Graphing and Optimization

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1 BARNMC_33886.QXD //7 :7 Page 74 Graphing and Optimization CHAPTER - First Derivative and Graphs - Second Derivative and Graphs -3 L Hôpital s Rule -4 Curve-Sketching Techniques - Absolute Maima and Minima -6 Optimization I N T R O D U C T I O N Since the derivative is associated with the slope of the graph of a function at a point, we might epect that it is also associated with other properties of a graph. As we will see in this and the net section, the derivative can tell us a great deal about the shape of the graph of a function. In addition, our investigation will lead to methods for finding absolute maimum and minimum values for functions that do not require graphing. Manufacturing companies can use these methods to find production levels that will minimize cost or maimize profit, pharmacologists can use them to find levels of drug dosages that will produce maimum sensitivity, and so on. Chapter Review Review Eercise 74

2 BARNMC_33886.QXD //7 :7 Page 7 Section - First Derivative and Graphs 7 Section - FIRST DERIVATIVE AND GRAPHS Increasing and Decreasing Functions Local Etrema First-Derivative Test Applications to Economics Increasing and Decreasing Functions Sign charts (Section 3-) will be used throughout this chapter. You will find it helpful to review the terminology and techniques for constructing sign charts now. Eplore & Discuss Figure shows the graph of y = f and a sign chart for f, where and f = 3-3 f = 3-3 = f() (, ) (, ) (, ) f () FIGURE Discuss the relationship between the graph of f and the sign of f over each interval on which f has a constant sign. Also, describe the behavior of the graph of f at each partition number for f. Slope positive y Slope a b c f FIGURE Slope negative As they are scanned from left to right, graphs of functions generally have rising and falling sections. If you scan the graph of f = 3-3 in Figure from left to right, you will see that On the interval - q, -, the graph of f is rising, f() is increasing,* and the slope of the graph is positive [f 7 ]. On the interval -,, the graph of f is falling, f is decreasing, and the slope of the graph is negative [f 6 ]. On the interval, q, the graph of f is rising, f is increasing, and the slope of the graph is positive [f 7 ]. At = - and =, the slope of the graph is [f = ]. In general, if f 7 (is positive) on the interval (a, b) (Fig. ), then f() increases and the graph of f rises as we move from left to right over the interval; if f 6 (is negative) on an interval (a, b), then f() decreases Ω and the graph of f falls as we move from left to right over the interval. We summarize these important results in Theorem. * Formally, we say that the function f is increasing on an interval (a, b) if f( ) 7 f( ) whenever a b, and f is decreasing on (a, b) if f( ) 6 f( ) whenever a b.

3 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization THEOREM INCREASING AND DECREASING FUNCTIONS For the interval (a, b), f f () Graph of f Eamples Increases Rises Decreases Ω Falls Ω Eplore & Discuss The graphs of f = and g = ƒ ƒ are shown in Figure 3. Both functions change from decreasing to increasing at =. Discuss the relationship between the graph of each function at = and the derivative of the function at =. f() g() (A) (B) FIGURE 3 EXAMPLE SOLUTION Finding Intervals on Which a Function Is Increasing or Decreasing Given the function f = 8 -, (A) Which values of correspond to horizontal tangent lines? (B) For which values of is f() increasing? Decreasing? (C) Sketch a graph of f. Add any horizontal tangent lines. (A) f = 8 - = = 4 Thus, a horizontal tangent line eists at = 4 only. (B) We will construct a sign chart for f to determine which values of make f 7 and which values make f 6. Recall from Section 3- that the partition numbers for a function are the points where the function is or discontinuous. Thus, when constructing a sign chart for f, we must locate all points where f = or f is discontinuous. From part (A), we know that f = 8 - = at = 4. Since f = 8 - is a polynomial, it is continuous for all. Thus, 4 is the only partition number. We construct a sign chart for the intervals - q, 4 and 4, q, using test numbers 3 and : f () f () (, 4) (4, ) 4 Increasing Decreasing Test Numbers f 3 -

4 BARNMC_33886.QXD //7 :7 Page 77 Section - First Derivative and Graphs 77 Hence, f() is increasing on - q, 4 and decreasing on 4, q. (C) f() f() f() increasing Horizontal tangent line f() decreasing MATCHED PROBLEM Repeat Eample for f = As Eample illustrates, the construction of a sign chart will play an important role in using the derivative to analyze and sketch the graph of a function f. The partition numbers for f are central to the construction of these sign charts and also to the analysis of the graph of y = f. We already know that if f c =, then the graph of y = f will have a horizontal tangent line at = c. But the partition numbers for f also include the numbers c such that f c does not eist.* There are two possibilities at this type of number: f(c) does not eist; or f(c) eists, but the slope of the tangent line at = c is undefined. DEFINITION Critical Values The values of in the domain of f where f = or where f does not eist are called the critical values of f. I N S I G H T The critical values of f are always in the domain of f and are also partition numbers for f, but f may have partition numbers that are not critical values. If f is a polynomial, then both the partition numbers for f and the critical values of f are the solutions of f =. We will illustrate the process for locating critical values with eamples. EXAMPLE SOLUTION Partition Numbers and Critical Values Find the critical values of f, the intervals on which f is increasing, and those on which f is decreasing, for f = + 3. Begin by finding the partition number for f : f = 3 =, only at = The partition number is in the domain of f, so is the only critical value of f. * We are assuming that f (c) does not eist at any point of discontinuity of f. There do eist functions f such that f is discontinuous at = c, yet f (c) eists. However, we do not consider such functions in this book.

5 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization f() The sign chart for f = 3 (partition number is ) is f () f () (, ) (, ) Increasing Increasing Test Numbers f œ FIGURE 4 The sign chart indicates that f() is increasing on - q, and, q. Since f is continuous at =, it follows that f() is increasing for all. The graph of f is shown in Figure 4. MATCHED PROBLEM Find the critical values of f, the intervals on which f is increasing, and those on which f is decreasing, for f = - 3. EXAMPLE 3 SOLUTION Partition Numbers and Critical Values Find the critical values of f, the intervals on which f is increasing, and those on which f is decreasing, for f = - >3. f = >3 = 3 - >3 To find partition numbers for f, we note that f is continuous for all, ecept for values of for which the denominator is ; that is, f does not eist and f is discontinuous at =. Since the numerator is the constant -, f Z for any value of. Thus, = is the only partition number for f. Since is in the domain of f, = is also the only critical value of f. When constructing the sign chart for f we use the abbreviation ND to note the fact that f is not defined at =. The sign chart for f = ->33 - >3 4 (partition number is ) is f() f () f () (, ) (, ) ND Decreasing Decreasing Test Numbers f FIGURE The sign chart indicates that f is decreasing on - q, and, q. Since f is continuous at =, it follows that f() is decreasing for all. Thus, a continuous function can be decreasing (or increasing) on an interval containing values of where f () does not eist. The graph of f is shown in Figure. Notice that the undefined derivative at = results in a vertical tangent line at =. In general, a vertical tangent will occur at c if f is continuous at c and if f () becomes larger and larger as approaches c. MATCHED PROBLEM 3 Find the critical values of f, the intervals on which f is increasing, and those on which f is decreasing, for f = + >3. EXAMPLE 4 Partition Numbers and Critical Values Find the critical values of f, the intervals on which f is increasing, and those on which f is decreasing, for f = -.

6 BARNMC_33886.QXD //7 :7 Page 79 f() SOLUTION f = Section - First Derivative and Graphs 79 - = - - f = = - - To find the partition numbers for f, note that f Z for any and f is not defined at =. Thus, = is the only partition number for f. However, = is not in the domain of f. Consequently, = is not a critical value of f. This function has no critical values. The sign chart for f = -> - (partition number is ) is f () f () (, ) (, ) ND Decreasing Decreasing Test Numbers f FIGURE 6 Thus, f is decreasing on - q, and, q. See the graph of f in Figure 6. MATCHED PROBLEM 4 EXAMPLE Find the critical values for f, the intervals on which f is increasing, and those on which f is decreasing, for f =. Partition Numbers and Critical Values Find the critical values of f, the intervals on which f is increasing, and those on which f is decreasing, for f = 8 ln -. SOLUTION The natural logarithm function ln is defined on, q, or 7, so f is defined only for 7. We have f = 8 ln -, 7 f = 8 - = 8 - Find a common denominator. Subtract numerators. = 8 - Factor numerator. f() 3 FIGURE 7 4 = Note that f = at - and at and f is discontinuous at. These are the partition numbers for f. Since the domain of f is, q, and - are not critical values. The remaining partition number,, is the only critical value for f. - + The sign chart for f =, 7 (partition number is ), is f () f () - +, 7 (, ) Increasing (, ) ND Decreasing Test Numbers f Thus, f is increasing on, and decreasing on, q. See the graph of f in Figure 7.

7 BARNMC_33886.QXD //7 :7 Page 8 8 CHAPTER Graphing and Optimization MATCHED PROBLEM Find the critical values of f, the intervals on which f is increasing, and those on which f is decreasing, for f = ln -. Eplore & Discuss 3 A student eamined the sign chart in Eample 4 and concluded that f = > - is decreasing for all ecept =. However, f = - 6 f3 =, which seems to indicate that f is increasing. Discuss the difference between the correct answer in Eample 4 and the student s answer. Eplain why the student s description of where f is decreasing is unacceptable. I N S I G H T Eample 4 illustrates two important ideas:. Do not assume that all partition numbers for the derivative f are critical values of the function f. To be a critical value, a partition number must also be in the domain of f.. The values for which a function is increasing or decreasing must always be epressed in terms of open intervals that are subsets of the domain of the function. Local Etrema When the graph of a continuous function changes from rising to falling, a high point, or local maimum, occurs; when the graph changes from falling to rising, a low point, or local minimum, occurs. In Figure 8, high points occur at c 3 and c 6, and low points occur at c and c 4. In general, we call f(c) a local maimum if there eists an interval (m, n) containing c such that f fc for all in m, n Note that this inequality need hold only for values of near c hence the use of the term local. y f() c c c 3 c 4 c c 6 c 7 FIGURE 8 The quantity f(c) is called a local minimum if there eists an interval (m, n) containing c such that f Ú fc for all in m, n The quantity f (c) is called a local etremum if it is either a local maimum or a local minimum. A point on a graph where a local etremum occurs is also called a turning point. Thus, in Figure 8 we see that local maima occur at c 3 and c 6, local minima occur at c and c 4, and all four values produce local etrema. Also, the local maimum fc 3 is not the highest point on the graph in Figure 8. Later in this chapter, we consider the problem of finding the highest and lowest points on a graph, or absolute etrema. For now, we are concerned only with locating local etrema. EXAMPLE 6 Analyzing a graph Use the graph of f in Figure 9 to find the intervals on which f is increasing, those on which f is decreasing, any local maima, and any local minima.

8 BARNMC_33886.QXD //7 :7 Page 8 Section - First Derivative and Graphs 8 f() FIGURE 9 SOLUTION The function f is increasing (the graph is rising) on - q, - and on 3, q and is decreasing (the graph is falling) on -, 3. Because the graph changes from rising to falling at = -, f- = 3 is a local maimum. Because the graph changes from falling to rising at = 3, f3 = - is a local minimum. MATCHED PROBLEM 6 Use the graph of g in Figure to find the intervals on which g is increasing, those on which g is decreasing, any local maima, and any local minima. g() FIGURE How can we locate local maima and minima if we are given the equation of a function and not its graph? The key is to eamine the critical values of the function. The local etrema of the function f in Figure 8 occur either at points where the derivative is ( c and c 3 ) or at points where the derivative does not eist ( c 4 and c 6 ). In other words, local etrema occur only at critical values of f. Theorem shows that this is true in general. THEOREM EXISTENCE OF LOCAL EXTREMA If f is continuous on the interval (a, b), c is a number in (a, b), and f(c) is a local etremum, then either f c = or f c does not eist (is not defined). Theorem states that a local etremum can occur only at a critical value, but it does not imply that every critical value produces a local etremum. In Figure 8, c and c are critical values (the slope is ), but the function does not have a local maimum or local minimum at either of these values. Our strategy for finding local etrema is now clear: We find all critical values of f and test each one to see if it produces a local maimum, a local minimum, or neither. First-Derivative Test If f eists on both sides of a critical value c, the sign of f can be used to determine whether the point (c, f(c)) is a local maimum, a local minimum, or

9 BARNMC_33886.QXD //7 :7 Page 8 8 CHAPTER Graphing and Optimization neither. The various possibilities are summarized in the following bo and are illustrated in Figure : f (c) : Horizontal tangent f() f() f() f() f(c) f(c) f(c) f(c) f () c (A) f(c) is a local minimum c (B) f(c) is a local maimum f () f () f () c (C) f(c) is neither a local maimum nor a local minimum c (D) f(c) is neither a local maimum nor a local minimum f (c) is not defined but f(c) is defined f() f() f() f() f(c) f(c) f(c) f(c) c (E) f(c) is a local minimum c (F) f(c) is a local maimum f () ND f () ND f () ND f () ND FIGURE Local etrema c (G) f(c) is neither a local maimum nor a local minimum c (H) f(c) is neither a local maimum nor a local minimum PROCEDURE First-Derivative Test for Local Etrema Let c be a critical value of f [f(c) is defined and either f c = or f c is not defined]. Construct a sign chart for f close to and on either side of c. Sign Chart f(c) f () f () ( m c n Decreasing Increasing ( f(c) is local minimum. If f () changes from negative to positive at c, then f(c) is a local minimum. f () ( m c n f () Increasing Decreasing ( f(c) is local maimum. If f () changes from positive to negative at c, then f(c) is a local maimum. f () f () ( m c n Decreasing Decreasing ( f(c) is not a local etremum. If f () does not change sign at c, then f(c) is neither a local maimum nor a local minimum. f () f () ( m c n Increasing Increasing ( f(c) is not a local etremum. If f () does not change sign at c, then f(c) is neither a local maimum nor a local minimum.

10 BARNMC_33886.QXD //7 :7 Page 83 Section - First Derivative and Graphs 83 EXAMPLE 7 Locating Local Etrema Given f = , (A) Find the critical values of f. (B) Find the local maima and minima. (C) Sketch the graph of f. SOLUTION (A) Find all numbers in the domain of f where f = or f does not eist. f = = = = = or = 3 f eists for all ; the critical values are = and = 3. (B) The easiest way to apply the first-derivative test for local maima and minima is to construct a sign chart for f for all. Partition numbers for f are = and = 3 (which also happen to be critical values of f). Sign chart for f = : f () f () (, ) (, 3) (3, ) Increasing Decreasing 3 Increasing Test Numbers f Local maimum Local minimum The sign chart indicates that f increases on - q,, has a local maimum at =, decreases on (, 3), has a local minimum at = 3, and increases on 3, q. These facts are summarized in the following table: f f() Graph of f (-q, ) + Increasing Rising = Local maimum Horizontal tangent (, 3) - Decreasing Falling = 3 Local minimum Horizontal tangent (3, q) + Increasing Rising (C) We sketch a graph of f, using the information from part (B) and point-by-point plotting. f() f() 6 Local maimum Local minimum 4

11 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization MATCHED PROBLEM 7 Given f = , (A) Find the critical values of f. (B) Find the local maima and minima. (C) Sketch a graph of f. How can you tell if you have found all the local etrema of a function? In general, this can be a difficult question to answer. However, in the case of a polynomial function, there is an easily determined upper it on the number of local etrema. Since the local etrema are the intercepts of the derivative, this it is a consequence of the number of intercepts of a polynomial. The relevant information is summarized in the following theorem, which is stated without proof: THEOREM 3 INTERCEPTS AND LOCAL EXTREMA OF POLYNOMIAL FUNCTIONS If f = a n n + a n - n - + Á + a + a, a n Z, is an nth-degree polynomial, then f has at most n intercepts and at most n - local etrema. Theorem 3 does not guarantee that every nth-degree polynomial has eactly n - local etrema; it says only that there can never be more than n - local etrema. For eample, the third-degree polynomial in Eample 7 has two local etrema, while the third-degree polynomial in Eample does not have any. Applications to Economics In addition to providing information for hand-sketching graphs, the derivative is an important tool for analyzing graphs and discussing the interplay between a function and its rate of change. The net two eamples illustrate this process in the contet of some applications to economics. EXAMPLE 8 Agricultural Eports and Imports Over the past several decades, the United States has eported more agricultural products than it has imported, maintaining a positive balance of trade in this area. However, the trade balance fluctuated considerably during that period.the graph in Figure approimates the rate of change of the balance of trade over a -year period, where B(t) is the balance of trade (in billions of dollars) and t is time (in years). B (t) Billions per year t Years FIGURE Rate of change of the balance of trade (A) Write a brief verbal description of the graph of y = Bt, including a discussion of any local etrema. (B) Sketch a possible graph of y = Bt.

12 BARNMC_33886.QXD //7 :7 Page 8 Section - First Derivative and Graphs 8 SOLUTION (A) The graph of the derivative y = B t contains the same essential information as a sign chart. That is, we see that B t is positive on (, 4), at t = 4, negative on (4, ), at t =, and positive on (, ). Hence, the trade balance increases for the first 4 years to a local maimum, decreases for the net 8 years to a local minimum, and then increases for the final 3 years. (B) Without additional information concerning the actual values of y = Bt, we cannot produce an accurate graph. However, we can sketch a possible graph that illustrates the important features, as shown in Figure 3. The absence of a scale on the vertical ais is a consequence of the lack of information about the values of B(t). B(t) Years FIGURE 3 Balance of trade t MATCHED PROBLEM 8 The graph in Figure 4 approimates the rate of change of the U.S. share of the total world production of motor vehicles over a -year period, where S(t) is the U.S. share (as a percentage) and t is time (in years). S (t) 3 Percent per year Years t FIGURE 4 (A) Write a brief verbal description of the graph of y = St, including a discussion of any local etrema. (B) Sketch a possible graph of y = St. EXAMPLE 9 Revenue Analysis The graph of the total revenue R() (in dollars) from the sale of bookcases is shown in Figure. (A) Write a brief verbal description of the graph of the marginal revenue function y = R, including a discussion of any intercepts. (B) Sketch a possible graph of y = R.

13 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization $4, R() $,, FIGURE Revenue SOLUTION (A) The graph of y = R indicates that R() increases on (, ), has a local maimum at =, and decreases on (,,). Consequently, the marginal revenue function R must be positive on (, ), at =, and negative on (,,). (B) A possible graph of y = R illustrating the information summarized in part (A) is shown in Figure 6. R (), FIGURE 6 Marginal revenue MATCHED PROBLEM 9 The graph of the total revenue R() (in dollars) from the sale of desks is shown in Figure 7. R() $6, $4, $, 4 6 8, FIGURE 7 (A) Write a brief verbal description of the graph of the marginal revenue function y = R, including a discussion of any intercepts. (B) Sketch a possible graph of y = R. Comparing Eamples 8 and 9, we see that we were able to obtain more information about the function from the graph of its derivative (Eample 8) than we were when the process was reversed (Eample 9). In the net section, we introduce some ideas that will enable us to etract additional information about the derivative from the graph of the function.

14 BARNMC_33886.QXD //7 :7 Page 87 Section - First Derivative and Graphs 87 Answers to Matched Problems. (A) Horizontal tangent line at = 3. (C) (B) Decreasing on - q, 3; increasing on 3, q f(). Partition number: = ; critical value: = ; decreasing for all 3. Partition number: = -; critical value: = -; increasing for all 4. Partition number: = ; no critical values; decreasing on - q, and, q. Partition number: = ; critical value: = ; increasing on, ; decreasing on, q 6. Increasing on -3, ; decreasing on - q, -3 and, q; local maimum at = ; local minimum at = (A) Critical values: =, = 4 (B) Local maimum at = ; local minimum at = 4 (C) f() 8. (A) The U.S. share of the world market decreases for 6 years to a local minimum, increases for the net years to a local maimum, and then decreases for the final 4 years. (B) S(t) t 9. (A) The marginal revenue is positive on (, 4), at = 4, and negative on (4,,). (B) R (), Eercise - A Problems 8 refer to the following graph of y = f: f() 3. Identify the intervals on which f Identify the intervals on which f 7.. Identify the coordinates of the points where f =. d a b c e f g h. Identify the intervals on which f() is increasing.. Identify the intervals on which f() is decreasing. 6. Identify the coordinates of the points where f does not eist. 7. Identify the coordinates of the points where f() has a local maimum. 8. Identify the coordinates of the points where f() has a local minimum.

15 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization In Problems 9 and, f() is continuous on - q, q and has critical values at = a, b, c, and d. Use the sign chart for f to determine whether f has a local maimum, a local minimum, or neither at each critical value. 4. f() 6 9. f () ND ND a b c d 6. f () ND ND. f() 6 a b c d 6 In Problems 8, match the graph of f with one of the sign charts a h in the figure.. f() 6 6. f() f() 7. f() f() 8. f()

16 BARNMC_33886.QXD //7 :7 Page 89 Section - First Derivative and Graphs 89 (a) (b) (c) (d) (e) (f) f () f () f () f () f () f () 3 ND 3 3 ND 3 3 ND 3 3. f = >3 36. f = 4>3-7 >3 In Problems 37 46, use a graphing calculator to approimate the critical values of f() to two decimal places. Find the intervals on which f() is increasing, the intervals on which f() is decreasing, and the local etrema. 37. f = f = f = f = f = ln f = 3 - >3-4>3 43. f = e f = e f = >3 + 4>3-46. f = ln - + In Problems 47 4, find the intervals on which f() is increasing and the intervals on which f() is decreasing. Then sketch the graph. Add horizontal tangent lines. (g) (h) f () f () ND 3 3 Figure for f = f = f = f = f = f = f = 4-8 f = B In Problems 9 36, find the intervals on which f() is increasing, the intervals on which f() is decreasing, and the local etrema. 9. f = - 4. f = f = f = f = f = f = f = f = f = f = - e - 3. f = ln - 3. f = 4 >3 - >3 3. f = - 9 >3 33. f = - ln 34. f = + e In Problems 6, f() is continuous on - q, q. Use the given information to sketch the graph of f.. 6. f () f () - - f() f() 3 -

17 BARNMC_33886.QXD //7 :7 Page 9 9 CHAPTER Graphing and Optimization 7. f () ND f 3 () f 4 () f() 8. f () ND f () f 6 () f() f- = 4, f =, f = -4; f - =, f =, f = ; f 7 on - q, - and, q; f 6 on -, and (, ) g () Figure (A) for g () 6. f- = -, f =, f = ; f - =, f = ; f 7 on - q, -, -,, and, q 6. f- =, f =, f = -; f - =, f =, f is not defined; f 7 on - q, - and, q; f 6 on -, and (, ) g 3 () g 4 () 6. f- =, f =, f = ; f - =, f =, f is not defined; f 7 on - q, - and (, ); f 6 on -, and, q Problems involve functions f f 6 and their derivatives, g g 6. Use the graphs shown in figures (A) and (B) to match each function with its derivative g j. f i f () f () g () g 6 () Figure (B) for 63 68

18 BARNMC_33886.QXD //7 :7 Page 9 Section - First Derivative and Graphs f f 7. f() 76. f() 6. f f 4 f 68. f 6 In Problems 69 74, use the given graph of y = f to find the intervals on which f is increasing, the intervals on which f is decreasing, and the local etrema. Sketch a possible graph of y = f. 77. f() 78. f() 69. f () 7. f () C In Problems 79 9, find the critical values, the intervals on which f() is increasing, the intervals on which f() is decreasing, and the local etrema. Do not graph. 79. f = f () 7. f () 8. f = f = f = f = f = f = f () 74. f () f = 3 - f = 3 - >3 + 4 f = 64 - >3 + 4 In Problems 7 78, use the given graph of y = f to find the intervals on which f 7, the intervals on which f 6, and the values of for which f =. Sketch a possible graph of y = f. 89. f = + 9. f = Let f = 3 + k, where k is a constant. Discuss the number of local etrema and the shape of the graph of f if (A) k 7 (B) k 6 (C) k = 9. Let f = 4 + k, where k is a constant. Discuss the number of local etrema and the shape of the graph of f if (A) k 7 (B) k 6 (C) k =

19 BARNMC_33886.QXD //7 :7 Page 9 9 CHAPTER Graphing and Optimization Applications 93. Profit analysis. The graph of the total profit P() (in dollars) from the sale of cordless electric screwdrivers is shown in the figure. 4,,, P(), (A) Write a brief verbal description of the graph of y = Bt, including a discussion of any local etrema. (B) Sketch a possible graph of y = Bt. 96. Price analysis. The graph in the figure approimates the rate of change of the price of eggs over a 7-month period, where E(t) is the price of a dozen eggs (in dollars) and t is time (in months).. E (t) 4, Figure for t (A) Write a brief verbal description of the graph of the marginal profit function y = P, including a discussion of any intercepts. (B) Sketch a possible graph of y = P. 94. Revenue analysis. The graph of the total revenue R() (in dollars) from the sale of cordless electric screwdrivers is shown in the figure. 4,,, R(),..3 Figure for 96 (A) Write a brief verbal description of the graph of y = Et, including a discussion of any local etrema. (B) Sketch a possible graph of y = Et. 97. Average cost. A manufacturer incurs the following costs in producing toasters in one day, for 6 6 : fied costs, $3; unit production cost, $ per toaster; equipment maintenance and repairs,. dollars. Thus, the cost of manufacturing toasters in one day is given by C = , Figure for 94 (A) Write a brief verbal description of the graph of the marginal revenue function y = R, including a discussion of any intercepts. (B) Sketch a possible graph of y = R. 9. Price analysis. The graph in the figure approimates the rate of change of the price of bacon over a 7-month period, where B(t) is the price of a pound of sliced bacon (in dollars) and t is time (in months) B (t) 3 7 Figure for 9 t (A) What is the average cost C per toaster if toasters are produced in one day? (B) Find the critical values of C, the intervals on which the average cost per toaster is decreasing, the intervals on which the average cost per toaster is increasing, and the local etrema. Do not graph. 98. Average cost. A manufacturer incurs the following costs in producing blenders in one day for 6 6 : fied costs, $4; unit production cost, $3 per blender; equipment maintenance and repairs,.8 dollars. (A) What is the average cost C per blender if blenders are produced in one day? (B) Find the critical values of C, the intervals on which the average cost per blender is decreasing, the intervals on which the average cost per blender is increasing, and the local etrema. Do not graph.

20 BARNMC_33886.QXD //7 :7 Page 93 Section - Second Derivative and Graphs Marginal analysis. Show that profit will be increasing over production intervals (a, b) for which marginal revenue is greater than marginal cost. [Hint: P = R - C]. Marginal analysis. Show that profit will be decreasing over production intervals (a, b) for which marginal revenue is less than marginal cost.. Medicine. A drug is injected into the bloodstream of a patient through the right arm. The concentration of the drug in the bloodstream of the left arm t hours after the injection is approimated by Ct =.8t t 6 t Find the critical values of C(t), the intervals on which the concentration of the drug is increasing, the intervals on which the concentration of the drug is decreasing, and the local etrema. Do not graph.. Medicine. The concentration C(t), in milligrams per cubic centimeter, of a particular drug in a patient s bloodstream is given by.3t Ct = t 6 t 6 + 6t + 9 where t is the number of hours after the drug is taken orally. Find the critical values of C(t), the intervals on which the concentration of the drug is increasing, the intervals on which the concentration of the drug is decreasing, and the local etrema. Do not graph. 3. Politics. Public awareness of a congressional candidate before and after a successful campaign was approimated by Pt = 8.4t t +. 6 t where t is time (in months) after the campaign started and P(t) is the fraction of people in the congressional district who could recall the candidate s (and later, congressman s) name. Find the critical values of P(t), the time intervals on which the fraction is increasing, the time intervals on which the fraction is decreasing, and the local etrema. Do not graph. Section - SECOND DERIVATIVE AND GRAPHS Using Concavity as a Graphing Tool Finding Inflection Points Analyzing Graphs Curve Sketching Point of Diminishing Returns In Section -, we saw that the derivative can be used when a graph is rising and falling. Now we want to see what the second derivative (the derivative of the derivative) can tell us about the shape of a graph. Using Concavity as a Graphing Tool Consider the functions f = and g = for in the interval, q. Since f = 7 for 6 6 q and g = 7 for 6 6 q both functions are increasing on, q.

21 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization Eplore & Discuss (A) Discuss the difference in the shapes of the graphs of f and g shown in Figure. f() g() (, ) (, ) (A) f() (B) g() FIGURE (B) Complete the following table, and discuss the relationship between the values of the derivatives of f and g and the shapes of their graphs:...7 f g We use the term concave upward to describe a graph that opens upward and concave downward to describe a graph that opens downward. Thus, the graph of f in Figure A is concave upward, and the graph of g in Figure B is concave downward. Finding a mathematical formulation of concavity will help us sketch and analyze graphs. It will be instructive to eamine the slopes of f and g at various points on their graphs (see Fig. ). We can make two observations about each graph:. Looking at the graph of f in Figure A, we see that f (the slope of the tangent line) is increasing and that the graph lies above each tangent line;. Looking at Figure B, we see that g is decreasing and that the graph lies below each tangent line. f() g() f () f (.7). f (.). f (.) g (). g (.7).6 g (.).7 g (.) (A) f() FIGURE (B) g() With these ideas in mind, we state the general definition of concavity. DEFINITION Concavity The graph of a function f is concave upward on the interval (a, b) if f is increasing on (a, b) and is concave downward on the interval (a, b) if f is decreasing on (a, b).

22 BARNMC_33886.QXD //7 :7 Page 9 Section - Second Derivative and Graphs 9 Geometrically, the graph is concave upward on (a, b) if it lies above its tangent lines in (a, b) and is concave downward on (a, b) if it lies below its tangent lines in (a, b). How can we determine when f is increasing or decreasing? In Section -, we used the derivative of a function to determine when that function is increasing or decreasing. Thus, to determine when the function f is increasing or decreasing, we use the derivative of f. The derivative of the derivative of a function is called the second derivative of the function. Various notations for the second derivative are given in the following bo: NOTATION Second Derivative For y = f, the second derivative of f, provided that it eists, is f = d d f Other notations for f are d y d y Returning to the functions f and g discussed at the beginning of this section, we have f = g = = > f = g = -> = f = d d = g = d d -> = - 4-3> = For 7, we see that f 7 ; thus, f is increasing and the graph of f is concave upward (see Fig. A). For 7, we also see that g 6 ; thus, g is decreasing and the graph of g is concave downward (see Fig. B). These ideas are summarized in the following bo: SUMMARY CONCAVITY For the interval (a, b), f () f () Graph of y = f() Eamples + Increasing Concave upward - Decreasing Concave downward I N S I G H T Be careful not to confuse concavity with falling and rising. A graph that is concave upward on an interval may be falling, rising, or both falling and rising on that interval. A similar statement holds for a graph that is concave downward. See Figure 3 on the net page.

23 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization f () > on (a, b) Concave upward f f f a b a b a b (A) f () is negative and increasing. Graph of f is falling. (B) f () increases from negative to positive. Graph of f falls, then rises. (C) f () is positive and increasing. Graph of f is rising. f () < on (a, b) Concave downward f f f a b a b a b (D) f () is positive and decreasing. Graph of f is rising. (E) f () decreases from positive to negative. Graph of f rises, then falls. FIGURE 3 Concavity (F) f () is negative and decreasing. Graph of f is falling. EXAMPLE Concavity of Graphs Determine the intervals on which the graph of each function is concave upward and the intervals on which it is concave downward. Sketch a graph of each function. (A) f = e (B) g = ln (C) h = 3 SOLUTION (A) f = e (B) g = ln (C) h = 3 f = e f = e g = g = - h = 3 h = 6 Since f 7 on - q, q, the graph of f = e [Fig. 4(A)] is concave upward on - q, q. The domain of g = ln Since h 6 when 6 is, q and g 6 on and h 7 when 7, this interval, so the graph the graph of h = 3 [Fig. of g = ln [Fig. 4(B)] 4(C)] is concave downward is concave downward on on - q, and concave, q. upward on, q. f () g() h() Concave upward 4 f () e (A) Concave upward for all. g() ln Concave downward (B) Concave downward for. 4 Concave downward h() 3 (C) Concavity changes at the origin. Concave upward FIGURE 4

24 BARNMC_33886.QXD //7 :7 Page 97 Section - Second Derivative and Graphs 97 MATCHED PROBLEM Determine the intervals on which the graph of each function is concave upward and the intervals on which it is cancave downward. Sketch a graph of each function. (A) f = -e - (B) g = ln (C) h = >3 Refer to Eample. The graphs of f = e and g = ln never change concavity. But the graph of h = 3 changes concavity at,. This point is called an inflection point. Eplore & Discuss Finding Inflection Points Discuss the relationship between the change in concavity of each of the following functions at = and the second derivative at and near : (A) f = 3 (B) g = 4>3 (C) h = 4 In general, an inflection point is a point on the graph of the function where the concavity changes (from upward to downward or from downward to upward). For the concavity to change at a point, f must change sign at that point. But in Section 3-, we saw that the partition numbers* identify the points where a function can change sign. Thus, we have the following theorem: THEOREM INFLECTION POINTS If y = f is continuous on (a, b) and has an inflection point at = c, then either f c = or f c does not eist. Note that inflection points can occur only at partition numbers of f, but not every partition number of f produces an inflection point. Two additional requirements must be satisfied for an inflection point to occur: A partition number c for f fl. f fl changes sign at c and. c is in the domain of f. Figure illustrates several typical cases. produces an inflection point for the graph of f only if c c c c f () f () f () f () ND (A) f (c) (B) f (c) (C) f (c) (D) f (c) is not defined FIGURE Inflection points If f c eists and f changes sign at = c, the tangent line at an inflection point (c, f(c)) will always lie below the graph on the side that is concave upward and above the graph on the side that is concave downward (see Fig. A, B, and C). *As we did with the first derivative, we assume that if f is discontinuous at c, then f c does not eist.

25 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization EXAMPLE Locating Inflection Points Find the inflection point(s) of f = SOLUTION Since inflection point(s) occur at values of where f changes sign, we construct a sign chart for f. We have f = f = f = 6 - = 6 - The sign chart for f = 6 - (partition number is ) is f () Graph of f (, ) Concave downward Inflection point (, ) Concave upward Test Numbers f From the sign chart, we see that the graph of f has an inflection point at =. That is, the point, f =, 3 f = 3-6 # + 9 # + = 3 is an inflection point on the graph of f. MATCHED PROBLEM Find the inflection point(s) of f = EXAMPLE 3 Locating Inflection Points Find the inflection point(s) of f = ln SOLUTION First we find the domain of f (a good first step for most calculus problems involving properties of the graph of a function). Since ln is defined only for 7, f is defined only for Use completing the square (Section -3). True for all. So the domain of f is - q, q. Now we find f and construct a sign chart for it. We have f = ln f = f = ( = =

26 BARNMC_33886.QXD //7 :7 Page 99 Section - Second Derivative and Graphs 99 = = The partition numbers for f are = and = 3. Sign chart for f f () (, ) (, 3) Concave downward Concave upward 3 (3, ) Concave downward Test Numbers f - 6 Inflection point Inflection point 4-6 The sign chart shows that the graph of f has inflection points at = and = 3. MATCHED PROBLEM 3 Find the inflection point(s) of f = ln - + I N S I G H T It is important to remember that the partition numbers for f are only candidates for inflection points. The function f must be defined at = c, and the second derivative must change sign at = c in order for the graph to have an inflection point at = c. For eample, consider f = 4 g = f = 4 3 g = - f = g = 3 In each case, = is a partition number for the second derivative, but neither the graph of f nor the graph of g has an inflection point at =. Function f does not have an inflection point at = because f does not change sign at = (see Fig. 6A). Function g does not have an inflection point at = because g() is not defined (see Fig. 6B). f() g() (A) f() 4 (B) g() FIGURE 6

27 BARNMC_33886.QXD //7 :7 Page 3 3 CHAPTER Graphing and Optimization Analyzing Graphs In the net eample, we combine increasing/decreasing properties with concavity properties to analyze the graph of a function. EXAMPLE 4 Analyzing a Graph Figure 7 shows the graph of the derivative of a function f. Use this graph to discuss the graph of f. Include a sketch of a possible graph of f. f () FIGURE 7 SOLUTION The sign of the derivative determines where the original function is increasing and decreasing, and the increasing/decreasing properties of the derivative determine the concavity of the original function. The relevant information obtained from the graph of f is summarized in Table, and a possible graph of f is shown in Figure 8. f() TABLE f œ () (Fig. 7) f() (Fig. 8) FIGURE 8 - q = = 6 6 = 6 6 q Negative and increasing Local maimum Negative and decreasing Local minimum Negative and increasing intercept Positive and increasing Decreasing and concave upward Inflection point Decreasing and concave downward Inflection point Decreasing and concave upward Local minimum Increasing and concave upward MATCHED PROBLEM 4 Figure 9 shows the graph of the derivative of a function f. Use this graph to discuss the graph of f. Include a sketch of a possible graph of f. f () FIGURE 9

28 BARNMC_33886.QXD //7 :7 Page 3 Curve Sketching Section - Second Derivative and Graphs 3 Graphing calculators and computers produce the graph of a function by plotting many points. Although the technology is quite accurate, important points on a plot many be difficult to identify. Using information gained from the function f and its derivatives, and plotting the important points intercepts, local etrema, and inflection points we can sketch by hand a very good representation of the graph of f. This graphing process is called curve sketching and is summarized net. PROCEDURE Graphing Strategy (First Version)* Step. Analyze f(). Find the domain and the intercepts. The intercepts are the solutions of f = and the y intercept is f. Step. Analyze f. Find the partition numbers for, and critical values of, f. Construct a sign chart for f, determine the intervals on which f is increasing and decreasing, and find local maima and minima. Step 3. Analyze f. Find the partition numbers for f. Construct a sign chart for f, determine the intervals on which the graph of f is concave upward and concave downward, and find inflection points. Step 4. Sketch the graph of f. Locate intercepts, local maima and minima, and inflection points. Sketch in what you know from steps 3. Plot additional points as needed and complete the sketch. Eample will illustrate the use of this strategy. EXAMPLE SOLUTION Using the Graphing Strategy Follow the graphing strategy and analyze the function State all the pertinent information and sketch the graph of f. Step. Analyze f(). Since f is a polynomial, its domain is - q. q. intercept: f = 4-3 = 3 - = =, y intercept: f = f = 4-3 Step. Analyze f. f = = Critical values of f(): and Partition numbers for f : and 3 Sign chart for f : f () f () (, ) (, w) (w, ) w Decreasing Decreasing Increasing Test Numbers f Local minimum *We will modify this summary in Section -4 to include some additional information about the graph of f.

29 BARNMC_33886.QXD //7 :7 Page 3 3 CHAPTER Graphing and Optimization Thus, f() is decreasing on - q, 3 is increasing on 3 and has a local minimum at = 3,, q,. Step 3. Analyze f. f = - = - Partition numbers for f : and Sign chart for f : f () Graph of f (, ) (, ) (, ) Concave upward Concave downward Concave upward Test Numbers f Inflection point Inflection point Thus, the graph of f is concave upward on - q, and, q, is concave downward on (, ), and has inflection points at = and =. Step 4. Sketch the graph of f. f() f() MATCHED PROBLEM Follow the graphing strategy and analyze the function f = State all the pertinent information and sketch the graph of f. I N S I G H T Refer to the solution to Eample. Combining the sign charts for f and f (Fig. ) partitions the real-number line into intervals on which neither f nor f changes sign. On each of these intervals, the graph of f() must have one of four basic shapes (see also Fig. 3, parts A, C, D, and F). This reduces sketching the graph of a function to plotting the points identified in the graphing strategy and connecting them with one of the basic shapes. f () Graph of f() (, ) f () Decreasing, concave upward (, ) Decreasing, concave downward (,.) Decreasing, concave upward. (., ) Increasing, concave upward Basic shape FIGURE

30 BARNMC_33886.QXD //7 :7 Page 33 Section - Second Derivative and Graphs 33 EXAMPLE 6 Using the Graphing Strategy Follow the graphing strategy and analyze the function f = 3 >3 - State all the pertinent information and sketch the graph of f. Round any decimal values to two decimal places. SOLUTION Step. Analyze f(). f = 3 >3 - Since p is defined for any and any positive p, the domain of f is - q, q. intercepts: Solve f = 3 >3 - = 3a >3-3 b = a - b = a - ba + b 3a >3 - A 3 ba>3 + A 3 b = The intercepts of f are y intercept: f =. Step. Analyze f. =, = a A 3 b 3 L 7., = a - A 3 b 3 L -7. f = >3 - = >3-4 = >3 - >3 + Critical values of f: = 3 = 8 and = - 3 = -8. Partition numbers for f: -8, 8 Sign chart for f : Again, a - b = a - ba + b f () (, 8) ( 8, 8) 8 8 (8, ) Increasing Decreasing Increasing Test Numbers f Local maimum Local minimum So f is increasing on - q, -8 and 8, q and decreasing on -8, 8, f-8 is a local maimum, and f(8) is a local minimum. Step 3. Analyze f. Partition number for f : f = >3 - f = 3 ->3 = 3 >3

31 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization Sign chart for f : f () (, ) ND Concave downward (, ) Concave upward Test Numbers f fl Inflection point So f is concave downward on - q,, is concave upward on, q, and has an inflection point at =. Step 4. Sketch the graph of f. 6 f() f () 3 / f MATCHED PROBLEM 6 Follow the graphing strategy and analyze the function f = 3 >3 -. State all the pertinent information and sketch the graph of f. Round any decimal values to two decimal places. Point of Diminishing Returns If a company decides to increase spending on advertising, it would epect sales to increase. At first, sales will increase at an increasing rate and then increase at a decreasing rate. The value of where the rate of change of sales changes from increasing to decreasing is called the point of diminishing returns. This is also the point where the rate of change has a maimum value. Money spent after this point may increase sales, but at a lower rate. The net eample illustrates this concept. EXAMPLE 7 Maimum Rate of Change Currently, a discount appliance store is selling large-screen television sets monthly. If the store invests $ thousand in an advertising campaign, the ad company estimates that sales will increase to N = When is rate of change of sales increasing and when is it decreasing? What is the point of diminishing returns and the maimum rate of change of sales? Graph N and N on the same coordinate system. SOLUTION The rate of change of sales with respect to advertising ependitures is N = 9-3 = 9 -

32 BARNMC_33886.QXD //7 :7 Page 3 Section - Second Derivative and Graphs 3 To determine when N is increasing and decreasing, we find N, the derivative of N : N = 8-3 = 36 - The information obtained by analyzing the signs of N and N is summarized in Table (sign charts are omitted). TABLE N () N () N () N() = Increasing Increasing, concave upward + Local maimum Inflection point - + Decreasing Increasing, concave downward Eamining Table, we see that N is increasing on (, 6) and decreasing on (6, 9). The point of diminishing returns is = 6 and the maimum rate of change is N 6 = 8. Note that N has a local maimum and N() has an inflection point at = 6. y y N() N () N () N () N () y N () Point of diminishing returns MATCHED PROBLEM 7 Repeat Eample 7 for N = Answers to Matched Problems. (A) Concave downward on - q, q f() f () e 3 (B) Concave upward on, q g() g() ln 3

33 BARNMC_33886.QXD //7 :7 Page CHAPTER Graphing and Optimization (C) Concave upward on - q, and concave downward on, q h() h() /3. The only inflection point is 3, f3 = 3, 8. f() 3. The inflection points are -, f- = -, ln 8 and 3, f3 = 3, ln f f() - q Positive and decreasing Increasing and concave downward = - Local minimum Inflection point Positive and increasing Increasing and concave upward = Local maimum Inflection point 6 6 Positive and decreasing Increasing and concave downward = intercept Local maimum 6 6 q Negative and decreasing Decreasing and concave downward. intercepts: -4, ; y intercept: f = Decreasing on - q, -3; increasing on -3, q; local minimum at = -3 Concave upward on - q, - and, q; concave downward on -, Inflection points at = - and = f() f() intercepts:, 7; y intercept: f() Decreasing on - q, and 8, q; increasing on, 8; local minimum: f() ; local maimum: f(8) 4 Concave downward on - q, and, q; no inflection points f() f()

34 BARNMC_33886.QXD //7 :8 Page 37 Section - Second Derivative and Graphs N is increasing on (, 8) and decreasing on (8, ). The point of diminishing returns is = 8 and the maimum rate of change is N 8 = 6. y N () y N() N () N () N () y N () 8 Point of diminishing returns Eercise - A. Use the graph of y = f to identify (A) Intervals on which the graph of f is concave upward (B) Intervals on which the graph of f is concave downward (C) Intervals on which f 6 (D) Intervals on which f 7 (E) Intervals on which f is increasing (F) Intervals on which f is decreasing (G) The coordinates of inflection points (H) The coordinates of local etrema for f a a b c d e f g h b c d f(). Use the graph of y = g to identify (A) Intervals on which the graph of g is concave upward (B) Intervals on which the graph of g is concave downward (C) Intervals on which g 6 (D) Intervals on which g 7 (E) Intervals on which g is increasing (F) Intervals on which g is decreasing (G) The coordinates of inflection points (H) The coordinates of local etrema for g g() e f h g In Problems 3 6, match the indicated conditions with one of the graphs (A) (D) shown in the figure. f() a b (A) f() a b (B) f() a b (C) f() a b (D) k = f 7 and f 7 on (a, b) 4. f 7 and f 6 on (a, b). f 6 and f 7 on (a, b) 6. f 6 and f 6 on (a, b) In Problems 7 8, find the indicated derivative for each function. 7. f for f = g for g = h for h = k for d y>d for y = - 8 >. d y>d for y = 3-4 >3 3. y for y = y for y = - 6. f for f = e - 6. f for f = e - 7. y for y = ln 8. y for y = ln In Problems 9 8, find the intervals on which the graph of f is concave upward, the intervals on which the graph of f is concave downward, and the inflection points. 9. f = f = 4 + 6

35 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization. f = f =, f =, f = -; f = f = f = f =, f = ; f 7 on - q, and, q; f 6 on (, );. f = ln f = ; f = ln f 7 on, q; f = 8e - e f 6 on - q, 8. f = e 3-9e 34. f- = -, f =, f = 4; f - =, f = ; In Problems 9 36, f() is continuous on - q, q. Use the given information to sketch the graph of f. f 7 on -, ; f 6 on - q, - and, q; f = ; f () f () f () f () f () f() f() f() ND 4 B f 7 on - q, ; f 6 on, q f- =, f = -, f = ; f =, f - and f are not defined; f 7 on (, ) and, q; f 6 on - q, - and -, ; f - and f are not defined; f 7 on -, ; f 6 on - q, - and, q f = -, f =, f = 4; f =, f =, f is not defined; f 7 on (, ) and (, ); f 6 on - q, and, q; f is not defined; f 7 on - q, ; f 6 on, q In Problems 37 8, summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of y = f f = f = f = f = f () ND f = f =. 4-3 f = f = f() f = f () f () ND ND f = f = f = f = 6-3. f = 3-4

36 BARNMC_33886.QXD //7 :8 Page 39 Section - Second Derivative and Graphs f = - e - f = - 3e - f = e. + 4e -. f = e. + e -. In Problems 63 7, apply steps 3 of the graphing strategy to f (). Use a graphing calculator to approimate (to two decimal places) intercepts, critical values, and coordinates of inflection points. Summarize all the pertinent information. 63. f = f = -4 + ln 64. f = f = - 3 ln 6. f = f = ln f = f = - ln f = In Problems 9 6, use the graph of y = f to discuss the graph of y = f. Organize your conclusions in a table (see Eample 4), and sketch a possible graph of y = f. 9. f () 6. f () f = f = f = C In Problems 7 74, assume that f is a polynomial. 7. Eplain how you can locate inflection points for the graph of y = f by eamining the graph of y = f. 7. Eplain how you can determine where f is increasing or decreasing by eamining the graph of y = f. 6. f () 6. f () 73. Eplain how you can locate local maima and minima for the graph of y = f by eamining the graph of y = f. 74. Eplain how you can locate local maima and minima for the graph of y = f by eamining the graph of y = f. Applications 7. Inflation. One commonly used measure of inflation is the annual rate of change of the Consumer Price Inde (CPI). A newspaper headline proclaims that the rate of change of inflation for consumer prices is increasing. What does this say about the shape of the graph of the CPI? 76. Inflation. Another commonly used measure of inflation is the annual rate of change of the Producer Price Inde (PPI). A government report states that the rate of change of inflation for producer prices is decreasing. What does this say about the shape of the graph of the PPI? 77. Cost analysis. A company manufactures a variety of lighting fitures at different locations. The total cost C() (in dollars) of producing desk lamps per week at plant A is shown in the figure. Discuss the graph of the marginal cost function C and interpret the graph of C () in terms of the efficiency of the production process at this plant. $, $, Figure for 77 C(), Production costs at plant A 78. Cost analysis. The company in Problem 77 produces the same lamp at another plant. The total cost C() (in dollars) of producing desk lamps per week at plant B is shown in the figure. Discuss the graph of the marginal cost function C and interpret the graph of C () in terms of the efficiency of the production process at

37 BARNMC_33886.QXD //7 :8 Page 3 3 CHAPTER Graphing and Optimization plant B. Compare the production processes at the two plants. $, C() It was found that the demand for the new hot dog is given approimately by P = 8 - ln where is the number of hot dogs (in thousands) that can be sold during one game at a price of $p. $, Figure for 78, Production costs at plant B 79. Revenue. The marketing research department of a computer company used a large city to test market the firm s new product. The department found that the relationship between price p (dollars per unit) and the demand (units per week) was given approimately by p =, Thus, weekly revenue can be approimated by R = p =, (A) Find the local etrema for the revenue function. (B) On which intervals is the graph of the revenue function concave upward? Concave downward? 8. Profit. Suppose that the cost equation for the company in Problem 79 is C = (A) Find the local etrema for the profit function. (B) On which intervals is the graph of the profit function concave upward? Concave downward? 8. Revenue. A cosmetics company is planning to introduce and promote a new lipstick line. After test marketing the new line in a carefully selected large city, the marketing research department found that the demand in that city is given approimately by p = e - where thousand lipsticks were sold per week at a price of $p each. (A) Find the local etrema for the revenue function. (B) On which intervals is the graph of the revenue function concave upward? Concave downward? 8. Revenue. A national food service runs food concessions for sporting events throughout the country. The company s marketing research department chose a particular football stadium to test market a new jumbo hot dog. (A) Find the local etrema for the revenue function. (B) On which intervals is the graph of the revenue function concave upward? Concave downward? 83. Production: point of diminishing returns. A T-shirt manufacturer is planning to epand its workforce. It estimates that the number of T-shirts produced by hiring new workers is given by T = When is the rate of change of T-shirt production increasing and when is it decreasing? What is the point of diminishing returns and the maimum rate of change of T-shirt production? Graph T and T on the same coordinate system. 84. Production: point of diminishing returns. A baseball cap manufacturer is planning to epand its workforce. It estimates that the number of baseball caps produced by hiring new workers is given by T = When is the rate of change of baseball cap production increasing and when is it decreasing? What is the point of diminishing returns and the maimum rate of change of baseball cap production? Graph T and T on the same coordinate system. 8. Advertising: point of diminishing returns. A company estimates that it will sell N() units of a product after spending $ thousand on advertising, as given by N = , 4 4 When is the rate of change of sales increasing and when is it decreasing? What is the point of diminishing returns and the maimum rate of change of sales? Graph N and N on the same coordinate system. 86. Advertising: point of diminishing returns. A company estimates that it will sell N() units of a product after spending $ thousand on advertising, as given by N = , 4 When is the rate of change of sales increasing and when is it decreasing? What is the point of diminishing returns and the maimum rate of change of sales? Graph N and N on the same coordinate system.

38 BARNMC_33886.QXD //7 :8 Page 3 Section -3 L Hôpital s Rule Advertising. An automobile dealer uses television advertising to promote car sales. On the basis of past records, the dealer arrived at the following data, where is the number of ads placed monthly and y is the number of cars sold that month: Number of Ads Number of Cars y (A) Enter the data in a graphing calculator and find a cubic regression equation for the number of cars sold monthly as a function of the number of ads. (B) How many ads should the dealer place each month to maimize the rate of change of sales with respect to the number of ads, and how many cars can the dealer epect to sell with this number of ads? Round answers to the nearest integer. 88. Advertising. A music store advertises on the radio to promote sales of CDs. The store manager used past records to determine the following data, where is the number of ads placed monthly and y is the number of CDs sold that month. Number of Ads Number of CDs y ,8 4,67,9 (A) Enter the data in a graphing calculator and find a cubic regression equation for the number of CDs sold monthly as a function of the number of ads. (B) How many ads should the store manager place each month to maimize the rate of change of sales with respect to the number of ads, and how many CDs can the manager epect to sell with this number of ads? Round answers to the nearest integer. 89. Population growth: bacteria. A drug that stimulates reproduction is introduced into a colony of bacteria. After t minutes, the number of bacteria is given approimately by Nt =, + 3t - t 3 t (A) When is the rate of growth, N t, increasing? Decreasing? (B) Find the inflection points for the graph of N. (C) Sketch the graphs of N and N on the same coordinate system. (D) What is the maimum rate of growth? 9. Drug sensitivity. One hour after milligrams of a particular drug are given to a person, the change in body temperature T(), in degrees Fahrenheit, is given by T = a - 9 b 6 The rate T at which T() changes with respect to the size of the dosage is called the sensitivity of the body to the dosage. (A) When is T increasing? Decreasing? (B) Where does the graph of T have inflection points? (C) Sketch the graphs of T and T on the same coordinate system. (D) What is the maimum value of T? 9. Learning. The time T (in minutes) it takes a person to learn a list of length n is Tn =.8n 3 -.n + 6n n Ú (A) When is the rate of change of T with respect to the length of the list increasing? Decreasing? (B) Where does the graph of T have inflection points? Graph T and T on the same coordinate system. (C) What is the minimum value of T n? Section -3 L HÔPITAL S RULE Introduction L Hôpital s Rule and the Indeterminate Form One-Sided Limits and Limits at q L Hôpital s Rule and the Indeterminate Form q> q Introduction The ability to evaluate a wide variety of different types of its is one of the skills that are necessary to apply the techniques of calculus successfully. Limits play a fundamental role in the development of the derivative and are an important

39 BARNMC_33886.QXD //7 :8 Page 3 3 CHAPTER Graphing and Optimization graphing tool. In order to deal effectively with graphs, we need to develop some additional methods for evaluating its. In this section, we discuss a powerful technique for evaluating its of quotients called L Hôpital s rule. The rule is named after the French mathematician Marquis de L Hôpital (66 74), who is generally credited with writing the first calculus tetbook. To use L Hôpital s rule, it is necessary to be familiar with the it properties of some basic functions. Figure reviews some its involving powers of that we have discussed earlier. The its in Figure are easily etended to functions of the form f = - c n and g = > - c n. In general, if n is an odd integer, its involving - c n or > - c n as approaches c or ; q behave, respectively, like the its of and > as approaches or ; q. If n is an even integer, its involving these epressions behave, respectively, like the its of and as approaches or ; q. y y y y 3 y 3 y 3 3 y 3 3 y B 3 B B (A) y 3 B 3 B B (B) y B B B (C) y B B Does not eist 3 B B B 3 B B (D) y FIGURE Limits involving powers of EXAMPLE Limits Involving Powers of c (A) Compare with in Figure. : - 4 = q : (B) Compare with in Figure. + 3 = - q : : - 4 (C) Compare with in Figure. : q = : q (D) Compare with in Figure. : - q 33 = -q : - q MATCHED PROBLEM Evaluate each it. 7 6 (A) (B) : : (C) (D) :- q + 3 : q 4

40 BARNMC_33886.QXD //7 :8 Page 33 Section -3 L Hôpital s Rule 33 Figure reviews some its of eponential and logarithmic functions. The its in Figure also generalize to other simple eponential and logarithmic forms. y y y 3 y ln y e y e 3 e B 3 e B e B (A) y e 3 e B 3 e B e B (B) y e FIGURE Limits involving eponential and logarithmic functions 3 ln B (C) y ln ln B EXAMPLE Limits Involving Eponential and Logarithmic Forms (A) Compare with e in Figure. : q e3 = q : q (B) Compare with in Figure. : q 4e- = : q e- (C) ln + 4 = q Compare with ln in Figure. : q : q (D) ln - = - q Compare with ln in Figure. : + : + MATCHED PROBLEM Evaluate each it. (A) - 6 e : - q (B) (C) ln + 4 (D) ln - + : - 4 :- q 3e : q Now that we have reviewed the it properties of some basic functions, we are ready to consider the main topic of this section: L Hôpital s rule a powerful tool for evaluating certain types of its. Eplore & Discuss It follows from Figure that Does : q e- = and ln + : q 3e- = ln + 3e - : q e - eist? If so, what is its value? Use graphical methods to support your answers. L Hôpital s Rule and the Indeterminate Form / Recall that the it f :c g is a > indeterminate form if f = and :c g = :c

41 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization The quotient property for its in Section 3- does not apply, since g =. :c If we are dealing with a > indeterminate form, the it may or may not eist, and we cannot tell which is true without further investigation. Each of the following is a indeterminate form: - 4 : - and e - e : - The first it can be evaluated by performing some algebraic simplifications, such as - 4 : - = - + : - = : + = 4 The second cannot. Instead, we turn to the powerful L Hôpital s rule, which we now state without proof. This rule can be used whenever a it is a indeterminate form. THEOREM L HÔPITAL S RULE FOR / INDETERMINATE FORMS: VERSION For c a real number, if f = and g =, then :c :c :c f g = :c f g provided that the second it eists or is + q or - q. The use of L Hôpital s rule is best illustrated through eamples. e - e EXAMPLE 3 L Hôpital s Rule Evaluate. : - SOLUTION Step. Check to see if L Hôpital s rule applies: e - e = e - e = and - = - = : : L Hôpital s rule does apply. Step. Apply L Hôpital s rule: form e - e : - = : e = : = e = e d d e - e d - d Apply L Hôpital s rule. e is continuous at. MATCHED PROBLEM 3 e - e 4 Evaluate :4-4.

42 BARNMC_33886.QXD //7 :8 Page 3 Section -3 L Hôpital s Rule 3 4 y y y I N S I G H T In L Hôpital s rule, the symbol f >g represents the derivative of f divided by the derivative of g, not the derivative of the quotient f>g. When applying L Hôpital s rule to a > indeterminate form, be certain that you differentiate the numerator and denominator separately. The functions y = e - e - and y = e FIGURE 3 of Eample 3 are different functions (see Fig. 3), but both functions have the same it e as approaches. Although y is undefined at =, the graph of y provides a check of the answer to Eample 3. ln + EXAMPLE 4 L Hôpital s Rule Evaluate. SOLUTION Step. Check to see if L Hôpital s rule applies: L Hôpital s rule does apply. Step. Apply L Hôpital s rule: : 4 ln + : = ln = and 4 = : form ln + : 4 = : ln + : 4 = : = : = : + = q d d ln d d Apply L Hôpital s rule. Multiply numerator and denominator by /4 3. Simplify. Apply Theorem in Section 3-3 and compare with Fig. (D). ln MATCHED PROBLEM 4 Evaluate. : - 3 ln EXAMPLE L Hôpital s Rule May Not Be Applicable Evaluate. : SOLUTION Step. Check to see if L Hôpital s rule applies: ln = ln =, but = Z : : L Hôpital s rule does not apply. Step. Evaluate by another method. The quotient property for its from Section 3- does apply, and we have

43 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization MATCHED PROBLEM Evaluate. Note that applying L Hôpital s rule would give us an incorrect result: : e ln : ln : I N S I G H T Z : ln = : = : d d ln d d ln = = > = : = As Eample illustrates, all its involving quotients are not > indeterminate forms. You must always check to see if L Hôpital s rule applies before you use it. EXAMPLE 6 Repeated Application of L Hôpital s Rule Evaluate SOLUTION Step. Check to see if L Hôpital s rule applies: L Hôpital s rule does apply. Step. Apply L Hôpital s rule: : e - - : = and e - - = : : form e - - = : d d d d e - - = : e - Since and : e : = - =, the new it obtained is also a > indeterminate form, and L Hôpital s rule can be applied again. Step 3. Apply L Hôpital s rule again: form : e - d d = : d d e - = : e = e = Thus, : e - - = : e - = : e = MATCHED PROBLEM 6 Evaluate: : e - - One-Sided Limits and Limits at q In addition to eamining the it as approaches c, we have discussed one-sided its and its at q in Chapter 3. L Hôpital s rule is valid in these cases also.

44 BARNMC_33886.QXD //7 :8 Page 37 Section -3 L Hôpital s Rule 37 THEOREM L HÔPITAL S RULE FOR / INDETERMINATE FORMS: VERSION (FOR ONE-SIDED LIMITS AND LIMITS AT INFINITY) The first version of L Hôpital s rule (Theorem ) remains valid if the symbol : c is replaced everywhere it occurs with one of the following symbols: : c + : c - : q : - q For eample, if f = and g =, then : q : q f : q g = f : q g provided that the second it eists or is + q or - q. Similar rules can be written for : c +, : c -, and : - q. ln EXAMPLE 7 L Hôpital s Rule for One-Sided Limits Evaluate. : + - SOLUTION Step. Check to see if L Hôpital s rule applies: L Hôpital s rule does apply. Step. Apply L Hôpital s rule: ln = and - + : : = + : + form ln - = : + = : + = : + - = q d ln d d - d > - Apply L Hôpital s rule. Simplify. The it as : + is q because > - has a vertical asymptote at (Theorem, Section 3-3) and - 7 for 7. = ln MATCHED PROBLEM 7 Evaluate. : - - ln + e - EXAMPLE 8 L Hôpital s Rule for Limits at Infinity Evaluate. SOLUTION Step. Check to see if L Hôpital s rule applies: : q ln + e- = ln + = ln = and : q e- = L Hôpital s rule does apply. : q e -

45 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization Step. Apply L Hôpital s rule: form ln + e - : q e - = : q = : q ln + e - : q e - = : q = d d 3ln + e- 4 + e - + = d d e- -e - > + e - -e - Apply L Hôpital s rule. Multiply numerator and denominator by -e. e - = : q ln + e MATCHED PROBLEM 8 Evaluate. : - q e L Hôpital s Rule and the Indeterminate Form In Section 3-3, we discussed techniques for evaluating its of rational functions such as () : q : q + : q Each of these its is an q> q indeterminate form. In general, if :c f = ; q and :c g = ;q, then f :c g is called an indeterminate form. Furthermore, : c can be replaced in all three its above with : c +, : c -, : q, or : - q. It can be shown that L Hôpital s rule also applies to these q> q indeterminate forms. THEOREM 3 L HÔPITAL S RULE FOR THE INDETERMINATE FORM : VERSION 3 Versions and of L Hôpital s rule for the indeterminate form > are also valid if the it of f and the it of g are both infinite; that is, both + q and - q are permissible for either it. Eplore & Discuss For eample, if :c + f = q and :c + g = -q, then L Hôpital s rule can be applied to :c + 3f>g4. Evaluate each of the its in () in two ways:. Use Theorem 4 in Section Use L Hôpital s rule. Given a choice, which method would you choose? Why? EXAMPLE 9 L Hôpital s Rule for the Indeterminate Form ln Evaluate. : q SOLUTION Step. Check to see if L Hôpital s rule applies: ln = q and : q : q = q L Hôpital s rule does apply.

46 BARNMC_33886.QXD //7 :8 Page 39 Section -3 L Hôpital s Rule 39 Step. Apply L Hôpital s rule: q q form : q ln = : q > = : q ln : q = : q = d ln d d d Apply L Hôpital s rule. Simplify. See Figure (D). ln MATCHED PROBLEM 9 Evaluate. : q EXAMPLE L Hôpital s Rule for the Indeterminate Form e Evaluate. : q SOLUTION Step. Check to see if L Hôpital s rule applies: L Hôpital s rule does apply. e = q and : q : q = q Step. Apply L Hôpital s rule: q q form : q e = : q d d e d d = : q Since : q e = q and : q = q, this it is an q> q indeterminate form and L Hôpital s rule can be applied again. Step 3. Apply L Hôpital s rule again: e q q form e : q = : q d d e d d e = : q = q Thus, e : q = : q e = e : q = q MATCHED PROBLEM Evaluate. : q e Eplore & Discuss 3 Let n be a positive integer. Eplain how L Hôpital s rule can be used to show that e : q n = q Does this imply that e increases more rapidly than any power of?

47 BARNMC_33886.QXD //7 :8 Page 3 3 CHAPTER Graphing and Optimization I N S I G H T The three versions of L Hôpital s rule cover a multitude of its far too many to remember case by case. Instead, we suggest you use the following pattern, common to all versions, as a memory aid:. All versions involve three its: f() g(), f(), and g().. The independent variable must behave the same in all three its. The acceptable behaviors are : c, : c +, : c -, : q, or : - q. ; q 3. The form of f() g() must be or and both f() and g() must ; q approach or both must approach ; q. Answers to Matched Problems. (A) q (B) q (C) (D) q. (A) q (B) (C) q (D) q 3. e 4 4. q q q EXERCISE -3 A Use L Hôpital s rule to find each it in Problems : 3-8 : - B : e : ln : : q : q : q ln In Problems 8, eplain why L Hôpital s rule does not apply. If the it eists, find it by other means : : - 4 Find each it in Problems 9 4. Note that L Hôpital s rule does not apply to every problem, and some problems will require more than one application of L Hôpital s rule. e : e 3 : : q ln : q 4 : q : -3 ln -.. : : ln - e - : : q + e - ln e 3 : ln + : ln + ln + 4. : + 3 : -. ln + ln + 6. : + : + e : : : : q : q e + e : q : q ln + 4e - e - e : C 43. Find. : + ln [Hint: Write ln = ln > Find. : + ln [Hint: Write ln = ln > - >.4 In Problems 4 48, n is a positive integer. Find each it. ln n : q ln : q n e : q n In Problems 49, show that the repeated application of L Hôpital s rule does not lead to a solution. Then use algebraic manipulation to evaluate each it. [Hint: For 7 and n 7, n n =. ] : q : - q e - 3 n : q e : : : q + 8 e 3 : q 3 + e - : - q + ln + e - : q ln + e : - q : q : e : 3

48 BARNMC_33886.QXD //7 :8 Page 3 Section -4 Curve-Sketching Techniques 3 Section -4 CURVE-SKETCHING TECHNIQUES Modifying the Graphing Strategy Using the Graphing Strategy Modeling Average Cost When we summarized the graphing strategy in Section -, we omitted one important topic: asymptotes. Polynomial functions do not have any asymptotes. Asymptotes of rational functions were discussed in Section 3-3, but what about all the other functions, such as logarithmic and eponential functions? Since investigating asymptotes always involves its, we can now use L Hôpital s rule (Section -3) as a tool for finding asymptotes of many different types of functions. Modifying the Graphing Strategy The first version of the graphing strategy in Section - made no mention of asymptotes. Including information about asymptotes produces the following (and final) version of the graphing strategy: PROCEDURE Graphing Strategy (Final Version) Step. Analyze f(). (A) Find the domain of f. (B) Find the intercepts. (C) Find asympotes. Step. Analyze f (). Find the partition numbers for, and critical values of, f (). Construct a sign chart for f (), determine the intervals on which f is increasing and decreasing, and find local maima and minima. Step 3. Analyze f (). Find the partition numbers of f (). Construct a sign chart for f (), determine the intervals on which the graph of f is concave upward and concave downward, and find inflection points. Step 4. Sketch the graph of f. Draw asymptotes and locate intercepts, local maima and minima, and inflection points. Sketch in what you know from steps 3. Plot additional points as needed and complete the sketch. Using the Graphing Strategy I N S I G H T From now on, you should always use the final version of the graphing strategy. If a function does not have any asymptotes, simply state this fact. We will illustrate the graphing strategy with several eamples: EXAMPLE SOLUTION Using the Graphing Strategy Use the graphing strategy to analyze the function f = - > -. State all the pertinent information and sketch the graph of f. Step. Analyze f(). (A) Domain: f = - - All real, ecept = (B) y intercept: f = - - = intercepts: Since a fraction is when its numerator is and its denominator is not, the intercept is =.

49 BARNMC_33886.QXD //7 :8 Page 3 3 CHAPTER Graphing and Optimization a m m (C) Horizontal asymptote: b n n = = Thus, the line y = is a horizontal asymptote. Vertical asymptote: The denominator is for =, and the numerator is not for this value. Therefore, the line = is a vertical asymptote. Step. Analyze f. f = Critical values of f(): None Partition number for f : = Sign chart for f : = - - f () f () (, ) (, ) ND Decreasing Decreasing Test Numbers f œ Thus, f() is decreasing on - q, and, q. There are no local etrema. Step 3. Analyze f. f = Partition number for f : = Sign chart for f : - 3 f () Graph of f (, ) ND Concave downward (, ) Concave upward Test Numbers f fl - 3 Thus, the graph of f is concave downward on - q, and concave upward on, q. Since f() is not defined, there is no inflection point at =, even though f changes sign at =. Step 4. Sketch the graph of f. Insert intercepts and asymptotes, and plot a few additional points (for functions with asymptotes, plotting additional points is often helpful). Then sketch the graph: - f() f () MATCHED PROBLEM Follow the graphing strategy and analyze the function f = > -. State all the pertinent information and sketch the graph of f.

50 BARNMC_33886.QXD //7 :8 Page 33 Section -4 Curve-Sketching Techniques 33 EXAMPLE Using the Graphing Strategy Use the graphing strategy to analyze the function g = - State all pertinent information and sketch the graph of g. SOLUTION Step. Analyze g(). (A) Domain: All real, ecept = (B) intercept: y intercept: Since is not in the domain of g, there is no y intercept. (C) Horizontal asymptote: y = (the ais) Vertical asymptote: The denominator of g() is at = and the numerator is not. So the line = (the y ais) is a vertical asymptote. Step. Analyze g. g = - = - = g = = = Critical values of g: = Partition numbers for g : =, = Sign chart for g : = = =. - 3 (, ) (, ) (, ) g () ND Function f() is decreasing on - q, and, q, is increasing on (, ), and has a local maimum at =. Step 3. Analyze g. g = g = = = = Partition numbers for g : =, = 3 =. Sign chart for g : (, ) (,.) (., ) g () ND. Function g() is concave downward on - q, and (,.), is concave upward on., q, and has an inflection point at =..

51 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization Step 4. Sketch the graph of g. Plot key points, note that the coordinate aes are asymptotes, and sketch the graph. g() Horizontal asymptote g() Vertical asymptote 3 MATCHED PROBLEM Use the graphing strategy to analyze the function h = State all pertinent information and sketch the graph of h. EXAMPLE 3 SOLUTION Graphing Strategy Follow the steps of the graphing strategy and analyze the function f = e. State all the pertinent information and sketch the graph of f. Step. Analyze f(): f = e. (A) Domain: All real numbers (B) y intercept: f = intercept: e = for = only, since e 7 for all. (C) Vertical asymptotes: None Horizontal asymptotes: We use tables to determine the nature of the graph of f as : q and : - q: : q f() ,64.66 : q : - q f() : Step. Analyze f : Critical value of f(): - Partition number for f : - Sign chart for f : d f = d e + e d d = e + e = e + (, ) (, ) f () f () Decreasing Increasing Test Numbers f œ - - e - Thus, f() decreases on - q, -, has a local minimum at = -, and increases on -, q. (Since e 7 for all, we do not have to evaluate e - to conclude that -e - 6 when using the test number -.)

52 BARNMC_33886.QXD //7 :8 Page 3 Section -4 Curve-Sketching Techniques 3 Step 3. Analyze f : f = e d d d d e = e + + e = e + Sign chart for f (partition number is -): (, ) (, ) f () Graph of f Concave downward Inflection point Concave upward -3 - Test Numbers f -e -3 e - Thus, the graph of f is concave downward on - q, -, has an inflection point at = -, and is concave upward on -, q. Step 4. Sketch the graph of f, using the information from steps to 3: f() 3 f() MATCHED PROBLEM 3 Analyze the function f = e -.. State all the pertinent information and sketch the graph of f. I N S I G H T Remember, if p is any real number, then e p 7. The sign of the eponent p does not matter. This is a useful fact to remember when you work with sign charts involving eponential forms. Eplore & Discuss Refer to the discussion of vertical asymptotes in the solution of Eample 3. We used tables of values to estimate its at infinity and determine horizontal asymptotes. In some cases, the functions involved in these its can be written in a form that allows us to use L Hôpital s rule. - q # form f = : - q : -q e Rewrite as a fraction. = - q> q form Apply L Hôpital s rule. = : -q -e - = : -q (-e ) = : -q e - Simplify. Property of e

53 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization Use algebraic manipulation and L Hôpital s rule to verify the value of each of the following its: (A) : q e-. = (B) : + (ln -.) = (C) ln = : + EXAMPLE 4 Graphing Strategy Let f = ln -.. Follow the steps in the graphing strategy and analyze this function. State all the pertinent information and sketch the graph of f. SOLUTION Step. Analyze f: f = ln -. = ln -.. (A) Domain:, q (B) y intercept: None [ f() is not defined.] intercept: Solve ln -. = ln -. = or = Discard, since is not in the domain of f. ln =. ln = a if and only if = e a. = e. intercept (C) Asymptotes: None. The following tables suggest the nature of the graph as : + and as : q:... : + f() : See Eplore & Discuss (B)., : q f() 8 4, 6,4, : q Step. Analyze f : f = d d ln + ln d d d -. d = + ln -. = ln Critical value of f(): Partition number for f : Sign chart for f : = + ln - (, ) (, ) Decreasing Increasing Test Numbers. f œ The function f() decreases on (, ), has a local minimum at =, and increases on, q.

54 BARNMC_33886.QXD //7 :8 Page 37 Section -4 Curve-Sketching Techniques 37 Step 3. Analyze f : d f = d ln + ln d d = + ln = + ln = ln = - ln = - = e - L.3679 Sign chart for f (partition number is e - ): (, e ) (e, ) Concave downward e Concave upward Test Numbers. f œ -.89 The graph of f() is concave downward on, e -, has an inflection point at = e -, and is concave upward on e -, q. Step 4. Sketch the graph of f, using the information from steps to 3:. f(). e - f() -.e - -. e. MATCHED PROBLEM 4 Analyze the function f = ln. State all pertinent information and sketch the graph of f. Modeling Average Cost EXAMPLE Average Cost Given the cost function C =, +., where is the number of items produced, use the graphing strategy to analyze the graph of the average cost function. State all the pertinent information and sketch the graph of the average cost function. Find the marginal cost function and graph it on the same set of coordinate aes. SOLUTION The average cost function is C =, +. =, +. Step. Analyze C. (A) Domain: Since negative values of do not make sense and C is not defined, the domain is the set of positive real numbers. (B) Intercepts: None a m m (C) Horizontal asymptote: b n n =. =.

55 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization Thus, there is no horizontal asymptote. Vertical asymptote: The line = is a vertical asymptote, since the denominator is and the numerator is not for =. Oblique asymptotes: If a graph approaches a line that is neither horizontal nor vertical as approaches q or - q, that line is called an oblique asymptote. If is a large positive number, then, is very small and That is, C =, +. L., 3C -.4 = : q : q = This implies that the graph of y = C approaches the line y =. as approaches q. That line is an oblique asymptote for the graph of y = C. [More generally, whenever f = n>d is a rational function for which the degree of n is more than the degree of d, we can use polynomial long division to write f = m + b + r>d, where the degree of r is less than the degree of d. The line y = m + b is then an oblique asymptote for the graph of y = f.] Step. Analyze C. C = -, +. =. -, Critical value for C: Partition numbers for C : and Sign chart for C : =. - + C () (, ) (, ) C() Decreasing Increasing Local minimum Test Numbers C œ -..8 Thus, C is decreasing on (, ), is increasing on, q, and has a local minimum at =. Step 3. Analyze C : C =, 3. C is positive for all positive ; therefore, the graph of y = C is concave upward on, q. Step 4. Sketch the graph of C. The graph of C is shown in Figure.

56 BARNMC_33886.QXD //7 :8 Page 39 Section -4 Curve-Sketching Techniques 39 y C (), C(). y. (oblique asymptote) Minimum average cost 3 4 FIGURE The marginal cost function is C =. The graph of this linear function is also shown in Figure. The graph in Figure illustrates an important principle in economics: The minimum average cost occurs when the average cost is equal to the marginal cost. MATCHED PROBLEM Given the cost function C =,6 +., where is the number of items produced, (A) Use the graphing strategy to analyze the graph of the average cost function. State all the pertinent information and sketch the graph of the average cost function. Find the marginal cost function and graph it on the same set of coordinate aes. Include any oblique asymptotes. (B) Find the minimum average cost. Answers to Matched Problems. Domain: All real, ecept = y intercept: f = ; intercept: Horizontal asymptote: y = -; vertical asymptote: = Increasing on - q, and, q Concave upward on - q, ; concave downward on, q f() f() Domain: All real, ecept = intercept: = = -.7 h() is not defined Vertical asymptote: = (the y ais) Horizontal asymptote: y = (the ais) Increasing on -., Decreasing on - q, -. and, q Local minimum at =. Concave upward on -., and, q

57 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization Concave downward on - q, -. Inflection point at = -. - h() h() 3. Domain: - q, q y intercept: f = intercept: = Horizontal asymptote: y = (the ais) Increasing on - q, Decreasing on, q Local maimum at = Concave downward on - q, 4 Concave upward on 4, q Inflection point at = 4 f() Domain:, q y intercept: None [f() is not defined] intercept: = Increasing on e -, q Decreasing on, e - Local minimum at = e - L.368 Concave upward on, q : q f() : q f().... : f() :. (A) Domain: Intercepts:, q None Vertical asymptote: = ; oblique asymptote: y =. Decreasing on (, 8); increasing on 8, q; local minimum at = 8 Concave upward on, q y 8 4 C ().,6 C(). y. (oblique asymptote) (B) Minimum average cost is 4 at = 8.

58 BARNMC_33886.QXD //7 :8 Page 33 Section -4 Curve-Sketching Techniques 33 Eercise -4 A. Use the graph of f to identify f() f () ND a b c d L M e f g h f () ND 4 4. Domain: All real ; f = -3; f = 3 : - q : q f () - (A) the intervals on which f 6 (B) the intervals on which f 7 (C) the intervals on which f() is increasing (D) the intervals on which f() is decreasing (E) the coordinate(s) of the point(s) where f() has a local maimum (F) the coordinate(s) of the point(s) where f() has a local minimum (G) the intervals on which f 6 (H) the intervals on which f 7 (I) the intervals on which the graph of f is concave upward (J) the intervals on which the graph of f is concave downward (K) the coordinate(s) of the inflection point(s) (L) the horizontal asymptote(s) (M) the vertical asymptote(s). Repeat Problem for the following graph of f: f() f () f () ND f() 3 ND. Domain: All real, ecept = -; f = q; f = -q; f = - : - + : - : q f () f () ND ND 4 6 a L b c de 6. Domain: All real, ecept = ; f = q; : - f = q; f = - + : : q -4 - In Problems 3, use the given information to sketch the graph of f. Assume that f is continuous on its domain and that all intercepts are included in the table of values. 3. Domain: All real ; : ; q f = f () f() - ND -4-4 f() - - f () ND

59 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization 7. Domain: All real, ecept = -; f-3 =, f- = 3, f = -, f = ; f 7 on - q, - and -, q; f 7 on - q, -; f 6 on -, q; vertical asymptote: = -; horizontal asymptote: y = 8. Domain: All real, ecept = ; f = -, f = ; f 6 on - q, and, q; f 6 on - q, ; f 7 on, q; vertical asymptote: = ; horizontal asymptote: y = - 9. Domain: All real, ecept = - and = ; f-3 = -, f =, f3 = ; f 6 on - q, - and, q; f 7 on -, ; f 6 on - q, - and -, ; f 7 on (, ) and, q; vertical asymptotes: = - and = ; horizontal asymptote: y =. Domain: All real, ecept = - and = ; f- =, f =, f = ; f 7 on - q, - and (, ); f 6 on -, and, q; f 7 on - q, -, -,, and, q; vertical asymptotes: = - and = ; horizontal asymptote: y = B In Problems 6, summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of y = f.. f = + 3. f = f = 4. f = f = + e f = 3 + 7e 7. f = e f = e f = ln -. f = ln + 4. f = - ln. f = ln f = 4. f = f = 6. f = f = 8. f = f = 3. f = f = + 3. f = f = f = f = 36. f = f = f = 4. f = f = 3 - e 4. f = - e 43. f = e f = e - 4. f = ln 46. f = ln 47. f = ln 48. f = 49. f =. f = f =. f = f = 4. f = f = 6. f = C In Problems 7 64, show that the line y = is an oblique asymptote for the graph of y = f, summarize all pertinent information obtained by applying the graphing strategy, and sketch the graph of y = f. 7. f = f = f = f = f = ln f = - 9 f = + 3 f = f = In Problems 6 7, summarize all pertinent information obtained by applying the graphing strategy, and sketch the graph of y = f. [Note: These rational functions are not reduced to lowest terms.] 6. f = f = f = f = f = f = f = f =

60 BARNMC_33886.QXD //7 :8 Page 333 Section -4 Curve-Sketching Techniques 333 Applications 73. Revenue. The marketing research department for a computer company used a large city to test market the firm s new product. The department found that the relationship between price p (dollars per unit) and demand (units sold per week) was given approimately by p =, Thus, weekly revenue can be approimated by R = p =, Graph the revenue function R. 74. Profit. Suppose that the cost function C() (in dollars) for the company in Problem 73 is (A) Write an equation for the profit P(). (B) Graph the profit function P. 7. Pollution. In Silicon Valley (California), a number of computer-related manufacturing firms were found to be contaminating underground water supplies with toic chemicals stored in leaking underground containers. A water quality control agency ordered the companies to take immediate corrective action and to contribute to a monetary pool for testing and cleanup of the underground contamination. Suppose that the required monetary pool (in millions of dollars) for the testing and cleanup is estimated to be given by P = C = where is the percentage (epressed as a decimal fraction) of the total contaminant removed. (A) Where is P() increasing? Decreasing? (B) Where is the graph of P concave upward? Downward? (C) Find any horizontal and vertical asymptotes. (D) Find the and y intercepts. (E) Sketch a graph of P. 76. Employee training. A company producing computer components has established that, on the average, a new employee can assemble N(t) components per day after t days of on-the-job training, as given by Nt = t t t Ú (A) Where is N(t) increasing? Decreasing? (B) Where is the graph of N concave upward? Downward? (C) Find any horizontal and vertical asymptotes. (D) Find the intercepts. (E) Sketch a graph of N. 77. Replacement time. An office copier has an initial price of $3,. A maintenance/service contract costs $3 for the first year and increases $ per year thereafter. It can be shown that the total cost of the copier (in dollars) after n years is given by Cn = 3, + n + n n Ú (A) Write an epression for the average cost per year, Cn, for n years. (B) Graph the average cost function found in part (A). (C) When is the average cost per year minimum? (This time is frequently referred to as the replacement time for this piece of equipment.) 78. Construction costs. The management of a manufacturing plant wishes to add a fenced-in rectangular storage yard of, square feet, using the plant building as one side of the yard (see the figure). If is the distance (in feet) from the building to the fence parallel to the building, show that the length of the fence required for the yard is given by L = +, 7 Figure for 78 Storage yard (A) Graph L. (B) What are the dimensions of the rectangle requiring the least amount of fencing? 79. Average and marginal costs. The total daily cost (in dollars) of producing park benches is given by C =, + +. (A) Sketch the graphs of the average cost function and the marginal cost function on the same set of coordinate aes. Include any oblique asymptotes. (B) Find the minimum average cost. 8. Average and marginal costs. The total daily cost (in dollars) of producing picnic tables is given by C = + +. (A) Sketch the graphs of the average cost function and the marginal cost function on the same set of coordinate aes. Include any oblique asymptotes. (B) Find the minimum average cost.

61 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization 8. Minimizing average costs. The data in the table give the total daily costs y (in dollars) of producing pepperoni pizzas at various production levels. Number of Pizzas Total Costs y ,4 3,4 (A) Enter the data into a graphing calculator and find a quadratic regression equation for the total costs. (B) Use the regression equation from part (A) to find the minimum average cost (to the nearest cent) and the corresponding production level (to the nearest integer). 8. Minimizing average costs. The data in the table give the total daily costs y (in dollars) of producing delue pizzas at various production levels. Number of Pizzas Total Costs y 9 7,,38,87 3,4 (A) Enter the data into a graphing calculator and find a quadratic regression equation for the total costs. (B) Use the regression equation from part (A) to find the minimum average cost (to the nearest cent) and the corresponding production level (to the nearest integer). 83. Medicine. A drug is injected into the bloodstream of a patient through her right arm. The concentration of the drug in the bloodstream of the left arm t hours after the injection is given by Graph C. 84. Physiology. In a study on the speed of muscle contraction in frogs under various loads, researchers W. O. Fems and J. Marsh found that the speed of contraction decreases with increasing loads. More precisely, they found that the relationship between speed of contraction, S (in centimeters per second), and load w (in grams) is given approimately by Graph S. Sw = 6 +.6w w 8. Psychology: retention. An eperiment on retention is conducted in a psychology class. Each student in the class is given one day to memorize the same list of 3 special characters. The lists are turned in at the end of the day, and for each succeeding day for 3 days, each student is asked to turn in a list of as many of the symbols as can be recalled. Averages are taken, and it is found that Nt = Ct =.4t t + t + t w Ú t Ú provides a good approimation of the average number N(t) of symbols retained after t days. Graph N. Section - ABSOLUTE MAXIMA AND MINIMA Absolute Maima and Minima Second Derivative and Etrema We are now ready to consider one of the most important applications of the derivative: the use of derivatives to find the absolute maimum or minimum value of a function. An economist may be interested in the price or production level of a commodity that will bring a maimum profit; a doctor may be interested in the time it takes for a drug to reach its maimum concentration in the bloodstream after an injection; and a city planner might be interested in the location of heavy industry in a city in order to produce minimum pollution in residential and business areas. In this section, we develop the procedures needed to find the absolute maimum and absolute minimum values of a function. Absolute Maima and Minima Recall that f(c) is a local maimum value if f fc for near c and a local minimum value if f Ú fc for near c. Now we are interested in finding the largest and the smallest values of f() throughout its domain.

62 BARNMC_33886.QXD //7 :8 Page 33 Section - Absolute Maima and Minima 33 DEFINITION Absolute Maima and Minima If fc Ú f for all in the domain of f, then f(c) is called the absolute maimum value of f. If fc f for all in the domain of f, then f() is called the absolute minimum value of f. Figure illustrates some typical eamples. f() f() f() f() f() 4 f() /3 (A) No absolute maimum or minimum One local maimum at One local minimum at (B) Absolute maimum at No absolute minimum FIGURE (C) Absolute minimum at No absolute maimum I N S I G H T If f(c) is the absolute maimum value of a function f, then f(c) is obviously a value of f. It is common practice to omit value and to refer to f(c) as the absolute maimum of f. In either usage, note that c is a value of in the domain of f where the absolute maimum occurs. It is incorrect to refer to c as the absolute maimum. Collectively, the absolute maimum and the absolute minimum are referred to as absolute etrema. Eplore & Discuss Functions f, g, and h, along with their graphs, are shown in Figure. f() g() h() (A) f() (B) g() FIGURE (C) h() (A) Which of these functions are continuous on 3-, 4? (B) Find the absolute maimum and the absolute minimum of each function on 3-, 4, if they eist, and the corresponding values of that produce these absolute etrema. (C) Suppose that a function p is continuous on 3-, 4 and satisfies p- = and p =. Sketch a possible graph for p. Does the function you graphed have an absolute maimum? An absolute minimum? Can you modify your sketch so that p does not have an absolute maimum or an absolute minimum on 3-, 4?

63 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization In many practical problems, the domain of a function is restricted because of practical or physical considerations. If the domain is restricted to some closed interval, as is often the case, then Theorem applies. THEOREM EXTREME VALUE THEOREM A function f that is continuous on a closed interval [a, b] has both an absolute maimum value and an absolute minimum value on that interval. It is important to understand that the absolute maimum and minimum values depend on both the function f and the interval [a, b]. Figure 3 illustrates four cases. f() f() f() f() Absolute maimum f() 4 Absolute maimum f() Absolute maimum f() Absolute maimum f() f() Absolute minimum f() 4 f(9) 73 Absolute minimum [ ] [ ] [ ] [ ] a b a 4 b a 4 b 8 a 3 b f(8) 78 Absolute minimum f(3) 73 f(9) Absolute minimum (A) [a, b] [, ] (B) [a, b] [4, ] (C) [a, b] [4, 8] (D) [a, b] [3, ] FIGURE 3 Absolute etrema for f = for various closed intervals In all four cases illustrated in Figure 3, the absolute maimum value and absolute minimum value occur at a critical value or an endpoint. This property is generalized in Theorem. Note that both the absolute maimum value and the absolute minimum value are unique, but each can occur at more than one point in the interval (Fig. 3D). THEOREM LOCATING ABSOLUTE EXTREMA Absolute etrema (if they eist) must always occur at critical values or at endpoints. Thus, to find the absolute maimum or minimum value of a continuous function on a closed interval, we simply identify the endpoints and the critical values in the interval, evaluate the function at each, and then choose the largest and smallest values out of this group. PROCEDURE Finding Absolute Etrema on a Closed Interval Step. Check to make certain that f is continuous over [a, b]. Step. Find the critical values in the interval (a, b). Step 3. Evaluate f at the endpoints a and b and at the critical values found in step. Step 4. The absolute maimum f() on [a, b] is the largest of the values found in step 3. Step. The absolute minimum f() on [a, b] is the smallest of the values found in step 3.

64 BARNMC_33886.QXD //7 :8 Page 337 Section - Absolute Maima and Minima 337 EXAMPLE Finding Absolute Etrema Find the absolute maimum and absolute minimum values of f = on each of the following intervals: (A) 3-6, 44 (B) 3-4, 4 (C) 3-, 4 SOLUTION (A) The function is continuous for all values of. f = = Thus, = -3 and = are critical values in the interval -6, 4. Evaluate f at the endpoints and critical values (-6, -3,, and 4), and choose the maimum and minimum from these: f-6 = -6 f-3 = f = - f4 = 69 Absolute minimum Absolute maimum (B) Interval: 3-4, 4 f() Absolute maimum - Absolute minimum - (C) Interval: 3-, 4 f() - Absolute maimum - Absolute minimum - The critical value = -3 is not included in this table, because it is not in the interval 3-, 4. MATCHED PROBLEM Find the absolute maimum and absolute minimum values of f = 3 - on each of the following intervals: (A) 3-, 4 (B) 3-3, 34 (C) 3-3, 4 Now, suppose that we want to find the absolute maimum or minimum value of a function that is continuous on an interval that is not closed. Since Theorem no longer applies, we cannot be certain that the absolute maimum or minimum value eists. Figure 4 illustrates several ways that functions can fail to have absolute etrema.

65 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization f() f() f() f() f() 8 3 f() (A) No absolute etrema on (, ): f() for all f() or for any 3 (B) No absolute etrema on (, ): 3 f() for (, ) f() 3 or for any (, ) FIGURE 4 Functions with no absolute etrema (C) No absolute etrema on (, ): Graph has vertical asymptotes at and In general, the best procedure to follow in searching for absolute etrema on an interval that is not of the form [a, b] is to sketch a graph of the function. However, many applications can be solved with a new tool that does not require any graphing. Second Derivative and Etrema The second derivative can be used to classify the local etrema of a function. Suppose that f is a function satisfying f c = and f c 7. First, note that if f c 7, it follows from the properties of its* that f 7 in some interval (m, n) containing c. Thus, the graph of f must be concave upward in this interval. But this implies that f is increasing in the interval. Since f c =, f must change from negative to positive at = c, and f(c) is a local minimum (see Fig. ). Reasoning in the same fashion, we conclude that if f c = and f c 6, then f(c) is a local maimum. Of course, it is possible that both f c = and f c =. In this case, the second derivative cannot be used to determine the shape of the graph around = c; fc may be a local minimum, a local maimum, or neither. f (c) f (c) f () f () f () f () f (c) f (c) m c n m c n (A) f (c) and f (c) implies f(c) is a local minimum (B) f (c) and f (c) implies f(c) is a local maimum FIGURE Second derivative and local etrema The sign of the second derivative thus provides a simple test for identifying local maima and minima. This test is most useful when we do not want to draw the graph of the function. If we are interested in drawing the graph and have already constructed the sign chart for f, the first-derivative test can be used to identify the local etrema. * Actually, we are assuming that f is continuous in an interval containing c. It is very unlikely that we will encounter a function for which f c eists but f is not continuous in an interval containing c.

66 BARNMC_33886.QXD //7 :8 Page 339 Section - Absolute Maima and Minima 339 RESULT Second-Derivative Test Let c be a critical value of f(). f c f c Graph of f is: f(c) Eample + Concave upward Local minimum - Concave downward Local maimum? Test does not apply EXAMPLE Testing Local Etrema Find the local maima and minima for each function. Use the second-derivative test when it applies. (A) (B) (C) f = f = e -. f = SOLUTION (A) Take first and second derivatives and find critical values: f = f = = f = 6 - = 6 - Critical values are = and = 3. (B) f = -6 6 f has a local maimum at =. f 3 = 6 7 f has a local minimum at = 3. f = e -. f = e -. + e = e f = e -. (-.) -. + e -. (-.) = e -. (.4 -.4) Critical value: = >. = f = e f has a local maimum at =. (C) f = f = = 3 - f = Critical values are = and =. f = f = The second-derivative test fails at both critical values, so the first-derivative test must be used. Sign chart for f = 3 - (partition numbers are and ): f () f () (, ) (, ) (, ) Decreasing Increasing Increasing Test Numbers f œ - - 8,33

67 BARNMC_33886.QXD //7 :8 Page CHAPTER Graphing and Optimization From the chart, we see that f() has a local minimum at = and does not have a local etremum at =. MATCHED PROBLEM Find the local maima and minima for each function. Use the second-derivative test when it applies. (A) f = (B) f = e - (C) f = I N S I G H T The second-derivative test does not apply if f c = or if f c is not defined. As Eample C illustrates, if f c =, then f(c) may or may not be a local etremum. Some other method, such as the first-derivative test, must be used when f c = or f c does not eist. The solution of many optimization problems involves searching for an absolute etremum. If the function in question has only one critical value, then the secondderivative test not only classifies the local etremum, but also guarantees that the local etremum is, in fact, the absolute etremum. THEOREM 3 SECOND-DERIVATIVE TEST FOR ABSOLUTE EXTREMUM Let f be continuous on an interval I with only one critical value c in I. If f c = and f c 7, then f(c) is the absolute minimum of f on I. ( I ) If f c = and f c 6, then f(c) is the absolute maimum of f on I. ( I ) Since the second-derivative test cannot be applied when f c = or f c does not eist, Theorem 3 makes no mention of these cases. EXAMPLE 3 Finding an Absolute Etremum on an Open Interval Find the absolute minimum value of each function on (, q). (A) (B) f = + 4 f = (ln ) - 3 ln SOLUTION (A) f = + 4 f = - 4 = - 4 = - + Critical values are = - and =. f = 8 3 The only critical value in the interval, q is =. Since f = 7, f = 4 is the absolute minimum value of f on, q.

68 BARNMC_33886.QXD //7 :8 Page 34 Section - Absolute Maima and Minima 34 (B) f = (ln ) - 3 ln f = ( ln ) - 3 = ln - 3 f = - ( ln - 3) = - ln Critical value is = e 3/. The only critical value in the interval, q is = e 3/. Since f e 3/ = /e 3 7, f(e 3/ ) = -. is the absolute minimum value of f on, q. MATCHED PROBLEM 3 Find the absolute maimum value of each function on, q. (A) f = - - (B) f = ln - Answers to Matched Problems. (A) Absolute maimum: f = 6; absolute minimum: f- = -6 (B) Absolute maimum: f- = 6; absolute minimum: f = -6 (C) Absolute maimum: f- = 6; absolute minimum: f = -. (A) f() is a local maimum; f(4) is a local minimum. (B) fln = - ln is a local minimum. (C) f() is a local minimum; there is no local etremum at =. 3. (A) f = - (B) f = ln - Eercise - A Problems refer to the graph of y = f() shown here. Find the absolute minimum and the absolute maimum over the indicated interval. f(). [, ]. [, 8] 3. [, 8] 4. [, ]. [, ] 6. [, 9] 7. [, 9] 8. [, ] 9. [, ]. [, 8] In Problems 6, find the absolute maimum and minimum, if either eists, for each function.. f = f = f = f = f = f = f = f = f = f =. f = f = f = B In Problems 7, find the indicated etremum of each function on the given interval. 7. Absolute minimum value on 3, q for f = Absolute maimum value on 3, q for f = Absolute maimum value on 3, q for f = 3-3 f = + f = f = Absolute minimum value on 3, q for f = Absolute minimum value on 3, q for f = Absolute minimum value on 3, q for f = Absolute maimum value on, q for f = 4-8 3

69 BARNMC_33886.QXD //7 :9 Page CHAPTER Graphing and Optimization 34. Absolute maimum value on (, q) for 48. Absolute minimum value on (, q) for f = Absolute maimum value on (, q) for f = Absolute minimum value on (, q) for f = Absolute maimum value on (, q) for f = Absolute maimum value on (, q) for f = Absolute minimum value on (, q) for f = Absolute minimum value on (, q) for f = Absolute minimum value on (, q) for f = e 4. Absolute maimum value on (, q) for f = 4 e 43. Absolute maimum value on (, q) for f = 3 e 44. Absolute minimum value on (, q) for f = e 4. Absolute maimum value on (, q) for f = - ln 46. Absolute minimum value on (, q) for f = 4 ln Absolute maimum value on (, q) for f = (3 - ln ) 49. Absolute maimum value on (, q) for. Absolute maimum value on (, q) for In Problems 6, find the absolute maimum and minimum, if either eists, for each function on the indicated intervals f = 3 (ln - ) f = ln(e - ) f = ln( e - ) f = (A) 3-, 4 (B) 3-, 34 (C) [, ]. f = (A) 3-3, 44 (B) 3-, 34 (C) 3-, 4 f = (A) [, 3] (B) [, 7] (C) [3, 6] 4. f = (A) 3-, 34 (B) [, ] (C) 3-3, 44. f = (A) 3-, 4 (B) [, 4] (C) 3-, 4 f = (A) 3-4, 44 (B) 3-, 4 (C) [, 3] In Problems 7 64, describe the graph of f at the given point relative to the eistence of a local maimum or minimum with one of the following phrases: Local maimum, Local minimum, Neither, or Unable to determine from the given information. Assume that f() is continuous on - q, q. 7. (, f()) if f = and f 7 8. (4, f(4)) if f 4 = and f , f-3 if f -3 = and f -3 = 6. -, f- if f - = and f (6, f(6)) if f 6 = and f 6 does not eist 6. (, f()) if f = and f does not eist 63. -, f- if f - = and f (, f()) if f = and f 7 Section -6 OPTIMIZATION Area and Perimeter Maimizing Revenue and Profit Inventory Control Now we want to use the calculus tools we have developed to solve optimization problems problems that involve finding the absolute maimum value or the absolute minimum value of a function. As you work through this section, note that the statement of the problem does not usually include the function that is to be optimized. Often, it is your responsibility to find the function and then to find its absolute etremum.

70 BARNMC_33886.QXD //7 :9 Page 343 Area and Perimeter Section -6 Optimization 343 The techniques used to solve optimization problems are best illustrated through eamples. EXAMPLE SOLUTION y $6 $ $ $ FIGURE Maimizing Area A homeowner has $3 to spend on building a fence around a rectangular garden. Three sides of the fence will be constructed with wire fencing at a cost of $ per linear foot. The fourth side is to be constructed with wood fencing at a cost of $6 per linear foot. Find the dimensions and the area of the largest garden that can be enclosed with $3 worth of fencing. To begin, we draw a figure (Fig. ), introduce variables, and look for relationships among the variables. Since we don t know the dimensions of the garden, the lengths of fencing are represented by the variables and y. The costs of the fencing materials are fied and are thus represented by constants. Now we look for relationships among the variables. The area of the garden is while the cost of the fencing is A = y C = y + + y + 6 = 8 + 4y The problem states that the homeowner has $3 to spend on fencing. We make the assumption that enclosing the largest area will use all of the money available for fencing. The problem has now been reduced to Maimize A = y subject to 8 + 4y = 3 Before we can use calculus techniques to find the maimum area A, we must epress A as a function of a single variable. We use the cost equation to einate one of the variables in the epression for the area (we choose to einate y either will work) y = 3 Now we consider the permissible values of. Because is one of the dimensions of a rectangle, must satisfy Ú 4y = 3-8 y = 8 - Length is always nonnegative. And because y = 8 - is also a dimension of a rectangle, y must satisfy y = 8 - Ú Width is always nonnegative. 8 Ú We summarize the preceding discussion by stating the following model for this optimization problem: Maimize A = 8 - for 4 Net, we find any critical values of A: A = y = 8 - = 8-4 Ú or 4 A = 8-4 = 8 = 4 = 8 4 = Critical value

71 BARNMC_33886.QXD //7 :9 Page CHAPTER Graphing and Optimization TABLE A() 8 4 MATCHED PROBLEM Since A() is continuous on [, 4], the absolute maimum value of A, if it eists, must occur at a critical value or an endpoint. Evaluating A at these values (Table ), we see that the maimum area is 8 when = and y = 8 - = 4 Finally, we must answer the questions posed in the problem. The dimensions of the garden with the maimum area of 8 square feet are feet by 4 feet, with one -foot side with wood fencing. Repeat Eample if the wood fencing costs $8 per linear foot and all other information remains the same. We summarize the steps we followed in the solution to Eample in the following bo: PROCEDURE Strategy for Solving Optimization Problems Step. Introduce variables, look for relationships among the variables, and construct a mathematical model of the form Maimize (or minimize) f() on the interval I Step. Find the critical values of f(). Step 3. Use the procedures developed in Section - to find the absolute maimum (or minimum) value of f() on the interval I and the value(s) of where this occurs. Step 4. Use the solution to the mathematical model to answer all the questions asked in the problem. EXAMPLE SOLUTION Minimizing Perimeter Refer to Eample.The homeowner judges that an area of 8 square feet for the garden is too small and decides to increase the area to, square feet. What is the minimum cost of building a fence that will enclose a garden with area, square feet? What are the dimensions of this garden? Assume that the cost of fencing remains unchanged. Refer to Figure in the solution for Eample. This time we want to minimize the cost of the fencing that will enclose, square feet. The problem can be epressed as Since and y represent distances, we know that Ú and y Ú. But neither variable can equal, because their product must be,. y =, Solve the area equation for y. y =,, C = = 8 +, The model for this problem is Minimize C = 8 + 4y subject to y =, Minimize C = 8 +, = 8 +, - Substitute for y in the cost equation. 7 for 7

72 BARNMC_33886.QXD //7 :9 Page 34 Section -6 Optimization 34 = 6 = The negative square root is discarded, since 7. We use the second derivative to determine the behavior at =. The second-derivative test shows that C() has a local minimum at =, and since = is the only critical value of 7, C() must be the absolute minimum value of C() for 7. When =, the cost is and C = 8 -, - = 8 -, = 8 =, =, 8 = 6 C = 8 -, - C = +, -3 =, 3 C =, 3 =.64 7 C = 8 +, y =, = + = $4 = The minimal cost for enclosing a,-square-foot garden is $4, and the dimensions are feet by feet, with one -foot side with wood fencing. MATCHED PROBLEM Repeat Eample if the homeowner wants to enclose an,8-square-foot garden and all other data remain unchanged. I N S I G H T The restrictions on the variables in the solutions of Eamples and are typical of problems involving areas or perimeters (or the cost of the perimeter): 8 + 4y = 3 y =, Cost of fencing (Eample ) Area of garden (Eample ) The equation in Eample restricts the values of to 4 or 3, 44 The endpoints are included in the interval for our convenience (a closed interval is easier to work with than an open one). The area function is defined at each endpoint, so it does no harm to include them. The equation in Eample restricts the values of to 7 or, q Neither endpoint can be included in this interval. We cannot include because the area is not defined when =, and we can never include q as an endpoint. Remember, q is not a number, but a symbol which denotes that the interval is unbounded.

73 BARNMC_33886.QXD //7 :9 Page CHAPTER Graphing and Optimization Maimizing Revenue and Profit EXAMPLE 3 SOLUTION Maimizing Revenue An office supply company sells mechanical pencils per year at $p per pencil.the price demand equation for these pencils is p = -.. What price should the company charge for the pencils to maimize revenue? What is the maimum revenue? Revenue = price * demand R = -. = -. Both price and demand must be nonnegative, so Ú and p = -. Ú Ú., Ú The mathematical model for this problem is Maimize R = -., R = = =. = Critical value. =, Use the second-derivative test for absolute etrema: Ma R = R, = $, When the demand is =,, the price is R = -. 6 for all -., = $ p = -. The company will realize a maimum revenue of $, when the price of a pencil is $. MATCHED PROBLEM 3 An office supply company sells heavy-duty paper shredders per year at $p per shredder. The price demand equation for these shredders is p = 3-3 What price should the company charge for the shredders to maimize revenue? What is the maimum revenue? EXAMPLE 4 SOLUTION Maimizing Profit The total annual cost of manufacturing mechanical pencils for the office supply company in Eample 3 is C =, + What is the company s maimum profit? What should the company charge for each pencil, and how many pencils should be produced? Using the revenue model in Eample 3, we have Profit = Revenue - Cost P = R - C = -. -, - = ,

74 BARNMC_33886.QXD //7 :9 Page 347 The mathematical model for profit is Maimize P = ,, (The restrictions on come from the revenue model in Eample 3.) P = 8 -. = 8 =. = 8. = 4, P = -. 6 for all Since = 4, is the only critical value and P 6, Ma P = P4, = $, Critical value Using the price demand equation from Eample 3 with = 4,, we find that p = -.4, = $6 p = -. Thus, a maimum profit of $, is realized when 4, pencils are manufactured annually and sold for $6 each. The results in Eamples 3 and 4 are illustrated in Figure. 3, R C Section -6 Optimization 347 Revenue Cost and revenue (dollars) Minimum cost,, Loss Profit Maimum profit Maimum revenue Profit Cost Loss,, Production (number of pencils per year) FIGURE I N S I G H T In Figure, notice that the maimum revenue and the maimum profit occur at different production levels. The profit is maimum when P = R - C = that is, when the marginal revenue is equal to the marginal cost. Notice that the slopes of the revenue function and the cost function are the same at this production level. MATCHED PROBLEM 4 The annual cost of manufacturing paper shredders for the office supply company in Matched Problem 3 is C = 9, + 3. What is the company s maimum profit? What should it charge for each shredder, and how many shredders should it produce?

75 BARNMC_33886.QXD //7 :9 Page CHAPTER Graphing and Optimization EXAMPLE Maimizing Profit The government decides to ta the company in Eample 4 $ for each pencil produced. Taking into account this additional cost, how many pencils should the company manufacture each week to maimize its weekly profit? What is the maimum weekly profit? How much should the company charge for the pencils to realize the maimum weekly profit? SOLUTION The ta of $ per unit changes the company s cost equation: C = original cost + ta =, + + =, + 4 The new profit function is P = R - C Thus, we must solve the following equation: Maimize P = ,, P = = = 3, = -. -, - 4 = , Critical value P = -. 6 for all Ma P = P3, = $4, Using the price demand equation (Eample 3) with = 3,, we find that p = -.3, = $7 p = -. Thus, the company s maimum profit is $4, when 3, pencils are produced and sold weekly at a price of $7. Even though the ta caused the company s cost to increase by $ per pencil, the price that the company should charge to maimize its profit increases by only $. The company must absorb the other $, with a resulting decrease of $7, in maimum profit. MATCHED PROBLEM The government decides to ta the office supply company in Matched Problem 4 $ for each shredder produced. Taking into account this additional cost, how many shredders should the company manufacture each week to maimize its weekly profit? What is the maimum weekly profit? How much should the company charge for the shredders to realize the maimum weekly profit? EXAMPLE 6 SOLUTION Maimizing Revenue When a management training company prices its seminar on management techniques at $4 per person,, people will attend the seminar. The company estimates that for each $ reduction in the price, an additional people will attend the seminar. How much should the company charge for the seminar in order to maimize its revenue? What is the maimum revenue? Let represent the number of $ price reductions. Then 4 - = price per customer, + = number of customers Revenue = price per customernumber of customers R = 4 - *, +

76 BARNMC_33886.QXD //7 :9 Page 349 Section -6 Optimization 349 TABLE R() 4, 4, 8 MATCHED PROBLEM 6 Eplore & Discuss Since price cannot be negative, we have 4 - Ú 4 Ú 8 Ú or 8 A negative value of would result in a price increase. Since the problem is stated in terms of price reductions, we must restrict so that Ú. Putting all this together, we have the following model: Maimize R = 4 -, + for 8 R = 4, + 3, - R = 3, - = 3, = = Critical value Since R() is continuous on the interval [, 8], we can determine the behavior of the graph by constructing a table. Table shows that R = $4, is the absolute maimum revenue. The price of attending the seminar at = is 4 - = $3. The company should charge $3 for the seminar in order to receive a maimum revenue of $4,. A walnut grower estimates from past records that if trees are planted per acre, each tree will average 6 pounds of nuts per year. If, for each additional tree planted per acre, the average yield per tree drops pounds, how many trees should be planted to maimize the yield per acre? What is the maimum yield? In Eample 6, letting be the number of $ price reductions produced a simple and direct solution to the problem. However, this is not the most obvious choice for a variable. Suppose that we proceed as follows: Let be the new price and let y be the attendance at that price level.then the total revenue is given by y. (A) Find y when = 4 and when = 39. Find the equation of the line through these two points. (B) Use the equation from part (A) to epress the revenue in terms of either or y, and use the epression you came up with to solve Eample 6. (C) Compare this method of solution to the one used in Eample 6 with respect to ease of comprehension and ease of computation. EXAMPLE 7 Maimizing Revenue After additional analysis, the management training company in Eample 6 decides that its estimate of attendance was too high. Its new estimate is that only additional people will attend the seminar for each $ decrease in price. All other information remains the same. How much should the company charge for the seminar now in order to maimize revenue? What is the new maimum revenue? SOLUTION TABLE 3 R() 4, 8 Under the new assumption, the model becomes Maimize R = 4 -, + 8 = 4, -, - R = -, - = -, = = - Critical value Note that = - is not in the interval [, 8]. Since R() is continuous on [, 8], we can use a table to find the absolute maimum revenue. Table 3 shows that the maimum revenue is R = $4,. The company should leave the price at $4. Any $ decreases in price will lower the revenue.

77 BARNMC_33886.QXD //7 :9 Page 3 3 CHAPTER Graphing and Optimization MATCHED PROBLEM 7 After further analysis, the walnut grower in Matched Problem 6 determines that each additional tree planted will reduce the average yield by 4 pounds. All other information remains the same. How many additional trees per acre should the grower plant now in order to maimize the yield? What is the new maimum yield? I N S I G H T The solution in Eample 7 is called an endpoint solution, because the optimal value occurs at the endpoint of an interval rather than at a critical value in the interior of the interval. It is always important to verify that the optimal value has been found. Inventory Control EXAMPLE 8 SOLUTION Inventory Control A recording company anticipates that there will be a demand for, copies of a certain compact disk (CD) during the net year. It costs the company $. to store a CD for one year. Each time it must make additional CDs, it costs $ to set up the equipment. How many CDs should the company make during each production run to minimize its total storage and setup costs? This type of problem is called an inventory control problem. One of the basic assumptions made in such problems is that the demand is uniform. For eample, if there are working days in a year, the daily demand would be,, = 8 CDs. The company could decide to produce all, CDs at the beginning of the year. This would certainly minimize the setup costs, but would result in very large storage costs. At the other etreme, the company could produce 8 CDs each day. This would minimize the storage costs, but would result in very large setup costs. Somewhere between these two etremes is the optimal solution that will minimize the total storage and setup costs. Let = number of CDs manufactured during each production run y = number of production runs It is easy to see that the total setup cost for the year is y, but what is the total storage cost? If the demand is uniform, the number of CDs in storage between production runs will decrease from to, and the average number in storage each day is. This result is illustrated in Figure 3. Number in storage Average number in storage First Second Third Fourth Production run FIGURE 3 Since it costs $. to store a CD for one year, the total storage cost is.> =. and the total cost is total cost = setup cost + storage cost C = y +.

78 BARNMC_33886.QXD //7 :9 Page 3 In order to write the total cost C as a function of one variable, we must find a relationship between and y. If the company produces CDs in each of y production runs, the total number of CDs produced is y. Thus, Certainly, must be at least and cannot eceed,. Therefore, we must solve the following equation: Thus, Minimize C = a, b +., C = 4,, - 4,, +. = C = - 4,, +. = 4,,. = 6,, = 4, y =, y =, +. Min C = C4, =, y =, 4, = Section -6 Optimization 3-4, is not a critical value, since,. C = 8,, 3 7 for H,, The company will minimize its total cost by making 4, CDs five times during the year. MATCHED PROBLEM 8 Repeat Eample 8 if it costs $ to set up a production run and $.4 to store a CD for one year. Answers to Matched Problems. The dimensions of the garden with the maimum area of 64 square feet are 6 feet by 4 feet, with one 6-foot side with wood fencing.. The minimal cost for enclosing a,8-square-foot garden is $48, and the dimensions are 3 feet by 6 feet, with one 3-foot side with wood fencing. 3. The company will realize a maimum revenue of $67, when the price of a shredder is $. 4. A maimum profit of $46,7 is realized when 4, shredders are manufactured annually and sold for $6 each.. A maimum profit of $378,7 is realized when 3,7 shredders are manufactured annually and sold for $7 each. 6. The maimum yield is, pounds per acre when additional trees are planted on each acre. 7. The maimum yield is, pounds when no additional trees are planted. 8. The company should produce, CDs four times a year.

79 BARNMC_33886.QXD //7 :9 Page 3 3 CHAPTER Graphing and Optimization Eercise -6 Preinary word problems:. How would you divide a -inch line so that the product of the two lengths is a maimum?. What quantity should be added to and subtracted from to produce the maimum product of the results? 3. Find two numbers whose difference is 3 and whose product is a minimum. 4. Find two positive numbers whose sum is 6 and whose product is a maimum.. Find the dimensions of a rectangle with perimeter centimeters that has maimum area. Find the maimum area. 6. Find the dimensions of a rectangle of area square centimeters that has the least perimeter. What is the perimeter? Problems 7 refer to a rectangular area enclosed by a fence that costs $B per foot. Discuss the eistence of a solution and the economical implications of each optimization problem. 7. Given a fied area, minimize the cost of the fencing. 8. Given a fied area, maimize the cost of the fencing. 9. Given a fied amount to spend on fencing, maimize the enclosed area.. Given a fied amount to spend on fencing, minimize the enclosed area.. Maimum revenue and profit. A company manufactures and sells videophones per week. The weekly price demand and cost equations are, respectively, p = -. and C =, + 3 (A) What price should the company charge for the phones, and how many phones should be produced to maimize the weekly revenue? What is the maimum weekly revenue? (B) What is the maimum weekly profit? How much should the company charge for the phones, and how many phones should be produced to realize the maimum weekly profit?. Maimum revenue and profit. A company manufactures and sells digital cameras per week. The weekly price demand and cost equations are, respectively, p = and C =, + 6 (A) What price should the company charge for the cameras, and how many cameras should be produced to maimize the weekly revenue? What is the maimum revenue? (B) What is the maimum weekly profit? How much should the company charge for the cameras, and how many cameras should be produced to realize the maimum weekly profit? 3. Maimum revenue and profit. A company manufactures and sells television sets per month. The monthly cost and price demand equations are C = 7, + 6 p = - 3 6, (A) Find the maimum revenue. (B) Find the maimum profit, the production level that will realize the maimum profit, and the price the company should charge for each television set. (C) If the government decides to ta the company $ for each set it produces, how many sets should the company manufacture each month to maimize its profit? What is the maimum profit? What should the company charge for each set? 4. Maimum revenue and profit. Repeat Problem 3 for C = 6, + 6 p = -,. Maimum profit. The following table contains price demand and total cost data for the production of radial arm saws, where p is the wholesale price (in dollars) of a saw for an annual demand of saws and C is the total cost (in dollars) of producing saws: p C 9 4 3,,,,8 6 8,, 9, (A) Find a quadratic regression equation for the price demand data, using as the independent variable. (B) Find a linear regression equation for the cost data, using as the independent variable. (C) What is the maimum profit? What is the wholesale price per saw that should be charged to realize the maimum profit? Round answers to the nearest dollar. 6. Maimum profit. The following table contains price demand and total cost data for the production of airbrushes, where p is the wholesale price (in dollars) of an airbrush for an annual demand of airbrushes and C is the total cost (in dollars) of producing airbrushes: p C,3 98 4, 3,3 84 7, 4, 67 9,,, (A) Find a quadratic regression equation for the price demand data, using as the independent variable. (B) Find a linear regression equation for the cost data, using as the independent variable. (C) What is the maimum profit? What is the wholesale price per airbrush that should be charged to realize the maimum profit? Round answers to the nearest dollar.

80 BARNMC_33886.QXD //7 :9 Page 33 Section -6 Optimization Maimum revenue. A deli sells 64 sandwiches per day at a price of $8 each. (A) A market survey shows that for every $. reduction in price, 4 more sandwiches will be sold. How much should the deli charge for a sandwich in order to maimize revenue? (B) A different market survey shows that for every $. reduction in the original $8 price, more sandwiches will be sold. Now how much should the deli charge for a sandwich in order to maimize revenue? 8. Maimum revenue. A university student center sells,6 cups of coffee per day at a price of $.4. (A) A market survey shows that for every $. reduction in price, more cups of coffee will be sold. How much should the student center charge for a cup of coffee in order to maimize revenue? (B) A different market survey shows that for every $. reduction in the original $.4 price, 6 more cups of coffee will be sold. Now how much should the student center charge for a cup of coffee in order to maimize revenue? 9. Car rental. A car rental agency rents cars per day at a rate of $3 per day. For each $ increase in rate, fewer cars are rented. At what rate should the cars be rented to produce the maimum income? What is the maimum income?. Rental income. A 3-room hotel in Las Vegas is filled to capacity every night at $8 a room. For each $ increase in rent, 3 fewer rooms are rented. If each rented room costs $ to service per day, how much should the management charge for each room to maimize gross profit? What is the maimum gross profit?. Agriculture. A commercial cherry grower estimates from past records that if 3 trees are planted per acre, each tree will yield an average of pounds of cherries per season. If, for each additional tree planted per acre (up to ), the average yield per tree is reduced by pound, how many trees should be planted per acre to obtain the maimum yield per acre? What is the maimum yield?. Agriculture. A commercial pear grower must decide on the optimum time to have fruit picked and sold. If the pears are picked now, they will bring 3 per pound, with each tree yielding an average of 6 pounds of salable pears. If the average yield per tree increases 6 pounds per tree per week for the net 4 weeks, but the price drops per pound per week, when should the pears be picked to realize the maimum return per tree? What is the maimum return? 3. Manufacturing. A candy bo is to be made out of a piece of cardboard that measures 8 by inches. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a rectangular bo. What size square should be cut from each corner to obtain a maimum volume? 4. Packaging. A parcel delivery service will deliver a package only if the length plus girth (distance around) does not eceed 8 inches. (A) Find the dimensions of a rectangular bo with square ends that satisfies the delivery service s restriction and has maimum volume. What is the maimum volume? (B) Find the dimensions (radius and height) of a cylindrical container that meets the delivery service s Length Girth Figure for 4 requirement and has maimum volume. What is the maimum volume?. Construction costs. A fence is to be built to enclose a rectangular area of 8 square feet. The fence along three sides is to be made of material that costs $6 per foot. The material for the fourth side costs $8 per foot. Find the dimensions of the rectangle that will allow the most economical fence to be built. 6. Construction costs. The owner of a retail lumber store wants to construct a fence to enclose an outdoor storage area adjacent to the store, using all of the store as part of one side of the area (see the figure). Find the dimensions that will enclose the largest area if (A) 4 feet of fencing material are used. (B) 4 feet of fencing material are used. Figure for 6 ft 7. Inventory control. A paint manufacturer has a uniform annual demand for 6, cans of automobile primer. It costs $4 to store one can of paint for one year and $ to set up the plant for production of the primer. How many times a year should the company produce this primer in order to minimize the total storage and setup costs? 8. Inventory control. A pharmacy has a uniform annual demand for bottles of a certain antibiotic. It costs $ to store one bottle for one year and $4 to place an order. How many times during the year should the pharmacy order the antibiotic in order to minimize the total storage and reorder costs? 9. Inventory control. A publishing company sells, copies of a certain book each year. It costs the company $ to store a book for one year. Each time it must print additional copies, it costs the company $, to set up the presses. How many books should the company produce during each printing in order to minimize its total storage and setup costs? 3. Operational costs. The cost per hour for fuel to run a train is v >4 dollars, where v is the speed of the train in miles per hour. (Note that the cost goes up as the square

81 BARNMC_33886.QXD //7 :9 Page CHAPTER Graphing and Optimization of the speed.) Other costs, including labor, are $3 per hour. How fast should the train travel on a 36-mile trip to minimize the total cost for the trip? 3. Construction costs. A freshwater pipeline is to be run from a source on the edge of a lake to a small resort community on an island miles offshore, as indicated in the figure. mi Island miles Figure for 3 (A) If it costs.4 times as much to lay the pipe in the lake as it does on land, what should be (in miles) to minimize the total cost of the project? (B) If it costs only. times as much to lay the pipe in the lake as it does on land, what should be to minimize the total cost of the project? [Note: Compare with Problem 36.] 3. Manufacturing costs. A manufacturer wants to produce cans that will hold ounces (approimately cubic inches) in the form of a right circular cylinder. Find the dimensions (radius of an end and height) of the can that will use the smallest amount of material. Assume that the circular ends are cut out of squares, with the corner portions wasted, and the sides are made from rectangles, with no waste. 33. Bacteria control. A lake used for recreational swimming is treated periodically to control harmful bacteria growth. Suppose that t days after a treatment, the concentration of bacteria per cubic centimeter is given by Ct = 3t - 4t + t 8 How many days after a treatment will the concentration be minimal? What is the minimum concentration? 34. Drug concentration. The concentration C(t), in milligrams per cubic centimeter, of a particular drug in a patient s bloodstream is given by.6t Ct = t + 4t + 4 where t is the number of hours after the drug is taken. How many hours after the drug is taken will the concentration be maimum? What is the maimum concentration? 3. Laboratory management. A laboratory uses white mice each year for eperimental purposes. It costs $4 to feed a mouse for one year. Each time mice are ordered from a supplier, there is a service charge of $ for processing the order. How many mice should be ordered each time to minimize the total cost of feeding the mice and of placing orders for the mice? 36. Bird flights. Some birds tend to avoid flights over large bodies of water during daylight hours. (Scientists speculate that more energy is required to fly over water than land, because air generally rises over land and falls over water during the day.) Suppose that an adult bird with this tendency is taken from its nesting area on the edge of a large lake to an island miles offshore and is then released (see the figure). mi Island Flight path miles Lake Nesting area Figure for 36 (A) If it takes.4 times as much energy to fly over water as land, how far up the shore (, in miles) should the bird head to minimize the total energy epended in returning to the nesting area? (B) If it takes only. times as much energy to fly over water as land, how far up the shore should the bird head to minimize the total energy epended in returning to the nesting area? [Note: Compare with Problem 3.] 37. Botany. If it is known from past eperiments that the height (in feet) of a certain plant after t months is given approimately by Ht = 4t > - t t how long, on the average, will it take a plant to reach its maimum height? What is the maimum height? 38. Pollution. Two heavily industrial areas are located miles apart, as indicated in the figure. If the concentration of particulate matter (in parts per million) decreases as the reciprocal of the square of the distance from the source, and if area A emits eight times the particulate matter as A, the concentration of particulate matter at any point between the two areas is given by C = 8k + k. 9., k 7 - How far from A will the concentration of particulate matter between the two areas be at a minimum? A A Figure for Politics. In a newly incorporated city, it is estimated that the voting population (in thousands) will increase according to Nt = 3 + t - t 3 t 8 where t is time in years. When will the rate of increase be most rapid? 4. Learning. A large grocery chain found that, on the average, a checker can recall P% of a given price list hours after starting work, as given approimately by P = At what time does the checker recall a maimum percentage? What is the maimum?

82 BARNMC_33886.QXD //7 :9 Page 3 Review Eercise 3 CHAPTER REVIEW Important Terms, Symbols, and Concepts - First Derivative and Graphs Increasing and decreasing properties of a function can be determined by eamining a sign chart for the derivative. A number c is a partition number for f () if f (c) = or f () is discontinuous at c. If c is also in the domain of f(), then c is a critical value. Increasing and decreasing properties and local etrema for f() can be determined by eamining the graph of f (). The first-derivative test is used to locate etrema of a function. - Second Derivative and Graphs The second derivative of a function f can be used to determine the concavity of the graph of f. Inflection points on a graph are points where the concavity changes. The concavity of the graph of f() can also be determined by an eamination of the graph of f (). A graphing strategy is used to organize the information obtained from the first and second derivatives. -3 L Hôpital s Rule Limits at infinity and infinite its involving powers of - c, e, and ln are reviewed. The first version of L Hôpital s rule applies to its involving the indeterminate form > as : c. You must always check that L Hôpital s rule applies. L Hôpital s rule can be applied more than once. L Hôpital s rule applies to one-sided its. L Hôpital s rule applies to its at infinity. L Hôpital s rule applies to its involving the indeterminate form q> q. -4 Curve-Sketching Techniques The graphing strategy first used in Section - is epanded to include horizontal and vertical asymptotes. If f() = n()>d() is a rational function with the degree of n() more than the degree of d(), then the graph of f() has an oblique asymptote of the form y = m + b. - Absolute Maima and Minima The steps involved in finding the absolute maimum and absolute minimum values of a continuous function on a closed interval are listed in a procedure. The second-derivative test for local etrema can be used to test critical values, but it does not work in all cases. If a function is continuous on an interval I and has only one critical value in I, the second-derivative test for absolute etrema can be used to find the absolute etrema, but it not does work in all cases. -6 Optimization The methods used to solve optimization problem are summarized and illustrated by eamples. Eamples E., p. 76 E., p. 77 E. 3, p. 78 E. 4, p. 78 E., p. 79 E. 6, p. 8 E. 7, p. 83 E., p. 96 E., p. 98 E. 3, p. 98 E. 4, p. 3 E., p. 3 E. 6, p. 33 E., p. 3 E., p. 33 E. 3, p. 34 E. 4, p. 3 E., p. 3 E. 6, p. 36 E. 7, p. 37 E. 8, p. 37 E. 9, p. 38 E., p. 39 E., p. 3 E., p. 33 E. 3, p. 34 E. 4, p. 36 E., p. 37 E.,.p. 337 E., p. 339 E. 3, p. 34 E., p. 343 E., p. 344 E. 3, p. 346 E. 4, p. 346 E., p. 348 E. 6, p. 348 E. 7, p. 349 E. 8, p. 3

83 BARNMC_33886.QXD //7 :9 Page CHAPTER Graphing and Optimization REVIEW EXERCISE Work through all the problems in this chapter review, and check your answers in the back of the book. Answers to all review problems are there, along with section numbers in italics to indicate where each type of problem is discussed. Where weaknesses show up, review appropriate sections in the tet. A Problems 8 refer to the graph of y = f that follows. Identify the points or intervals on the ais that produce the indicated behavior.. f() is increasing.. f 6 3. The graph of f is concave downward. f() a c c c 3 c 4 c c 6 c 7 b 4. Local minima. Absolute maima 6. f appears to be. 7. f does not eist. 8. Inflection points In Problems 9 and, use the given information to sketch the graph of f. Assume that f is continuous on its domain and that all intercepts are included in the information given. 9. Domain: All real f() f = f = f = f = 8. f = f =. f = f = - e -. f = 3 ln Find each it in Problems 3 3. e : : ln( + ). 6. : - e : q : + ln( + ) : + e + e - - : ln : q ln( + 6) ln( + 6) : q ln( + 3) : ln( + 3) 33. Use the graph of y = f f () shown here to discuss the graph of y = f. Organize your conclusions in a table (see Eample 4, Section -). Sketch a possible graph of y = f. f () ND Figure for 33 and 34 f () ND 34. Refer to the proceding graph of y = f. Which of the following could be the graph of y = f? (A) f () (B) f (). Domain: All real ; f- =, f =, f = ; f = ; f 6 on - q, ; f 7 on, q; f - =, f = ; f 6 on - q, - and, q; f 7 on -, ; f = ; f = : -q : q. Find f for f = (C) f (). Find y for y = In Problems 3, summarize all the pertinent information obtained by applying the final version of the graphing strategy (Section 4) to f, and sketch the graph of f. 3. f =

84 BARNMC_33886.QXD //7 :9 Page 37 Applications Use the second-derivative test to find any local etrema for f = Find the absolute maimum and absolute minimum, if either eists, for y = f = Find the absolute minimum, if it eists, for y = f = Find the absolute maimum value, if it eists, for f = - ln Find the absolute maimum value, if it eists, for f = e Let y = f be a polynomial function with local minima at = a and = b, a 6 b. Must f have at least one local maimum between a and b? Justify your answer. 4. The derivative of f = - is f = - -. Since f 6 for Z, is it correct to say that f() is decreasing for all ecept =? Eplain. 4. Discuss the difference between a partition number for f and a critical value of f(), and illustrate with eamples. C 43. Find the absolute maimum for f if Graph f and f on the same coordinate system for Find two positive numbers whose product is 4 and whose sum is a minimum. What is the minimum sum? In Problems 4 and 46, apply the graphing strategy and summarize the pertinent information. Round any approimate values to two decimal places f = f = f = Find the absolute maimum value, if it eists, for f() = e Find the absolute maimum value, if it eists, for f() = ln e 7 APPLICATIONS 49. Price analysis. The graph in the figure approimates the rate of change of the price of tomatoes over a 6-month period, where p(t) is the price of a pound of tomatoes and t is time (in months). (A) Write a brief verbal description of the graph of y = pt, including a discussion of local etrema and inflection points. (B) Sketch a possible graph of y = pt p (t) Figure for 49. Maimum revenue and profit. A company manufactures and sells electric stoves per month. The monthly cost and price demand equations are, respectively, C = 3 +, p = -., (A) Find the maimum revenue. (B) How many stoves should the company manufacture each month to maimize its profit? What is the maimum monthly profit? How much should the company charge for each stove? (C) If the government decides to ta the company $ for each stove it produces, how many stoves should the company manufacture each month to maimize its profit? What is the maimum monthly profit? How much should the company charge for each stove?. Construction. A fence is to be built to enclose a rectangular area. The fence along three sides is to be made of material that costs $ per foot. The material for the fourth side costs $ per foot. (A) If the area is, square feet, find the dimensions of the rectangle that will allow the most economical fence to be built. (B) If $3, is available for the fencing, find the dimensions of the rectangle that will enclose the most area.. Rental income. A -room hotel in Fresno is filled to capacity every night at a rate of $4 per room. For each $ increase in the nightly rate, 4 fewer rooms are rented. If each rented room costs $8 a day to service, how much should the management charge per room in order to maimize gross profit? What is the maimum gross profit? 3. Inventory control. A computer store sells 7, boes of floppy disks annually. It costs the store $. to store a bo of disks for one year. Each time it reorders disks, the store must pay a $. service charge for processing the order. How many times during the year should the store order disks to minimize the total storage and reorder costs?

f'(x) = x 4 (2)(x - 6)(1) + (x - 6) 2 (4x 3 ) f'(x) = (x - 2) -1/3 = x 2 ; domain of f: (-, ) f'(x) = (x2 + 1)4x! 2x 2 (2x) 4x f'(x) =

f'(x) = x 4 (2)(x - 6)(1) + (x - 6) 2 (4x 3 ) f'(x) = (x - 2) -1/3 = x 2 ; domain of f: (-, ) f'(x) = (x2 + 1)4x! 2x 2 (2x) 4x f'(x) = 85. f() = 4 ( - 6) 2 f'() = 4 (2)( - 6)(1) + ( - 6) 2 (4 3 ) = 2 3 ( - 6)[ + 2( - 6)] = 2 3 ( - 6)(3-12) = 6 3 ( - 4)( - 6) Thus, the critical values are = 0, = 4, and = 6. Now we construct the sign chart

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