APPLICATIONS OF DIFFERENTIATION

Transcription

1 4 APPLICATIONS OF DIFFERENTIATION Calculus reveals all the important aspects of graphs of functions. Members of the famil of functions f c sin are illustrated. We have alread investigated some of the applications of derivatives, but now that we know the differentiation rules we are in a better position to pursue the applications of differentiation in greater depth. Here we learn how derivatives affect the shape of a graph of a function and, in particular, how the help us locate maimum and minimum values of functions. Man practical problems require us to minimize a cost or maimize an area or somehow find the best possible outcome of a situation. In particular, we will be able to investigate the optimal shape of a can and to eplain the location of rainbows in the sk. 7

2 4. MAXIMUM AND MINIMUM VALUES f(d) Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) wa of doing something. Here are eamples of such problems that we will solve in this chapter: What is the shape of a can that minimizes manufacturing costs? What is the maimum acceleration of a space shuttle? (This is an important question to the astronauts who have to withstand the effects of acceleration.) What is the radius of a contracted windpipe that epels air most rapidl during a cough? At what angle should blood vessels branch so as to minimize the energ epended b the heart in pumping blood? These problems can be reduced to finding the maimum or minimum values of a function. Let s first eplain eactl what we mean b maimum and minimum values. DEFINITION A function f has an absolute maimum (or global maimum) at c if f c f for all in D, where D is the domain of f. The number f c is called the maimum value of f on D. Similarl, f has an absolute minimum at c if f c f for all in D and the number f c is called the minimum value of f on D. The maimum and minimum values of f are called the etreme values of f. f(a) a b c d e FIGURE Minimum value f(a), maimum value f(d) Figure shows the graph of a function f with absolute maimum at d and absolute minimum at a. Note that d, f d is the highest point on the graph and a, f a is the lowest point. If we consider onl values of near b [for instance, if we restrict our attention to the interval a, c], then f b is the largest of those values of f and is called a local maimum value of f. Likewise, f c is called a local minimum value of f because f c f for near c [in the interval b, d, for instance]. The function f also has a local minimum at e. In general, we have the following definition. = DEFINITION A function f has a local maimum (or relative maimum) at c if f c f when is near c. [This means that f c f for all in some open interval containing c.] Similarl, f has a local minimum at c if f c f when is near c. FIGURE Minimum value, no maimum = FIGURE 3 No minimum, no maimum EXAMPLE The function f cos takes on its (local and absolute) maimum value of infinitel man times, since cos n for an integer n and cos for all. Likewise, cosn is its minimum value, where n is an integer. M EXAMPLE If f, then f f because for all. Therefore f is the absolute (and local) minimum value of f. This corresponds to the fact that the origin is the lowest point on the parabola. (See Figure.) However, there is no highest point on the parabola and so this function has no maimum value. EXAMPLE 3 From the graph of the function f 3, shown in Figure 3, we see that this function has neither an absolute maimum value nor an absolute minimum value. In fact, it has no local etreme values either. M M 7

3 7 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION V EXAMPLE 4 The graph of the function (_, 37) =3$-6 +8 f (, 5) _ (3, _7) FIGURE 4 is shown in Figure 4. You can see that f 5 is a local maimum, whereas the absolute maimum is f 37. (This absolute maimum is not a local maimum because it occurs at an endpoint.) Also, f is a local minimum and f 3 7 is both a local and an absolute minimum. Note that f has neither a local nor an absolute maimum at 4. M We have seen that some functions have etreme values, whereas others do not. The following theorem gives conditions under which a function is guaranteed to possess etreme values. 3 THE EXTREME VALUE THEOREM If f is continuous on a closed interval a, b, then f attains an absolute maimum value f c and an absolute minimum value f d at some numbers c and d in a, b. The Etreme Value Theorem is illustrated in Figure 5. Note that an etreme value can be taken on more than once. Although the Etreme Value Theorem is intuitivel ver plausible, it is difficult to prove and so we omit the proof. FIGURE 5 a c d b a c d=b a c d c b Figures 6 and 7 show that a function need not possess etreme values if either hpothesis (continuit or closed interval) is omitted from the Etreme Value Theorem. 3 FIGURE 6 This function has minimum value f()=, but no maimum value. FIGURE 7 This continuous function g has no maimum or minimum. The function f whose graph is shown in Figure 6 is defined on the closed interval [, ] but has no maimum value. (Notice that the range of f is [, 3). The function takes on values arbitraril close to 3, but never actuall attains the value 3.) This does not contradict the Etreme Value Theorem because f is not continuous. [Nonetheless, a discontinuous function could have maimum and minimum values. See Eercise 3(b).]

4 SECTION 4. MAXIMUM AND MINIMUM VALUES 73 c d FIGURE 8 {c, f(c)} {d, f(d)} The function t shown in Figure 7 is continuous on the open interval (, ) but has neither a maimum nor a minimum value. [The range of t is,. The function takes on arbitraril large values.] This does not contradict the Etreme Value Theorem because the interval (, ) is not closed. The Etreme Value Theorem sas that a continuous function on a closed interval has a maimum value and a minimum value, but it does not tell us how to find these etreme values. We start b looking for local etreme values. Figure 8 shows the graph of a function f with a local maimum at c and a local minimum at d. It appears that at the maimum and minimum points the tangent lines are horizontal and therefore each has slope. We know that the derivative is the slope of the tangent line, so it appears that f c and f d. The following theorem sas that this is alwas true for differentiable functions. N Fermat s Theorem is named after Pierre Fermat (6 665), a French lawer who took up mathematics as a hobb. Despite his amateur status, Fermat was one of the two inventors of analtic geometr (Descartes was the other). His methods for finding tangents to curves and maimum and minimum values (before the invention of its and derivatives) made him a forerunner of Newton in the creation of differential calculus. 4 FERMAT S THEOREM If f has a local maimum or minimum at c, and if f c eists, then f c. PROOF Suppose, for the sake of definiteness, that f has a local maimum at c. Then, according to Definition, f c f if is sufficientl close to c. This implies that if h is sufficientl close to, with h being positive or negative, then f c f c h and therefore 5 f c h f c We can divide both sides of an inequalit b a positive number. Thus, if h and h is sufficientl small, we have f c h f c h Taking the right-hand it of both sides of this inequalit (using Theorem.3.), we get hl But since f c eists, we have f c h f c h hl f c h l f c h f c h h l f c h f c h and so we have shown that f c. If h, then the direction of the inequalit (5) is reversed when we divide b h: f c h f c h So, taking the left-hand it, we have h f c h l f c h f c h f c h f c h l h

5 74 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION FIGURE 9 If ƒ=, then fª()= but ƒ has no maimum or minimum. = = FIGURE If ƒ=, then f()= is a minimum value, but fª() does not eist. We have shown that f c and also that f c. Since both of these inequalities must be true, the onl possibilit is that f c. We have proved Fermat s Theorem for the case of a local maimum. The case of a local minimum can be proved in a similar manner, or we could use Eercise 76 to deduce it from the case we have just proved (see Eercise 77). The following eamples caution us against reading too much into Fermat s Theorem. We can t epect to locate etreme values simpl b setting f and solving for. EXAMPLE 5 If f 3, then f 3, so f. But f has no maimum or minimum at, as ou can see from its graph in Figure 9. (Or observe that 3 for but 3 for.) The fact that f simpl means that the curve 3 has a horizontal tangent at,. Instead of having a maimum or minimum at,, the curve crosses its horizontal tangent there. M f EXAMPLE 6 The function has its (local and absolute) minimum value at, but that value can t be found b setting f because, as was shown in Eample 5 in Section.8, f does not eist. (See Figure.) M WARNING Eamples 5 and 6 show that we must be careful when using Fermat s Theorem. Eample 5 demonstrates that even when f c there need not be a maimum or minimum at c. (In other words, the converse of Fermat s Theorem is false in general.) Furthermore, there ma be an etreme value even when f c does not eist (as in Eample 6). Fermat s Theorem does suggest that we should at least start looking for etreme values of f at the numbers c where f c or where f c does not eist. Such numbers are given a special name. 6 DEFINITION A critical number of a function f is a number c in the domain of f such that either f c or f c does not eist. M N Figure shows a graph of the function f in Eample 7. It supports our answer because there is a horizontal tangent when.5 and a vertical tangent when. 3.5 _.5 5 _ FIGURE V EXAMPLE 7 Find the critical numbers of f SOLUTION The Product Rule gives f 35 4 ( ) 35 [The same result could be obtained b first writing f ] Therefore f if 8, that is, 3, and f does not eist when. Thus the 3 critical numbers are and. M In terms of critical numbers, Fermat s Theorem can be rephrased as follows (compare Definition 6 with Theorem 4): 7 If f has a local maimum or minimum at c, then c is a critical number of f.

6 SECTION 4. MAXIMUM AND MINIMUM VALUES 75 To find an absolute maimum or minimum of a continuous function on a closed interval, we note that either it is local [in which case it occurs at a critical number b (7)] or it occurs at an endpoint of the interval. Thus the following three-step procedure alwas works. THE CLOSED INTERVAL METHOD To find the absolute maimum and minimum values of a continuous function f on a closed interval a, b:. Find the values of f at the critical numbers of f in a, b.. Find the values of f at the endpoints of the interval. 3. The largest of the values from Steps and is the absolute maimum value; the smallest of these values is the absolute minimum value. V EXAMPLE 8 Find the absolute maimum and minimum values of the function f = -3 + (4, 7) SOLUTION Since f is continuous on [, 4], we can use the Closed Interval Method: f 3 3 f Since f eists for all, the onl critical numbers of f occur when f, that is, or. Notice that each of these critical numbers lies in the interval (, 4). The values of f at these critical numbers are f f 3 The values of f at the endpoints of the interval are 5 5 FIGURE (, _3) 3 4 f ( ) 8 f 4 7 Comparing these four numbers, we see that the absolute maimum value is f 4 7 and the absolute minimum value is f 3. Note that in this eample the absolute maimum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of f is sketched in Figure. M If ou have a graphing calculator or a computer with graphing software, it is possible to estimate maimum and minimum values ver easil. But, as the net eample shows, calculus is needed to find the eact values. 8 EXAMPLE 9 (a) Use a graphing device to estimate the absolute minimum and maimum values of the function f sin,. (b) Use calculus to find the eact minimum and maimum values. _ FIGURE 3 π SOLUTION (a) Figure 3 shows a graph of f in the viewing rectangle, b, 8. B moving the cursor close to the maimum point, we see that the -coordinates don t change ver much in the vicinit of the maimum. The absolute maimum value is about 6.97 and it occurs when 5.. Similarl, b moving the cursor close to the minimum point, we see that the absolute minimum value is about.68 and it occurs when.. It is

7 76 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION possible to get more accurate estimates b zooming in toward the maimum and minimum points, but instead let s use calculus. (b) The function f sin is continuous on,. Since f cos, we have f when cos and this occurs when 3 or 53. The values of f at these critical points are f 3 3 sin 3 3 s and The values of f f at the endpoints are sin s f and f 6.8 Comparing these four numbers and using the Closed Interval Method, we see that the absolute minimum value is f 3 3 s3 and the absolute maimum value is f s3. The values from part (a) serve as a check on our work. M EXAMPLE The Hubble Space Telescope was deploed on April 4, 99, b the space shuttle Discover. A model for the velocit of the shuttle during this mission, from liftoff at t until the solid rocket boosters were jettisoned at t 6 s, is given b vt.3t 3.99t 3.6t 3.83 (in feet per second). Using this model, estimate the absolute maimum and minimum values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters. SOLUTION We are asked for the etreme values not of the given velocit function, but rather of the acceleration function. So we first need to differentiate to find the acceleration: NASA at vt d dt.3t 3.99t 3.6t t.858t 3.6 We now appl the Closed Interval Method to the continuous function a on the interval t 6. Its derivative is at.78t.858 The onl critical number occurs when at : t Evaluating at at the critical number and at the endpoints, we have a 3.6 at.5 a So the maimum acceleration is about 6.87 fts and the minimum acceleration is about.5 fts. M

8 SECTION 4. MAXIMUM AND MINIMUM VALUES EXERCISES. Eplain the difference between an absolute minimum and a local minimum.. Suppose f is a continuous function defined on a closed interval a, b. (a) What theorem guarantees the eistence of an absolute maimum value and an absolute minimum value for f? (b) What steps would ou take to find those maimum and minimum values? 3 4 For each of the numbers a, b, c, d, r, and s, state whether the function whose graph is shown has an absolute maimum or minimum, a local maimum or minimum, or neither a maimum nor a minimum Use the graph to state the absolute and local maimum and minimum values of the function Sketch the graph of a function f that is continuous on [, 5] and has the given properties. 7. Absolute minimum at, absolute maimum at 3, local minimum at 4 8. Absolute minimum at, absolute maimum at 5, local maimum at, local minimum at 4 9. Absolute maimum at 5, absolute minimum at, local maimum at 3, local minima at and 4. f has no local maimum or minimum, but and 4 are critical numbers. a b c d r s =ƒ a b c d r s (a) Sketch the graph of a function that has a local maimum at and is differentiable at. (b) Sketch the graph of a function that has a local maimum at and is continuous but not differentiable at. = (c) Sketch the graph of a function that has a local maimum at and is not continuous at.. (a) Sketch the graph of a function on [, ] that has an absolute maimum but no local maimum. (b) Sketch the graph of a function on [, ] that has a local maimum but no absolute maimum. 3. (a) Sketch the graph of a function on [, ] that has an absolute maimum but no absolute minimum. (b) Sketch the graph of a function on [, ] that is discontinuous but has both an absolute maimum and an absolute minimum. 4. (a) Sketch the graph of a function that has two local maima, one local minimum, and no absolute minimum. (b) Sketch the graph of a function that has three local minima, two local maima, and seven critical numbers. 5 8 Sketch the graph of f b hand and use our sketch to find the absolute and local maimum and minimum values of f. (Use the graphs and transformations of Sections. and.3.) 5. f 8 3, 6. f 3, 5 7. f, 8. f, 9. f,. f,. f, 3. f, 3. f ln, 4. f t cos t, f s f e f 4 f t 3 if if 3 if if 9 44 Find the critical numbers of the function. 9. f f 3 3. f f st 3t 4 4t 3 6t 34. tt 3t t 36. hp p p 4

9 78 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 37. ht t 34 t t s f cos sin 4. t 4 tan 43. f e f ln ; A formula for the derivative of a function f is given. How man critical numbers does f have? f 5e. sin Find the absolute maimum and absolute minimum values of f on the given interval. 47. f 3 5, 48. f 3 3, 5. f 3 6 9, 5. f 4 3, 5. f 3,, 53. f,, 54. f 4, f t ts4 t, 56. f t s 3 t 8 t, 57. f t cos t sin t, 58. f t t cott, 59. f e 8, 6. f ln, If a and b are positive numbers, find the maimum value of f a b,. ; 64. Use a graph to estimate the critical numbers of correct to one decimal place. ; (a) Use a graph to estimate the absolute maimum and minimum values of the function to two decimal places. (b) Use calculus to find the eact maimum and minimum values. 65. F 45 4 f ln, f e e, 4, 4,, 8, 4 [, ] f 3 3, f 5 3, 66. f e 3,, 3, f 3 3,, 3, 4, 74,, 4 t 3 3 f cos, f s f cos, 69. Between C and 3C, the volume V (in cubic centimeters) of kg of water at a temperature T is given approimatel b the formula V T.8543T.679T 3 Find the temperature at which water has its maimum densit. 7. An object with weight W is dragged along a horizontal plane b a force acting along a rope attached to the object. If the rope makes an angle with the plane, then the magnitude of the force is F where is a positive constant called the coefficient of friction and where. Show that F is minimized when tan. W sin cos 7. A model for the US average price of a pound of white sugar from 993 to 3 is given b the function St.337t 5.937t t 3.369t.4458t.474 where t is measured in ears since August of 993. Estimate the times when sugar was cheapest and most epensive during the period ; 7. On Ma 7, 99, the space shuttle Endeavour was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table gives the velocit data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. Event Time (s) Velocit (fts) Launch Begin roll maneuver 85 End roll maneuver 5 39 Throttle to 89% 447 Throttle to 67% 3 74 Throttle to 4% Maimum dnamic pressure Solid rocket booster separation 5 45 (a) Use a graphing calculator or computer to find the cubic polnomial that best models the velocit of the shuttle for the time interval t, 5. Then graph this polnomial. (b) Find a model for the acceleration of the shuttle and use it to estimate the maimum and minimum values of the acceleration during the first 5 seconds.

10 APPLIED PROJECT THE CALCULUS OF RAINBOWS When a foreign object lodged in the trachea (windpipe) forces a person to cough, the diaphragm thrusts upward causing an increase in pressure in the lungs. This is accompanied b a contraction of the trachea, making a narrower channel for the epelled air to flow through. For a given amount of air to escape in a fied time, it must move faster through the narrower channel than the wider one. The greater the velocit of the airstream, the greater the force on the foreign object. X ras show that the radius of the circular tracheal tube contracts to about two-thirds of its normal radius during a cough. According to a mathematical model of coughing, the velocit v of the airstream is related to the radius r of the trachea b the equation vr kr rr r r r where k is a constant and r is the normal radius of the trachea. The restriction on r is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than r is prevented (otherwise the person would suffocate). (a) Determine the value of r in the interval [ r, r ] at which v has an absolute maimum. How does this compare with eperimental evidence? (b) What is the absolute maimum value of v on the interval? (c) Sketch the graph of v on the interval, r. 74. Show that 5 is a critical number of the function but t does not have a local etreme value at Prove that the function t 5 3 f 5 has neither a local maimum nor a local minimum. 76. If f has a minimum value at c, show that the function t f has a maimum value at c. 77. Prove Fermat s Theorem for the case in which f has a local minimum at c. 78. A cubic function is a polnomial of degree 3; that is, it has the form f a 3 b c d, where a. (a) Show that a cubic function can have two, one, or no critical number(s). Give eamples and sketches to illustrate the three possibilities. (b) How man local etreme values can a cubic function have? å A from sun to observer A P P L I E D P R O J E C T å O C Formation of the primar rainbow B D(å ) THE CALCULUS OF RAINBOWS Rainbows are created when raindrops scatter sunlight. The have fascinated mankind since ancient times and have inspired attempts at scientific eplanation since the time of Aristotle. In this project we use the ideas of Descartes and Newton to eplain the shape, location, and colors of rainbows.. The figure shows a ra of sunlight entering a spherical raindrop at A. Some of the light is reflected, but the line AB shows the path of the part that enters the drop. Notice that the light is refracted toward the normal line AO and in fact Snell s Law sas that sin k sin, where is the angle of incidence, is the angle of refraction, and k 4 3 is the inde of refraction for water. At B some of the light passes through the drop and is refracted into the air, but the line BC shows the part that is reflected. (The angle of incidence equals the angle of reflection.) When the ra reaches C, part of it is reflected, but for the time being we are more interested in the part that leaves the raindrop at C. (Notice that it is refracted awa from the normal line.) The angle of deviation D is the amount of clockwise rotation that the ra has undergone during this three-stage process. Thus ras from sun ras from sun 4 observer 38 Show that the minimum value of the deviation is D 38 and occurs when. The significance of the minimum deviation is that when we have D, so. This means that man ras with become deviated b approimatel the same amount. It is the concentration of ras coming from near the direction of minimum deviation that creates the brightness of the primar rainbow. The figure at the left shows that the angle of elevation from the observer up to the highest point on the rainbow is (This angle is called the rainbow angle.) D D Problem eplains the location of the primar rainbow, but how do we eplain the colors? Sunlight comprises a range of wavelengths, from the red range through orange, ellow,

11 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION to observer from sun D å å A Formation of the secondar rainbow C B green, blue, indigo, and violet. As Newton discovered in his prism eperiments of 666, the inde of refraction is different for each color. (The effect is called dispersion.) For red light the refractive inde is k.338 whereas for violet light it is k B repeating the calculation of Problem for these values of k, show that the rainbow angle is about 4.3 for the red bow and 4.6 for the violet bow. So the rainbow reall consists of seven individual bows corresponding to the seven colors. 3. Perhaps ou have seen a fainter secondar rainbow above the primar bow. That results from the part of a ra that enters a raindrop and is refracted at A, reflected twice (at B and C), and refracted as it leaves the drop at D (see the figure). This time the deviation angle D is the total amount of counterclockwise rotation that the ra undergoes in this four-stage process. Show that D 6 and D has a minimum value when cos k 8 Taking k 4 3, show that the minimum deviation is about 9 and so the rainbow angle for the secondar rainbow is about 5, as shown in the figure. C. Donald Ahrens Show that the colors in the secondar rainbow appear in the opposite order from those in the primar rainbow. 4. THE MEAN VALUE THEOREM We will see that man of the results of this chapter depend on one central fact, which is called the Mean Value Theorem. But to arrive at the Mean Value Theorem we first need the following result. N Rolle s Theorem was first published in 69 b the French mathematician Michel Rolle (65 79) in a book entitled Méthode pour résoudre les égalitéz. He was a vocal critic of the methods of his da and attacked calculus as being a collection of ingenious fallacies. Later, however, he became convinced of the essential correctness of the methods of calculus. ROLLE S THEOREM Let f be a function that satisfies the following three hpotheses:. f is continuous on the closed interval a, b.. f is differentiable on the open interval a, b. 3. f a f b Then there is a number c in a, b such that f c.

12 SECTION 4. THE MEAN VALUE THEOREM 8 Before giving the proof let s take a look at the graphs of some tpical functions that satisf the three hpotheses. Figure shows the graphs of four such functions. In each case it appears that there is at least one point c, f c on the graph where the tangent is horizontal and therefore f c. Thus Rolle s Theorem is plausible. a c c b a c b a c c b a c b (a) (b) (c) (d) FIGURE N Take cases PROOF There are three cases: CASE I N f k, a constant Then f, so the number c can be taken to be an number in a, b. CASE II N f f a for some in a, b [as in Figure (b) or (c)] B the Etreme Value Theorem (which we can appl b hpothesis ), f has a maimum value somewhere in a, b. Since f a f b, it must attain this maimum value at a number c in the open interval a, b. Then f has a local maimum at c and, b hpothesis, f is differentiable at c. Therefore f c b Fermat s Theorem. CASE III N f f a for some in a, b [as in Figure (c) or (d)] B the Etreme Value Theorem, f has a minimum value in a, b and, since f a f b, it attains this minimum value at a number c in a, b. Again f c b Fermat s Theorem. EXAMPLE Let s appl Rolle s Theorem to the position function s f t of a moving object. If the object is in the same place at two different instants t a and t b, then f a f b. Rolle s Theorem sas that there is some instant of time t c between a and b when f c ; that is, the velocit is. (In particular, ou can see that this is true when a ball is thrown directl upward.) M M N Figure shows a graph of the function f 3 discussed in Eample. Rolle s Theorem shows that, no matter how much we enlarge the viewing rectangle, we can never find a second -intercept. _ 3 EXAMPLE Prove that the equation 3 has eactl one real root. SOLUTION First we use the Intermediate Value Theorem (.5.) to show that a root eists. Let f 3. Then f and f. Since f is a polnomial, it is continuous, so the Intermediate Value Theorem states that there is a number c between and such that f c. Thus the given equation has a root. To show that the equation has no other real root, we use Rolle s Theorem and argue b contradiction. Suppose that it had two roots a and b. Then f a f b and, since f is a polnomial, it is differentiable on a, b and continuous on a, b. Thus, b Rolle s Theorem, there is a number c between a and b such that f c. But f 3 for all FIGURE _3 (since ) so f can never be. This gives a contradiction. Therefore the equation can t have two real roots. M

13 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Our main use of Rolle s Theorem is in proving the following important theorem, which was first stated b another French mathematician, Joseph-Louis Lagrange. N The Mean Value Theorem is an eample of what is called an eistence theorem. Like the Intermediate Value Theorem, the Etreme Value Theorem, and Rolle s Theorem, it guarantees that there eists a number with a certain propert, but it doesn t tell us how to find the number. THE MEAN VALUE THEOREM Let f be a function that satisfies the following hpotheses:. f is continuous on the closed interval a, b.. f is differentiable on the open interval a, b. Then there is a number c in a, b such that f c f b f a b a or, equivalentl, f b f a f cb a Before proving this theorem, we can see that it is reasonable b interpreting it geometricall. Figures 3 and 4 show the points Aa, f a and Bb, f b on the graphs of two differentiable functions. The slope of the secant line AB is 3 m AB f b f a b a which is the same epression as on the right side of Equation. Since f c is the slope of the tangent line at the point c, f c, the Mean Value Theorem, in the form given b Equation, sas that there is at least one point Pc, f c on the graph where the slope of the tangent line is the same as the slope of the secant line AB. In other words, there is a point P where the tangent line is parallel to the secant line AB. P{c, f(c)} P B A{a, f(a)} A P B{b, f(b)} a c b a c c b FIGURE 3 FIGURE 4 PROOF We appl Rolle s Theorem to a new function h defined as the difference between f and the function whose graph is the secant line AB. Using Equation 3, we see that the equation of the line AB can be written as or as f a f a f b f a b a f b f a b a a a

14 SECTION 4. THE MEAN VALUE THEOREM 83 A FIGURE 5 ƒ h() =ƒ f(b)-f(a) f(a)+ (-a) b-a LAGRANGE AND THE MEAN VALUE THEOREM The Mean Value Theorem was first formulated b Joseph-Louis Lagrange (736 83), born in Ital of a French father and an Italian mother. He was a child prodig and became a professor in Turin at the tender age of 9. Lagrange made great contributions to number theor, theor of functions, theor of equations, and analtical and celestial mechanics. In particular, he applied calculus to the analsis of the stabilit of the solar sstem. At the invitation of Frederick the Great, he succeeded Euler at the Berlin Academ and, when Frederick died, Lagrange accepted King Louis XVI s invitation to Paris, where he was given apartments in the Louvre and became a professor at the Ecole Poltechnique. Despite all the trappings of luur and fame, he was a kind and quiet man, living onl for science. B So, as shown in Figure 5, 4 First we must verif that h satisfies the three hpotheses of Rolle s Theorem.. The function h is continuous on a, b because it is the sum of f and a first-degree polnomial, both of which are continuous.. The function h is differentiable on a, b because both f and the first-degree polnomial are differentiable. In fact, we can compute h directl from Equation 4: 3. (Note that f a and f b f ab a are constants.) Therefore, ha hb. h f f a h f ha f a f a hb f b f a f b f a f b f a Since h satisfies the hpotheses of Rolle s Theorem, that theorem sas there is a number c in a, b such that hc. Therefore hc f c f b f a b a f b f a b a f b f a b a f b f a b a b a f b f a b a a a a and so f c f b f a b a M = - B V EXAMPLE 3 To illustrate the Mean Value Theorem with a specific function, let s consider f 3, a, b. Since f is a polnomial, it is continuous and differentiable for all, so it is certainl continuous on, and differentiable on,. Therefore, b the Mean Value Theorem, there is a number c in, such that f f f c Now f 6, f, and f 3, so this equation becomes O FIGURE 6 c 6 3c 6c which gives c 4 3, that is, c s3. But c must lie in,, so c s3. Figure 6 illustrates this calculation: The tangent line at this value of c is parallel to the secant line OB. M V EXAMPLE 4 If an object moves in a straight line with position function s f t, then the average velocit between t a and t b is f b f a b a

15 84 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION and the velocit at t c is f c. Thus the Mean Value Theorem (in the form of Equation ) tells us that at some time t c between a and b the instantaneous velocit f c is equal to that average velocit. For instance, if a car traveled 8 km in hours, then the speedometer must have read 9 kmh at least once. In general, the Mean Value Theorem can be interpreted as saing that there is a number at which the instantaneous rate of change is equal to the average rate of change over an interval. M The main significance of the Mean Value Theorem is that it enables us to obtain information about a function from information about its derivative. The net eample provides an instance of this principle. V EXAMPLE 5 Suppose that f 3 and f 5 for all values of. How large can f possibl be? SOLUTION We are given that f is differentiable (and therefore continuous) everwhere. In particular, we can appl the Mean Value Theorem on the interval,. There eists a number c such that f f f c so f f f c 3 f c We are given that f 5 for all, so in particular we know that f c 5. Multipling both sides of this inequalit b, we have f c, so f 3 f c 3 7 The largest possible value for f is 7. M The Mean Value Theorem can be used to establish some of the basic facts of differential calculus. One of these basic facts is the following theorem. Others will be found in the following sections. 5 THEOREM If f for all in an interval a, b, then f is constant on a, b. PROOF Let and be an two numbers in a, b with. Since f is differentiable on a, b, it must be differentiable on, and continuous on,. B appling the Mean Value Theorem to f on the interval,, we get a number c such that c and 6 f f f c Since f for all, we have f c, and so Equation 6 becomes f f or f f Therefore f has the same value at an two numbers and in a, b. This means that f is constant on a, b. M 7 COROLLARY If f t for all in an interval a, b, then f t is constant on a, b; that is, f t c where c is a constant.

16 SECTION 4. THE MEAN VALUE THEOREM 85 PROOF Let F f t. Then F f t for all in a, b. Thus, b Theorem 5, F is constant; that is, f t is constant. M NOTE Care must be taken in appling Theorem 5. Let f The domain of f is D and f for all in D. But f is obviousl not a constant function. This does not contradict Theorem 5 because D is not an interval. Notice that f is constant on the interval, and also on the interval,. EXAMPLE 6 Prove the identit tan cot. if if SOLUTION Although calculus isn t needed to prove this identit, the proof using calculus is quite simple. If f tan cot, then f for all values of. Therefore f C, a constant. To determine the value of C, we put [because we can evaluate f eactl]. Then C f tan cot 4 4 Thus tan cot. M 4. EXERCISES 4 Verif that the function satisfies the three hpotheses of Rolle s Theorem on the given interval. Then find all numbers c that satisf the conclusion of Rolle s Theorem... f 3 6, 3. f 5 3, f s 3, 4. f cos,, 9 8, 78, 3, 3 5. Let f 3. Show that f f but there is no number c in, such that f c. Wh does this not contradict Rolle s Theorem? 6. Let f tan. Show that f f but there is no number c in, such that f c. Wh does this not contradict Rolle s Theorem? 7. Use the graph of f to estimate the values of c that satisf the conclusion of the Mean Value Theorem for the interval, 8. =ƒ 8. Use the graph of f given in Eercise 7 to estimate the values of c that satisf the conclusion of the Mean Value Theorem for the interval, 7.

17 86 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION ; 9. (a) Graph the function f 4 in the viewing rectangle, b,. (b) Graph the secant line that passes through the points, 5 and 8, 8.5 on the same screen with f. (c) Find the number c that satisfies the conclusion of the Mean Value Theorem for this function f and the interval, 8. Then graph the tangent line at the point c, f c and notice that it is parallel to the secant line. ;. (a) In the viewing rectangle 3, 3 b 5, 5, graph the function f 3 and its secant line through the points, 4 and, 4. Use the graph to estimate the -coordinates of the points where the tangent line is parallel to the secant line. (b) Find the eact values of the numbers c that satisf the conclusion of the Mean Value Theorem for the interval, and compare with our answers to part (a). 4 Verif that the function satisfies the hpotheses of the Mean Value Theorem on the given interval. Then find all numbers c that satisf the conclusion of the Mean Value Theorem.. f 3 5,,. f 3,, 3. f e,, 3 4. f,, 4 5. Let f 3. Show that there is no value of c in, 4 such that f 4 f f c4. Wh does this not contradict the Mean Value Theorem? 6. Let f. Show that there is no value of c such that f 3 f f c3. Wh does this not contradict the Mean Value Theorem? 7. Show that the equation has eactl one real root. 8. Show that the equation sin has eactl one real root. 9. Show that the equation 3 5 c has at most one root in the interval,.. Show that the equation 4 4 c has at most two real roots.. (a) Show that a polnomial of degree 3 has at most three real roots. (b) Show that a polnomial of degree n has at most n real roots.. (a) Suppose that f is differentiable on and has two roots. Show that f has at least one root. (b) Suppose f is twice differentiable on and has three roots. Show that f has at least one real root. (c) Can ou generalize parts (a) and (b)? 3. If f and f for 4, how small can f 4 possibl be? 4. Suppose that 3 f 5 for all values of. Show that 8 f 8 f Does there eist a function f such that f, f 4, and f for all? 6. Suppose that f and t are continuous on a, b and differentiable on a, b. Suppose also that f a ta and f t for a b. Prove that f b tb. [Hint: Appl the Mean Value Theorem to the function h f t.] 7. Show that s if. 8. Suppose f is an odd function and is differentiable everwhere. Prove that for ever positive number b, there eists a number c in b, b such that f c f bb. 9. Use the Mean Value Theorem to prove the inequalit sin a sin b a b 3. If f c (c a constant) for all, use Corollar 7 to show that f c d for some constant d. 3. Let f and Show that f t for all in their domains. Can we conclude from Corollar 7 that f t is constant? 3. Use the method of Eample 6 to prove the identit 33. Prove the identit t sin cos arcsin arctan s 34. At : PM a car s speedometer reads 3 mih. At : PM it reads 5 mih. Show that at some time between : and : the acceleration is eactl mih. 35. Two runners start a race at the same time and finish in a tie. Prove that at some time during the race the have the same speed. [Hint: Consider f t tt ht, where t and h are the position functions of the two runners.] 36. A number a is called a fied point of a function f if f a a. Prove that if f for all real numbers, then f has at most one fied point. if if for all a and b

18 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH D Man of the applications of calculus depend on our abilit to deduce facts about a function f from information concerning its derivatives. Because f represents the slope of the curve f at the point, f, it tells us the direction in which the curve proceeds at each point. So it is reasonable to epect that information about f will provide us with information about f. A B FIGURE N Let s abbreviate the name of this test to the I/D Test. C WHAT DOES f SAY ABOUT f? To see how the derivative of f can tell us where a function is increasing or decreasing, look at Figure. (Increasing functions and decreasing functions were defined in Section..) Between A and B and between C and D, the tangent lines have positive slope and so f. Between B and C, the tangent lines have negative slope and so f. Thus it appears that f increases when f is positive and decreases when f is negative. To prove that this is alwas the case, we use the Mean Value Theorem. INCREASING/DECREASING TEST (a) If f on an interval, then f is increasing on that interval. (b) If f on an interval, then f is decreasing on that interval. PROOF (a) Let and be an two numbers in the interval with. According to the definition of an increasing function (page ) we have to show that f f. Because we are given that f, we know that f is differentiable on,. So, b the Mean Value Theorem there is a number c between and such that f f f c Now f c b assumption and because. Thus the right side of Equation is positive, and so f f or f f This shows that f is increasing. Part (b) is proved similarl. V EXAMPLE Find where the function where it is decreasing. f is increasing and M SOLUTION f 3 4 To use the ID Test we have to know where f and where f. This depends on the signs of the three factors of f, namel,,, and. We divide the real line into intervals whose endpoints are the critical numbers,, and and arrange our work in a chart. A plus sign indicates that the given epression is positive, and a minus sign indicates that it is negative. The last column of the chart gives the

19 88 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION conclusion based on the ID Test. For instance, f for, so f is decreasing on (, ). (It would also be true to sa that f is decreasing on the closed interval,.) Interval f f _ 3 _3 decreasing on (, ) increasing on (, ) decreasing on (, ) increasing on (, ) FIGURE The graph of f shown in Figure confirms the information in the chart. M Recall from Section 4. that if f has a local maimum or minimum at c, then c must be a critical number of f (b Fermat s Theorem), but not ever critical number gives rise to a maimum or a minimum. We therefore need a test that will tell us whether or not f has a local maimum or minimum at a critical number. You can see from Figure that f 5 is a local maimum value of f because f increases on, and decreases on,. Or, in terms of derivatives, f for and f for. In other words, the sign of f changes from positive to negative at. This observation is the basis of the following test. THE FIRST DERIVATIVE TEST Suppose that c is a critical number of a continuous function f. (a) If f changes from positive to negative at c, then f has a local maimum at c. (b) If f changes from negative to positive at c, then f has a local minimum at c. (c) If f does not change sign at c (for eample, if f is positive on both sides of c or negative on both sides), then f has no local maimum or minimum at c. The First Derivative Test is a consequence of the ID Test. In part (a), for instance, since the sign of f changes from positive to negative at c, f is increasing to the left of c and decreasing to the right of c. It follows that f has a local maimum at c. It is eas to remember the First Derivative Test b visualizing diagrams such as those in Figure 3. fª()> fª()< fª()< fª()> fª()> fª()> fª()< fª()< c c c c (a) Local maimum (b) Local minimum (c) No maimum or minimum (d) No maimum or minimum FIGURE 3

20 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 89 V EXAMPLE Find the local minimum and maimum values of the function f in Eample. SOLUTION From the chart in the solution to Eample we see that f changes from negative to positive at, so f is a local minimum value b the First Derivative Test. Similarl, f changes from negative to positive at, so f 7 is also a local minimum value. As previousl noted, f 5 is a local maimum value because f changes from positive to negative at. EXAMPLE 3 Find the local maimum and minimum values of the function M t sin SOLUTION To find the critical numbers of t, we differentiate: t cos So t when cos. The solutions of this equation are 3 and 43. Because t is differentiable everwhere, the onl critical numbers are 3 and 43 and so we analze t in the following table. N The + signs in the table come from the fact that t when cos. From the graph of cos, this is true in the indicated intervals. Interval t cos t increasing on, 3 decreasing on 3, 43 increasing on 43, Because t changes from positive to negative at 3, the First Derivative Test tells us that there is a local maimum at 3 and the local maimum value is t3 3 Likewise, t changes from negative to positive at 43 and so t sin 3 sin s3 3 s s3 4 s is a local minimum value. The graph of t in Figure 4 supports our conclusion. 6 FIGURE 4 =+ sin π M

21 9 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION WHAT DOES f SAY ABOUT f? Figure 5 shows the graphs of two increasing functions on a, b. Both graphs join point A to point B but the look different because the bend in different directions. How can we distinguish between these two tpes of behavior? In Figure 6 tangents to these curves have been drawn at several points. In (a) the curve lies above the tangents and f is called concave upward on a, b. In (b) the curve lies below the tangents and t is called concave downward on a, b. B B f g A A a b a b FIGURE 5 (a) (b) B B f g A A FIGURE 6 (a) Concave upward (b) Concave downward DEFINITION If the graph of f lies above all of its tangents on an interval I, then it is called concave upward on I. If the graph of f lies below all of its tangents on I, it is called concave downward on I. Figure 7 shows the graph of a function that is concave upward (abbreviated CU) on the intervals b, c, d, e, and e, p and concave downward (CD) on the intervals a, b, c, d, and p, q. B C D P a b c d e p q FIGURE 7 CD CU CD CU CU CD Let s see how the second derivative helps determine the intervals of concavit. Looking at Figure 6(a), ou can see that, going from left to right, the slope of the tangent increases.

22 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 9 This means that the derivative f is an increasing function and therefore its derivative f is positive. Likewise, in Figure 6(b) the slope of the tangent decreases from left to right, so f decreases and therefore f is negative. This reasoning can be reversed and suggests that the following theorem is true. A proof is given in Appendi F with the help of the Mean Value Theorem. CONCAVITY TEST (a) If f for all in I, then the graph of f is concave upward on I. (b) If f for all in I, then the graph of f is concave downward on I. EXAMPLE 4 Figure 8 shows a population graph for Cprian honebees raised in an apiar. How does the rate of population increase change over time? When is this rate highest? Over what intervals is P concave upward or concave downward? P 8 Number of bees (in thousands) 6 4 FIGURE Time (in weeks) 8 t SOLUTION B looking at the slope of the curve as t increases, we see that the rate of increase of the population is initiall ver small, then gets larger until it reaches a maimum at about t weeks, and decreases as the population begins to level off. As the population approaches its maimum value of about 75, (called the carring capacit), the rate of increase, Pt, approaches. The curve appears to be concave upward on (, ) and concave downward on (, 8). M In Eample 4, the population curve changed from concave upward to concave downward at approimatel the point (, 38,). This point is called an inflection point of the curve. The significance of this point is that the rate of population increase has its maimum value there. In general, an inflection point is a point where a curve changes its direction of concavit. DEFINITION A point P on a curve f is called an inflection point if f is continuous there and the curve changes from concave upward to concave downward or from concave downward to concave upward at P. For instance, in Figure 7, B, C, D, and P are the points of inflection. Notice that if a curve has a tangent at a point of inflection, then the curve crosses its tangent there. In view of the Concavit Test, there is a point of inflection at an point where the second derivative changes sign.

23 9 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION V EXAMPLE 5 Sketch a possible graph of a function f that satisfies the following conditions: i f on,, iii f, l f on, ii f on, and,, f l f on, - =_ FIGURE 9 SOLUTION Condition (i) tells us that f is increasing on, and decreasing on,. Condition (ii) sas that f is concave upward on, and,, and concave downward on,. From condition (iii) we know that the graph of f has two horizontal asmptotes: and. We first draw the horizontal asmptote as a dashed line (see Figure 9). We then draw the graph of f approaching this asmptote at the far left, increasing to its maimum point at and decreasing toward the -ais at the far right. We also make sure that the graph has inflection points when and. Notice that we made the curve bend upward for and, and bend downward when is between and. M f Another application of the second derivative is the following test for maimum and minimum values. It is a consequence of the Concavit Test. fª(c)= P f(c) ƒ THE SECOND DERIVATIVE TEST Suppose f is continuous near c. (a) If f c and f c, then f has a local minimum at c. (b) If f c and f c, then f has a local maimum at c. FIGURE f (c)>, f is concave upward c For instance, part (a) is true because f near c and so f is concave upward near c. This means that the graph of f lies above its horizontal tangent at c and so f has a local minimum at c. (See Figure.) V EXAMPLE 6 Discuss the curve with respect to concavit, points of inflection, and local maima and minima. Use this information to sketch the curve. SOLUTION If f 4 4 3, then f f 4 To find the critical numbers we set f and obtain and 3. To use the Second Derivative Test we evaluate f at these critical numbers: f f 3 36 Since f 3 and f 3, f 3 7 is a local minimum. Since f, the Second Derivative Test gives no information about the critical number. But since f for and also for 3, the First Derivative Test tells us that f does not have a local maimum or minimum at. [In fact, the epression for f shows that f decreases to the left of 3 and increases to the right of 3.]

24 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 93 =$-4 Since f when or, we divide the real line into intervals with these numbers as endpoints and complete the following chart. inflection points (, ) 3 Interval f Concavit (, ) upward (, ) downward (, ) upward FIGURE (, _6) (3, _7) The point, is an inflection point since the curve changes from concave upward to concave downward there. Also,, 6 is an inflection point since the curve changes from concave downward to concave upward there. Using the local minimum, the intervals of concavit, and the inflection points, we sketch the curve in Figure. M NOTE The Second Derivative Test is inconclusive when f c. In other words, at such a point there might be a maimum, there might be a minimum, or there might be neither (as in Eample 6). This test also fails when f c does not eist. In such cases the First Derivative Test must be used. In fact, even when both tests appl, the First Derivative Test is often the easier one to use. N Tr reproducing the graph in Figure with a graphing calculator or computer. Some machines produce the complete graph, some produce onl the portion to the right of the -ais, and some produce onl the portion between and 6. For an eplanation and cure, see Eample 7 in Section.4. An equivalent epression that gives the correct graph is FIGURE (4, %?#) =@?#(6-)!?# EXAMPLE 7 Sketch the graph of the function f SOLUTION You can use the differentiation rules to check that the first two derivatives are Since f when 4 and f does not eist when or 6, the critical numbers are, 4, and 6. To find the local etreme values we use the First Derivative Test. Since f changes from negative to positive at, f is a local minimum. Since f changes from positive to negative at 4, f 4 53 is a local maimum. The sign of f does not change at 6, so there is no minimum or maimum there. (The Second Derivative Test could be used at 4, but not at or 6 since f does not eist at either of these numbers.) Looking at the epression for f and noting that 43 for all, we have f for and for 6 and f for 6. So f is concave downward on, and, 6 and concave upward on 6,, and the onl inflection point is 6,. The graph is sketched in Figure. Note that the curve has vertical tangents at, and 6, because as l and as l6. M f f l f Interval f f decreasing on (, ) increasing on (, 4) decreasing on (4, 6) decreasing on (6, ) EXAMPLE 8 Use the first and second derivatives of f e, together with asmptotes, to sketch its graph. SOLUTION Notice that the domain of f is, so we check for vertical asmptotes b computing the left and right its as l. As l, we know that t l,

25 94 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION so e e t l tl and this shows that is a vertical asmptote. As l, we have t l, so e e t l tl TEC In Module 4.3 ou can practice using graphical information about f to determine the shape of the graph of f. As l, we have l and so e e l This shows that is a horizontal asmptote. Now let s compute the derivative. The Chain Rule gives f e Since e and for all, we have f for all. Thus f is decreasing on, and on,. There is no critical number, so the function has no maimum or minimum. The second derivative is f e e 4 e 4 Since e and 4, we have f when and when f. So the curve is concave downward on (, and concave upward on ( ), ) and on,. The inflection point is (, e ). To sketch the graph of f we first draw the horizontal asmptote (as a dashed line), together with the parts of the curve near the asmptotes in a preinar sketch [Figure 3(a)]. These parts reflect the information concerning its and the fact that f is decreasing on both, and,. Notice that we have indicated that f l as l even though f does not eist. In Figure 3(b) we finish the sketch b incorporating the information concerning concavit and the inflection point. In Figure 3(c) we check our work with a graphing device. = 4 = inflection point = _3 3 (a) Preinar sketch (b) Finished sketch (c) Computer confirmation FIGURE 3 M

26 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH EXERCISES Use the given graph of f to find the following. (a) The open intervals on which f is increasing. (b) The open intervals on which f is decreasing. (c) The open intervals on which f is concave upward. (d) The open intervals on which f is concave downward. (e) The coordinates of the points of inflection. (c) On what intervals is f concave upward or concave downward? Eplain. (d) What are the -coordinates of the inflection points of f? Wh? =fª() Suppose ou are given a formula for a function f. (a) How do ou determine where f is increasing or decreasing? (b) How do ou determine where the graph of f is concave upward or concave downward? (c) How do ou locate inflection points? 4. (a) State the First Derivative Test. (b) State the Second Derivative Test. Under what circumstances is it inconclusive? What do ou do if it fails? 5 6 The graph of the derivative f of a function f is shown. (a) On what intervals is f increasing or decreasing? (b) At what values of does f have a local maimum or minimum? 5. =fª() The graph of the second derivative f of a function f is shown. State the -coordinates of the inflection points of f. Give reasons for our answers. 8. The graph of the first derivative f of a function f is shown. (a) On what intervals is f increasing? Eplain. (b) At what values of does f have a local maimum or minimum? Eplain. 6. =f () =fª() (a) Find the intervals on which f is increasing or decreasing. (b) Find the local maimum and minimum values of f. (c) Find the intervals of concavit and the inflection points. 9. f f f cos sin, 5. f e e 6. f ln 7. f 4 3 f 3 f sin cos, f ln s 9 Find the local maimum and minimum values of f using both the First and Second Derivative Tests. Which method do ou prefer? 9. f. f f s. (a) Find the critical numbers of f 4 3. (b) What does the Second Derivative Test tell ou about the behavior of f at these critical numbers? (c) What does the First Derivative Test tell ou? 3. Suppose f is continuous on,. (a) If f and f 5, what can ou sa about f? (b) If f 6 and f 6, what can ou sa about f? 4 9 Sketch the graph of a function that satisfies all of the given conditions. 8. f s e 4. f for all, vertical asmptote, f if or 3, f if 3 5. f f f 4, f if or 4, f if or 4, f if 3, f if or 3

27 96 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 6. f f, f if, f if, f if, f if, inflection point, f f l f 7. f if, f if,,, f if 8. f if, f if,, f, f f, l f if 3, f if 3 9. f and f for all 3. Suppose f3, f 3, and f and f for all. (a) Sketch a possible graph for f. (b) How man solutions does the equation f have? Wh? (c) Is it possible that f 3? Wh? 3 3 The graph of the derivative f of a continuous function f is shown. (a) On what intervals is f increasing or decreasing? (b) At what values of does f have a local maimum or minimum? (c) On what intervals is f concave upward or downward? (d) State the -coordinate(s) of the point(s) of inflection. (e) Assuming that f, sketch a graph of f. 3. =fª() _ (d) Use the information from parts (a) (c) to sketch the graph. Check our work with a graphing device if ou have one. 33. f f t h h f cos cos, 44. ft t cos t, 45 5 (a) Find the vertical and horizontal asmptotes. (b) Find the intervals of increase or decrease. (c) Find the local maimum and minimum values. (d) Find the intervals of concavit and the inflection points. (e) Use the information from parts (a) (d) to sketch the graph of f f tan, 5. A s 3 C 3 4 f s f e t 53. Suppose the derivative of a function f is f On what interval is f increasing? 54. Use the methods of this section to sketch the curve 3 3a a 3, where a is a positive constant. What do the members of this famil of curves have in common? How do the differ from each other? f 46. f 49. f ln ln 5. f 5. 3 f 3 B 3 3 f ln 4 7 e e f e arctan 3. _ =fª() ; (a) Use a graph of f to estimate the maimum and minimum values. Then find the eact values. (b) Estimate the value of at which f increases most rapidl. Then find the eact value. 55. f s 56. f e (a) Find the intervals of increase or decrease. (b) Find the local maimum and minimum values. (c) Find the intervals of concavit and the inflection points. ; (a) Use a graph of f to give a rough estimate of the intervals of concavit and the coordinates of the points of inflection. (b) Use a graph of f to give better estimates. 57. f cos cos,

28 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 97 CAS Estimate the intervals of concavit to one decimal place b using a computer algebra sstem to compute and graph f. 59. f s 6. A graph of a population of east cells in a new laborator culture as a function of time is shown. (a) Describe how the rate of population increase varies. (b) When is this rate highest? (c) On what intervals is the population function concave upward or downward? (d) Estimate the coordinates of the inflection point. 6. Let f t be the temperature at time t where ou live and suppose that at time t 3 ou feel uncomfortabl hot. How do ou feel about the given data in each case? (a) f 3, f 3 4 (b) f 3, f 3 4 (c) f 3, f 3 4 (d) f 3, f f 3 4 Number of east cells f tan Time (in hours) 63. Let Kt be a measure of the knowledge ou gain b studing for a test for t hours. Which do ou think is larger, K8 K7 or K3 K? Is the graph of K concave upward or concave downward? Wh? Coffee is being poured into the mug shown in the figure at a constant rate (measured in volume per unit time). Sketch a rough graph of the depth of the coffee in the mug as a function of time. Account for the shape of the graph in terms of concavit. What is the significance of the inflection point? ; 65. A drug response curve describes the level of medication in the bloodstream after a drug is administered. A surge function St At p e kt is often used to model the response curve, reflecting an initial surge in the drug level and then a more gradual decline. If, for a particular drug, A., p 4, k.7, and t is measured in minutes, estimate the times corresponding to the inflection points and eplain their significance. If ou have a graphing device, use it to graph the drug response curve. 66. The famil of bell-shaped curves occurs in probabilit and statistics, where it is called the normal densit function. The constant is called the mean and the positive constant is called the standard deviation. For simplicit, let s scale the function so as to remove the factor ) and let s analze the special case where. So we stud the function (s (a) Find the asmptote, maimum value, and inflection points of f. (b) What role does pla in the shape of the curve? ; (c) Illustrate b graphing four members of this famil on the same screen. 67. Find a cubic function f a 3 b c d that has a local maimum value of 3 at and a local minimum value of at. 68. For what values of the numbers a and b does the function f ae b have the maimum value f? s e f e 69. Show that the curve has three points of inflection and the all lie on one straight line. 7. Show that the curves e and e touch the curve e sin at its inflection points. 7. Suppose f is differentiable on an interval I and f for all numbers in I ecept for a single number c. Prove that f is increasing on the entire interval I Assume that all of the functions are twice differentiable and the second derivatives are never. 7. (a) If f and t are concave upward on I, show that f t is concave upward on I. (b) If f is positive and concave upward on I, show that the function t f is concave upward on I. 73. (a) If f and t are positive, increasing, concave upward functions on I, show that the product function ft is concave upward on I. (b) Show that part (a) remains true if f and t are both decreasing.

29 98 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (c) Suppose f is increasing and t is decreasing. Show, b giving three eamples, that ft ma be concave upward, concave downward, or linear. Wh doesn t the argument in parts (a) and (b) work in this case? 74. Suppose f and t are both concave upward on,. Under what condition on f will the composite function h f t be concave upward? 75. Show that tan for. [Hint: Show that f tan is increasing on,.] 76. (a) Show that e for. (b) Deduce that e for. (c) Use mathematical induction to prove that for and an positive integer n, e n! n! 77. Show that a cubic function (a third-degree polnomial) alwas has eactl one point of inflection. If its graph has three -intercepts,, and 3, show that the -coordinate of the inflection point is 33. ; 78. For what values of c does the polnomial P 4 c 3 have two inflection points? One inflection point? None? Illustrate b graphing P for several values of c. How does the graph change as c decreases? 79. Prove that if c, f c is a point of inflection of the graph of f and f eists in an open interval that contains c, then f c. [Hint: Appl the First Derivative Test and Fermat s Theorem to the function t f.] 8. Show that if f 4, then f, but, is not an inflection point of the graph of f. t 8. Show that the function has an inflection point at, but t does not eist. 8. Suppose that f is continuous and f c f c, but f c. Does f have a local maimum or minimum at c? Does f have a point of inflection at c? 83. The three cases in the First Derivative Test cover the situations one commonl encounters but do not ehaust all possibilities. Consider the functions f, t, and h whose values at are all and, for, f t 4 sin 4 sin h 4 sin (a) Show that is a critical number of all three functions but their derivatives change sign infinitel often on both sides of. (b) Show that f has neither a local maimum nor a local minimum at, t has a local minimum, and h has a local maimum. 4.4 INDETERMINATE FORMS AND L HOSPITAL S RULE Suppose we are tring to analze the behavior of the function Although F is not defined when, we need to know how F behaves near. In particular, we would like to know the value of the it F l In computing this it we can t appl Law 5 of its (the it of a quotient is the quotient of the its, see Section.3) because the it of the denominator is. In fact, although the it in () eists, its value is not obvious because both numerator and denominator approach and is not defined. In general, if we have a it of the form where both f l and tl as la, then this it ma or ma not eist and is called an indeterminate form of tpe. We met some its of this tpe in Chapter. For la ln ln f t

30 SECTION 4.4 INDETERMINATE FORMS AND L HOSPITAL S RULE 99 rational functions, we can cancel common factors: l l We used a geometric argument to show that l sin l But these methods do not work for its such as (), so in this section we introduce a sstematic method, known as l Hospital s Rule, for the evaluation of indeterminate forms. Another situation in which a it is not obvious occurs when we look for a horizontal asmptote of F and need to evaluate the it l ln It isn t obvious how to evaluate this it because both numerator and denominator become large as l. There is a struggle between numerator and denominator. If the numerator wins, the it will be ; if the denominator wins, the answer will be. Or there ma be some compromise, in which case the answer ma be some finite positive number. In general, if we have a it of the form la f t where both f l (or ) and tl (or ), then the it ma or ma not eist and is called an indeterminate form of tpe. We saw in Section.6 that this tpe of it can be evaluated for certain functions, including rational functions, b dividing numerator and denominator b the highest power of that occurs in the denominator. For instance, l l This method does not work for its such as (), but l Hospital s Rule also applies to this tpe of indeterminate form. L HOSPITAL L Hospital s Rule is named after a French nobleman, the Marquis de l Hospital (66 74), but was discovered b a Swiss mathematician, John Bernoulli ( ). You might sometimes see l Hospital spelled as l Hôpital, but he spelled his own name l Hospital, as was common in the 7th centur. See Eercise 77 for the eample that the Marquis used to illustrate his rule. See the project on page 37 for further historical details. L HOSPITAL S RULE Suppose f and t are differentiable and t on an open interval I that contains a (ecept possibl at a). Suppose that or that f la f la (In other words, we have an indeterminate form of tpe or.) Then la f t and and la f t if the it on the right side eists (or is or ). t la t la

31 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION f g NOTE L Hospital s Rule sas that the it of a quotient of functions is equal to the it of the quotient of their derivatives, provided that the given conditions are satisfied. It is especiall important to verif the conditions regarding the its of f and t before using l Hospital s Rule. a a FIGURE =m (-a) =m (-a) N Figure suggests visuall wh l Hospital s Rule might be true. The first graph shows two differentiable functions f and t, each of which approaches as la. If we were to zoom in toward the point a,, the graphs would start to look almost linear. But if the functions actuall were linear, as in the second graph, then their ratio would be which is the ratio of their derivatives. This suggests that la m a m m a m f t la f t NOTE L Hospital s Rule is also valid for one-sided its and for its at infinit or negative infinit; that is, la can be replaced b an of the smbols la, la, l, or l. NOTE 3 For the special case in which f a ta, f and t are continuous, and ta, it is eas to see wh l Hospital s Rule is true. In fact, using the alternative form of the definition of a derivative, we have la f t f a ta la It is more difficult to prove the general version of l Hospital s Rule. See Appendi F. ln V EXAMPLE Find. l SOLUTION Since ln ln l we can appl l Hospital s Rule: la la f f a t ta f f a a t ta a la and f t la f f a a t ta a l Notice that when using l Hospital s Rule we differentiate the numerator and denominator separatel. We do not use the Quotient Rule. l ln l d ln d d d l l M N The graph of the function of Eample is shown in Figure. We have noticed previousl that eponential functions grow far more rapidl than power functions, so the result of Eample is not unepected. See also Eercise 69. = e l EXAMPLE Calculate. SOLUTION We have and l l e, so l Hospital s Rule gives l e l d d e d d l Since e l and l as l, the it on the right side is also indeterminate, but a second application of l Hospital s Rule gives e FIGURE l e l e e l M

32 SECTION 4.4 INDETERMINATE FORMS AND L HOSPITAL S RULE 3 N The graph of the function of Eample 3 is shown in Figure 3. We have discussed previousl the slow growth of logarithms, so it isn t surprising that this ratio approaches as l. See also Eercise 7. ln V EXAMPLE 3 Calculate. ls 3 SOLUTION Since ln l and s 3 l as l, l Hospital s Rule applies: ln ls 3 l 3 3 _ = ln Œ, Notice that the it on the right side is now indeterminate of tpe. But instead of appling l Hospital s Rule a second time as we did in Eample, we simplif the epression and see that a second application is unnecessar: ln ls 3 l 3 3 l 3 s 3 M FIGURE 3 N The graph in Figure 4 gives visual confirmation of the result of Eample 4. If we were to zoom in too far, however, we would get an inaccurate graph because tan is close to when is small. See Eercise 38(d) in Section.. tan EXAMPLE 4 Find. (See Eercise 38 in Section..) l 3 SOLUTION Noting that both tan l and 3 l as l, we use l Hospital s Rule: tan sec l 3 l 3 Since the it on the right side is still indeterminate of tpe, we appl l Hospital s Rule again: sec l 3 sec tan l 6 Because l sec, we simplif the calculation b writing tan - = _ FIGURE 4 sec tan l 6 3 tan sec l l 3 tan l We can evaluate this last it either b using l Hospital s Rule a third time or b writing tan as sin cos and making use of our knowledge of trigonometric its. Putting together all the steps, we get tan sec l 3 l 3 sec tan l 6 3 tan l 3 sec l 3 M sin EXAMPLE 5 Find. l cos SOLUTION If we blindl attempted to use l Hospital s Rule, we would get sin cos l cos l sin This is wrong! Although the numerator sin l as l, notice that the denominator cos does not approach, so l Hospital s Rule can t be applied here.

33 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION The required it is, in fact, eas to find because the function is continuous at the denominator is nonzero there: l sin cos sin cos Eample 5 shows what can go wrong if ou use l Hospital s Rule without thinking. Other its can be found using l Hospital s Rule but are more easil found b other methods. (See Eamples 3 and 5 in Section.3, Eample 3 in Section.6, and the discussion at the beginning of this section.) So when evaluating an it, ou should consider other methods before using l Hospital s Rule. and M INDETERMINATE PRODUCTS If la f and la t (or ), then it isn t clear what the value of la f t, if an, will be. There is a struggle between f and t. If f wins, the answer will be ; if t wins, the answer will be (or ). Or there ma be a compromise where the answer is a finite nonzero number. This kind of it is called an indeterminate form of tpe. We can deal with it b writing the product ft as a quotient: ft f t or ft This converts the given it into an indeterminate form of tpe or so that we can use l Hospital s Rule. t f N Figure 5 shows the graph of the function in Eample 6. Notice that the function is undefined at ; the graph approaches the origin but never quite reaches it. = ln V EXAMPLE 6 Evaluate ln. l SOLUTION The given it is indeterminate because, as l, the first factor approaches while the second factor ln approaches. Writing, we have l as l, so l Hospital s Rule gives ln ln l l l l M NOTE In solving Eample 6 another possible option would have been to write ln l l ln FIGURE 5 This gives an indeterminate form of the tpe, but if we appl l Hospital s Rule we get a more complicated epression than the one we started with. In general, when we rewrite an indeterminate product, we tr to choose the option that leads to the simpler it. INDETERMINATE DIFFERENCES If la f and la t, then the it f t la is called an indeterminate form of tpe. Again there is a contest between f and t. Will the answer be ( f wins) or will it be ( t wins) or will the compromise on a finite number? To find out, we tr to convert the difference into a quotient (for instance, b using

34 SECTION 4.4 INDETERMINATE FORMS AND L HOSPITAL S RULE 33 a common denominator, or rationalization, or factoring out a common factor) so that we have an indeterminate form of tpe or. EXAMPLE 7 Compute sec tan. SOLUTION First notice that sec l and tan l as l, so the it is indeterminate. Here we use a common denominator: sec tan sin l l cos cos Note that the use of l Hospital s Rule is justified because sin l and cos l as l. M INDETERMINATE POWERS Several indeterminate forms arise from the it. f and t tpe la. f and t tpe la 3. f and t tpe la l la la la sin cos l cos l sin f t la Each of these three cases can be treated either b taking the natural logarithm: let f t, then ln t ln f or b writing the function as an eponential: (Recall that both of these methods were used in differentiating such functions.) In either method we are led to the indeterminate product t ln f, which is of tpe. EXAMPLE 8 Calculate sin 4cot. SOLUTION First notice that as l, we have sin 4l and cot l, so the given it is indeterminate. Let Then l so l Hospital s Rule gives ln sin 4 ln l l tan So far we have computed the it of f t t ln f e sin 4 cot ln ln sin 4 cot cot ln sin 4 l 4 cos 4 sin 4 sec 4 ln, but what we want is the it of. To find this

35 34 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION we use the fact that e ln : sin l 4cot e ln e 4 l l M N The graph of the function,, is shown in Figure 6. Notice that although is not defined, the values of the function approach as l. This confirms the result of Eample 9. EXAMPLE 9 Find. l SOLUTION Notice that this it is indeterminate since for an but for an. We could proceed as in Eample 8 or b writing the function as an eponential: e ln e ln In Eample 6 we used l Hospital s Rule to show that _ FIGURE 6 Therefore ln l e ln e l l M 4.4 EXERCISES 4 Given that which of the following its are indeterminate forms? For those that are not an indeterminate form, evaluate the it where possible.. (a) (c) (e) (b) (d). (a) f p (b) (c) 3. (a) f p (b) (c) f la la l a la la p 4. (a) f t (b) f (c) la (d) p f (e) p q (f) la f t h la p p l a q pq p la p q la t la la la q la la p la f h la f p hp l a p q la p h la la q sp 5 64 Find the it. Use l Hospital s Rule where appropriate. If there is a more elementar method, consider using it. If l Hospital s Rule doesn t appl, eplain wh l l 5 cos 9.. l sin e t.. tan p l tan q ln l s ln l 9... tl e l 3 t 3 e l 6 l a l b sin 4 l tan 5 e 3t tl t sin l csc l ln ln l l ln sin. l e 3

36 SECTION 4.4 INDETERMINATE FORMS AND L HOSPITAL S RULE 35 tanh l tan 5 t 3 t t l t sin l 9. cos l 3. sin l cos ln l cos sin l tan sin l 3 ln l cos m cos n l l l tan 4 s s l cos ; Use a graph to estimate the value of the it. Then use l Hospital s Rule to find the eact value. l ; Illustrate l Hospital s Rule b graphing both f t and f t near to see that these ratios have the same it as l. Also calculate the eact value of the it. 67. f e, 68. f sin, t 3 4 t sec l l 3 a a a l cos l 39. sin 4. l 4 e e l sin cos ln a la lne e a e l 69. Prove that e l n for an positive integer n. This shows that the eponential function approaches infinit faster than an power of. 4. cot sin ln tan 46. l 47. l ln ln l 3 e l l (s ) l l l l l sin ln l tan sec l 4 tan l csc cot l lcot e l tan l l a b ln ln l l e tan 4 cot l l 7. Prove that ln l p for an number p. This shows that the logarithmic function approaches more slowl than an power of. 7. What happens if ou tr to use l Hospital s Rule to evaluate l s Evaluate the it using another method. 7. If an object with mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into account, is v mt c e ctm where t is the acceleration due to gravit and c is a positive constant. (In Chapter 9 we will be able to deduce this equation from the assumption that the air resistance is proportional to the speed of the object; c is the proportionalit constant.) (a) Calculate t l v. What is the meaning of this it? (b) For fied t, use l Hospital s Rule to calculate c l v. What can ou conclude about the velocit of a falling object in a vacuum?

37 36 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 73. If an initial amount A of mone is invested at an interest rate r compounded n times a ear, the value of the investment after t ears is A A r nnt If we let nl, we refer to the continuous compounding of interest. Use l Hospital s Rule to show that if interest is compounded continuousl, then the amount after t ears is A A e rt 74. If a metal ball with mass m is projected in water and the force of resistance is proportional to the square of the velocit, then the distance the ball travels in time t is st m c ln cosh tc mt where c is a positive constant. Find cl st. 75. If an electrostatic field E acts on a liquid or a gaseous polar dielectric, the net dipole moment P per unit volume is PE e E e E e E e E E Show that El PE. 76. A metal cable has radius r and is covered b insulation, so that the distance from the center of the cable to the eterior of the insulation is R. The velocit v of an electrical impulse in the cable is v c R r ln R r where c is a positive constant. Find the following its and interpret our answers. (a) v (b) v Rlr rl 77. The first appearance in print of l Hospital s Rule was in the book Analse des Infiniment Petits published b the Marquis de l Hospital in 696. This was the first calculus tetbook ever published and the eample that the Marquis used in that book to illustrate his rule was to find the it of the function sa 3 4 as 3 aa a s 4 a 3 as approaches a, where a. (At that time it was common to write aa instead of a.) Solve this problem. 78. The figure shows a sector of a circle with central angle. Let A be the area of the segment between the chord PR and the arc PR. Let B be the area of the triangle PQR. Find l. 79. If f is continuous, f, and f 7, evaluate 8. For what values of a and b is the following equation true? 8. If f is continuous, use l Hospital s Rule to show that Eplain the meaning of this equation with the aid of a diagram. 8. If f is continuous, show that 83. Let ; 84. Let hl O l hl l f 3 f 5 sin 3 a b f h f h h f h f f h h f e (a) Use the definition of derivative to compute f. (b) Show that f has derivatives of all orders that are defined on. [Hint: First show b induction that there is a polnomial p n and a nonnegative integer k n such that f n p nf k n for.] f P Q B( ) A( ) if if if if f f (a) Show that f is continuous at. (b) Investigate graphicall whether f is differentiable at b zooming in several times toward the point, on the graph of f. (c) Show that f is not differentiable at. How can ou reconcile this fact with the appearance of the graphs in part (b)? R

38 SECTION 4.5 SUMMARY OF CURVE SKETCHING 37 Thomas Fisher Rare Book Librar W R I T I N G P R O J E C T The Internet is another source of information for this project. Click on Histor of Mathematics for a list of reliable websites. THE ORIGINS OF L HOSPITAL S RULE L Hospital s Rule was first published in 696 in the Marquis de l Hospital s calculus tetbook Analse des Infiniment Petits, but the rule was discovered in 694 b the Swiss mathematician John (Johann) Bernoulli. The eplanation is that these two mathematicians had entered into a curious business arrangement whereb the Marquis de l Hospital bought the rights to Bernoulli s mathematical discoveries. The details, including a translation of l Hospital s letter to Bernoulli proposing the arrangement, can be found in the book b Eves []. Write a report on the historical and mathematical origins of l Hospital s Rule. Start b providing brief biographical details of both men (the dictionar edited b Gillispie [] is a good source) and outline the business deal between them. Then give l Hospital s statement of his rule, which is found in Struik s sourcebook [4] and more briefl in the book of Katz [3]. Notice that l Hospital and Bernoulli formulated the rule geometricall and gave the answer in terms of differentials. Compare their statement with the version of l Hospital s Rule given in Section 4.4 and show that the two statements are essentiall the same.. Howard Eves, In Mathematical Circles (Volume : Quadrants III and IV) (Boston: Prindle, Weber and Schmidt, 969), pp... C. C. Gillispie, ed., Dictionar of Scientific Biograph (New York: Scribner s, 974). See the article on Johann Bernoulli b E. A. Fellmann and J. O. Fleckenstein in Volume II and the article on the Marquis de l Hospital b Abraham Robinson in Volume VIII. 3. Victor Katz, A Histor of Mathematics: An Introduction (New York: HarperCollins, 993), p D. J. Struik, ed., A Sourcebook in Mathematics, 8 (Princeton, NJ: Princeton Universit Press, 969), pp = _ 4 _ FIGURE 8 = FIGURE SUMMARY OF CURVE SKETCHING So far we have been concerned with some particular aspects of curve sketching: domain, range, and smmetr in Chapter ; its, continuit, and asmptotes in Chapter ; derivatives and tangents in Chapters and 3; and etreme values, intervals of increase and decrease, concavit, points of inflection, and l Hospital s Rule in this chapter. It is now time to put all of this information together to sketch graphs that reveal the important features of functions. You might ask: Wh don t we just use a graphing calculator or computer to graph a curve? Wh do we need to use calculus? It s true that modern technolog is capable of producing ver accurate graphs. But even the best graphing devices have to be used intelligentl. We saw in Section.4 that it is etremel important to choose an appropriate viewing rectangle to avoid getting a misleading graph. (See especiall Eamples, 3, 4, and 5 in that section.) The use of calculus enables us to discover the most interesting aspects of graphs and in man cases to calculate maimum and minimum points and inflection points eactl instead of approimatel. For instance, Figure shows the graph of f At first glance it seems reasonable: It has the same shape as cubic curves like 3, and it appears to have no maimum or minimum point. But if ou compute the derivative, ou will see that there is a maimum when.75 and a minimum when. Indeed, if we zoom in to this portion of the graph, we see that behavior ehibited in Figure. Without calculus, we could easil have overlooked it. In the net section we will graph functions b using the interaction between calculus and graphing devices. In this section we draw graphs b first considering the following

39 38 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION information. We don t assume that ou have a graphing device, but if ou do have one ou should use it as a check on our work. GUIDELINES FOR SKETCHING A CURVE (a) Even function: reflectional smmetr (b) Odd function: rotational smmetr FIGURE 3 The following checklist is intended as a guide to sketching a curve f b hand. Not ever item is relevant to ever function. (For instance, a given curve might not have an asmptote or possess smmetr.) But the guidelines provide all the information ou need to make a sketch that displas the most important aspects of the function. A. Domain It s often useful to start b determining the domain D of f, that is, the set of values of for which f is defined. B. Intercepts The -intercept is f and this tells us where the curve intersects the -ais. To find the -intercepts, we set and solve for. (You can omit this step if the equation is difficult to solve.) C. Smmetr (i) If f f for all in D, that is, the equation of the curve is unchanged when is replaced b, then f is an even function and the curve is smmetric about the -ais. This means that our work is cut in half. If we know what the curve looks like for, then we need onl reflect about the -ais to obtain the complete curve [see Figure 3(a)]. Here are some eamples:, and cos. (ii) If f f for all in D, then f is an odd function and the curve is smmetric about the origin. Again we can obtain the complete curve if we know what it looks like for. [Rotate 8 about the origin; see Figure 3(b).] Some simple eamples of odd functions are, 3, 5, and sin. (iii) If f p f for all in D, where p is a positive constant, then f is called a periodic function and the smallest such number p is called the period. For instance, sin has period and tan has period. If we know what the graph looks like in an interval of length p, then we can use translation to sketch the entire graph (see Figure 4)., 4, FIGURE 4 Periodic function: translational smmetr a-p a a+p a+p D. Asmptotes (i) Horizontal Asmptotes. Recall from Section.6 that if either l f L or l f L, then the line L is a horizontal asmptote of the curve f. If it turns out that l f (or ), then we do not have an asmptote to the right, but that is still useful information for sketching the curve. (ii) Vertical Asmptotes. Recall from Section. that the line a is a vertical asmptote if at least one of the following statements is true: f f la la f f la la

40 SECTION 4.5 SUMMARY OF CURVE SKETCHING 39 (For rational functions ou can locate the vertical asmptotes b equating the denominator to after canceling an common factors. But for other functions this method does not appl.) Furthermore, in sketching the curve it is ver useful to know eactl which of the statements in () is true. If f a is not defined but a is an endpoint of the domain of f, then ou should compute la f or la f, whether or not this it is infinite. (iii) Slant Asmptotes. These are discussed at the end of this section. E. Intervals of Increase or Decrease Use the I/D Test. Compute f and find the intervals on which f is positive ( f is increasing) and the intervals on which f is negative ( f is decreasing). F. Local Maimum and Minimum Values Find the critical numbers of f [the numbers c where f c or f c does not eist]. Then use the First Derivative Test. If f changes from positive to negative at a critical number c, then f c is a local maimum. If f changes from negative to positive at c, then f c is a local minimum. Although it is usuall preferable to use the First Derivative Test, ou can use the Second Derivative Test if f c and f c. Then f c implies that f c is a local minimum, whereas f c implies that f c is a local maimum. G. Concavit and Points of Inflection Compute f and use the Concavit Test. The curve is concave upward where f and concave downward where f. Inflection points occur where the direction of concavit changes. H. Sketch the Curve Using the information in items A G, draw the graph. Sketch the asmptotes as dashed lines. Plot the intercepts, maimum and minimum points, and inflection points. Then make the curve pass through these points, rising and falling according to E, with concavit according to G, and approaching the asmptotes. If additional accurac is desired near an point, ou can compute the value of the derivative there. The tangent indicates the direction in which the curve proceeds. EXAMPLE Use the guidelines to sketch the curve V. A. The domain is,,, B. The - and -intercepts are both. C. Since f f, the function f is even. The curve is smmetric about the -ais. D. l l = Therefore the line is a horizontal asmptote. Since the denominator is when, we compute the following its: =_ = l l FIGURE 5 Preinar sketch N We have shown the curve approaching its horizontal asmptote from above in Figure 5. This is confirmed b the intervals of increase and decrease. l l Therefore the lines and are vertical asmptotes. This information about its and asmptotes enables us to draw the preinar sketch in Figure 5, showing the parts of the curve near the asmptotes.

41 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION E. f 4 4 = =_ = FIGURE 6 Finished sketch of = - Since f when and f when, f is increasing on, and, and decreasing on, and,. F. The onl critical number is. Since f changes from positive to negative at, f is a local maimum b the First Derivative Test. G. Since 4 for all, we have and. Thus the curve is concave upward on the intervals, and, and concave downward on,. It has no point of inflection since and are not in the domain of f. H. Using the information in E G, we finish the sketch in Figure 6. M EXAMPLE Sketch the graph of f. s A. Domain, B. The - and -intercepts are both. C. Smmetr: None D. Since ls E. f f f &? &? there is no horizontal asmptote. Since s l as l and f is alwas positive, we have l s and so the line is a vertical asmptote. &? f s (s ) We see that f when (notice that 4 3 is not in the domain of f ), so the onl critical number is. Since f when and f when, f is decreasing on, and increasing on,. F. Since f and f changes from negative to positive at, f is a local (and absolute) minimum b the First Derivative Test G. f =_ FIGURE 7 = œ + Note that the denominator is alwas positive. The numerator is the quadratic 3 8 8, which is alwas positive because its discriminant is b 4ac 3, which is negative, and the coefficient of is positive. Thus f for all in the domain of f, which means that f is concave upward on, and there is no point of inflection. H. The curve is sketched in Figure 7. M

42 SECTION 4.5 SUMMARY OF CURVE SKETCHING 3 (_, _/e) FIGURE 8 = V EXAMPLE 3 Sketch the graph of f e. A. The domain is. B. The - and -intercepts are both. C. Smmetr: None D. Because both and e become large as l, we have l e. As l, however, e l and so we have an indeterminate product that requires the use of l Hospital s Rule: E. Thus the -ais is a horizontal asmptote. Since e is alwas positive, we see that f when, and f when. So f is increasing on, and decreasing on,. F. Because f and f changes from negative to positive at, f e is a local (and absolute) minimum. G. Since f if and f if, f is concave upward on, and concave downward on,. The inflection point is, e. H. We use this information to sketch the curve in Figure 8. M EXAMPLE 4 Sketch the graph of f cos. sin A. The domain is. B. The -intercept is f. The -intercepts occur when cos, that is, n, where n is an integer. C. f is neither even nor odd, but f f for all and so f is periodic and has period. Thus, in what follows, we need to consider onl and then etend the curve b translation in part H. D. Asmptotes: None e l l e l f e e e e e l f e e e E. f sin sin cos cos sin sin sin Thus f when sin &? sin &? So f is increasing on 76, 6 and decreasing on, 76 and 6,. F. From part E and the First Derivative Test, we see that the local minimum value is f 76 s3 and the local maimum value is f 6 s3. G. If we use the Quotient Rule again and simplif, we get cos sin f sin 3 Because sin 3 and sin for all, we know that f when cos, that is, 3. So f is concave upward on, 3 and concave downward on, and 3,. The inflection points are, and 3,.

43 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION π 6, œ 3 H. The graph of the function restricted to is shown in Figure 9. Then we etend it, using periodicit, to the complete graph in Figure. π π 3π π _π π π 3π FIGURE 9 7π - 6, œ 3 FIGURE M EXAMPLE 5 Sketch the graph of ln4. A. The domain is 4 4, B. The -intercept is f ln 4. To find the -intercept we set ln4 We know that ln, so we have 4? 3 and therefore the -intercepts are s3. C. Since f f, f is even and the curve is smmetric about the -ais. D. We look for vertical asmptotes at the endpoints of the domain. Since 4 l as l and also as l, we have ln4 l ln4 l Thus the lines and are vertical asmptotes. =_ (, ln 4) {_œ 3, } {œ 3, } = E. Since f when and f when, f is increasing on, and decreasing on,. F. The onl critical number is. Since f changes from positive to negative at, f ln 4 is a local maimum b the First Derivative Test. G. f 4 f FIGURE =ln(4 - ) Since f for all, the curve is concave downward on, and has no inflection point. H. Using this information, we sketch the curve in Figure. M SLANT ASYMPTOTES Some curves have asmptotes that are oblique, that is, neither horizontal nor vertical. If f m b l then the line m b is called a slant asmptote because the vertical distance

44 SECTION 4.5 SUMMARY OF CURVE SKETCHING 33 =ƒ ƒ-(m+b) =m+b between the curve f and the line m b approaches, as in Figure. (A similar situation eists if we let l.) For rational functions, slant asmptotes occur when the degree of the numerator is one more than the degree of the denominator. In such a case the equation of the slant asmptote can be found b long division as in the following eample. EXAMPLE 6 Sketch the graph of f 3 V. FIGURE A. The domain is,. B. The - and -intercepts are both. C. Since f f, f is odd and its graph is smmetric about the origin. D. Since is never, there is no vertical asmptote. Since f l as l and f l as l, there is no horizontal asmptote. But long division gives f 3 f l as l So the line is a slant asmptote. E. f Since f for all (ecept ), f is increasing on,. F. Although f, f does not change sign at, so there is no local maimum or minimum. G. f = + Since f when or s3, we set up the following chart: 3œ 3 _œ 3, _ 4 œ 3, 3œ 3 4 inflection points Interval 3 3 f f s3 CU on (, s3 ) s3 CD on (s3, ) s3 CU on (, s3 ) s3 CD on (s3, ) = FIGURE 3 The points of inflection are (s3, 3 4s3 ),,, and (s3, 3 4s3 ). H. The graph of f is sketched in Figure 3. M

45 34 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 5 Use the guidelines of this section to sketch the curve EXERCISES e 3 e 5. tan 53. In the theor of relativit, the mass of a particle is s s s s s s s s where m is the rest mass of the particle, m is the mass when the particle moves with speed v relative to the observer, and c is the speed of light. Sketch the graph of m as a function of v. 54. In the theor of relativit, the energ of a particle is E sm c 4 h c where is the rest mass of the particle, is its wave length, and h is Planck s constant. Sketch the graph of E as a function of. What does the graph sa about the energ? m m s v c 55. The figure shows a beam of length L embedded in concrete walls. If a constant load W is distributed evenl along its length, the beam takes the shape of the deflection curve W 4EI 4 WL EI 3 WL 4EI where E and I are positive constants. ( E is Young s modulus of elasticit and I is the moment of inertia of a cross-section of the beam.) Sketch the graph of the deflection curve. W m 9. s 3 3. s sin sin 3 3. cos 33. tan, 34. tan, 35. sin, 36. sec tan, 37. sin 38. cos 39. e sin 4. e sin, 4. e e 46. ln ln 44. e 47. lnsin 48. e e ln sin cos 49. e 5. 3e 56. Coulomb s Law states that the force of attraction between two charged particles is directl proportional to the product of the charges and inversel proportional to the square of the distance between them. The figure shows particles with charge located at positions and on a coordinate line and a particle with charge at a position between them. It follows from Coulomb s Law that the net force acting on the middle particle is F k k where k is a positive constant. Sketch the graph of the net force function. What does the graph sa about the force? + _ L +

46 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS Find an equation of the slant asmptote. Do not sketch the curve Use the guidelines of this section to sketch the curve. In guideline D find an equation of the slant asmptote e Show that the curve tan has two slant asmptotes: and. Use this fact to help sketch the curve. 68. Show that the curve s 4 has two slant asmptotes: and. Use this fact to help sketch the curve. 69. Show that the lines ba and ba are slant asmptotes of the hperbola a b. 7. Let f 3. Show that f l This shows that the graph of f approaches the graph of, and we sa that the curve f is asmptotic to the parabola. Use this fact to help sketch the graph of f. 7. Discuss the asmptotic behavior of f 4 in the same manner as in Eercise 7. Then use our results to help sketch the graph of f. 7. Use the asmptotic behavior of f cos to sketch its graph without going through the curve-sketching procedure of this section. 4.6 N If ou have not alread read Section.4, ou should do so now. In particular, it eplains how to avoid some of the pitfalls of graphing devices b choosing appropriate viewing rectangles. 4, _5 5 _ FIGURE =ƒ _3 FIGURE =ƒ _5 GRAPHING WITH CALCULUS AND CALCULATORS The method we used to sketch curves in the preceding section was a culmination of much of our stud of differential calculus. The graph was the final object that we produced. In this section our point of view is completel different. Here we start with a graph produced b a graphing calculator or computer and then we refine it. We use calculus to make sure that we reveal all the important aspects of the curve. And with the use of graphing devices we can tackle curves that would be far too complicated to consider without technolog. The theme is the interaction between calculus and calculators. EXAMPLE Graph the polnomial f Use the graphs of f and f to estimate all maimum and minimum points and intervals of concavit. SOLUTION If we specif a domain but not a range, man graphing devices will deduce a suitable range from the values computed. Figure shows the plot from one such device if we specif that 5 5. Although this viewing rectangle is useful for showing that the asmptotic behavior (or end behavior) is the same as for 6, it is obviousl hiding some finer detail. So we change to the viewing rectangle 3, b 5, shown in Figure. From this graph it appears that there is an absolute minimum value of about 5.33 when.6 (b using the cursor) and f is decreasing on,.6 and increasing on.6,. Also, there appears to be a horizontal tangent at the origin and inflection points when and when is somewhere between and. Now let s tr to confirm these impressions using calculus. We differentiate and get f f

47 36 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION =fª() _3 _5 FIGURE 3 =ƒ FIGURE 4 _3 =f () _3 FIGURE 5 3!* When we graph f in Figure 3 we see that f changes from negative to positive when.6; this confirms (b the First Derivative Test) the minimum value that we found earlier. But, perhaps to our surprise, we also notice that f changes from positive to negative when and from negative to positive when.35. This means that f has a local maimum at and a local minimum when.35, but these were hidden in Figure. Indeed, if we now zoom in toward the origin in Figure 4, we see what we missed before: a local maimum value of when and a local minimum value of about. when.35. What about concavit and inflection points? From Figures and 4 there appear to be inflection points when is a little to the left of and when is a little to the right of. But it s difficult to determine inflection points from the graph of f, so we graph the second derivative f in Figure 5. We see that f changes from positive to negative when.3 and from negative to positive when.9. So, correct to two decimal places, f is concave upward on,.3 and.9, and concave downward on.3,.9. The inflection points are.3,.8 and.9,.5. We have discovered that no single graph reveals all the important features of this polnomial. But Figures and 4, when taken together, do provide an accurate picture. M V EXAMPLE Draw the graph of the function in a viewing rectangle that contains all the important features of the function. Estimate the maimum and minimum values and the intervals of concavit. Then use calculus to find these quantities eactl. SOLUTION Figure 6, produced b a computer with automatic scaling, is a disaster. Some graphing calculators use, b, as the default viewing rectangle, so let s tr it. We get the graph shown in Figure 7; it s a major improvement. The -ais appears to be a vertical asmptote and indeed it is because Figure 7 also allows us to estimate the -intercepts: about.5 and 6.5. The eact values are obtained b using the quadratic formula to solve the equation 7 3 ; we get (7 s37 ). f l =ƒ _5 5 FIGURE 6 =ƒ FIGURE 7 =ƒ = 5 FIGURE 8 To get a better look at horizontal asmptotes, we change to the viewing rectangle, b 5, in Figure 8. It appears that is the horizontal asmptote and this is easil confirmed: 7 3 l l 7 3

48 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 37 _3 To estimate the minimum value we zoom in to the viewing rectangle 3, b 4, in Figure 9. The cursor indicates that the absolute minimum value is about 3. when.9, and we see that the function decreases on,.9 and, and increases on.9,. The eact values are obtained b differentiating: =ƒ FIGURE 9 _4 f This shows that f when 6 and when 6 7 f and when. The eact minimum value is f ( 6 7) Figure 9 also shows that an inflection point occurs somewhere between and. We could estimate it much more accuratel using the graph of the second derivative, but in this case it s just as eas to find eact values. Since f (7 9 4 we see that f when 9. So is concave upward on ( 9 7 f 7, ) and, and concave downward on (, 9 7). The inflection point is ( 9 7, 7 7 ). The analsis using the first two derivatives shows that Figures 7 and 8 displa all the major aspects of the curve. M =ƒ _ 3 V EXAMPLE 3 Graph the function f. 4 4 SOLUTION Drawing on our eperience with a rational function in Eample, let s start b graphing f in the viewing rectangle, b,. From Figure we have the feeling that we are going to have to zoom in to see some finer detail and also zoom out to see the larger picture. But, as a guide to intelligent zooming, let s first take a close look at the epression for f. Because of the factors and 4 4 in the denominator, we epect and 4 to be the vertical asmptotes. Indeed FIGURE _ l and l _ 3 4 FIGURE To find the horizontal asmptotes we divide numerator and denominator b 6 : This shows that f l as l, so the -ais is a horizontal asmptote. It is also ver useful to consider the behavior of the graph near the -intercepts using an analsis like that in Eample in Section.6. Since is positive, f does not change sign at and so its graph doesn t cross the -ais at. But, because of the factor 3, the graph does cross the -ais at and has a horizontal tangent there. Putting all this information together, but without using derivatives, we see that the curve has to look something like the one in Figure.

49 38 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Now that we know what to look for, we zoom in (several times) to produce the graphs in Figures and 3 and zoom out (several times) to get Figure _ =ƒ _.5 FIGURE =ƒ _.5.5 _. FIGURE 3 FIGURE 4 =ƒ We can read from these graphs that the absolute minimum is about. and occurs when. There is also a local maimum. when.3 and a local minimum when.5. These graphs also show three inflection points near 35, 5, and and two between and. To estimate the inflection points closel we would need to graph f, but to compute f b hand is an unreasonable chore. If ou have a computer algebra sstem, then it s eas to do (see Eercise 5). We have seen that, for this particular function, three graphs (Figures, 3, and 4) are necessar to conve all the useful information. The onl wa to displa all these features of the function on a single graph is to draw it b hand. Despite the eaggerations and distortions, Figure does manage to summarize the essential nature of the function. M N The famil of functions f sin sin c where c is a constant, occurs in applications to frequenc modulation (FM) snthesis. A sine wave is modulated b a wave with a different frequenc sin c. The case where c is studied in Eample 4. Eercise 5 eplores another special case. EXAMPLE 4 Graph the function f sin sin. For, estimate all maimum and minimum values, intervals of increase and decrease, and inflection points correct to one decimal place. f SOLUTION We first note that f is periodic with period. Also, f is odd and for all. So the choice of a viewing rectangle is not a problem for this function: We start with, b.,.. (See Figure 5.).. =ƒ π π =fª() _. FIGURE 5 _. FIGURE 6 It appears that there are three local maimum values and two local minimum values in that window. To confirm this and locate them more accuratel, we calculate that and graph both f and f in Figure 6. f cos sin cos

50 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 39. f π f _. FIGURE 7 Using zoom-in and the First Derivative Test, we find the following values to one decimal place. Intervals of increase:,.6,.,.6,.,.5 Intervals of decrease:.6,.,.6,.,.5, Local maimum values: f.6, f.6, f.5 Local minimum values: f..94, f..94 The second derivative is f cos sin sin 4 sin cos sin Graphing both f and f in Figure 7, we obtain the following approimate values:. Concave upward on:.8,.3,.8,.3 Concave downward on:,.8,.3,.8,.3, Inflection points:,,.8,.97,.3,.97,.8,.97,.3,.97 _π FIGURE 8 _. π Having checked that Figure 5 does indeed represent f accuratel for, we can state that the etended graph in Figure 8 represents f accuratel for. M Our final eample is concerned with families of functions. As discussed in Section.4, this means that the functions in the famil are related to each other b a formula that contains one or more arbitrar constants. Each value of the constant gives rise to a member of the famil and the idea is to see how the graph of the function changes as the constant changes. _5 4 = ++ FIGURE 9 c= FIGURE c= _ = +- _5 4 V EXAMPLE 5 How does the graph of f c var as c varies? SOLUTION The graphs in Figures 9 and (the special cases c and c ) show two ver different-looking curves. Before drawing an more graphs, let s see what members of this famil have in common. Since for an value of c, the all have the -ais as a horizontal asmptote. A vertical asmptote will occur when c. Solving this quadratic equation, we get s c. When c, there is no vertical asmptote (as in Figure 9). When c, the graph has a single vertical asmptote because l l c l When c, there are two vertical asmptotes: s c (as in Figure ). Now we compute the derivative: f c This shows that f when (if c ), f when, and

51 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION TEC See an animation of Figure in Visual 4.6. f when. For c, this means that f increases on, and decreases on,. For c, there is an absolute maimum value f c. For c, f c is a local maimum value and the intervals of increase and decrease are interrupted at the vertical asmptotes. Figure is a slide show displaing five members of the famil, all graphed in the viewing rectangle 5, 4 b,. As predicted, c is the value at which a transition takes place from two vertical asmptotes to one, and then to none. As c increases from, we see that the maimum point becomes lower; this is eplained b the fact that c l as cl. As c decreases from, the vertical asmptotes become more widel separated because the distance between them is s c, which becomes large as cl. Again, the maimum point approaches the -ais because c l as cl. c=_ c= c= c= c=3 FIGURE The famil of functions ƒ=/( ++c) There is clearl no inflection point when c. For c we calculate that f c c 3 and deduce that inflection points occur when s3c 3. So the inflection points become more spread out as c increases and this seems plausible from the last two parts of Figure. M 4.6 ; EXERCISES 8 Produce graphs of f that reveal all the important aspects of the curve. In particular, ou should use graphs of f and f to estimate the intervals of increase and decrease, etreme values, intervals of concavit, and inflection points... f f f f f f tan 5 cos 7. f 4 7 cos, 8. f e Produce graphs of f that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and intervals of concavit, and use calculus to find these intervals eactl. 9.. (a) Graph the function. (b) Use l Hospital s Rule to eplain the behavior as l. (c) Estimate the minimum value and intervals of concavit. Then use calculus to find the eact values.. f 8 3 f f ln. f e

52 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 3 CAS CAS CAS CAS 3 4 Sketch the graph b hand using asmptotes and intercepts, but not derivatives. Then use our sketch as a guide to producing graphs (with a graphing device) that displa the major features of the curve. Use these graphs to estimate the maimum and minimum values If f is the function considered in Eample 3, use a computer algebra sstem to calculate f and then graph it to confirm that all the maimum and minimum values are as given in the eample. Calculate f and use it to estimate the intervals of concavit and inflection points. 6. If f is the function of Eercise 4, find f and f and use their graphs to estimate the intervals of increase and decrease and concavit of f. 7 Use a computer algebra sstem to graph f and to find f and f. Use graphs of these derivatives to estimate the intervals of increase and decrease, etreme values, intervals of concavit, and inflection points of f. 9. f s 5 sin,. f e arctan e. f. f e 3 4 (a) Graph the function. (b) Eplain the shape of the graph b computing the it as l or as l. (c) Estimate the maimum and minimum values and then use calculus to find the eact values. (d) Use a graph of f to estimate the -coordinates of the inflection points. 3. f s 7. f 8. f f f In Eample 4 we considered a member of the famil of functions f sin sin c that occur in FM snthesis. Here we investigate the function with c 3. Start b graphing f in the viewing rectangle, b.,.. How man local maimum points do ou see? The graph has more than are visible to the naked ee. To discover the hidden maimum and minimum points ou will need to eamine the graph of f ver carefull. In fact, it helps to look at the graph of f at e tan f sin sin the same time. Find all the maimum and minimum values and inflection points. Then graph f in the viewing rectangle, b.,. and comment on smmetr Describe how the graph of f varies as c varies. Graph several members of the famil to illustrate the trends that ou discover. In particular, ou should investigate how maimum and minimum points and inflection points move when c changes. You should also identif an transitional values of c at which the basic shape of the curve changes. 6. f 3 c 7. f 4 c 8. f s c 9. f e c 3. f ln c f c c 3. f 33. f c sin c 34. The famil of functions f t Ce at e bt, where a, b, and C are positive numbers and b a, has been used to model the concentration of a drug injected into the bloodstream at time t. Graph several members of this famil. What do the have in common? For fied values of C and a, discover graphicall what happens as b increases. Then use calculus to prove what ou have discovered. 35. Investigate the famil of curves given b f e c, where c is a real number. Start b computing the its as l. Identif an transitional values of c where the basic shape changes. What happens to the maimum or minimum points and inflection points as c changes? Illustrate b graphing several members of the famil. 36. Investigate the famil of curves given b the equation f 4 c. Start b determining the transitional value of c at which the number of inflection points changes. Then graph several members of the famil to see what shapes are possible. There is another transitional value of c at which the number of critical numbers changes. Tr to discover it graphicall. Then prove what ou have discovered. 37. (a) Investigate the famil of polnomials given b the equation f c 4. For what values of c does the curve have minimum points? (b) Show that the minimum and maimum points of ever curve in the famil lie on the parabola. Illustrate b graphing this parabola and several members of the famil. 38. (a) Investigate the famil of polnomials given b the equation f 3 c. For what values of c does the curve have maimum and minimum points? (b) Show that the minimum and maimum points of ever curve in the famil lie on the curve 3. Illustrate b graphing this curve and several members of the famil. 3.

53 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 4.7 OPTIMIZATION PROBLEMS The methods we have learned in this chapter for finding etreme values have practical applications in man areas of life. A businessperson wants to minimize costs and maimize profits. A traveler wants to minimize transportation time. Fermat s Principle in optics states that light follows the path that takes the least time. In this section and the net we solve such problems as maimizing areas, volumes, and profits and minimizing distances, times, and costs. In solving such practical problems the greatest challenge is often to convert the word problem into a mathematical optimization problem b setting up the function that is to be maimized or minimized. Let s recall the problem-solving principles discussed on page 76 and adapt them to this situation: STEPS IN SOLVING OPTIMIZATION PROBLEMS. Understand the Problem The first step is to read the problem carefull until it is clearl understood. Ask ourself: What is the unknown? What are the given quantities? What are the given conditions?. Draw a Diagram In most problems it is useful to draw a diagram and identif the given and required quantities on the diagram. 3. Introduce Notation Assign a smbol to the quantit that is to be maimized or minimized (let s call it Q for now). Also select smbols a, b, c,...,, for other unknown quantities and label the diagram with these smbols. It ma help to use initials as suggestive smbols for eample, A for area, h for height, t for time. 4. Epress Q in terms of some of the other smbols from Step If Q has been epressed as a function of more than one variable in Step 4, use the given information to find relationships (in the form of equations) among these variables. Then use these equations to einate all but one of the variables in the epression for Q. Thus Q will be epressed as a function of one variable, sa, Q f. Write the domain of this function. 6. Use the methods of Sections 4. and 4.3 to find the absolute maimum or minimum value of f. In particular, if the domain of f is a closed interval, then the Closed Interval Method in Section 4. can be used. N Understand the problem N Analog: Tr special cases N Draw diagrams EXAMPLE A farmer has 4 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area? SOLUTION In order to get a feeling for what is happening in this problem, let s eperiment with some special cases. Figure (not to scale) shows three possible was of laing out the 4 ft of fencing Area= =, ft@ FIGURE Area=7 =7, ft@ Area= 4=4, ft@

54 SECTION 4.7 OPTIMIZATION PROBLEMS 33 N Introduce notation We see that when we tr shallow, wide fields or deep, narrow fields, we get relativel small areas. It seems plausible that there is some intermediate configuration that produces the largest area. Figure illustrates the general case. We wish to maimize the area A of the rectangle. Let and be the depth and width of the rectangle (in feet). Then we epress A in terms of and : A A We want to epress A as a function of just one variable, so we einate b epressing it in terms of. To do this we use the given information that the total length of the fencing is 4 ft. Thus 4 FIGURE From this equation we have 4, which gives A 4 4 Note that and (otherwise A ). So the function that we wish to maimize is A 4 The derivative is A 4 4, so to find the critical numbers we solve the equation 4 4 which gives 6. The maimum value of A must occur either at this critical number or at an endpoint of the interval. Since A, A6 7,, and A, the Closed Interval Method gives the maimum value as A6 7,. [Alternativel, we could have observed that A 4 for all, so A is alwas concave downward and the local maimum at 6 must be an absolute maimum.] Thus the rectangular field should be 6 ft deep and ft wide. M h V EXAMPLE A clindrical can is to be made to hold L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. FIGURE 3 r πr SOLUTION Draw the diagram as in Figure 3, where r is the radius and h the height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the clinder (top, bottom, and sides). From Figure 4 we see that the sides are made from a rectangular sheet with dimensions r and h. So the surface area is r A r rh h To einate h we use the fact that the volume is given as L, which we take to be 3 cm. Thus r h Area {πr@} FIGURE 4 Area (πr)h which gives h r. Substitution of this into the epression for A gives A r r r r r

55 34 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Therefore the function that we want to minimize is Ar r r r To find the critical numbers, we differentiate: Ar 4r 4r 3 5 r r N In the Applied Project on page 333 we investigate the most economical shape for a can b taking into account other manufacturing costs. =A(r) Then Ar when r 3 5, so the onl critical number is r s 3 5. Since the domain of A is,, we can t use the argument of Eample concerning endpoints. But we can observe that Ar for r s 3 5 and Ar for r s 3 5, so A is decreasing for all r to the left of the critical number and increasing for all r to the right. Thus r s 3 5 must give rise to an absolute minimum. [Alternativel, we could argue that Arl as rl and Arl as rl, so there must be a minimum value of Ar, which must occur at the critical number. See Figure 5.] The value of h corresponding to r s 3 5 is FIGURE 5 r h r Thus, to minimize the cost of the can, the radius should be s 3 5 cm and the height should be equal to twice the radius, namel, the diameter. r M NOTE The argument used in Eample to justif the absolute minimum is a variant of the First Derivative Test (which applies onl to local maimum or minimum values) and is stated here for future reference. TEC Module 4.7 takes ou through si additional optimization problems, including animations of the phsical situations. FIRST DERIVATIVE TEST FOR ABSOLUTE EXTREME VALUES Suppose that c is a critical number of a continuous function f defined on an interval. (a) If f for all c and f for all c, then f c is the absolute maimum value of f. (b) If f for all c and f for all c, then f c is the absolute minimum value of f. NOTE An alternative method for solving optimization problems is to use implicit differentiation. Let s look at Eample again to illustrate the method. We work with the same equations A r rh r h but instead of einating h, we differentiate both equations implicitl with respect to r: A 4r h rh rh r h The minimum occurs at a critical number, so we set A, simplif, and arrive at the equations r h rh h rh and subtraction gives r h, or h r.

56 SECTION 4.7 OPTIMIZATION PROBLEMS 35 V EXAMPLE 3 Find the point on the parabola that is closest to the point, 4. SOLUTION The distance between the point, 4 and the point, is d s 4 (, 4) = (See Figure 6.) But if, lies on the parabola, then, so the epression for d becomes d s( ) 4 (, ) 3 4 (Alternativel, we could have substituted s to get d in terms of alone.) Instead of minimizing d, we minimize its square: d f ( ) 4 FIGURE 6 (You should convince ourself that the minimum of d occurs at the same point as the minimum of d, but d is easier to work with.) Differentiating, we obtain f ( ) so f when. Observe that f when and f when, so b the First Derivative Test for Absolute Etreme Values, the absolute minimum occurs when. (Or we could simpl sa that because of the geometric nature of the problem, it s obvious that there is a closest point but not a farthest point.) The corresponding value of is. Thus the point on closest to, 4 is,. M A 3 km C D 8 km EXAMPLE 4 A man launches his boat from point A on a bank of a straight river, 3 km wide, and wants to reach point B, 8 km downstream on the opposite bank, as quickl as possible (see Figure 7). He could row his boat directl across the river to point C and then run to B, or he could row directl to B, or he could row to some point D between C and B and then run to B. If he can row 6 kmh and run 8 kmh, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared with the speed at which the man rows.) SOLUTION If we let be the distance from C to D, then the running distance is DB 8 and the Pthagorean Theorem gives the rowing distance as. We use the equation AD s 9 B time distance rate Then the rowing time is s 9 6 and the running time is 8 8, so the total time T as a function of is FIGURE 7 T s The domain of this function T is, 8. Notice that if, he rows to C and if 8, he rows directl to B. The derivative of T is T 6s 9 8

57 36 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Thus, using the fact that, we have T &? 6s 9 8 &? 4 3s 9 &? &? 7 8 &? 9 s7 T =T() 4 6 FIGURE 8 The onl critical number is 9s7. To see whether the minimum occurs at this critical number or at an endpoint of the domain, 8, we evaluate T at all three points: T.5 T 9 s7 s T8 s Since the smallest of these values of T occurs when 9s7, the absolute minimum value of T must occur there. Figure 8 illustrates this calculation b showing the graph of T. Thus the man should land the boat at a point 9s7 km ( 3.4 km) downstream from his starting point. M V EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle of radius r. _r FIGURE 9 (, ) r SOLUTION Let s take the semicircle to be the upper half of the circle r with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the -ais as shown in Figure 9. Let, be the verte that lies in the first quadrant. Then the rectangle has sides of lengths and, so its area is A To einate we use the fact that, lies on the circle r and so sr. Thus A sr The domain of this function is r. Its derivative is A sr r sr sr which is when r, that is, rs (since ). This value of gives a maimum value of A since A and Ar. Therefore the area of the largest inscribed rectangle is A r s r s r r r FIGURE r r cos r sin SOLUTION A simpler solution is possible if we think of using an angle as a variable. Let be the angle shown in Figure. Then the area of the rectangle is A r cos r sin r sin cos r sin

58 SECTION 4.7 OPTIMIZATION PROBLEMS 37 We know that sin has a maimum value of and it occurs when. So A has a maimum value of r and it occurs when. Notice that this trigonometric solution doesn t involve differentiation. In fact, we didn t need to use calculus at all. 4 M APPLICATIONS TO BUSINESS AND ECONOMICS In Section 3.7 we introduced the idea of marginal cost. Recall that if C, the cost function, is the cost of producing units of a certain product, then the marginal cost is the rate of change of C with respect to. In other words, the marginal cost function is the derivative, C, of the cost function. Now let s consider marketing. Let p be the price per unit that the compan can charge if it sells units. Then p is called the demand function (or price function) and we would epect it to be a decreasing function of. If units are sold and the price per unit is p, then the total revenue is and R is called the revenue function. The derivative R of the revenue function is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold. If units are sold, then the total profit is and P is called the profit function. The marginal profit function is P, the derivative of the profit function. In Eercises ou are asked to use the marginal cost, revenue, and profit functions to minimize costs and maimize revenues and profits. V EXAMPLE 6 A store has been selling DVD burners a week at $35 each. A market surve indicates that for each $ rebate offered to buers, the number of units sold will increase b a week. Find the demand function and the revenue function. How large a rebate should the store offer to maimize its revenue? SOLUTION If is the number of DVD burners sold per week, then the weekl increase in sales is. For each increase of units sold, the price is decreased b $. So for each additional unit sold, the decrease in price will be and the demand function is The revenue function is R p P R C p R p 45 Since R 45, we see that R when 45. This value of gives an absolute maimum b the First Derivative Test (or simpl b observing that the graph of R is a parabola that opens downward). The corresponding price is p and the rebate is Therefore, to maimize revenue, the store should offer a rebate of $5. M

59 38 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 4.7 EXERCISES. Consider the following problem: Find two numbers whose sum is 3 and whose product is a maimum. (a) Make a table of values, like the following one, so that the sum of the numbers in the first two columns is alwas 3. On the basis of the evidence in our table, estimate the answer to the problem. First number Second number Product (b) Use calculus to solve the problem and compare with our answer to part (a).. Find two numbers whose difference is and whose product is a minimum. 3. Find two positive numbers whose product is and whose sum is a minimum. 4. Find a positive number such that the sum of the number and its reciprocal is as small as possible. 5. Find the dimensions of a rectangle with perimeter m whose area is as large as possible. 6. Find the dimensions of a rectangle with area m whose perimeter is as small as possible. 7. A model used for the ield Y of an agricultural crop as a function of the nitrogen level N in the soil (measured in appropriate units) is Y kn N where k is a positive constant. What nitrogen level gives the best ield? 8. The rate in mg carbonm 3 h at which photosnthesis takes place for a species of phtoplankton is modeled b the function P I I I 4 where I is the light intensit (measured in thousands of footcandles). For what light intensit is P a maimum? 9. Consider the following problem: A farmer with 75 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens? (a) Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these configurations. Does it appear that there is a maimum area? If so, estimate it. (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with our smbols. (c) Write an epression for the total area. (d) Use the given information to write an equation that relates the variables. (e) Use part (d) to write the total area as a function of one variable. (f) Finish solving the problem and compare the answer with our estimate in part (a).. Consider the following problem: A bo with an open top is to be constructed from a square piece of cardboard, 3 ft wide, b cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a bo can have. (a) Draw several diagrams to illustrate the situation, some short boes with large bases and some tall boes with small bases. Find the volumes of several such boes. Does it appear that there is a maimum volume? If so, estimate it. (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with our smbols. (c) Write an epression for the volume. (d) Use the given information to write an equation that relates the variables. (e) Use part (d) to write the volume as a function of one variable. (f) Finish solving the problem and compare the answer with our estimate in part (a).. A farmer wants to fence an area of.5 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. How can he do this so as to minimize the cost of the fence?. A bo with a square base and open top must have a volume of 3 3, cm. Find the dimensions of the bo that minimize the amount of material used. 3. If cm of material is available to make a bo with a square base and an open top, find the largest possible volume of the bo. 4. A rectangular storage container with an open top is to have a 3 volume of m. The length of its base is twice the width. Material for the base costs $ per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container. 5. Do Eercise 4 assuming the container has a lid that is made from the same material as the sides. 6. (a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square. (b) Show that of all the rectangles with a given perimeter, the one with greatest area is a square. 7. Find the point on the line 4 7 that is closest to the origin. 8. Find the point on the line 6 9 that is closest to the point 3,. 9. Find the points on the ellipse 4 4 that are farthest awa from the point,.

60 SECTION 4.7 OPTIMIZATION PROBLEMS 39 ;. Find, correct to two decimal places, the coordinates of the point on the curve tan that is closest to the point,.. Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius r.. Find the area of the largest rectangle that can be inscribed in the ellipse a b. 3. Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle. 4. Find the dimensions of the rectangle of largest area that has its base on the -ais and its other two vertices above the -ais and ling on the parabola Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r. 6. Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3 cm and 4 cm if two sides of the rectangle lie along the legs. 7. A right circular clinder is inscribed in a sphere of radius r. Find the largest possible volume of such a clinder. 8. A right circular clinder is inscribed in a cone with height h and base radius r. Find the largest possible volume of such a clinder. 9. A right circular clinder is inscribed in a sphere of radius r. Find the largest possible surface area of such a clinder. 3. A Norman window has the shape of a rectangle surmounted b a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See Eercise 56 on page 3.) If the perimeter of the window is 3 ft, find the dimensions of the window so that the greatest possible amount of light is admitted. 3. The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is fied at 384 cm, find the dimensions of the poster with the smallest area. 3. A poster is to have an area of 8 in with -inch margins at the bottom and sides and a -inch margin at the top. What dimensions will give the largest printed area? 33. A piece of wire m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maimum? (b) A minimum? 34. Answer Eercise 33 if one piece is bent into a square and the other into a circle. 35. A clindrical can without a top is made to contain V cm 3 of liquid. Find the dimensions that will minimize the cost of the metal to make the can. 36. A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest lad- der that will reach from the ground over the fence to the wall of the building? 37. A cone-shaped drinking cup is made from a circular piece of paper of radius R b cutting out a sector and joining the edges CA and CB. Find the maimum capacit of such a cup. 38. A cone-shaped paper drinking cup is to be made to hold 7 cm 3 of water. Find the height and radius of the cup that will use the smallest amount of paper. 39. A cone with height h is inscribed in a larger cone with height H so that its verte is at the center of the base of the larger cone. Show that the inner cone has maimum volume when h 3 H. 4. An object with weight W is dragged along a horizontal plane b a force acting along a rope attached to the object. If the rope makes an angle with a plane, then the magnitude of the force is W F where is a constant called the coefficient of friction. For what value of is F smallest? A 4. If a resistor of R ohms is connected across a batter of E volts with internal resistance r ohms, then the power (in watts) in the eternal resistor is P sin cos E R R r If E and r are fied but R varies, what is the maimum value of the power? 4. For a fish swimming at a speed v relative to the water, the energ ependiture per unit time is proportional to v 3. It is believed that migrating fish tr to minimize the total energ required to swim a fied distance. If the fish are swimming against a current u u v, then the time required to swim a distance L is Lv u and the total energ E required to swim the distance is given b L Ev av 3 v u where a is the proportionalit constant. (a) Determine the value of v that minimizes E. (b) Sketch the graph of E. C R Note: This result has been verified eperimentall; migrating fish swim against a current at a speed 5% greater than the current speed. B

61 33 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 43. In a beehive, each cell is a regular heagonal prism, open at one end with a trihedral angle at the other end as in the figure. It is believed that bees form their cells in such a wa as to minimize the surface area for a given volume, thus using the least amount of wa in cell construction. Eamination of these cells has shown that the measure of the ape angle is amazingl consistent. Based on the geometr of the cell, it can be shown that the surface area S is given b S 6sh 3 s cot (3s s3) csc where s, the length of the sides of the heagon, and h, the height, are constants. (a) Calculate. (b) What angle should the bees prefer? (c) Determine the minimum surface area of the cell (in terms of s and h). Note: Actual measurements of the angle in beehives have been made, and the measures of these angles seldom differ from the calculated value b more than. dsd 47. An oil refiner is located on the north bank of a straight river that is km wide. A pipeline is to be constructed from the refiner to storage tanks located on the south bank of the river 6 km east of the refiner. The cost of laing pipe is $4,km over land to a point P on the north bank and $8,km under the river to the tanks. To minimize the cost of the pipeline, where should P be located? ; 48. Suppose the refiner in Eercise 47 is located km north of the river. Where should P be located? 49. The illumination of an object b a light source is directl proportional to the strength of the source and inversel proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed ft apart, where should an object be placed on the line between the sources so as to receive the least illumination? 5. Find an equation of the line through the point 3, 5 that cuts off the least area from the first quadrant. rear of cell trihedral angle 5. Let a and b be positive numbers. Find the length of the shortest line segment that is cut off b the first quadrant and passes through the point a, b. b 44. A boat leaves a dock at : PM and travels due south at a speed of kmh. Another boat has been heading due east at 5 kmh and reaches the same dock at 3: PM. At what time were the two boats closest together? 45. Solve the problem in Eample 4 if the river is 5 km wide and point B is onl 5 km downstream from A. 46. A woman at a point A on the shore of a circular lake with radius mi wants to arrive at the point C diametricall opposite A on the other side of the lake in the shortest possible time. She can walk at the rate of 4 mih and row a boat at mih. How should she proceed? s h front of cell B 5. At which points on the curve does the tangent line have the largest slope? 53. (a) If C is the cost of producing units of a commodit, then the average cost per unit is c C. Show that if the average cost is a minimum, then the marginal cost equals the average cost. (b) If C 6, 4 3, in dollars, find (i) the cost, average cost, and marginal cost at a production level of units; (ii) the production level that will minimize the average cost; and (iii) the minimum average cost. 54. (a) Show that if the profit P is a maimum, then the marginal revenue equals the marginal cost. (b) If C 6, is the cost function and p 7 7 is the demand function, find the production level that will maimize profit. 55. A baseball team plas in a stadium that holds 55, spectators. With ticket prices at $, the average attendance had been 7,. When ticket prices were lowered to $8, the average attendance rose to 33,. (a) Find the demand function, assuming that it is linear. (b) How should ticket prices be set to maimize revenue? A C 56. During the summer months Terr makes and sells necklaces on the beach. Last summer he sold the necklaces for $ each and his sales averaged per da. When he increased the price b $, he found that the average decreased b two sales per da. (a) Find the demand function, assuming that it is linear. (b) If the material for each necklace costs Terr $6, what should the selling price be to maimize his profit?

62 SECTION 4.7 OPTIMIZATION PROBLEMS 33 CAS 57. A manufacturer has been selling television sets a week at $45 each. A market surve indicates that for each $ rebate offered to the buer, the number of sets sold will increase b per week. (a) Find the demand function. (b) How large a rebate should the compan offer the buer in order to maimize its revenue? (c) If its weekl cost function is C 68, 5, how should the manufacturer set the size of the rebate in order to maimize its profit? 58. The manager of a -unit apartment comple knows from eperience that all units will be occupied if the rent is $8 per month. A market surve suggests that, on average, one additional unit will remain vacant for each $ increase in rent. What rent should the manager charge to maimize revenue? 59. Show that of all the isosceles triangles with a given perimeter, the one with the greatest area is equilateral. 6. The frame for a kite is to be made from si pieces of wood. The four eterior pieces have been cut with the lengths indicated in the figure. To maimize the area of the kite, how long should the diagonal pieces be? a a ; 6. A point P needs to be located somewhere on the line AD so that the total length L of cables linking P to the points A, B, and C is minimized (see the figure). Epress L as a function of AP and use the graphs of L and dld to estimate the minimum value. b b this consumption G. Using the graph, estimate the speed at which G has its minimum value. 63. Let v be the velocit of light in air and v the velocit of light in water. According to Fermat s Principle, a ra of light will travel from a point A in the air to a point B in the water b a path ACB that minimizes the time taken. Show that where (the angle of incidence) and (the angle of refraction) are as shown. This equation is known as Snell s Law. 64. Two vertical poles PQ and ST are secured b a rope PRS going from the top of the first pole to a point R on the ground between the poles and then to the top of the second pole as in the figure. Show that the shortest length of such a rope occurs when. A c 4 6 P sin v sin v C B S A P 5 m Q R T B m 3 m D 6. The graph shows the fuel consumption c of a car (measured in gallons per hour) as a function of the speed v of the car. At ver low speeds the engine runs inefficientl, so initiall c decreases as the speed increases. But at high speeds the fuel consumption increases. You can see that cv is minimized for this car when v 3 mih. However, for fuel efficienc, what must be minimized is not the consumption in gallons per hour but rather the fuel consumption in gallons per mile. Let s call C 65. The upper right-hand corner of a piece of paper, in. b 8 in., as in the figure, is folded over to the bottom edge. How would ou fold it so as to minimize the length of the fold? In other words, how would ou choose to minimize? 8

63 33 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 66. A steel pipe is being carried down a hallwa 9 ft wide. At the end of the hall there is a right-angled turn into a narrower hallwa 6 ft wide. What is the length of the longest pipe that can be carried horizontall around the corner? observer stand so as to maimize the angle ee b the painting?) subtended at his h 6 d An observer stands at a point P, one unit awa from a track. Two runners start at the point S in the figure and run along the track. One runner runs three times as fast as the other. Find the maimum value of the observer s angle of sight between the runners. [Hint: Maimize tan.] 68. A rain gutter is to be constructed from a metal sheet of width 3 cm b bending up one-third of the sheet on each side through an angle. How should be chosen so that the gutter will carr the maimum amount of water? P S cm cm cm 69. Where should the point P be chosen on the line segment AB so as to maimize the angle? 7. Find the maimum area of a rectangle that can be circumscribed about a given rectangle with length L and width W. [Hint: Epress the area as a function of an angle.] 7. The blood vascular sstem consists of blood vessels (arteries, arterioles, capillaries, and veins) that conve blood from the heart to the organs and back to the heart. This sstem should work so as to minimize the energ epended b the heart in pumping the blood. In particular, this energ is reduced when the resistance of the blood is lowered. One of Poiseuille s Laws gives the resistance R of the blood as A where L is the length of the blood vessel, r is the radius, and C is a positive constant determined b the viscosit of the blood. (Poiseuille established this law eperimentall, but it also follows from Equation 8.4..) The figure shows a main blood vessel with radius r branching at an angle into a smaller vessel with radius vascular branching r B R C L r 4 a r r C b B 3 P A 7. A painting in an art galler has height h and is hung so that its lower edge is a distance d above the ee of an observer (as in the figure). How far from the wall should the observer stand to get the best view? (In other words, where should the 5 Manfred Cage / Peter Arnold