. Since c is between 1 and 3, c is positive, so c = 2. Our primar use of the Intermediate Value Theorem is in the construction of sign diagrams, as in Section 2.4, since it guarantees us that polnomial functions are alwas positive (+) or alwas negative ( ) on intervals which do not contain an of its zeros. The general algorithm for polnomials is given below. Steps for Constructing a Sign Diagram for a Polnomial Function Suppose f is a polnomial function. 1. Find the zeros of f and place them on the number line with the number 0 above them. 2. Choose a real number, called a test value, in each of the intervals determined in step Determine the sign of f() for each test value in step 2, and write that sign above the corresponding interval. Eample Construct a sign diagram for f() = 3 ( 3) 2 ( + 2) ( ). Use it to give a rough sketch of the graph of = f(). Solution. First, we find the zeros of f b solving 3 ( 3) 2 ( + 2) ( ) = 0. We get = 0, = 3 and = 2. (The equation = 0 produces no real solutions.) These three points divide the real number line into four intervals: (, 2), ( 2, 0), (0, 3) and (3, ). We select the test values = 3, = 1, = 1 and = 4. We find f( 3) is (+), f( 1) is ( ) and f(1) is (+) as is f(4). Wherever f is (+), its graph is above the -ais; wherever f is ( ), its graph is below the -ais. The -intercepts of the graph of f are ( 2, 0), (0, 0) and (3, 0). Knowing f is smooth and continuous allows us to sketch its graph. (+) 0 ( ) 0 (+) 0 (+) A sketch of = f() A couple of notes about the Eample are in order. First, note that we purposefull did not label the -ais in the sketch of the graph of = f(). This is because the sign diagram gives us the zeros and the relative position of the graph - it doesn t give us an information as to how high or low the graph stras from the -ais. Furthermore, as we have mentioned earlier in the tet, without Calculus, the values of the relative maimum and minimum can onl be found approimatel using a calculator. If we took the time to find the leading term of f, we would find it to be 8. Looking

9 3.1 Graphs of Polnomials 243 at the end behavior of f, we notice that it matches the end behavior of = 8. This is no accident, as we find out in the net theorem. Theorem 3.2. End Behavior for Polnomial Functions: The end behavior of a polnomial f() = a n n +a n 1 n a 2 2 +a 1 +a 0 with a n 0 matches the end behavior of = a n n. To see wh Theorem 3.2 is true, let s first look at a specific eample. Consider f() = If we wish to eamine end behavior, we look to see the behavior of f as ±. Since we re concerned with s far down the -ais, we are far awa from = 0 so can rewrite f() for these values of as ( f() = ) As becomes unbounded (in either direction), the terms and , as the table below indicates become closer and closer to In other words, as ±, f() 4 3 ( ) = 4 3, which is the leading term of f. The formal proof of Theorem 3.2 works in much the same wa. Factoring out the leading term leaves ( f() = a n n 1 + a n 1 a n a 2 a n n 2 + a 1 a n n 1 + a ) 0 a n n As ±, an term with an in the denominator becomes closer and closer to 0, and we have f() a n n. Geometricall, Theorem 3.2 sas that if we graph = f() using a graphing calculator, and continue to zoom out, the graph of it and its leading term become indistinguishable. Below are the graphs of = (the thicker line) and = 4 3 (the thinner line) in two different windows. A view close to the origin. A zoomed out view.

10 244 Polnomial Functions Let s return to the function in Eample 3.1.5, f() = 3 ( 3) 2 (+2) ( ), whose sign diagram and graph are reproduced below for reference. Theorem 3.2 tells us that the end behavior is the same as that of its leading term 8. This tells us that the graph of = f() starts and ends above the -ais. In other words, f() is (+) as ±, and as a result, we no longer need to evaluate f at the test values = 3 and = 4. Is there a wa to eliminate the need to evaluate f at the other test values? What we would reall need to know is how the function behaves near its zeros - does it cross through the -ais at these points, as it does at = 2 and = 0, or does it simpl touch and rebound like it does at = 3. From the sign diagram, the graph of f will cross the -ais whenever the signs on either side of the zero switch (like the do at = 2 and = 0); it will touch when the signs are the same on either side of the zero (as is the case with = 3). What we need to determine is the reason behind whether or not the sign change occurs. (+) 3 0 ( ) 0 (+) 0 (+) A sketch of = f() Fortunatel, f was given to us in factored form: f() = 3 ( 3) 2 ( + 2). When we attempt to determine the sign of f( 4), we are attempting to find the sign of the number ( 4) 3 ( 7) 2 ( 2), which works out to be ( )(+)( ) which is (+). If we move to the other side of = 2, and find the sign of f( 1), we are determining the sign of ( 1) 3 ( 4) 2 (+1), which is ( )(+)(+) which gives us the ( ). Notice that signs of the first two factors in both epressions are the same in f( 4) and f( 1). The onl factor which switches sign is the third factor, ( + 2), precisel the factor which gave us the zero = 2. If we move to the other side of 0 and look closel at f(1), we get the sign pattern (+1) 3 ( 2) 2 (+3) or (+)(+)(+) and we note that, once again, going from f( 1) to f(1), the onl factor which changed sign was the first factor, 3, which corresponds to the zero = 0. Finall, to find f(4), we substitute to get (+4) 3 (+2) 2 (+5) which is (+)(+)(+) or (+). The sign didn t change for the middle factor ( 3) 2. Even though this is the factor which corresponds to the zero = 3, the fact that the quantit is squared kept the sign of the middle factor the same on either side of 3. If we look back at the eponents on the factors ( + 2) and 3, we see that the are both odd, so as we substitute values to the left and right of the corresponding zeros, the signs of the corresponding factors change which results in the sign of the function value changing. This is the ke to the behavior of the function near the zeros. We need a definition and then a theorem. Definition 3.3. Suppose f is a polnomial function and m is a natural number. If ( c) m is a factor of f() but ( c) m+1 is not, then we sa = c is a zero of multiplicit m. Hence, rewriting f() = 3 ( 3) 2 ( + 2) as f() = ( 0) 3 ( 3) 2 ( ( 2)) 1, we see that = 0 is a zero of multiplicit 3, = 3 is a zero of multiplicit 2 and = 2 is a zero of multiplicit 1.

11 3.1 Graphs of Polnomials 245 Theorem 3.3. The Role of Multiplicit: Suppose f is a polnomial function and = c is a zero of multiplicit m. If m is even, the graph of = f() touches and rebounds from the -ais at (c, 0). If m is odd, the graph of = f() crosses through the -ais at (c, 0). Our last eample shows how end behavior and multiplicit allow us to sketch a decent graph without appealing to a sign diagram. Eample Sketch the graph of f() = 3(2 1)( + 1) 2 using end behavior and the multiplicit of its zeros. Solution. The end behavior of the graph of f will match that of its leading term. To find the leading term, we multipl b the leading terms of each factor to get ( 3)(2)() 2 = 6 3. This tells us that the graph will start above the -ais, in Quadrant II, and finish below the -ais, in Quadrant IV. Net, we find the zeros of f. Fortunatel for us, f is factored. 15 Setting each factor equal to zero gives is = 1 2 and = 1 as zeros. To find the multiplicit of = 1 2 we note that it corresponds to the factor (2 1). This isn t strictl in the form required in Definition 3.3. If we factor out the 2, however, we get (2 1) = 2 ( 1 2), and we see that the multiplicit of = 1 2 is 1. Since 1 is an odd number, we know from Theorem 3.3 that the graph of f will cross through the -ais at ( 1 2, 0). Since the zero = 1 corresponds to the factor ( + 1) 2 = ( ( 1)) 2, we find its multiplicit to be 2 which is an even number. As such, the graph of f will touch and rebound from the -ais at ( 1, 0). Though we re not asked to, we can find the -intercept b finding f(0) = 3(2(0) 1)(0 + 1) 2 = 3. Thus (0, 3) is an additional point on the graph. Putting this together gives us the graph below. 15 Obtaining the factored form of a polnomial is the main focus of the net few sections.

12 246 Polnomial Functions Eercises In Eercises 1-10, find the degree, the leading term, the leading coefficient, the constant term and the end behavior of the given polnomial. 1. f() = g() = q(r) = 1 16r 4 4. Z(b) = 42b b 3 5. f() = π s(t) = 4.9t 2 + v 0 t + s 0 7. P () = ( 1)( 2)( 3)( 4) 8. p(t) = t 2 (3 5t)(t 2 + t + 4) 9. f() = 2 3 ( + 1)( + 2) G(t) = 4(t 2) 2 ( t In Eercises 11-20, find the real zeros of the given polnomial and their corresponding multiplicities. Use this information along with a sign chart to provide a rough sketch of the graph of the polnomial. Compare our answer with the result from a graphing utilit. 11. a() = ( + 2) g() = ( + 2) f() = 2( 2) 2 ( + 1) 14. g() = (2 + 1) 2 ( 3) 15. F () = 3 ( + 2) P () = ( 1)( 2)( 3)( 4) 17. Q() = ( + 5) 2 ( 3) h() = 2 ( 2) 2 ( + 2) H(t) = (3 t)(t 2 + 1) 20. Z(b) = b(42 b 2 ) In Eercises 21-26, given the pair of functions f and g, sketch the graph of = g() b starting with the graph of = f() and using transformations. Track at least three points of our choice through the transformations. State the domain and range of g. 21. f() = 3, g() = ( + 2) f() = 4, g() = ( + 2) f() = 4, g() = 2 3( 1) f() = 5, g() = f() = 5, g() = ( + 1) f() = 6, g() = Use the Intermediate Value Theorem to prove that f() = has a real zero in each of the following intervals: [ 4, 3], [0, 1] and [2, 3]. 28. Rework Eample assuming the bo is to be made from an 8.5 inch b 11 inch sheet of paper. Using scissors and tape, construct the bo. Are ou surprised? Consider decorating the bo and presenting it to our instructor. If done well enough, mabe our instructor will issue ou some bonus points. Or mabe not. )

13 3.1 Graphs of Polnomials 247 In Eercises 29-31, suppose the revenue R, in thousands of dollars, from producing and selling hundred LCD TVs is given b R() = for Use a graphing utilit to graph = R() and determine the number of TVs which should be sold to maimize revenue. What is the maimum revenue? 30. Assume that the cost, in thousands of dollars, to produce hundred LCD TVs is given b C() = for 0. Find and simplif an epression for the profit function P (). (Remember: Profit = Revenue - Cost.) 31. Use a graphing utilit to graph = P () and determine the number of TVs which should be sold to maimize profit. What is the maimum profit? 32. While developing their newest game, Sasquatch Attack!, the makers of the PortaBo (from Eample 2.1.5) revised their cost function and now use C() = , for 0. As before, C() is the cost to make PortaBo Game Sstems. Market research indicates that the demand function p() = remains unchanged. Use a graphing utilit to find the production level that maimizes the profit made b producing and selling PortaBo game sstems. 33. According to US Postal regulations, a rectangular shipping bo must satisf the inequalit Length + Girth 130 inches for Parcel Post and Length + Girth 108 inches for other services. 17 Let s assume we have a closed rectangular bo with a square face of side length as drawn below. The length is the longest side and is clearl labeled. The girth is the distance around the bo in the other two dimensions so in our case it is the sum of the four sides of the square, 4. (a) Assuming that we ll be mailing a bo via Parcel Post where Length + Girth = 130 inches, epress the length of the bo in terms of and then epress the volume V of the bo in terms of. (b) Find the dimensions of the bo of maimum volume that can be shipped via Parcel Post. (c) Repeat parts 33a and 33b if the bo is shipped using other services. length 17 See here for details.

14 248 Polnomial Functions 34. We now revisit the data set from Eercise 6b in Section 2.5. In that eercise, ou were given a chart of the number of hours of dalight the get on the 21 st of each month in Fairbanks, Alaska based on the 2009 sunrise and sunset data found on the U.S. Naval Observator website. We let = 1 represent Januar 21, 2009, = 2 represent Februar 21, 2009, and so on. The chart is given again for reference. Month Number Hours of Dalight Find cubic (third degree) and quartic (fourth degree) polnomials which model this data and comment on the goodness of fit for each. What can we sa about using either model to make predictions about the ear 2020? (Hint: Think about the end behavior of polnomials.) Use the models to see how man hours of dalight the got on our birthda and then check the website to see how accurate the models are. Knowing that Sasquatch are largel nocturnal, what das of the ear according to our models are going to allow for at least 14 hours of darkness for field research on the elusive creatures? 35. An electric circuit is built with a variable resistor installed. For each of the following resistance values (measured in kilo-ohms, kω), the corresponding power to the load (measured in milliwatts, mw ) is given in the table below. 18 Resistance: (kω) Power: (mw ) (a) Make a scatter diagram of the data using the Resistance as the independent variable and Power as the dependent variable. (b) Use our calculator to find quadratic (2nd degree), cubic (3rd degree) and quartic (4th degree) regression models for the data and judge the reasonableness of each. (c) For each of the models found above, find the predicted maimum power that can be delivered to the load. What is the corresponding resistance value? (d) Discuss with our classmates the limitations of these models - in particular, discuss the end behavior of each. 36. Show that the end behavior of a linear function f() = m + b is as it should be according to the results we ve established in the section for polnomials of odd degree. 19 (That is, show that the graph of a linear function is up on one side and down on the other just like the graph of = a n n for odd numbers n.) 18 The authors wish to thank Don Anthan and Ken White of Lakeland Communit College for devising this problem and generating the accompaning data set. 19 Remember, to be a linear function, m 0.

15 3.1 Graphs of Polnomials There is one subtlet about the role of multiplicit that we need to discuss further; specificall we need to see how the graph crosses the -ais at a zero of odd multiplicit. In the section, we deliberatel ecluded the function f() = from the discussion of the end behavior of f() = n for odd numbers n and we said at the time that it was due to the fact that f() = didn t fit the pattern we were tring to establish. You just showed in the previous eercise that the end behavior of a linear function behaves like ever other polnomial of odd degree, so what doesn t f() = do that g() = 3 does? It s the flattening for values of near zero. It is this local behavior that will distinguish between a zero of multiplicit 1 and one of higher odd multiplicit. Look again closel at the graphs of a() = ( + 2) 2 and F () = 3 ( + 2) 2 from Eercise Discuss with our classmates how the graphs are fundamentall different at the origin. It might help to use a graphing calculator to zoom in on the origin to see the different crossing behavior. Also compare the behavior of a() = ( + 2) 2 to that of g() = ( + 2) 3 near the point ( 2, 0). What do ou predict will happen at the zeros of f() = ( 1)( 2) 2 ( 3) 3 ( 4) 4 ( 5) 5? 38. Here are a few other questions for ou to discuss with our classmates. (a) How man local etrema could a polnomial of degree n have? How few local etrema can it have? (b) Could a polnomial have two local maima but no local minima? (c) If a polnomial has two local maima and two local minima, can it be of odd degree? Can it be of even degree? (d) Can a polnomial have local etrema without having an real zeros? (e) Wh must ever polnomial of odd degree have at least one real zero? (f) Can a polnomial have two distinct real zeros and no local etrema? (g) Can an -intercept ield a local etrema? Can it ield an absolute etrema? (h) If the -intercept ields an absolute minimum, what can we sa about the degree of the polnomial and the sign of the leading coefficient?

16 250 Polnomial Functions Answers 1. f() = Degree 2 Leading term 3 2 Leading coefficient 3 Constant term 4 As, f() As, f() 2. g() = Degree 5 Leading term 3 5 Leading coefficient 3 Constant term 1 As, g() As, g() 3. q(r) = 1 16r 4 Degree 4 Leading term 16r 4 Leading coefficient 16 Constant term 1 As r, q(r) As r, q(r) 4. Z(b) = 42b b 3 Degree 3 Leading term b 3 Leading coefficient 1 Constant term 0 As b, Z(b) As b, Z(b) 5. f() = π Degree 17 Leading term 3 17 Leading coefficient 3 Constant term 1 3 As, f() As, f() 6. s(t) = 4.9t 2 + v 0 t + s 0 Degree 2 Leading term 4.9t 2 Leading coefficient 4.9 Constant term s 0 As t, s(t) As t, s(t) 7. P () = ( 1)( 2)( 3)( 4) Degree 4 Leading term 4 Leading coefficient 1 Constant term 24 As, P () As, P () 8. p(t) = t 2 (3 5t)(t 2 + t + 4) Degree 5 Leading term 5t 5 Leading coefficient 5 Constant term 0 As t, p(t) As t, p(t)

17 3.1 Graphs of Polnomials f() = 2 3 ( + 1)( + 2) 2 Degree 6 Leading term 2 6 Leading coefficient 2 Constant term 0 As, f() As, f() 10. G(t) = 4(t 2) 2 ( t + 1 ) 2 Degree 3 Leading term 4t 3 Leading coefficient 4 Constant term 8 As t, G(t) As t, G(t) 11. a() = ( + 2) 2 = 0 multiplicit 1 = 2 multiplicit g() = ( + 2) 3 = 0 multiplicit 1 = 2 multiplicit f() = 2( 2) 2 ( + 1) = 2 multiplicit 2 = 1 multiplicit g() = (2 + 1) 2 ( 3) = 1 2 multiplicit 2 = 3 multiplicit

18 252 Polnomial Functions 15. F () = 3 ( + 2) 2 = 0 multiplicit 3 = 2 multiplicit P () = ( 1)( 2)( 3)( 4) = 1 multiplicit 1 = 2 multiplicit 1 = 3 multiplicit 1 = 4 multiplicit Q() = ( + 5) 2 ( 3) 4 = 5 multiplicit 2 = 3 multiplicit f() = 2 ( 2) 2 ( + 2) 2 = 2 multiplicit 2 = 0 multiplicit 2 = 2 multiplicit H(t) = (3 t) ( t ) = 3 multiplicit Z(b) = b(42 b 2 ) b = 42 multiplicit 1 b = 0 multiplicit 1 b = 42 multiplicit t b

19 3.1 Graphs of Polnomials g() = ( + 2) domain: (, ) range: (, ) 22. g() = ( + 2) domain: (, ) range: [1, ) g() = 2 3( 1) 4 domain: (, ) range: (, 2] 24. g() = 5 3 domain: (, ) range: (, )

20 254 Polnomial Functions 25. g() = ( + 1) domain: (, ) range: (, ) 26. g() = 8 6 domain: (, ) range: (, 8] We have f( 4) = 23, f( 3) = 5, f(0) = 5, f(1) = 3, f(2) = 5 and f(3) = 5 so the Intermediate Value Theorem tells us that f() = has real zeros in the intervals [ 4, 3], [0, 1] and [2, 3]. 28. V () = (8.5 2)(11 2) = , 0 < < Volume is maimized when 1.58, so the dimensions of the bo with maimum volume are: height 1.58 inches, width 5.34 inches, and depth 7.84 inches. The maimum volume is cubic inches. 29. The calculator gives the location of the absolute maimum (rounded to three decimal places) as and Since represents the number of TVs sold in hundreds, = corresponds to TVs. Since we can t sell half of a TV, we compare R(6.30) and R(6.31) , so selling 631 TVs results in a (slightl) higher revenue. Since represents the revenue in thousands of dollars, the maimum revenue is $1,115, P () = R() C() = , The calculator gives the location of the absolute maimum (rounded to three decimal places) as and Since represents the number of TVs sold in hundreds, = corresponds to TVs. Since we can t sell 0.7 of a TV, we compare P (3.89) and P (3.90) , so selling 390 TVs results in a (slightl) higher revenue. Since represents the revenue in thousands of dollars, the maimum revenue is $35, Making and selling 71 PortaBos ields a maimized profit of $

21 3.1 Graphs of Polnomials (a) Our ultimate goal is to maimize the volume, so we ll start with the maimum Length + Girth of 130. This means the length is The volume of a rectangular bo is alwas length width height so we get V () = 2 (130 4) = (b) Graphing = V () on [0, 33] [0, 21000] shows a maimum at (21.67, ) so the dimensions of the bo with maimum volume are 21.67in in in. for a volume of in. 3. (c) If we start with Length + Girth = 108 then the length is and the volume is V () = Graphing = V () on [0, 27] [0, 11700] shows a maimum at (18.00, ) so the dimensions of the bo with maimum volume are 18.00in in. 36in. for a volume of in. 3. (Calculus will confirm that the measurements which maimize the volume are eactl 18in. b 18in. b 36in., however, as I m sure ou are aware b now, we treat all calculator results as approimations and list them as such.) 34. The cubic regression model is p 3 () = It has R 2 = which isn t bad. The graph of = p 3 () in the viewing window [ 1, 13] [0, 24] along with the scatter plot is shown below on the left. Notice that p 3 hits the -ais at about = making this a bad model for future predictions. To use the model to approimate the number of hours of sunlight on our birthda, ou ll have to figure out what decimal value of is close enough to our birthda and then plug it into the model. M (Jeff s) birthda is Jul 31 which is 10 das after Jul 21 ( = 7). Assuming 30 das in a month, I think = 7.33 should work for m birthda and p 3 (7.33) The website sas there will be about hours of dalight that da. To have 14 hours of darkness we need 10 hours of dalight. We see that p 3 (1.96) 10 and p 3 (10.05) 10 so it seems reasonable to sa that we ll have at least 14 hours of darkness from December 21, 2008 ( = 0) to Februar 21, 2009 ( = 2) and then again from October 21,2009 ( = 10) to December 21, 2009 ( = 12). The quartic regression model is p 4 () = It has R 2 = which is good. The graph of = p 4 () in the viewing window [ 1, 15] [0, 35] along with the scatter plot is shown below on the right. Notice that p 4 (15) is above 24 making this a bad model as well for future predictions. However, p 4 (7.33) making it much better at predicting the hours of dalight on Jul 31 (m birthda). This model sas we ll have at least 14 hours of darkness from December 21, 2008 ( = 0) to about March 1, 2009 ( = 2.30) and then again from October 10, 2009 ( = 9.667) to December 21, 2009 ( = 12). = p 3 () = p 4 ()

22 256 Polnomial Functions 35. (a) The scatter plot is shown below with each of the three regression models. (b) The quadratic model is P 2 () = with R 2 = The cubic model is P 3 () = with R 2 = The quartic model is P 4 () = with R 2 = (c) The maimums predicted b the three models are P 2 (5.737) 1.648, P 3 (4.232) and P 4 (3.784) 1.630, respectivel. = P 2 () = P 3 () = P 4 ()