Applications of Derivatives

Transcription

1 Chapter Applications of Derivatives. Etrema of Functions on Intervals Maimum and Minimum Values of a Function Relative Etrema and Critical Numbers Finding the Etrema on a Closed Interval Maimum and Minimum Values of a Function In Chapter, we studied several rules for finding the derivative of a function. In this section, we use derivatives to find the maimum and minimum values of a differentiable function. Such values have important consequences. The information of knowing how to maimize returns or minimize costs is important to know. What ou have been obliged to discover b ourself leaves a path in our mind which ou can use again when the need arises. - G.C. Lichtenberg Definition Maimum and Minimum Values of a Function Let f be a function that is defined on an interval I containing c.. f(c) istheminimum value of f on I if f(c) f() for all in I.. f(c) isthemaimum value of f on I if f(c) f() for all in I. The maimum and minimum values of f on I are called the etreme values or etrema of f on I. In the net eample, we use the graph to identif an etrema. It is possible for a function not to have a maimum or minimum value. Eample Finding the Etrema of a Function Find the etrema of f() =( ) + in the indicated interval. a) [, ] b) [, ) 5 Figure a In [, ], the maimum value is f() = 5 and the minimum value is f() =. 5 Solution a) The graph of = f() in[, ] is given in Figure a. The highest point is (, 5) and the lowest point is (, ). Then the maimum value of f is f() = 5 and the minimum value is f() =. b) In Figure b, we see the graph of = f() in[, ). The point (, 5) is not the highest point since (, 5) does not belong to the graph of = f() in[, ). Figure b In [, ), f has no maimum value and the minimum value is f() =. 7

2 8 CHAPTER. APPLICATIONS OF DERIVATIVES There is no highest point in [, ) because given an point we can alwas find a higher point. In other words, none of the following values of f() f(.9) = (.9 ) +=.6 f(.99) = (.99 ) +=.960. f() = ( ) +, if < is the maimum value of f() on[, ). Then f has no maimum value on [, ). However, the point (, ) is the lowest point on the graph. Hence, the minimum value of f on [, ) is f() =. Tr This Find the etrema of f() = + in the indicated interval, see Figure c. a) [, 0] b) (, ) c) (0, ) d) (0, ] Figure c The graph of f() = +. Eample Finding the Etrema of a Function Find the etrema of in the interval [, ] g() = ( ) + if = if = 5 Figure The graph has no minimum value in [, ], but the maimum value is 5. Solution In Figure, the point (, 5) is the highest point on the graph of = g(). Then the maimum value of g on [, ] is g() = 5. Note, the point (, ) does not lie on the graph of = g() andcannotbethe lowest point on the graph. We cannot claim that a certain point on the graph is the lowest point since there will be another point on the graph that is lower. Hence, g() hasno minimum value on [, ]. Tr This Find the etrema of g() =( ) + where lies in [, ] and =. In particular, g() is undefined. The net theorem states a condition when a function is guaranteed to have etreme values. A proof of the theorem is omitted since it is beond the scope of this book. Theorem. The Etreme Value Theorem A continuous function on a closed interval [a, b] is guaranteed to have a maimum value and a minimum value on [a, b].

3 .. EXTREMA OF FUNCTIONS ON INTERVALS 9 Relative Etrema and Critical Numbers A function f has a relative maimum value at point P (c, f(c)) if P is higher than nearb points, or geometricall P is at the top of a hill. Similarl, f has a relative minimum value at P if P is lower than nearb points, or P is at the bottom of a valle. As seen in Figure, the graph of = f() hasarelative maimum value at the point (, 5), or f() = 5 is a relative maimum value of f. Similarl, f has relative minimum values at (0, 0) and (, 7). That is, f(0) = 0 and f() = 7 are relative minimum values of f. f 5 Definition Definition of Relative Etrema Let f be a function.. f(c) isarelative maimum value of f if there is an open interval I containing c for which f(c) isthemaimumvalueoff on I.. f(c) isarelative minimum value of f if there is an open interval I containing c for which f(c) istheminimumvalueoff on I. The relative maimum values and relative minimum values of f are called the relative etreme values, or relative etrema of f. The relative maimum values and relative minimum values are also called relative maima and relative minima, respectivel. 7 Figure Relative minimum values are f(0) = 0 and f() = 7, and the relative maimum value is f() = 5. Clearl, an etreme value of a function is a relative etreme value. However, a relative etreme value is not necessaril an etreme value. In Figure, f(0) = 0 and f() = 5 are relative etreme values of f but the are not the minimum and maimum values of f. In Eample, we will see a particular case of a general phenomena. That is, if f(c) is a relative etreme value then f (c) is either zero or undefined. Eample The Derivative at a Relative Etrema f Determine the relative etreme values f(c) off, andevaluatef (c). a) f() = + b) f() = / See the graphs in Figures a and b. Solution a) In Figure a, we see that the point (0, ) is a relative maimum point. Then f(0) = is a relative maimum value of f. Similarl, the point (, ) is a local minimum point. Thus, f() = is a relative minimum value of f. The derivative of f is f () = 6 =( ). Substituting =0and =intof (), we obtain b) Note, the derivative of f is f (0) = 0 and f () = 0. f () = / =. Using Figure b, we see that f(0) = is the maimum value of f. Then f (0) is undefined since it evaluates to the epression 0. Finall, from the graph we see that there is no relative minimum value as seen in Figure b. Figure a f(0) = is the relative maimum value, and f() = is the relative minimum value of f. f Figure b f(0) = is the maimum value of f.

4 0 CHAPTER. APPLICATIONS OF DERIVATIVES f Tr This For each relative etreme value f(c), evaluate f (c). See Figures 5a and 5b. / a) f() = b) f() = / Figure 5a f() = f Figure 5b / f() = / In Eample, we have seen that the values of the derivative at the relative etrema are either zero or undefined. These -values are called critical numbers. Definition The Definition of a Critical Number Let f be a function which is defined at a number c. If f (c) =0orf (c) doesnoteist, then c is a critical number of f. Eample Finding the Critical Numbers Find the critical numbers of each function. a) f() = 6 + b) g() = / ( +) Solution a) Evaluate and factor the derivative f as follows: f () = = 6( 6) = 6( )( +). Then f () =0eactlwhen = or =. Thus, the critical numbers are =, ; see local etrema in Figure 6a. f Figure 6a: The critical numbers of f are and. b) Appl the product rule and factor g (). g() = / ( +) g () = / ( +)+ / g () = / [( +)+] g 5 +8 () =

5 .. EXTREMA OF FUNCTIONS ON INTERVALS Note, g () =0ifandonlif 5 +8=0. Then = 8 is a critical number. 5 Moreover, the derivative g (0) is undefined since it evaluates to an undefined epres-.thus, = 0 is also a critical number. sion 8 0 Hence, the critical numbers are 8 and 0; see the relative etrema in Figure 6b. 5 g 8 5 Figure 6b: The critical numbers of g are 0 and 8/5. Tr This Find the critical numbers of each function. a) f() = + b) f() = c) f() = + A first step in finding the etreme values of a differentiable function is to find the critical numbers. To illustrate this point, in Figure 7 we see that f has relative minimum and maimum values at C and C, respectivel. B Theorem., theabcissasofc and C are critical numbers of f. Thus, the relative etrema can be found from the list of critical numbers. However, we caution the student that there are critical numbers that do not provide etreme values. In an case, the set of critical numbers is a short list that we could use to find the relative etrema. C f C Figure 7 The graph of f has relative etreme values at the critical points C and C.

6 CHAPTER. APPLICATIONS OF DERIVATIVES Theorem. The Relative Etrema Occur Eactl at Critical Points Let f be a function. If f(c) is a relative etreme value, then c is a critical number of f. Proof If f (c) does not eist, then the theorem is true and there would be nothing else to prove. So, we assume f (c) eists and we have to prove f (c) =0. Iff (c) = 0,then either f (c) > 0orf (c) < 0. Suppose f f(c + h) f(c) (c) = lim > 0. h 0 h B Eercise 7 in Chapter, there eists an open interval ( δ, δ) withδ>0suchthat f(c + h) f(c) h whenever δ <h<δwith h = 0. If we multipl both sides of the previous inequalit with a positive h that is less than δ, thenweobtain > 0 f(c + h) f(c) h h > h 0 f(c + h) f(c) > 0 f(c + h) > f(c). This implies f(c) is not a relative maimum value of f. On the other hand if we multipl the same inequalit with a negative h with δ <h<0, then we reverse the direction of the inequalit and find f(c + h) f(c) h h < h 0 f(c + h) f(c) < 0 f(c + h) < f(c). This shows f(c) is not a relative minimum value of f. Since f(c) is a relative etreme value of f, we have a contradiction. Similarl, if f (c) < 0 then we would reach the same contradiction. Hence, we conclude f (c) =0. Finding the Etrema on a Closed Interval Recall, the etrema of a continuous function f defined on a closed interval [a, b] are guaranteed to eist because of Theorem.. As a consequence of Theorem., the following guidelines can be used to find the etrema of f on [a, b]. Guidelines for Finding the Etrema of a Continuous Function f on a Closed Interval [a, b]. Find the critical numbers c of f in (a, b).. Evaluate f(c) at the critical numbers c in (a, b).. Evaluate f(a) andf(b).. The largest value in steps - is the maimum value of f on [a, b]. The smallest value in steps - is the minimum value of f on [a, b].

7 .. EXTREMA OF FUNCTIONS ON INTERVALS Eample 5 Finding the Etrema on a Closed Interval Find the etrema of f() = on [0, ], as in Figure 8 Solution The derivative of f is f () =. The critical numbers of f are found as follows: = 0 = = = ±. The onl critical number in the open interval (0, ) is =,and f() = () =. The values of f at the endpoints of [0, ] are f(0) = and f() = () = 6. 6 Figure 8 On [0, ], the maimum, value of f is f() = 6, and the minimum value is f() =. From the list below f(), critical number 0, endpoint, endpoint 6 we conclude the maimum value of f is 6, and the minimum value is. Tr This 5 Find the etrema of the given function on [0, ]. f() = Eample 6 Find the Etrema on [a, b] Find the etrema of g() = / on the interval [, ], see Figure 9. Solution The derivative of g is g () = = 6/. Note, = 0 is a critical number of g. Ifg () =0,then 6 / = 0 / = = ± 7 ±0. Then = ±/ 7 are critical numbers on (, ). Moreover, we find g ± / = 7 7 = 7 =. Figure 9 On [, ], the maimum value of g is achieved at =, and the minimum value at the critical numbers ±/ 7.

8 CHAPTER. APPLICATIONS OF DERIVATIVES In addition, at the endpoints of [, ] we have g( ) = and g() = 0. Now, identif the largest and smallest values of g in the table g() 0, critical number 0 ±/ 7, critical numbers /( ) 0.8, endpoint., endpoint 0. Hence, the maimum value of g is g( ) = and the minimum value is g(±/ 7) = = 9. Tr This 6 Find the etrema of g() = / on [0, 8]. Π Π cossin Π 6 5 Π 6 Figure 0 The maimum value is h (π/6) = / and the minimum value is h (5π/6) = /. Π Eample 7 Finding the Etrema on a Closed Interval Find the etrema of h() =cos +sin on [ π, π], see Figure 0. Solution The derivative of h is h () = sin +cos = sin +( sin ) Since cos = sin = ( sin +sin ) h () = ( sin )(sin +) If h () =0for on the interval ( π, π), then Moreover, we find π h 6 5π h 6 h π sin = = π 6, 5π 6 or sin = or = π. = cos π 6 +sinπ = + =, = cos 5π 6 +sin5π = =, and = cos π +sin( π) =0. The values of h at the endpoints of [ π, π] are h ( π) = cos( π)+sin( π) = +0= h (π) = cosπ +sinπ = +0=.

9 .. EXTREMA OF FUNCTIONS ON INTERVALS 5 The values of h at the critical numbers and endpoints are listed below: h() π/6, critical number /.6 5π/6, critical number /.6 π/, critical number 0 π, endpoint π, endpoint Identif the largest and smallest values of h in the table. Hence, the maimum value of h is h(π/6) =,andtheminimumvalueish(5π/6) =. Tr This 7 Find the etrema of f() =cos +sin on [0,π].. Check-It Out. Find the critical numbers of f() =. Find the critical numbers of g(t) =sint t on the open interval (0, π).. Find the etrema of f() = on [0, ]. True or False. If false, eplain or show an eample that shows it is false.. The number f() is the minimum value of f() =.. If f(c) is a relative etreme value of f, thenf (c) =0.. If c is a critical number of a function f, thenf(c) is a relative etreme value of f.. The critical numbers of f() = are = ±. 5. The critical number of f() =( a)( b) is =(a + b)/. 6. The critical numbers of =sin(π) on the open interval (0, ) are =,. 7. g() = has a critical number. 8. The etrema of f() = on [0, ] are 0 and The etrema of f() =sin on [0,π] are π/ and π. 0. The critical numbers of f(t) = t are t =0,. t + Eercises for Section. In Eercises -, use the graph to find the etrema of the function in the indicated interval.. f() = ( ) +5,[, ]. f() = ( ) +5,(, ) 5 5 For No. For No.

10 6 CHAPTER. APPLICATIONS OF DERIVATIVES. g() = ( ) +5 if = if =, [0, ] 5 For No. For No. ( ) + if is in (, ) (, ). g() = if =, [, ] In Eercises 5-8, find the value of the derivative at each relative etremum. 5. f() = 6. f() = For No. 5 For No f() = / 8. f() =8 / / 7 For No. 7 For No. 8 7 In Eercises 9-8, find the critical numbers of the function. 9. f() = f() = g() = 5/ /. g() = / ( ). f() =. f() = g() =cos +, 0 <<π 6. g() = 6sin +,0<<π 7. f() =sin cos, 0<<π 8. f() = sin sin, 0<<π

11 .. EXTREMA OF FUNCTIONS ON INTERVALS 7 In Eercises 9-, find the etrema of the function in the indicated closed interval. 9. f() =( ), [, ] 0. f() = (5 + ), [, 0]. f() = 7 +5,[, ]. f() = +,[, ]. g() = ,[ 6, ]. g() = + +,[.5, 0.5] 5. g() =( ) ( +),[ /, ] 6. g() =( +) ( ), [, /] 7. f() =( ),[0, ] 8. f() =, [, ] 9. f() =cos, [0, π] 0. f() = sin π, [0, ]. f(t) = t t,[, ]. h(t) =,[0, ] t + t +. k(t) =6sint t, [0,π]. M(s) =sins +coss, [0,π/] Applications 5. The Path of Least Cost A plumbing project involves installing pvc pipes from A to C to B, see figure below. Along the horizontal, the installation costs $ per foot. Along the diagonal, the cost is $ per foot due to etra labor. Find the minimum cost of the plumbing project. 0 A C B 0 P 00 For No. 5 For No Staking Two Antennas Two antennas are 00 feet apart, and their heights from the ground are 50 feet and 75 feet. Suppose a cable is connected from point P to the top of each antenna, see above figure. Where should P be located so that the least amount of cable is used? 7. Generating a Right Circular Clinder The perimeter of a rectangle is 6 feet. If the rectangle is revolved about one of it sides, a right circular clinder is generated. Find the maimum volume of the clinder. See figure below. A0, B,0 C, For No. 7 For No Bring out a Calculator AlinesegmenthasendpointsA(0, ) and B(,0) where 0. A second line segment has endpoints B(, 0) and C(, ). Find if two times the length of the first segment plus the length of the second segment is the minimum. See figure above.

12 8 CHAPTER. APPLICATIONS OF DERIVATIVES. The Mean Value Theorem Rolle s Theorem The Mean Value Theorem Michel Rolle (65-79), afrenchmathematician, published Rolle s Theorem in 69. His theorem plas an important role in the proofs of several calculus theorems. Rolle s Theorem The above-named theorem is a basic result that makes possible the application of derivatives to finding the etreme values of a differentiable function. Also, Rolle s Theorem establishes a connection between critical numbers and the functional values of a differentiable function, see Figure. c, f c a b c, f c Figure Rolle s Theorem: If f(a) = f(b), then f (c) =0forsomec in (a, b). Theorem. Rolle s Theorem Let f be a function that is continuous on [a, b] and differentiable on (a, b). If f(a) =f(b), then there eists a number c in (a, b) satisfing f (c) =0. Proof We consider three cases. Case If f() =f(a) for all in [a, b], then f is a constant function. Thus, f (c) =0 for all c in (a, b). Case Suppose f( 0) >f(a) for some 0 in (a, b). Recall, the Etreme Value Theorem assures that f has a maimum value f(c) where c is some number c in [a, b]. Since f( 0) >f(a) =f(b), we find c = a, b. Then f(c) is a relative etreme value of f. B Theorem., c is a critical number of f in (a, b). Since f is differentiable in (a, b), we have f (c) =0. Case Suppose f( 0) <f(a) for some 0 in (a, b). We consider the minimum value f(c) of f. Similarl, as in Case we conclude that c lies in (a, b) andf (c) =0. Eample Illustrating Rolle s Theorem Show Rolle s Theorem applies to the function on the indicated interval: f() = / /, [0, ] Then find all the numbers c that satisf Rolle s Theorem. Solution To show Rolle s Theorem applies we have to verif the following: a) f is continuous on [0, ], b) f is differentiable on (0, ), and c) f(0) = f(). We establish the validit of statements a), b), andc) as follows.

13 .. THE MEAN VALUE THEOREM 9 a) The radical function = and the composite function =( ) are continuous everwhere, see page and Theorem.0. Then the difference function f() = / / is continuous on [0, ]. b) Appling the power rule, we find 0.7 f () = / / = ( ) () / Since f () isundefinedonlwhen =0,f is differentiable on (0, ). c) Clearl, f() = / / =0andf(0) = 0. Thus, Rolle s Theorem applies. Using (), we find that f (c) =0implies c =0. Hence, c =/ istheonlnumberin(0, ) that satisfies Rolle s Theorem, see Figure. Figure The values of f at the endpoints of [0, ] are f(0) = f() = 0. Rolle s Theorem guarantees the eistence of c =/ inthe interval (0, ) such that f (c) =0. Tr This Find all numbers c satisfing Rolle s Theorem. a) f() =9, [, 0] b) f() =sin, [π/, 5π/] Man functions satisf the hpothesis of Rolle s Theorem. These include polnomial, rational, and trigonometric functions provided these functions are defined on the interval [a, b] in Rolle s Theorem. Recall, polnomial, rational, and trigonometric functions are differentiable in their domains of definition, see Sections. Eample Appling Rolle s Theorem Appl Rolle s Theorem and the Intermediate Value Theorem to show that the graph of f() = has eactl one -intercept. Solution We find two functional values of f with opposite signs: f() = 5 +()+=5 and f( ) = ( ) 5 +( ) + = B the Intermediate Value Theorem (see page 5), we can find a number 0 in (, ) such that f( 0)=0. Then( 0, 0) is an -intercept of the graph of f. Suppose (, 0) is another -intercept of the graph f and = 0. Then f( )= f( 0) = 0. Appling Rolle s Theorem, there eists a real number c between and 0 such that f (c) = 0. But this is impossible for the derivative has no zero, i.e., f () =5 += 0 for all. Hence, the graph of f has eactl one -intercept, namel, ( 0, 0). Tr This Show the graph of f() = 9 has eactl one -intercept using Rolle s Theorem and the Intermediate Value Theorem. Figure Asketchof the graph of f() =

14 50 CHAPTER. APPLICATIONS OF DERIVATIVES The Mean Value Theorem was first proved b Joseph-Louis Lagrange (76-8). At the oung age of 9, Lagrange became a professor in Turin. The Mean Value Theorem The following theorem is a generalization of Rolle s Theorem. The Mean Value Theorem can be applied to classif increasing or decreasing functions in terms of the derivative, as we will see in Section.. b, f b Theorem. The Mean Value Theorem Let f be a function that is continuous on [a, b] and differentiable on (a, b). Then there eists a number c in (a, b) satisfing a, f a f (c) = f(b) f(a). b a a c b Figure Mean Value Theorem: The line joining (a, f(a)) to (b, f(b)) is parallel to a tangent line at some point (c, f(c)) where a<c<b. Proof The slope of the secant line joining A(a, f(a)) to B(b, f(b)) is An equation of the secant line is m = f(b) f(a). b a s() =m( a)+f(a). Let g() be the difference f() and s() as defined below: g() =f() (m( a)+f(a)). Then g(a) =0andg(b) =0becauseofthedefinitionofm. Observe, g () = d d [f()] d [m( a)+f(a)] d = f () m. Appling Rolle s Theorem, there eists a number c in (a, b) satisfing g (c) = 0 f (c) m = 0. Hence, we obtain f f(b) f(a) (c) =m =. b a This completes the proof of the Mean value Theorem. In other words, the Mean Value Theorem implies that the average rate of change of a function f over [a, b] is equal to the rate of change of f at some number c in (a, b). Geometricall, the line joining the end points (a, f(a)) and (b, f(b)) is parallel to a tangent line at some point (c, f(c)) where a<c<b, see Figure. Eample Illustrating the Mean Value Theorem Find the values of c that satisf the Mean Value Theorem for the given function in the indicated interval. f() = +, [, 5]

15 .. THE MEAN VALUE THEOREM 5 Solution If c satisfies the Mean Value Theorem, then f (c) = f(5) f( ) 5+ = 6 ( ) 8 = 96 8 f (c) =. 5,6 5 5 Net, solve the equation f () =. = 5 = 0 ( +5)( ) = 0 = 5, Note, c must lie in the open interval (, 5). Thus, the values of c satisfing the Mean Value Theorem are c =andc = 5/, as shown in Figure 5., Figure 5 The tangent lines at the points where = 5/ and = are parallel to the secant line that contains the endpoints (, ) and (5, 6). Tr This Find the values of c that satisf the Mean Value Theorem for the function in the interval [0, ]. h() = + Note, the values of c in the Mean Value Theorem belong to the open interval (a, b). The theorem does not indicate how man such c s there are. The theorem simpl states the eistence of such a number. Eample An Application of the Mean Value Theorem A police officer clocks the speed of a certain car at 65 mph. Five minutes later another police officer finds the same car going at 60 mph. If the police officers are eight miles apart, eplain wh the car was speeding at 96 mph at some point between the two police officers. Solution Let t be a fraction of an hour after the first officer clocked the car s speed. Denote b s(t) the corresponding distance in miles between the first officer and the car at time t. Note, five minutes is equivalent to / of an hour. Then s(0) = 0 and s =8. Suppose s(t) is a differentiable function of t. Then the Mean Value Theorem applies to s(t) on[0, ]. Thus, there eists a number c in (0, ) such that s (c) = s(/) s(0) / 0 = 8 0 / =96mph. Hence, at some point between the two police officers the car s speed was eactl 96 mph.

16 5 CHAPTER. APPLICATIONS OF DERIVATIVES Tr This Johnn made a 00-mile trip in hours. Show that at some point in the trip Johnn was driving at 75 mph.. Check-It Out. Find the values of c that satisf Rolle s Theorem for f() = +in[, ].. Find the values of c that satisf the Mean Value Theorem for f() = in [0, ].. Use Rolle s Theorem to eplain wh f() = cos + has onl one -intercept. True or False. If false, eplain or show an eample that shows it is false.. For the function f() = 6 +9 in[0, ], the value of c that satisfies Rolle s Theorem is c =.. Rolle s Theorem applies to f() =( ) / in the interval [0, ].. For the function h(t) = t t in [, ], the value of c that satisfies the Mean Value Theorem is c =/.. If = f() is a differentiable function and f(a) =f(b), then there is eactl one c in (a, b) satisfing f (c) =0. 5. Suppose a differentiable function = f() has -intercepts (a, 0) and (b, 0), a = b. Then = f() has a critical number c in (a, b). 6. A police officer spotted a car traveling at 55 mph. After minutes, a state trooper clocked the same car at 60 mph. If the officer and trooper are 5 miles apart, then the car s speed at some point in between was eactl 00 mph. 7. The function f() =sin(π) + satisfies Rolle s Theorem in the interval [0, /]. 8. For f() =/ in [, ], there eists a number c in (, ) such that f (c) = f() f( ). ( ) 9. For f() =/ in [, ], there eists a number c in (, ) such that f (c) = f() f( ). ( ) 0. If a car can accelerate from 60 mph to 70 mph in minute, then the car s acceleration at some instant is eactl 600 mph per hour. Eercises for Section. In Eercises -8, determine if Rolle s theorem applies to the function in the indicated interval. If it does, find the values of c that satisf Rolle s Theorem.. f() =+6,[0, 6]. f() =, [0, /]. s(t) =t t +,[0, ]. g(t) =t 6t +t, [, ] 5. T (θ) =sinθ, [π/6, 5π/6] 6. T (θ) =tanθ +cotθ, [π/6,π/] 7. p(w) =sinw,[, ] 8. h(α) =cos[(α α +5)π], [, ]

17 .. THE MEAN VALUE THEOREM 5 In Eercises 9-6, find the values of c that satisf the Mean Value Theorem. 9. g(t) =t t + t, [0, ] 0. f() = +,[, 0]. R(s) =s s +s, [, ]. C(w) =w(w ) + w, [0, ]. f() = +,[, ]. f() =,[0, ] + 5. A(h) =h h,[0, ] 6. p(t) =t + t,[0, ] Applications of the Mean Value Theorem 7. Let = f() be a function such that f (a) = for an a. Prove that the equation f(w) =w has at most one solution w. 8. Let = g(t) be a function such that for some nonzero constant k we have g (t) = k for an t. Prove g() =k has at most one solution. 9. Arithmetic Mean Verif that the value of c that satisfies the Mean Value Theorem for f() = on the interval [a, b] isc =(a + b)/. 0. Geometric Mean Let a, b > 0 be positive numbers. Show that the value of c that satisfies the Mean Value Theorem for r() =/ on the interval [a, b] isc = ab.. Let f be a differentiable function such that f() = and 0 f () forall. Find a maimum possible value for f(6).. Let p be a differentiable function such that p(0) = and p () for all. Find a minimum possible value for p().. If a = b, showthat sin b sin a b a.. If =, show that tan tan. 5. Let = f() be differentiable function that satisfies f() = and f() =. Show that there eists a number c in the open interval (, ) such that the tangent line to the graph of f at the point (c, f(c)) passes through the origin. 6. Show that = A + has three distinct zeros if A>, and has eactl one zero if A<. 7. Driving in an interstate Jimmie went from one eit to another eit in 80 seconds. If the eits are miles apart, eplain wh Jimmie s speed was eactl 90 mph at some point between the two eits. 8. Let = f() becontinuouson[a, b] and differentiable on (a, b). If f () = 0 for all in (a, b), then f(s) =f(t) for all s, t in [a, b].

18 5 CHAPTER. APPLICATIONS OF DERIVATIVES. Increasing and Decreasing Functions Increasing and Decreasing Functions The First Derivative Test Increasing and Decreasing Functions In this section, we will use the derivative to identif the relative etrema of a differentiable function. First, we need a few preliminaries. Definition Increasing, Decreasing, and Constant Functions Let f be a function on an interval I.. f is increasing on I if f( ) <f( ) whenever < and, belong to I. a b Figure a The function f is increasing on the open interval (a, b).. f is decreasing on I if f( ) >f( ) whenever < and, belong to I.. f is constant on I if f( )=f( ) for all, in I. Geometricall, a function f is increasing on an interval I if its graph is rising as moves to the right in I. In Figure a, f is increasing on an open interval (a, b). Likewise, f is decreasing on I if its graph is falling as moves to the right in I. In Figure b, f is decreasing on (b, c). The derivative can tell us if a differentiable function is increasing or decreasing. In Figure c, we see f is increasing on (a, b), the tangent lines have positive slopes for in (a, b), and consequentl f () > 0. Similarl, f is decreasing on (b, c), the tangent lines have negative slopes, and f () < 0 for in (b, c). The net theorem summarizes these results. b c Figure b The function f is decreasing on (b, c). a b c Figure c A function is increasing, decreasing, or constant according to whether its derivative is positive, negative, or zero, respectivel. d Theorem.5 Increasing and Decreasing Test Suppose f is a continuous function on [a, b], and differentiable on (a, b).. If f () > 0 for all in (a, b), then f is increasing on [a, b].. If f () < 0 for all in (a, b), then f is decreasing on [a, b]. If f () = 0 for all in (a, b), then f is constant on [a, b].

19 .. INCREASING AND DECREASING FUNCTIONS 55 Proof Suppose f () > 0 for all in (a, b). If a < b, thenbthemeanvalue Theorem there eists a number c such that <c< and f (c) = f() f(). Since f (c) and are positive, f( ) f( )isalsopositive.thus, f( ) >f( ) and consequentl f is increasing on [a, b]. The remaining two cases are proved similarl, see Eercises 59 and 6 at the end of the section. The above theorem implies that we have to solve inequalities such as f () > or f () < 0 to determine where a function is increasing or decreasing. In the net eample, we use a standard method to solve such inequalities. The method involves finding the critical numbers and determining the signs of f () off the critical numbers. This method is described in more details after Tr This. Eample Identif Where f is Increasing or Decreasing Find the open intervals where f() = is increasing or decreasing. Solution Since the derivative is given b f () = +6=( )( ) the critical numbers are =,. If we delete the critical numbers from the number line R, we obtain a union of open intervals: R, =,,,. Choose a number in I =,,sa =. Determine the sign of f ( ); in fact f ( ) > 0. Then f () > 0 for all in I. Thus, f is increasing on I b Theorem.5. Repeat this procedure and choose test values and from I =, and I =,., 07, 5 Then determine the sign of f ( i)whichisthesignoff on the test open interval I i.the following table summarizes the results. Test Open Interval,,, Test Value i = = = Sign of f ( i) f ( ) > 0 f () < 0 f () > 0 Appl Theorem.5 Increasing Decreasing Increasing Hence, f is increasing on (, ), increasing on (, ), and decreasing on (, ). The graph of f is shown in Figure. Figure f is increasing on (, )and(, ), and decreasing on (, ). The sign of f ( ) will be the same as the sign of f () foranothertestvalue in I.For if f ( )andf ( )haveoppositesignswith and in I,thenbtheIntermediuateValue Theorem there eist a number 0 in I for which f ( 0 )=0;butthisisacontradictionsinceI does not contain an critical number of f.

20 56 CHAPTER. APPLICATIONS OF DERIVATIVES Tr This Find the open intervals on which the function is increasing or decreasing. a) s(t) = 6t +96t +0 b) p() = 6 + The guidelines below are for students to help them develop a strateg for finding open intervals on which a function is increasing or decreasing. Guidelines for finding open intervals where a function is increasing or decreasing Let f be a differentiable function on an open interval (a, b).. Find the critical numbers i in (a, b) and arrange in ascending order: a< < < < n <b.. Form the open intervals I =(a, ), I =(, ),..., I n+ =( n,b). Select an test value i in I i.. Appl Theorem.5: If f ( i) > 0thenf is increasing on I i. If f ( i) < 0thenf is decreasing on I i. In the above, we allow for a = or b =. The First Derivative Test The sign of the slope of a tangent line is an indicator of whether a function is locall increasing or decreasing. This observation is the basis of the First Derivative Test. However, we need to introduce some terminologies. For our purposes, think of the function g in the net theorem as the first or second derivative of a function f. Definition 5 AFunctionChangingitsSignataNumber Let g be a function, and let c be a number., 07. g changes from positive to negative at c if there are open intervals (b, c) and(c, d) such that g is positive on (b, c) andg is negative on (c, d)., 5. g changes from negative to positive at c if there are open intervals (b, c) and(c, d) such that g is negative on (b, c) andg is positive on (c, d).. The sign of g stas constant about c if there are open intervals (b, c) and(c, d) such that either g is negative on (b, c) and(c, d), or g is positive on (b, c) and(c, d). Figure The derivative changes sign at a local etrema. For instance, see Figure, the sign of the derivative f changes from positive to negative at =. Also, f changes its sign from negative to positive at =. Moreover, we see that f has relative maima f( ) and relative minima f( ). The First Derivative Test states that sign changes of f are indicators of the local etrema of a function.

21 .. INCREASING AND DECREASING FUNCTIONS 57 Theorem.6 The First Derivative Test Let f be a differentiable function, and let c be a critical number of f.. If f changes from positive to negative at c, thenf(c) is a relative maimum of f.. If f changes from negative to positive at c, thenf(c) is a relative minimum of f.. If the sign of f stas constant about c, thenf(c) is not a relative etremum of f. Proof and Suppose f changes from positive to negative at c. Chooseb and d such that f () > 0 for all in (b, c) f () < 0 for all in (c, d). Then f is increasing on (b, c) andf is decreasing on (c, d) b Theorem.5. Since f is continuous at c, f is increasing on (b, c] andf is decreasing on [c, d). Thus, f(c) isthe maimum value of f on the (b, d). Hence, f(c) is a relative maimum value of f. The proof of the second part is similar. We leave the proof as an eercise for the student to prove, see Eercise 6. If the sign of f stas constant at c, then there eist numbers c,c such that either f is positive on (c,c)and(c, c ), or f is negative on (c,c)and(c, c ). Appling Theorem.5 and the continuit of f at c, weobtainthatf is either increasing or decreasing on (c,c ). Hence, f(c) is neither a relative maimum nor relative minimum value of f. Eample Appling the First Derivative Test Find the relative etrema of g() = / /. Solution The derivative of g is g () = / / = / /. Then = 0 is a critical number of g since g (0) is undefined and g(0) is defined. To find another critical number, suppose g () =0. Then / = 0 = = 8. Thus, the critical numbers are =0,. Consider the open intervals 8 I =(, 0), I = 0,, I = 8 8, following the guidelines after Eample. Findtestvalues i in I i,andevaluateg ( i)as follows: Test Open Interval I =(, 0) I = 0, I 8 =, 8 Test Value i = =/9 = Sign of g ( i) g ( ) > 0 g 9 > 0 g () < 0 Positive Positive Negative Note, g changes from positive to negative at. Appling the First Derivative Test, g has 8 a relative maimum value at =. Since the sign of 8 g stas constant at =0,g does not have a relative etreme value at = 0. Hence, the relative maimum value of g is / / g = = =. The graph of g is shown in Figure. Rel. ma. 8, Figure g has a relative maima at = since 8 g changes from positive to negative at =. 8

22 58 CHAPTER. APPLICATIONS OF DERIVATIVES Tr This Find the relative etrema of g(t) = t/ 5 t5/. Eample Using the First Derivative Test Find the relative etrema of h() = sin +cos in the open interval (0, π), see Figure 5. Solution The derivative is h () = cos sin. Π Π To find the critical numbers, we find the zeros of the derivative. cos sin = 0 cos = sin = tan Figure 5 The relative etrema of h are = ±. = π, π. Then we find test values i in the open intervals I i where I = 0, π π, I =, π π, I =, π. following the guidelines after Eample. Test Open Interval 0, π π, π π, π Test Value i = π 6 = π = π Sign of h ( i) h ( )= h ( )= h ( )= Positive Negative Positive Hence, b the First Derivative Test, the relative maimum value of h is π h = sin π +cosπ = and the relative minimum value of h is π h = sin π +cosπ =. Tr This Find the relative etrema of k() =cos sin where 0 <<π.

23 .. INCREASING AND DECREASING FUNCTIONS 59 Eample Using the First Derivative Test Find the relative etrema of m() = +. Solution Evaluate m () as follows: m () = = ( 6) = ( )( +)( +) Factoring The critical numbers of m are = ±. Note, = 0 is not a critical number for m(0) is undefined. Form the open intervals using the critical numbers: I =(, ), I =(, 0), I =(0, ), I =(, ). Select certain test values i in I i and evaluate m ( i). Test Open Interval (, ) (, 0) (0, ) (, ) Test Value i = = = = Value of m ( i) m ( ) > 0 m ( ) < 0 m ( ) < 0 h ( ) > 0 Positive Negative Negative Positive We appl the First Derivative Test. Since m changes from positive to negative at =, the relative maimum value of m is m ( ) = ( ) + = = Similarl, m changes from negative to positive at =. Hence, the relative minimum value of m is m () = =. The graph of m is shown in Figure 6.,6 0 Figure 6 Relative maima and relative minima,6 Tr This Find the relative etrema of R() = Eample 5 The Longest Rod Through a Corner Find the length of the longest rod that can be carried horizontall from one hall onto the other hall, as in Figure 7. Suppose the halls are feet wide and the meet at a right angle. Solution The length of the longest rod that can be carried horizontall through the halls is the minimum value of L + L, as seen Figure 7. Using right triangle trigonometr, we obtain L = L + L =secα +cscα, 0 <α< π. The derivative of L with respect to α is L (α) = secα tan α cscα cot α sin α = cos α cos α sin α = sin α cos α sin. α cos α L Α L Α Figure 7 L and L are the hpotenuse of right triangles with angle α.

24 60 CHAPTER. APPLICATIONS OF DERIVATIVES Then the critical numbers of L in (0, π ) must satisf sin α cos α =0 or tanα =. Thus, the critical number is α = π. Consider the open intervals I = 0, π π, I =, π. The sign of L (α) in the above open intervals are described below. Test Open Interval I = 0, π I = π, π Test Value α i α = π α 6 = π Sign of L (α i) L π 6. L π. Negative Positive L Α Note, L is decreasing on (0,π/) and L is increasing on (π/,π/) b Theorem.5. Thus,theminimumvalueofL is π π π L =sec +csc =8. feet. L Α Hence, the length of the longest possible rod is 8 feet. Figure 8 Find the minimum sum of L + L given the widths of d and d. Tr This 5 As in Eample 5 but suppose the widths of the halls are ard and ards, see Figure 8. Find the length of the longest rod that can be carried horizontall from one hall onto the other hall.. Check-It Out. Find the open intervals on which f() = + 0 is increasing or decreasing.. Find the relative etrema of s(t) =t +t t + 7 b using the First Derivative Test.. Find the relative etrema of =sin()+cos() where 0 <<π.. Find the maimum product of two positive numbers and where + =. True or False. If false, eplain or show an eample that shows it is false.. If f () > 0 for in ( 0, 0), then f is increasing on ( 0, 0).. If f () < 0 for in (, ), then f is decreasing on [, ].. If f() =( ) 5,thenthesignoff () stasconstantabout.. If = f() iscontinuouson(0, 6) and the sign of f () changes from negative to positive at, then f() is a relative minimum of f. 5. If f () > 0 for in (, 0) and f () < 0 for in (0, ), then f(0) is a relative maimum of f. 6. If f () =( ) ( +),thenf is increasing on (, ). 7. If f () = ( ) 5,thenf () > 0 for in (, ). 8. If g(t) = t(t ), then g(/) is a relative minimum of g. 9. The function s(t) =cos(t)+t 5 has a relative maimum at t = π/6. 0. The minimum sum + of two positive numbers and for which =istwo.

25 .. INCREASING AND DECREASING FUNCTIONS 6 Eercises for Section. In Eercises -0, determine the open intervals on which the function is increasing or decreasing.. f() =5( )( +). g() = ( +). =( ) ( ). =( )( +5) 0( +5) 5. p() = 5 6. r() = S(t) = t t +t 6 8. g(t) = t +t 6t = t +6t +t 0. =6t t t +. R() = +. m() = + 9. P () = +. N() = 5. s(θ) =sin (θ), 0 <θ<π 6. t(θ) =cos (θ), 0 <θ<π 7. =sin() cos(), 0 <<π 8. =cos()+sin(), 0 <<π 9. f() =sin() cos(), 0 <<π 0. g() =cos() sin(),0<<π In Eercises -6, determine all the relative etreme values of the function. Appl the First Derivative Test.. f() =( ). g() =( )( +). h() = (t) =(t +) (t ) 5. (t) =(t ) (t +) 6. = f() = M() = + 9. L() = N() = +. k() = 8. f() = / + /. f() = ( ). f() = 8 5. f() = + if if > 6. f() = if + if > 7. f() =( +) / 8. f() = f() = / ( ) / 0. f() = ( ). g(t) =+sin (t), 0 <t<π. g(t) =cos (t), π <t<π. v(θ) =sin(θ)+ cos(θ), 0 <θ<π. A(w) =sin(w)+w π, π <w<π 5. f(α) =cos (α) sin(α), 0 <α<π 6. g(β) =sin (β)+cos(β), 0 <β<π Applications 7. The difference between two numbers is one. Find the minimum product of two such numbers. 8. What is the minimum sum of two positive numbers whose product is one? 9. Find the maimum area of a rectangle that has a perimeter of feet. What if the perimeter is p feet?

26 6 CHAPTER. APPLICATIONS OF DERIVATIVES 50. Find the minimum perimeter of a rectangle that has an area of 6 square inches. What if the area is A square inches? 5. Let f() be the square of the distance between the point (0, ) and a point (, ) on the parabola =. a) Find open intervals on which = f() is increasing or decreasing. 6 b) Which points on the parabola are closest to (0, )? 5. Let D() be the square of the distance between the point (, 0) and a point (, ) on the curve =. a) Find open intervals on which = D() is increasing or decreasing. b) Which point on the graph of = is nearest to (, 0)? 5. A rectangle is bounded b the -and -ais and the line + =, see figure below. Find the dimensions of the rectangle that has the maimum area. 5. A right triangle is bounded b the - and-ais and a line that passes through the point (, ). Find the dimensions of such a triangle that has the minimum area. Figure for No.5, Figure for No.5 Theor and Proofs 55. Find and sketch the graph of a function = p() thatsatisfies a) p() =, p () = 0, p () > 0if>, and p () < 0when< b) p() =, p () = 0, p () > 0if = 56. Sketch the graph of a differentiable function m = v(t) thatsatisfies a) v() = 5, v( ) = 5, v ( ) = v () = 0, v (t) > 0if t <, and v (t) < 0when t > b) v() = v( ) =, v () = v ( ) = 0, v (t) > 0ift<, v (t) < 0 if <t<0, v (t) > 0if0<t<, and v (t) < 0ift> 57. Let f() =( a)( b). Prove = f() is decreasing on the open interval a+b c, a+b+c where c = a ab + b 58. Let p() =( a) ( b), a<b. Prove = p() is increasing on a, a+b and (b, ). 59. Let f be continuous on [a, b] and differentiable on (a, b). Suppose f () < 0 for all in (a, b). Prove f is decreasing on [a, b]. Hint: Mean Value Theorem. 60. Suppose f is continuous on [a, b]. If f () > 0 for in (a, b), prove f is increasing on [a, b]. 6. Let f be continuous on [a, b] and differentiable on (a, b). If f () = 0 for all in (a, b), prove f is a constant function on [a, b]. 6. Let c be a critical number of a differentiable function f. Iff changes from negative to positive at c, prove f(c) is a local minimum of f. Odd Ball Problems 6. Find a rd degree polnomial that has a relative maimum point at (, ) and a relative minimum point at (, ). 6. Find a rd degree polnomial that has a relative maimum point at (, 5) and a relative minimum point at (, ). 65. If 0 < π, prove tan tan (). 66. If 0 π, prove cos cos.

27 .. CONCAVITY OF THE GRAPHS OF FUNCTIONS 6. Concavit of the Graphs of Functions Concavit Points of Inflection The Second Derivative Test Concavit In Section., we discussed how the sign of the derivative f determines whether f is increasing or decreasing. Also, we studied the First Derivative Test which is a criterion for finding the relative etrema of f. Inthissection, wewillseehowthesignofthesecond derivative f determines whether the graph of f is curving upward or curving downward. In addition, we discuss the Second Derivative Test which is another criterion for finding the relative etrema of f. Definition 6 Concavit of a Graph Let f be a differentiable function on an open interval I. The graph of f is concave upward if the derivative f is an increasing function on I. Likewise, the graph of f is concave downward if f is decreasing on I. For simplicit, we sa f is concave upward or downward on an interval I if the graph of f is concave or downward on I, respectivel. Consider the graphs of f and p in Figures a and a, respectivel. Note, f and p are increasing since the slopes of the tangent lines to the graphs of f and p are increasing. B definition, f and p are concave upward. Similarl, we see the graphs of g and q in Figures b and b. Since g and q are decreasing functions, g and q are concave downward. The graph of f is concave upward The graph of g is concave downward Figure a. f is increasing. Figure b. g is decreasing. The graph of p is concave upward The graph of q is concave downward Figure a. p is increasing. Figure b. q is decreasing. Moreover, we claim f is concave upward if and onl if the graph of f lies above all its tangent lines. Also, f is concave downward if and onl if the graph of f lies below all its tangent lines. For a proof, see Theorem A.6, page.

28 6 CHAPTER. APPLICATIONS OF DERIVATIVES The net theorem is a straightforward wa to determine concavit. The proof follows directl from Definition 6 and Theorem.5 in Section.. f concave downward concave upward Theorem.7 Concavit Test Let f be twice differentiable on an open interval I.. If f () > 0 for all in I, then the graph of f is concave upward on I.. If f () < 0 for all in I, the graph of f is concave downward on I. In Figure, we see the graphs of f() = + and f () =6. Sincef () > 0 on (0, ), f is concave upward on (0, ) b Theorem.7. Likewise, since f () < 0on (, 0), f is concave downward on (, 0). f ''6 Eample Determining Concavit f '' is positive Find the open intervals where the graph of f '' is negative is concave upward or concave downward. f() = + Solution We obtain the derivatives as follows: Figure The concavit of the graph of f is determined b the signs of f. f () = 6 f () = = ( ). Then =0, are the solutions to f () =0. Wetestthesignsoff () ontheopen intervals I 0 =(, 0), I =(0, ), and I =(, ). Test open interval (, 0) (0, ) (, ) Test value i = = = Value of f ( i) f ( )= f ( )= f ( )= Sign of f on I i Positive Negative Positive Appling Theorem.7, f is concave upward on (, 0) and (, ), and concave downward on (0, ). The graph of f is shown in Figure. Tr This Figure A sketch of the graph of f() = + Find the open intervals on which the graph of = + is concave upward or concave downward. Eample Determining Concavit on Open Intervals Discuss the concavit of the graph of f() = +. Solution Appling the General Power Rule,we obtain f () = ( +) () = 8 ( +).

29 .. CONCAVITY OF THE GRAPHS OF FUNCTIONS 65 Then b the Quotient Rule, we find f () = ( +) ( 8) + 8(( +)()) ( +) = 8( +) ( +)+ ( +) Factor out 8( +) = 8( +) ( +) f () = ( )( +) ( +). Simplif Note, f () = 0 precisel when = ±. We test the signs of f () on(, ), (, ), and (, ) as follows: Test open interval (, ) (, ) (, ) Test value i 0 = 0 =0 0 = Value of f ( i) f ( ) 0. f ( )= / f ( ) 0. Sign of f on test open interval Positive Negative Positive B Theorem.7, f is concave upward (, ) and (, ), and concave downward on (, ). The graph of f is shown in Figure 5.,.5,.5 Figure 5. The graph of f is concave upward on (, ) and (, ), and concave downward on (, ). Tr This Find the open intervals on which the graph of = downward. + is concave upward or concave Eample Determining Concavit on Open Intervals Discuss the concavit of the graph of g() = Solution Then The first derivative is, as shown in Figure 6. g () = ( ) () ( ) Quotient rule = + ( ). Simplif g () = ( ) () ( + )[( )()] ( ) Quotient rule = ( ) ( ) ( + ) ( ) Factor out ( ) = ( +) ( ). Simplif Then =0istheonlsolutiontog () =0,Note,g () isundefinedwhen = ±. Net, we obtain the sign of g () on the open intervals determined b =0, ±. Test open interval (, ) (, 0) (0, ) (, ) Test value i = = 0.5 =0.5 = Value of g ( i) g ( ).0 g ( ) 7.7 g ( ) 7.7 g ( ).0 Sign of g () Negative Positive Negative Positive Figure 6 Appl the Concavit Test to determine the concavit of the graph of g.

30 66 CHAPTER. APPLICATIONS OF DERIVATIVES Hence, b Theorem.7, g is concave upward on (, 0) and (, ), and concave downward on (, ) and (0, ). Tr This Find the open intervals on which the graph of is concave upward or concave downward. = Points of Inflection A point on a graph where concavit changes is called a point of inflection. In Figure 7, we see three points of inflection P, Q, andr. Observe, the concavit to the nearb left side is different from the concavit to the nearb right side of a point of inflection. P Q R Figure 7 Points of Inflection at P, Q, and R. Definition 7 Point of Inflection Let f be continuous at c. Then(c, f(c)) is a point of inflection of the graph of f if there are open intervals (a, c) and(c, b) such that either a) f is concave upward on (a, c) and concave downward on (c, b), or b) f is concave downward on (a, c) and concave upward on (c, b). For brevit, we ma write inflection point of f. Apointofinflectionoff is not necessaril a critical point of f. However, an inflection point is a critical point of the derivative f as the net theorem shows.

31 .. CONCAVITY OF THE GRAPHS OF FUNCTIONS 67 Theorem.8 Necessar Condition for a Point of Inflection If (c, f(c)) is a point of inflection of f, thenf (c) =0orf (c) doesnoteist. Proof If f (c) does not eist, then the theorem is true. Suppose f (c) eists.thenwe appl the continuit of f at c and Definition 7. That is, there eist intervals (a, c] and [c, b) suchthat a) f is increasing on (a, c] and decreasing on [c, b), or b) f is decreasing on (a, c] and increasing on [c, b). Thus, f (c) is a relative etreme value of f. Hence, f (c) = 0 b Theorem. in page. Eample Finding Points of Inflection Find the points of inflection and discuss the concavit of the graph of f() = +. Solution The first two derivatives of f are given b f () = +6 f () = + = ( +). Then f () =0onlif =0or =. We determine the signs of f on I =(, ), I =(, 0), I =(0, ). Test open interval I =(, ) I =(, 0) I =(0, ) Test value i = = / = Value of f ( i) f ( ) = f ( /) = f () = Sign of f Positive Negative Positive Appling Theorem.7, f is concave upward on (, ) and (0, ), and concave downward on (, 0). Hence, b Definition 7 the points of inflection of f are (0,f( )) = (0, ) and (,f( )) = (, ). See the graph of f in Figure 8. Figure 8 A sketch of the graph of f() = +. Tr This Find the points of inflection and discuss the concavit of =. Eample 5 Finding Points of Inflection Discuss the concavit and find the points of inflection of f() = 5/ 8/ +.

32 68 CHAPTER. APPLICATIONS OF DERIVATIVES Solution The first two derivatives are f () = 0 / 8 5/ 0,, f () = 0 9 / 0 9 / = 0( ) 9. Figure 9 At a point of inflection (c, f(c)), either f (c) =0 or f (c) isundefined. Note, f () =0if =,andf () isundefinedif = 0. Net, determine the signs of f () on the open intervals I =(, 0), I =(0, ), I =(, ). The results are as follows: Test open interval I =(, 0) I =(0, ) I =(, ) Test value i = =0.5 = Value of f ( i) f ( ) 8.8 f ( ).8 f ( ).5 Sign of f Negative Positive Negative B Theorem.7, f is concave upward on (0, ), and concave downward on (, 0) and (, ). Since the direction of concavit changes at =0and =,thepointsofinflection of f are (0,f(0)) = (0, ) and (,f()) = (, ). See the graph of f in Figure 9. Concave Upward Tr This 5 Find the points of inflection of the graph of = 5/ 5 /. Include a discussion of the concavit of the graph. c Figure 0a If f (c) =0andf (c) > 0, then f(c) is a relative minimum value of f. Concave Downward Observe, if f (c) =0orf (c) isundefined,wecannotconcludethat(c, f(c)) is a point of inflection. For eample, let g() = and h() = /. Since g () = and h () = are increasing functions, the graphs of g and h are concave upward on (, ) b Definition 6. Thus, g and h do not have points of inflections even though g (0) = 0 and h (0) is undefined for g () = and h () = 9. The Second Derivative Test The above named-test uses the second derivative to identif certain relative etreme values of a function. We describe a geometric interpretation of the test. If f (c) =0andf is concave upward on an open interval I containing c as in Figure 0a, then f(c) is a relative minima of f. Similarl, if f is concave downward on I, thenf(c) is a relative maima of f, see Figure 0b. c Theorem.9 The Second Derivative Test Figure 0b If f (c) =0andf (c) < 0, then f(c) is a relative maimum value of f. Let f be a function and suppose f (c) =0. a) If f (c) > 0, then f(c) is a relative minimum value of f. b) If f (c) < 0, then f(c) is a relative maimum value of f.

33 .. CONCAVITY OF THE GRAPHS OF FUNCTIONS 69 Proof Suppose f (c) > 0. f f () f (c) f () (c) =lim =lim c c c c > 0 Then there eists an open interval (a, b) containingc such that f () > 0whenever is in (a, b) and = c. c Thus, f () < 0 for all in (a, c), and f () > 0 for all in (c, b). Note, f is continuous at c since f (c) eists.thenf is decreasing on (a, c] and decreasing on [c, b). Hence, f(c) is a relative minimum of f b the First Derivative Test in page 5. The proof of part b) is left as an eercise. Note, the Second Derivative Test is inconclusive if f (c) = 0. In such a case, an alternative is to appl the First Derivative Test to identif an relative etrema. Eample 6 Appling the Second Derivative Test Find the relative etrema of f() = 0 5. Solution First, we determine the derivative of f. f () = = ( ) () Then the solutions to f () = 0 are =0, ±. The second derivative of f is f () = d d = 6 f () = ( ). We appl the Second Derivative Test with c =0, ±. c 0 Sign of f (c) f ( ) > 0 f (0) = 0 f () < 0 Conclusion Relative minimum Test fails Relative maimum Hence, a local maimum value of f is and a local minimum value is f() = 8 () =. 0 f( ) = 8 ( ) =. 0 Note, the Second Derivative Test fails when = 0. However, using () weclaimthatthe sign of f () stasconstantat =0. Thus,f(0) is not a relative etrema of f b the First Derivative Test, see page 57. The graph of f is given in Figure Figure The point (,.) is a relative maimum, and (,.) is a relative minimum of f. Tr This 6 Appl the Second Derivative Test to find the relative etrema of s(t) =8t t +5.

34 70 CHAPTER. APPLICATIONS OF DERIVATIVES Eample 7 Appling the Second Derivative Test Find the relative etrema of g() =sin +cos(), 0 <<π. Solution The derivative is Π 6 Π 5 Π 6 Π Π g () = cos sin() = cos sin cos Since sin =sin cos = cos( sin). Figure The graph of g() =sin +cos() in 0 <<π. Then the critical numbers in (0, π) satisf either Thus, the critical numbers are The second derivative is cos =0or sin =. = π, π, π 6, 5π 6. g () = d [ cos sin()] d = sin cos(). We appl the Second Derivative Test as follows: π 6 c Value of g (c) g ( π )= 6 g ( π )= g ( 5π )= 6 g ( π )=6 Conclusion Rel ma Rel min Rel ma Rel min Hence, the relative maimum value of g() is π π g = 5π 6 = g 6. and the relative minimum values are π π g =andg =. 5π 6 π The graph of g is in Figure. Tr This 7 Use the Second Derivative Test to find the relative etrema of f() =cos() cos, π << 5π.

35 .. CONCAVITY OF THE GRAPHS OF FUNCTIONS 7. Check-It Out. Find the open intervals on which the graph of f() = + + is concave upward or concave downward.. Find the points of inflection of the graph of p() = Use the Second Derivative Test to find the relative etrema of s(t) = t t. True or False. If false, eplain or show an eample that shows it is false.. The graph of p() = is concave upward on (, ).. The graph of T () =sin is concave downward on (0,π).. A point of inflection of the graph of = is (0, 0).. A point of inflection of the graph of = is (0, 0). 5. If (c, f(c)) is a point of inflection of = f(), then f (c) =0orf (c) isundefined. 6. If g (c) =0,then(c, g(c)) is a point of inflection of = g(). 7. If R (c) =0andR (c) > 0, then R(c) is relative minimum value of = R(). 8. If = M() is concave upward on (a, b) and concave downward on (b, c), then (b, M(b)) is a point of inflection of the graph of M. 9. If = p() isapolnomialsuchthatp () < 0on(a, b) andp () > 0on(b, c), then (b, p(b)) is a point of inflection of = p(). 0. It is possible that a point of inflection of = f() is also a relative etrema of f. Eercises for Section. In Eercises -, find the open intervals where the graph is concave upward or concave downward. Also, identif the inflection points.. =. = Figure for No. Figure for No.

36 7 CHAPTER. APPLICATIONS OF DERIVATIVES. p(t) =t t +t +. =sint, 0 t π Π Π Π Π t t Figure for No. Figure for No. In Eercises 5-, find the open intervals where the graph of the function is concave upward or downward. Then determine the points of inflection. 5. p() = p() = g() = / / 8. h() = / + / 9. f() = 5/ 5 / 0. g() = 7/ /. =+cos (), 0 <<π. = +sin (), 0 <<π. =sin()+cos(), 0 <<π. =cos() sin(), 0 <<π 5. f() = P (t) = t t m() =. s(t) = + 6. f() = R(w) = 0. n() = 96w w + + t w. T (w) = t w In Eercises -8, find the relative etrema b using the Second Derivative Test when applicable.. f() = +5. f() = + 5. g() = f() = C(w) =(w +6)(00 w ) 8. A(w) =(w +5)(69 w ) 9. g() =sin(), 0 π 0. f() =cos()+, π 6 5π 6. L(α) =tan(α) α, π <α< π. R(β) =β +cot(β), 0 <β<π. = 8t t f() = +. Q(p) = 0p p g() = 9 7. K() = / / 8. M(c) =8c / c /

37 .. CONCAVITY OF THE GRAPHS OF FUNCTIONS 7 In Eercises 9-, sketch a graph of a function with the given properties. There are several possible answers. Then find a formula = f() for such a function. 9. f() =, f () = 0, and f () > 0 for << 0. f() =, f () = 0, and f () < 0 for <<. f(0) =, f (0) = 0, f () < 0 for <, and f () > 0 for >. f() = 5, f( ) = 5, f (±) = 0, f () > 0 for <0, and f () < 0 for >0 Applications. Find the points on the graph of = that are closest to the origin.. Find the shortest distance between the origin and the graph of =. 5. Find the shortest distance between the parabola = and the point (, 0). 6. Find the point on the graph of =sin that is nearest to the point ( π, 0). 7. A rectangle is bounded b the and -aes and the graph of = in the first quadrant, see figure below. Find the smallest possible perimeter of such a rectangle. Figure for No. 7 Figure for No Find the minimum length of the line segment that is tangent to the graph of = and whose endpoints lie on the positive - and -aes, see figure above. 9. The diagonals of a rectangle contain the origin and two vertices of the rectangle lie on the branches of the graph of =. Find the minimum perimeter of such a rectangle. 50. Find the points on the graph of = that are nearest the origin.

38 7 CHAPTER. APPLICATIONS OF DERIVATIVES Prove, or disprove b finding a counter eample 5. If f () is a twice differentiable concave upward function, prove or disprove that f () > 0 for all. 5. If f() is a concave upward function, prove or disprove that =(f()) is concave upward. 5. If f and f are positive functions on the interval (0, ), prove that = f( ) is concave upward on (0, ). Is the converse true? 5. Find conditions on f and g so that = f(g()) is concave upward.

39 .5. LIMITS AT INFINITY 75.5 Limits at Infinit Limits at Infinit Horizontal Asmptotes Infinite Limits at Infinit Limits at Infinit In the section, we stud the end behavior of a function. Specificall, we analze the limit of a function f() as increases or decreases without bound. To illustrate, we list several values of a function for large : + f() = f() The above values suggest that f() approaches as increases without bound, and we write lim f() =. Limit at Positive Infinit Similarl, as decreases without bound the table of values indicate f() approaches, and we write lim f() =. Limit at Negative Infinit In Figure, we see the graph of f is getting closer to the horizontal line =as. Also, the graph of f is getting nearer to the line = as. For clarit, we point out that does not represent a number. Eample Finding a Limit at Infinit Complete the table below where f() = f() Figure As,wefind f() approaches. As,wesee f() approaches. Then estimate the limits at and : a) lim + b) lim + Solution Using a calculator, we find f() Observe, f() approaches as or.thenwefind lim f() = and lim f() =. Tr This Estimate the limits a) lim + b) lim + b constructing a table of values for f() = +.

40 76 CHAPTER. APPLICATIONS OF DERIVATIVES LΕ L LΕ f approaches L approaches For eactness, we will need the definition of a limit at infinit. M Figure a For >M, the graph of = f() lies between the lines = L + and = L. Definition 8 Limits at Infinit Let L be a real number and let f be a function.. The statement lim f() =L means that for each ε>0 there eists a positive number M>0suchthat f() L <εwhenever >M.. The statement lim f() = L means that for each ε > 0 there eists a negative number N<0suchthat f approaches L f() L <εwhenever <N. approaches N LΕ L LΕ Figure b For <N, the graph of = f() lies between the lines = L + and = L. In words, the definition of a limit at infinit implies that f() iswithinεunits of L for all >M (where M depends on ε). Geometricall, the graph of = f() for >M lies between the horizontal lines = L + ε and = L ε, as shown in Figure a. Similar statements can be made for a limit at minus infinit (or negative infinit), see Figure b. Not all functions have limits at ±. The graph of =sin oscillates infinitel man times between the lines = and =, see Figure c. Then it is impossible to find apositivenumberm > 0 such that the graph of =sin for >Mlies between the horizontal lines Y = L ±.Thus,theinfinitelimit lim sin does not eist. Eample Proving a Limit at Infinit limsin does not eist. Similarl, Π Π Π Π Prove the following limit b using Definition 8: lim =0 Solution Let f() =, ε>0, and L = 0. Appling the notation in Definition 8, we analze f() L = 0 = <ε The above inequalit is equivalent to ε <. Figure c sin does not eist lim lim sin does not eist Let M = ε.thus, f() L <εwhenever >M. Hence, b Definition 8 we have proved lim =0. Tr This Use Definition 8 to prove lim =0.

41 .5. LIMITS AT INFINITY 77 Several properties of limits etend to limits at infinit. Such properties of limits are discussed in Section.. Forinstance,if limf() and lim g() botheistthen. lim (f()+g()) = lim f()+ lim g(). lim (f() g()) = lim f() lim g(). lim (f()g()) = lim f() lim g() f() lim f(). lim g() = lim g() provided lim g() = 0,and 5. lim (cf()) = c lim f() where c is an real number. The above formulas remain true when isreplaced b. The net theorem is useful for evaluating limits at infinit of rational functions. A proof of the theorem is given in Appendi A., page. Theorem.0 Basic Rules for Limits at Infinit If r>0 is a positive rational number and c is a real number, then c. lim =0,and r. lim c r = 0 provided r is well-defined for all <0. Eample Finding Limits at Infinit Evaluate the limits: a) lim 0 b) lim + 5 Solution a) Appling the limit of a difference, we find lim 0 = lim lim 0 = 0 B Theorem.0 =. b) Likewise, appling the limit of a sum we obtain lim + 5 = lim + lim 5 = + 0 Theorem.0 =. Tr This Evaluate the limits. a) lim 0 0 b) lim c) lim

42 78 CHAPTER. APPLICATIONS OF DERIVATIVES Horizontal Asmptotes When the eist, the limits at infinit are useful in describing the end behavior of the graph of a function. Definition 9 Horizontal Asmptote Aline = L is a horizontal asmptote of the graph of a function = f() ifeither lim f() =L or lim f() =L. Since the limits in Definition 9 involve or, a function ma have at most two distinct horizontal asmptotes. Eample Finding the Horizontal Asmptotes f 5 9 Evaluate the limits 5 a) lim +9 b) lim + Then find the horizontal asmptotes of the graph of f() = Solution Evaluate the limit as follows: 5 lim +9 5 = lim +9 Figure The line = is a horizontal asmptote. Similarl, as we obtain 5 = lim + 9 = 0 +0 = B Theorem.0 5 lim +9 = lim =. Thus, the line = Finall, we have is the horizontal asmptote of the graph of f as shown in Figure. lim + = lim + =0. Tr This Evaluate the limits a) lim 8 b) lim 5 8 Then find the horizontal asmptotes of = 8.

43 .5. LIMITS AT INFINITY 79 For the rational functions, the limit at infinit is determined b the leading coefficients in the numerator and denominator, as discussed below. Summar for Finding Limits at Infinit of Rational Functions Let p() and q() be polnomials of degree n and m, respectivel. Suppose a and b are the leading coefficients of p() and q(), respectivel.. If n = m, then. If n<m,then p() lim q() = a b = lim p() q(). p() lim q() =0= lim p() q().. If n>m,thelimitdoesnoteistas or. In an case, a rational function has at most one horizontal asmptote. The proof of the above summar is left as an eercise at the end of this section. For eample, the polnomials in the numerator and denominator in the rational function R() = are of both of degree two. Thus, as or,weobtainthatr() /5, which is the quotient of the leading coefficients. In Figure, the line = 5 r Figure : R() = is the horizontal asmptote of the graph of R. However, for a non-rational function its graph ma possibl have two distinct horizontal asmptotes. Eample 5 AGraphWithTwoHorizontalAsmptotes Evaluate the limits: a) lim +9 b) lim +9

44 80 CHAPTER. APPLICATIONS OF DERIVATIVES Solution a) We rewrite as follows: 9 +9 Appling Theorem.0, we obtain = +9 = + 9 lim +9 = lim b) Since = whenever <0, we rewrite = =. Figure 5 The lines = ± are horizontal asmptotes. +9 Thus, b Theorem.0 we obtain lim +9 = = lim = = =. In particular, the lines = ± are the horizontal asmptotes of the graph of = as seen in Figure Tr This 5 Evaluate each limit. a) lim + b) lim 9 + In evaluating limits at infinit involving trigonometric functions, the Squeeze Theorem as eplained in Eercise 5 will be helpful to know. Eample 6 Limits at Infinit with Trigonometric Functions Evaluate each limit: +cos a) lim +cos b) lim Solution Adding, weobtain For all, we recall If we divide b >0, then However, if <0then cos. +cos +. () +cos +cos +. +.

45 .5. LIMITS AT INFINITY 8 Using the leading coefficients, we find + lim == lim cos and + lim == lim. Hence, b the Squeeze Theorem we obtain +cos lim == lim +cos. 0 As seen in Figure 6, the line = is the horizontal asmptote of = +cos. Tr This 6 Evaluate the following limits. +sin a) lim + +sin b) lim + Figure 6 The graph of = +cos and the horizontal asmptote =. Eample 7 The Limit of a Population of Bacteria A nutrient is needed to help grow a population of bacteria. The number of bacteria is modeled b the function N(t) = t 00 + t where t 0 is in hours. Determine the number of bacteria when t =,t =8andt =7 hr. Then find the limit of N(t) ast. Nt 0000 t Nt t Solution Using a calculator, the number of bacteria are N() = () =0, () 8 7 t N(8) = (8) =, (8) N(7) = (7) =, (7) Figure 7 As t increases without bound, N(t) approaches 5,000. We evaluate the limit as follows: lim t t 00 + t t t = lim t t = =5, 000 Thus, N(t) 5, 000 bacteria as t. In Figure 7, we see a graph of the population versus time.

46 8 CHAPTER. APPLICATIONS OF DERIVATIVES Tr This 7 The population of a bacteria is approimated b N(t) =0, t t Find the number of bacteria after t = hours. Also, find the limit of N(t) ast. Infinite Limits at Infinit If f() increases without bound as, wesaf() hasaninfinitelimitatinfinit and write f() =. Similarl, if f() decreases without bound as,then lim f() has a negative infinite limit at infinit and we write lim below describes the notion of an infinite limit at infinit. f() =. Thedefinition Definition 0 Infinite Limits at Infinit Let = f() be a function.. We write lim f() = if for each positive number M>0there eists a positive number N>0suchthat f() >M whenever >N.. The statement lim f() = means that for each negative number M<0there eists a positive number N>0 such that f() <M whenever >N. We have similar statements for lim f() = and lim f() =. In evaluating infinite limits, the following observations are useful. If L>0 is a positive number, lim f() =L, and lim g() =, then lim f()g() = () and g() lim =. (5) f() The proofs of these observations are left as eercises at the end of the section. Eample 8 Evaluating Infinite Limits at Infinit Evaluate the limits at infinit. a) lim Solution b) lim 0 5 a) Since >when >, it follows that increases without bound as.then lim =.

47 .5. LIMITS AT INFINITY 8 b) Factor the monomial with the highest eponent: 0 5= 5 5. B part a), as. Moreover, we find lim 5 5 = 0 0= f 0 5 Appling observation () in the previous page, we obtain lim 0 5 = lim lim = = Note, the few values of f() = 0 5shownbelow Figure 8 As increases without bound, f() increases without bound f() 56, 5 7, 995, 899, 995 support the fact that f() increases without bound as. behavior of the graph of f is seen in Figure 8. The right-end Tr This 8 Evaluate the limits at infinit. a) lim b) lim + Eample 9 Limits at Infinit of a Rational Function 5 Find the limits at infinit. + a) lim 5 + b) lim + 7 Solution a) The degree of the denominator in + is smaller than the degree of the numerator. In such a case, multipl and divide b / d where d is the degree of the denominator. + / / = +/ /. Appling Theorem.0 and observation (5), we find + lim +/ = lim / 0 Figure 9a The graph of = +. = +0 0 = = In Figure 9a, we see the graph of = +.

48 8 CHAPTER. APPLICATIONS OF DERIVATIVES b) The degree of the denominator in 5 + is smaller than the degree of the numerator. + We follows the strateg applied in part a). Then we multipl and divide b /. Then lim / / = +/ +/ / = lim +/ Figure 9b The graph of = = = = The graph of = 5 + is shown in Figure 9b. + Tr This 9 Evaluate the limits at infinit. a) lim + b) lim +.5 Check-It Out. Evaluate the limits. 00 a) lim + c) lim + b) lim 6 + d) lim. Find the horizontal asmptote of f() =.. Find a positive number N>0suchthat < whenever >N. True or False. If false, revise the statement to make it true. f(). If lim f() = lim g() =, then lim g() =.. If lim f() = lim g() =, then lim f()+ lim g() =.. If lim. lim f() =0. and lim sin does not eist g() =, then lim 5. lim sin cos = 6. lim = (f()g()) = lim ( ) = 8. lim + + = 9. The graph of a function ma have two horizontal asmptotes. 0. The graph of f() = + has no horizontal asmptote. +

49 Eercises for Section.5.5. LIMITS AT INFINITY 85 In Eercises -8, use the given graph to evaluate the limit. a) lim f() b) lim f(). f() = +. f() = + +. f() = +. f() = For No. For No. For No. For No f() = 6. f() = + 7. f() = sin 8. f() = +cos For No. 5 For No. 6 For No. 7 For No. 8 In Eercises 9-, evaluate the limit. If applicable, indicate if the limit is or. 9. lim + 0. lim 6. lim +7. lim lim + 8. lim +. lim lim + 7. lim +sin 0. lim cos. lim lim lim 5. lim ( ) cos π 5. lim 8. lim +cos. lim sin lim lim + 0. lim lim sin 6. lim sin(π) 9. lim. lim sin

50 86 CHAPTER. APPLICATIONS OF DERIVATIVES cos. lim sin. lim π Encounters with the indeterminate form. In Eercises 5-0, evaluate the infinite limits. 5. lim + 8. lim lim lim +8 In Eercises -, determine the horizontal asmptotes of the function. 7. lim + 0. lim +. f() = h(t) = +. s(t) = t t + In Eercises 5-8, sketch the graph of a function that satisfies the given statements.. R() =+ sin 5. f(0) = 0, lim f() =, lim f() = f() =, lim f() =0 6. f(0) =, lim 7. f(0) = 0, lim f() =π/, lim 8. f(0) =, f() =, f( ) = 0, lim Applications f() = π/ f() =, lim f() = 9. During the spring season the proportion of adults with sinus problems due to pollen is P (t) = t t + where t is the number of weeks past March, the first da of spring. Determine the limit of P (t) ast. What is a practical interpretation of the limit? 50. The probabilit that a certain injur will heal in hours is given b P () = where is the number + milligrams of an ointment that should be administered right after the injur. Evaluate lim P (). 5. A compan has to spend $0,000 to set-up a manufacturing equipment. After the set-up, it will cost the compan $5 to manufacture each item. a) Find the total cost C() of manufacturing items with the set-up cost included. b) Find the limit at infinit of the average cost per item, i.e., the limit of C() as. 5. In a given da, the amount of nitrogen dioide in the air is modeled b A(t) = 50t +5( ) t + where t is the number of hours past midnight, and A(t) is measured using the pollutant standard inde (PSI ). a) Find the amount of nitrogen dioide in the air at AM. b) Find the limit of A(t) ast. Theor and Writing Proofs 5. Let p() andq() be nonzero polnomials of degree n and m, respectivel. Suppose a and b are the leading coefficients of p() andq(), respectivel. p() a) If n = m, prove lim q() = a b = lim p() q(). p() b) If n<m,show lim q() =0= lim p() q(). 5. Suppose h() f() g() for all in (a, ), and lim h() = lim g() =L. Prove lim f() =L. 55. Let L be a nonzero real number, lim f() =L, and lim g() =. Prove lim f()g() =sign(l) where sign(l) denotesthesignofl.

51 .6. SUMMARY OF CURVE SKETCHING 87.6 Summar of Curve Sketching Curve Sketching Techniques We use the following features to sketch a detailed graph of a function. For a review of the features, we list the pertinent sections in parentheses. Domain and range Smmetr -intercepts and -intercepts Continuit (Section.) Vertical and horizontal asmptotes (Sections.5 and.5) Differentiabilit (Section.) Relative etrema (Section.) Concavit (Section.) Points of inflection (Section.) Limits at infinit (Section.5) A graphing utilit is helpful in obtaining different parts or sections of the graph of a function. For eample, the viewing windows in Figure a and b show certain parts of the graph of f() = 6. However, the second viewing window shows a holistic view of the graph. The guidelines listed below should help the student determine important features of the graph. Consequentl help her or him to choose a good viewing window. Not ever feature is relevant to ever function. For instance, a graph ma have no asmptotes and the range ma not be determined before the relative etrema are identified. 0 f Figure a 5 f Guidelines For Sketching the Graph of a Function. Determine the domain, and if possible the range.. Find the intercepts, smmetr, and horizontal and vertical asmptotes.. Find the -values for which f () andf () are either zero or undefined.. Determine the relative etrema and points of inflection from step. Figure b Viewing windows for the graph of f() = 6. Eample Sketching the Graph of a Polnomial Function Sketch the graph of p() = Solution Then we find The first three features of the graph follow directl. lim Domain: (, ) -intercept: (0, ) Infinite limits: p() =, lim p() = First derivative: p () = +6 =( ) and b the product rule we obtain Second derivative: p () = ( ) + ( ) = ( )(( ) + ) = ( )( ) 0, 0, Figure The graph of,0 p() =

52 88 CHAPTER. APPLICATIONS OF DERIVATIVES Then the critical numbers and the zeros of p () determine the following test open intervals: Critical numbers: =0, Test open intervals: (, 0), (0, ), (, ), (, ) We list the signs of p () andp () on the test open intervals, determine whether the graph of f is increasing or decreasing, and discuss concavit. Also, we list all relative etrema and points of inflection. p () p () Propert of the Graph (, 0) + Decreasing, concave upward (0, ) + + Increasing, concave upward (, ) + Increasing, concave downward (, ) + + Increasing, concave upward =0 0 + Relative Minimum = + 0 Point of Inflection = 0 0 Point of inflection Appling the Second Derivative Test, the relative minimum value of p is p(0) =. Since the sign of p () changesat =and =,then(, ) and (, 0) are inflection points. In Figure, the graph implies that the range is [, ). Tr This Sketch the graph of F (w) =w 5 5w.Applthesketchingguidelinesinpage87. Eample The Graph of a Radical Function Sketch the graph of f() =.,, 8 Figure The graph of f() =. Solution The first few features of the graph are Appling the quotient rule, we find Similarl, we obtain Domain: [, ) Limit at infinit: Vertical asmptote: = First derivative: f () = lim f() = = ( ) / Second derivative: f () = ( )/ ( ) ( ) = ( ) ( ) ( ) 5/ = ( ) 5/ Then the critical numbers and the test open intervals are as follows: Critical number: = Test open interval: (, ), (, ), (, ) Note f () = 0

53 .6. SUMMARY OF CURVE SKETCHING 89 We summarize the signs of f and f on the test open intervals. From which, we find where f is increasing, decreasing, and concave upward or downward. Also, we identif the relative etrema b the Second Derivative Test and find the points of inflection. f () f () Propert of the Graph (, ) + Decreasing, concave upward (, ) + + Increasing, concave upward (, ) + Increasing, concave downward = 0 + Relative Minimum = + 0 Point of inflection The relative minimum value is f() =. Since the sign of f () changesat =,the point (, ) is an inflection point. From the graph in Figure, we conclude the range of f is [, ). Tr This Sketch the graph of R() = +.Appltheguidelinesinpage87. Eample The Graph of a Rational Function Sketch the graph of f() = 9. Solution The following features can be computed directl: Domain: { : = ±} Smmetr: With respect to the -ais Intercepts: (±, 0), 0, 9 Vertical asmptotes: = ± Limits at infinit: f() =, lim f() = Horizontal asmptote: Appling the quotient rule, we obtain lim = st derivative: f () = (9 )() ( )( ) (9 ) = Differentiating again, we find 6 (9 ) nd derivative: f () = (9 ) (6) (6)( (9 )) (9 ) = 6(9 )((9 )+ ) (9 ) = 8( +) (9 ) The following test open intervals are determined from the critical numbers and the domain. Critical number: =0 Test open intervals: (, ), (, 0), (0, ), (, )

54 90 CHAPTER. APPLICATIONS OF DERIVATIVES Figure Important features of a graph are determined b using calculus. We determine the signs of f and f, and where f is increasing or decreasing. Also, we discuss concavit and identif the relative etrema b using the Second Derivative Test. f () f () Propert of the Graph (, ) Decreasing, concave downward (, 0) + Decreasing, concave upward (0, ) + + Increasing, concave upward (, ) + Increasing, concave downward =0 0 + Relative Minimum The relative minimum value of f is f(0) = 9. Finall, the range is (, ) 9, as seen from the graph of f in Figure. Tr This Sketch the graph of g(t) = t t.usetheguidelinesinpage87. Eample Sketching the Graph of a Radical Function Sketch the graph of f() = / ( +5). Usetheguidelinesinpage Solution The following features of the graph follow directl.,,6 5 0,0 0 Figure 5 The graph of f() = / ( +5) lim Domain: (, ) Intercepts: (0, 0), ( 5, 0) Limits at infinit: f() =, Appling the product rule, we obtain lim f() = First derivative: f () = / + 5( +) ( +5)=. Differentiating again, b the quotient rule we find Second derivative: f () = 5 5( +) 9 = 0( ) 9 /. We list the test open intervals determined b the critical numbers and the the zero of f (), i.e., =. Critical number: =, 0 Test open intervals: (, ), (, 0), (0, ), (, ) We tabulate the signs of f and f on the test open intervals. We determine where f is increasing or decreasing, and discuss concavit. Also, we identif the relative etrema and the points of inflection. f () f () Propert of the Graph (, ) + Increasing, concave downward (, 0) Decreasing, concave downward (0, ) + Increasing, concave downward (, ) + + Increasing, concave upward = 0 Relative Maimum =0 undefined undefined Relative Minimum = + 0 Point of inflection

55 .6. SUMMARY OF CURVE SKETCHING 9 We have a local maimum value at = b the Second Derivative Test. A relative maimum value is f( ) =. Note, the sign of f () changes from negative to positive at = 0. Thus, b the First Derivative Test a relative minimum value is f(0) = 0. Since the sign of f () changesat =thepoint(, 6) is an inflection point. In Figure 5, we see the graph of f and the range is (, ). Tr This Sketch the graph of H(t) =6 t t /.Applthesketchingguidelinesinpage87. Eample 5 The Graph of a Trigonometric Functions Sketch the graph of T () = sin +cos. Solution First, we sketch the graph for in [0, π]. Since cos andsin(kπ) =0 where k is an integer, we obtain Domain: (, ) Π, -intercepts: (0, 0), (π, 0), (π, 0) Appling the quotient rule and the identit cos +sin =,wefind Π Π First derivative: T () = ( + cos )cos (sin )( sin ) ( + cos ) Π, = cos + ( + cos ) Figure 6a Second derivative: T () = ( + cos ) ( sin) ( cos +)( sin( + cos )) ( + cos ) Π, Π, = = sin( + cos )( ( + cos )+(cos +)) ( + cos ) sin(cos ) ( + cos ) Note, =0,π,π satisf T () = 0. Together with the critical numbers, we list the test open intervals: Critical numbers: = π, π Test open intervals: 0, π π,,π, π, π π,, π We tabulate the signs of T () andt () on the test open intervals. Then determine where T is increasing or decreasing, and discuss concavit. T () T () Propert of the Graph (0, π/) + Increasing, concave downward (π/,π) Decreasing, concave downward (π, π/) + Decreasing, concave upward (π/, π) + + Increasing, concave upward =π/ 0 Relative Maimum = π 0 Point of Inflection =π/ 0 + Relative Minimum Π Π, Π Π, Figure 6b Appl the periodicit to etend the graph of T () = sin +cos.

56 9 CHAPTER. APPLICATIONS OF DERIVATIVES B the Second Derivative Test, a local maimum value is T ( π )=, and a relative minimum is T ( π )=. Since the sign of T () changesat = π, (π, 0) is a point of inflection. The graph of T in [0, π] is given in Figure 6a. Since T ( +π) =T (), we etend the graph to [ π, π], see Figure 6b. Using the etrema, we obtain that the range is [, ]. Tr This 5 Sketch the graph of N() = cos +sin.appltheguidelinesinpage87. Eample 6 Sketching the Graph of a Rational Function Sketch the graph of f() = Solution. The first few features of the graph can be obtained directl., 6, 6 Domain: { : = ± } Vertical asmptotes: = ± Smmetr: with respect to the origin Intercept: (0, 0) Limits at infinit: f() =, lim f() = Appling the quotient rule, we find lim st derivative: f () = ( )( ) ( 8) ( ) Figure 7 The graph of f() =. = (( )+8 ) ( ) f () = ( ) ( ) = ( ) Critical numbers: = 0, ± Differentiating again, we obtain nd derivative: f () = ( ) (6 6 ) ( )( 6( )) ( ) = ( )(6 6 )+6( ) ( ) = ( ) f () = 8 +6 = ( +) ( ) ( ) We consider the test open intervals determined b the critical numbers, the zeros of f (), and the domain: Test open intervals:,,,,, 0, 0,,,,,

57 .6. SUMMARY OF CURVE SKETCHING 9 Below, we list the signs of f and f on the test open intervals. Then we identif an relative etrema b the Second Derivative Test. We describe concavit and the points of inflection, too. f () f () Propert of the Graph ( ) + Decreasing, concave upward (, ) + + Increasing, concave upward (, 0) + Increasing, concave downward (0, ) + + Increasing, concave upward (, ) + Increasing, concave downward, ) Decreasing, concave downward ( = 0 Relative maimum = Point of inflection = 0 + Relative minimum The relative maimum value of f is f f = 6 and the relative minimum value is =.Note,thesignoff changes at =0. Thus,(0, 0) is an inflection point. 6 Finall, the range of f is (, ) as seen in Figure 7. Tr This 6 Sketch the graph of h(t) = t using the guidelines in page 87. t The graph in Eample 6 has a slant asmptote or oblique asmptote. Precisel, aline = m + b where m = 0 is a slant asmptote of the graph of a function f if either or lim [f() (m + b)] = 0 lim [f() (m + b)] = 0. If R() =p()/q() is a rational function and the degree of p() is one more than the degree of q(), then the graph of = R() hasaslantasmptote. Tofindtheslant asmptote, use long division to epress R() as a sum of linear function and another rational function. For eample, using long division we find Then we obtain Thus, = f() = lim f() = + /. / = lim = 0 is a slant asmptote of the graph of f, see Figure 8. Figure 8 The line = is a slant asmptote of the graph of f() =..6 Check-It Out. Find the domain, asmptotes, and the smmetr of the graph of R() =.. Find the domain, range, and open intervals where the graph of f() = + is increasing, decreasing, concave upward, or concave downward.. Find the critical numbers and open intervals in [0, π] where g(t) =sint +cost is increasing or decreasing.

58 9 CHAPTER. APPLICATIONS OF DERIVATIVES True or False. If false, eplain or show an eample that shows it is false. In Eercises -6, use the graph below to answer the problems. Figure for nos -6: The graph of f() =. f () > 0on(0, ). f is increasing on (, 0). lim f() =. The range of f is (, ). 5. The graph has no slant asmptote. 6. (0, 0) is a local maimum point. 7. The domain of = sin is { : = kπ where k is an integer }. +cos 8. The graph of = sin is smmetric about the -ais. 9. The graph of = tan is smmetric about the origin. 0. If f ( 0)=0,then f( 0) is a relative etreme value of f. Eercises for Section.6 In Eercises -, an approimate sketch of the graph of a function is given. Find the relative etrema, open intervals where the function is increasing or decreasing, limits at infinit, and range.. f() =6 +. g() = 6 For No. For No.

59 .6. SUMMARY OF CURVE SKETCHING 95. p(t) =t 5/ 5t /. C(t) =sint +cost, 0 t π t Π Π t For No. For No. In Eercises 5-, sketch the graph of the function. Find the intercepts, relative etrema, points of inflection, asmptotes, and the range. 5. g(t) = t t 7. f() = 9 9. r(w) = 6. h(t) = 8t t 8. R() = w w 0. C(w) = w w +. p() =8 / /. F () = 9 6 / /. =. = 5 0 In Eercises 5-8, sketch the graph of the function over the indicated interval. Find the coordinates of the local etrema and points of inflection. 5. T () = sin 5cosθ,[0, π] 6. S(θ) = cos sin θ 7. R() = tan, π, π 8. f(θ) =tanθ 8θ,,[0, π] π, π In Eercises 9-, find a function with the given asmptotes. There are several possible answers. 9. Vertical asmptote =, horizontal asmptote = 0. Vertical asmptotes = ±, no horizontal asmptote. Vertical asmptote =0,slantasmptote = +. Vertical asmptote =, slant asmptotes =, = In Eercises -6, the graphs of = f () and = f () are given. Sketch the graph of f if the graph passes through the indicated point P.. P (0, 0). P (0, 0) f '' f ' f ' f '' For No. For No.

60 96 CHAPTER. APPLICATIONS OF DERIVATIVES 5. P (0, ) 6. P (, 0) f '' f ' f ' f '' For No. 5 For No. 6 v v s't 6.5 Figure for 7 t Applications 7. Free-falling Object A ball is thrown from a certain height s 0 above the ground. The graph shown is that of the velocit function v(t) in ft/sec over [0 sec, 6.5 sec], which is the time interval when the ball is above the ground. If v() = 0, find s 0 and the maimum height reached b the ball. 8. FM Radio Sketch the graph and find all relative etrema of F (θ) =cos(θ +sinθ) overthe interval [0,π]. A function such as F is used in frequenc modulation FM snthesis. 9. Fourier sine series Find all the relative etrema of S(θ) = sin θ + sin θ over the interval [0,π], 6 and sketch the graph of S. The function S is called a Fourier sine series and has applications in signal processing. 0. Harmonic oscillator Find the critical numbers of the function = cos(t) +cos(t) over the interval [0,π]. Moreover, show that the function satisfies the differential equation +6 = 9 cos t. Such a differential equation is said to model the motion of a weight that is attached to a spring. Theor and Writing Proofs. Let = a + b + c be a quadratic function where a = 0. Use calculus to determine the coordinates of the verte of the parabola. When is the graph concave upward? Concave downward?. Determine the local etrema and points of inflection of the cubic polnomial c() = + k where k is a constant.. The signs of the first two derivatives of = G(t) on certain open intervals are listed below. If the graph of G passes through the points A(, ), B(, ), and (, ), then sketch a possible graph of G. Open interval G (t) G (t) (, ) + (, ) (, ) + (, ) + +. The derivative of a polnomial = p() isp () =( ) ( ). Find the relative etrema and points of inflection of the graph of p. 5. Let p be a polnomial with an even degree n. Prove that p has at least one relative etreme value.

61 .7. OPTIMIZATION PROBLEMS 97.7 Optimization Problems Solving Optimization Problems The applications of derivatives in finding minimum and maimum values are useful in solving optimization problems. For instance, we ma want to know the maimum area, shortest distance, maimum profit, and least cost among others. In the process, we have to convert word problems into equations as the following eamples illustrate. Eample Maimizing the Area of a Rectangle A homeowner plans to build a rectangular enclosure along the side of her house using 500 feet of fencing material, see Figure a. Find the dimensions of the enclosure that has the largest area assuming no fencing material is needed along the side of the house. Solution The area A of the rectangle is A =. Primar equation Wall Since there are 500 feet of fencing material and onl three sides of the rectangle require fencing, we have + = 500. Secondar equation Substituting into the primar equation, we obtain =500. (6) A = = (500 ) =500. Since and must be nonnegative, using (6) wefindthat 50. Then the feasible domain of A as a function of is [0, 50]. The graph of A as a function of is seen in Figure b. Now, we find the critical numbers of A in (0, 50). da d =500 = 0 Set the derivative to zero. = 5 Critical number Net, evaluate A at the critical number and at the endpoints of [0, 5] A 0, 50 0 Thus, the maimum value of A occurs when = 5 ft. Correspondingl, we get =50 ft b using (6). Therefore, the dimensions of the rectangular enclosure with the largest possible area are 5 ft b 50 ft. Figure a A rectangular enclosure where the dotted side represents the side that uses no fencing material. A 5, 50 Figure b Asketchof the graph of A =500 in the feasible domain [0, 50]. In Eample, with the given 500 feet of fencing we must have realized that there were infinitel man was to construct a rectangular enclosure. For instance, we list a few possible dimensions of such rectangular enclosures with their corresponding areas. Dimensions Area 50 ft b 50 ft b 00 ft 0, 000 ft 80 ft b 80 ft b 0 ft 7, 00 ft 50 ft b 50 ft b 00 ft 0, 000 ft

62 98 CHAPTER. APPLICATIONS OF DERIVATIVES Area 0,000 ft Area 7,00 ft Area 0,000 ft Figure Rectangular enclosures with the same perimeters but different areas. Tr This Find the dimensions of the rectangle with the maimum possible area that has a perimeter of 00 feet. The following guidelines are useful in solving optimization problems, as illustrated in Eample. Guidelines for Solving Optimization Problems. Identif all quantities and draw a picture of the problem. A quantit could be a variable or constant such as area, perimeter, etc.. Write a primar equation that involves the variable to be optimized, i.e., maimized or minimized.. Epress the quantit to be optimized as a function of one variable. A substitution and a secondar equation ma be needed to do this.. Find the feasible domain, i.e., the set of numbers for which the optimization problem makes sense. 5. Find the maimum or minimum value of the quantit to be optimized using the calculus techniques in Sections. through.6. f, Eample Minimizing the Distance from a Point to a Graph,0 Find the point on the graph of f() = that is closest to the point (, 0). What is the minimum distance between the graph of f and (, 0)? Figure a Find the point (, ) on the graph of f() = that is nearest to (, 0). Solution Let d be the distance between (, 0) and a point (, ) on the graph of f, see Figure a. d = ( ) +( 0) = ( +)+ Primar equation Since (, ) lies on the graph of f() =,wecansubstitute = Secondar equation into the primar equation. Then we write d() = ( +)+( ) = +.

63 .7. OPTIMIZATION PROBLEMS 99 Using the domain of f() =, we find that the feasible domain of d() is[0, ). Appling the Chain Rule, we obtain d () = d + = d ( +) / ( ) = +. d Now, if d () = 0 then = 0, = Critical number Using the First Derivative Test, we find that the minimum value of d occurs when =/ as seen in Figure b. Hence, the minimum distance between the graph of f and (, 0) is d(/) = +=. Finall, the point on the graph of f() = that is closest to (, 0) is,f(/) =,, or,. Figure b The shortest distance between (, 0) and the graph of f is d(/) = /. Tr This Find the point on the graph of f() = that is closest to the point (, 0). Eample Maimizing the Area of an Inscribed Rectangle A rectangle is inscribed in the region bounded b the semicircle = and the -ais, see Figure a. Find the length and width of the rectangle with the maimum possible area. P, Solution Let P (, ) be a verte of a rectangle such that P lies on the semicircle in the first quadrant, see Figure a. Since and are the length and width of the rectangle, respectivel, the area of the rectangle is Since we obtain A =. Primar equation =, A =. Secondar equation A feasible domain of A is [0, ]. Using the Product Rule, we obtain da d = + = + = +( ) Factor Figure a Inscribing a rectangle in a semicircle with radius. da d = ( ) Simplif

64 00 CHAPTER. APPLICATIONS OF DERIVATIVES If da/d =0,then A, Figure b The maimum possible value of A is and this occurs when = /. =0 or = Critical number The values of A at the critical number and the endpoints of [0, ] are A(0) = 0, A( /) =, and A() = 0. Thus, the maimum value of A occurs when = /, see Figure b. Hence, the length and width of the rectangle with the maimum area are length = = = and width = = = = =. Tr This A rectangle is inscribed in a region bounded b the semicircle = 6 and the -ais. Find the length and width of the rectangle that has the maimum possible area. A ft E 5 ft Eample Minimizing the Total Length A -feet high post and a 5-feet high post are 0 feet apart, see Figure 5a. The posts are to be supported b two wires staked at a common point C in the horizontal ground between the posts. Where should the wires be staked to minimize the sum of the lengths of the wires? Solution Let = BC and = CD be the lengths of the line segments shown in Figure 5a. B the Pthagorean Theorem, we obtain AC = 9+ and CE = 5 +. Then the sum L of the lengths of the wires is B C D BD0 ft Figure 5a A -ft post and a 5-ft post are supported b wires that are staked at point C. L = AC + CE = Primar equation Since the posts are 0 feet apart, we have = 0. Secondar equation Substituting into the primar equation, we find L = (0 ) = See Figure 5b. Since and 0 are nonnegative, let [0, 0] be the feasible domain of L. UsingtheChainRule,wefind dl d =

65 .7. OPTIMIZATION PROBLEMS 0 If dl/d =0,then = = Then cross-multipl and set one side of the equation to zero. 5 L = 0 ( 5)( +5) = 0 = 5/ Critical number in [0, 0] Figure 5b The minimum total length L occurs when =.75 ft. The values of L at the endpoints of [0, 0] and the critical number are L(0)., L(.75).8, and L(0) 5.. Hence, the wires must be staked at a point in the ground that is =.75 ft from the -ft post. Tr This If the posts in Eample are 6 feet apart, then where should the wires be staked to minimize the sum of the lengths of the wires? Eample 5 Using the Least Material Find the radius and height of a -liter can with the least surface area. Assume the can is a right circular clinder that has a cover at the top and bottom. Solution Let r and h be the radius and height of the can, respectivel. Since the can has a top and bottom, its surface area A is A =πr +πrh. Primar equation Since the volume is liter or equivalentl 000 cm,wefind Substituting πr h =000. h = 000 πr into the primar equation, we obtain A =πr 000 +πr πr Secondar equation =πr r Since r must be positive, the feasible domain of A is (0, ) as in Figure 6b. Then If da/dr =0,then da dr πr 000 = 0 =πr 000 r = πr 000 r. 500 r = π = 5 π π Critical number Figure 6a A right circular clinder with atopandbottom. Letr be the radius and h the height A AΠr 000r 500 Π Figure 6b The graph of the surface area A as a function of its radius r. r

66 0 CHAPTER. APPLICATIONS OF DERIVATIVES Appling the First Derivative Test, we find that the minimum surface area A occurs at the critical number. Hence, the radius and height of the can with the least possible surface area are radius = r = 5 π 5. cm, and π height = h = 000 πr = = 0 π π π π π 0.8 cm. Tr This 5 Find the radius and height of a -liter can with the least surface area. Assume the can is a right circular clinder that has a cover at the bottom but is open at the top..7 Check-It Out. Gar wants to build a fence on two adjacent lots, as shown in Figure. No fencing is needed along the boundar represented b the dotted lines. If 00 feet of fencing material is available, then find and as indicated in the figure if the area enclosed is the maimum. Figure for No.. The length from the left-most side to the right-most side is, and the width is.. Find the minimum value of p if the primar equation is p = +, secondar equation is =, and the feasible domain is >0. True or False. If false, eplain or show an eample that shows it is false.. The distance between (a, b) and a point (, ) on the parabola = is ( + a) +( + b).. If a right circular clinder has a cover at the top and bottom, then its surface area is S =πrh +πr where r and h are the radius and height, respectivel.. If the perimeter of a rectangle is feet, then the area of such a rectangle is ft.. If the sum of two numbers is 0, then the maimum possible product of two such numbers is If the area of a rectangle is square inches, then the dimensions of such a rectangle are inches b inches.

67 .7. OPTIMIZATION PROBLEMS 0 6. The two positive numbers that have a product of 5 and the minimum possible sum are 5 and If two negative numbers differ b, then the minimum product of two such numbers does not eist. 8. A bo with a square base and open top has a surface area of 00 square inches. Then the shape of the bo that has the maimum volume is a cube, i.e., the height of the bo is the same as side of the base. 9. A rectangle is inscribed in a semicircle with radius, see figure below. If the indicated radial line makes an angle θ with the positive -ais where 0 <θ<π/, then the area of the rectangle is sin θ. Θ Figure for No. 9 and 0 0. In the above figure, the minimum perimeter of the inscribed rectangle is. Eercises for Section.7 In Eercises -6, optimize the indicated variable given the primar and secondar equations, and feasible domain.. Minimize D, primar equation D =(.5) +, secondar equation =, feasible domain <<.. Minimize D, primar equation D = ( ) +, secondar equation = +, feasible domain <<.. Maimize A, primar equation A = l w, secondar equation l +w = 0, feasible domain 0 w 0.. Maimize A, primar equation A = bh, secondar equation b + h = 7, feasible domain 0 b Minimize S, primar equation S =πrh + πr, secondar equation r h = π, feasible domain 0 <r<. 6. Maimize V, primar equation V = πr h, secondar equation πrh +πr =π, feasible domain 0 <r<. Optimization Problems 7. Find the point on the graph of f() = that is nearest the given point. a) (9, 0) b) (, 0) 8. Find the point on the graph of f() = that is closest to the indicated point. a) (, 0) b) (, 0) 9. Find the number which eceeds its square b the greatest value. 0. Find the positive number that eceeds its cube b the greatest amount.

68 0 CHAPTER. APPLICATIONS OF DERIVATIVES. A homeowner plans to fence a rectangular region that must have an area of 0,000 square feet. In addition, the homeowner plans to subdivide the region into three equal rectangles b inserting two smaller fences, as seen below. Find the dimensions of the rectangular region that will use the least amount of fence. h r Figure for No.. Figure for No... A window consists of a rectangle and a semicircle that is mounted on top of the rectangle, see figure above. Let h be the height from the base of the rectangle to the base of the semicircle. If the perimeter of the window is 0 feet, find the radius of the semicircle so that the window has the maimum area.. From an aluminum square sheet, inches on each side, a bo with an open top is to be made b cutting off small squares from each corner and bending up the sides. How large a square should be cut from each corner so that the bo has the maimum possible volume? Figure for No.. The sum of the perimeters of a square and an equilateral triangle is feet. Find the length of the side of the square that minimizes the sum of the areas of the square and triangle. 5. A right triangle in the first quadrant is bounded b coordinate aes, and a line through point (, ), as seen below. Find the coordinates of the vertices if the triangle has the minimum area. Does the triangle have the minimum perimeter?, Figure for No. 5 Figure for No Find the area of the largest isosceles triangle that can be inscribed in a circle with radius units. 7. Find the dimensions of the right circular clinder with a volume of 0 cubic inches and that has the minimum surface area. Assume the clinder has a top and bottom.

69 .7. OPTIMIZATION PROBLEMS A track and field has the shape of a rectangle and two semicircles attached to the sides of the rectangle. The perimeter of the track and field is 00 meters. Find the length of the rectangle and the radius of the semicircle if the rectangular region has the maimum possible area. 9. Two cities A and B are at a distance of 6 miles from each other, see figure below. Let = PC denote the distance from P to C. The cities plan to build a recreation center C along a certain highwa. Suppose the perpendicular distances from A and B to the highwa are miles and 7 miles, respectivel. Find the value of that minimizes the sum AC + CB of the distances. The figure is not drawn to scale. ( # $ % km B P C! &&' " A Figure for No. 9 Figure for No A forest ranger must travel from point A to point C through some point P, see figure above. There is a road from P to C that runs from west to east. The point B on the road nearest A is km awa, and the distance from B to C is 0 km. The forest ranger can travel diagonall from A to P at the speed of 5 kph, and drive faster from P to C at 0 kph. How far from B should P be located so that the forest ranger travels the least amount of time from A to P to B?. Mark plans to make a rectangular garden using feet of fencing material. He figures that onl three sides of the garden will be fenced and the fourth side will be left open without a fence. What is the maimum area of the garden that that he can enclose?. If the base b and perimeter p of a triangle are fied, then determine the remaining two sides of the triangle with the maimum area.. A crew is needed to service a manufacturing plant. It has been found that the number of hours needed to do the job is 5 + 6/ where is the size of the crew. If each worker is paid $6 per hour and the use of an equipment is $60 per hour, find the most economical size of the crew.. A field trip will cost each student $60 if 00 students join the trip. If more than 00 students join, then the cost of the trip per student will be reduced b 5 cents times the number of students in ecess of 00. How man students should join the trip to realize the largest gross income? 5. A contractor wishes to find the dimensions of a rectangular piece of land where she can build a house that will have an area of,000 square feet. Moreover, the land must have 50 feet of space from the eterior of the house to the front and back propert lines, and 7.5 feet of space from the eterior to the left and right propert boundaries. Find the dimensions of the land with the least area on which the house can be built. 6. What are the dimensions of the rectangle with the greatest area that can be inscribed in a triangle with base 0 ft and height 5 ft, if one side of the rectangle lies on the base of the triangle? See figure below. Figure for No Find the minimum distance between a point (p, q) andaline = m + b.

70 06 CHAPTER. APPLICATIONS OF DERIVATIVES 8. A right circular cone is made b cutting a sector from a unit circle, see figure below. Let θ be the central angle of the sector. Find θ if the cone has the maimum volume.! Figure for No. 8 Figure for No A rain gutter is made b folding up the edges of an 8-inch metal sheet, see figure above. Suppose each of the left and right edges of the gutter are 6 inches long. Find the angle that the edges must be folded so that the gutter holds the maimum volume. 0. Find the volume of the largest circular cone that can be inscribed in a sphere of radius r.. A -b-6 rectangular card is folded in such a wa that verte A is placed at some point B on the opposite longer side, see figure below. Let c be the length of the crease that is formed. Find the minimum value of c. B c R A Figure for No. Figure for No.. A rectangle region R in the first quadrant has a verte at the origin, the opposite verte is a point in the parabola =( ), and two sides that lie in the coordinate aes, see figure above. Find the maimum area of such a rectangle.. A rope of length L is formed into a sector of a circle. Find the central angle of the sector that gives the maimum area of the sector. Odd Ball Problems. Determine the constant C if G() = + C has a relative minimum at =. 5. Determine the constant K if the function H(t) =t + K t has points of inflection at t = ± 6. Find the maimum and minimum values of f() =A cos()+bcos assuming B A > Find the maimum and minimum values of f() =A cos()+bcos assuming A >B>0. 8. A circle of radius is centered at (0, ). A second circle of radius lies in the first quadrant, and is tangent to the first circle and -ais. In the figure above, we see a third circle that is tangent to the first and second circles. If the center of the third circle lies in the -ais, find its radius. What is the limit of the common point between the first and third circles as the radius of the third circle increases to?

71 .8. NEWTON S METHOD 07.8 Newton s Method Aproimating a Real Zero of a Function ASufficientConditionforConvergence Aproimating a Real Zero of a Function The solutions of f() = 0 are not eas to solve. In fact not all solutions are necessaril real numbers. If c is a real zero of f, we can approimate c to an degree of precision under certain conditions. We will discuss and appl Newton s method to approimate c. The first step is to choose an arbitrar number that is approimatel c, as in Figure a. The graph of f will help ou choose.thetangentlineat(,f( )) is given b f( ) = f ( )( ) c = f ( )( )+f( ). If (, 0) is the -intercept of the tangent line, then 0 = f ( )( )+f( ) Figure a = f() f ( ) if f ( ) = 0 Now, we have two approimations for c, namel, and. We continue the process. Similarl, we can find the -intercept (, 0) of the tangent line at (,f( )). In general, the nth approimation to c is, f n+ = n f(n) f ( n) if f ( n) = 0 The repeated applications of the above procedure is called Newton s Method., f c ASummarofNewton smethod Let f be a differentiable function such that f(c) = 0. Step. Find an initial estimate of c. Step. The (n +) st approimation is n+ = n f(n) f ( n). Step. If n+ n is less than the desired accurac, let n+ be the final estimate of c. Otherwise, return to Step and find a new approimation. Under certain conditions, the approimations become more precise as n increases. Figure b Newton s Method finds sequential approimations,,,... to an -intercept of the graph of a function. Eample Approimating Appl Newton s Method to approimate the real root of = 0 as follows: use = and estimate. f Solution formula Let f() =. Since f () =, the iterative process is given b the n+ = n f(n) f ( n) = n n. n Figure The tangent line at the point where =has -intercept =.5.

72 08 CHAPTER. APPLICATIONS OF DERIVATIVES Since =,wefind = = =.5. Then the third estimate is = =.5 (.5) (.5).95 Continuing the process, we obtain =.89. The first iteration of Newton s Method is shown in Figure. Using a calculator, we find.8907 which agrees with up to four decimal places Tr This Repeat Eample but approimate the real root of =0..5, f.5 Eample Finding a Zero of a Function The function f() = 5 + has a zero in [, ], see Figure. Appl Newton s Method with =.5 tofind n+ if n+ n < Then find the value of f( n+). Solution Since f () =5 +, the iterative process is given b the formula n+ = n f(n) f ( n) = n 5 n n + n 5 n n +. Using a calculator with =.5, we obtain the following estimates Figure The tangent line to the graph of f at the point with =.5 has -intercept =.95. We stop iterating since < Then the zero of f() in[, ] is approimatel Finall, satisfies f( ) Tr This The function g() =cos has a zero in [0, ]. Appl Newton s Method to find n+ if =and n+ n <

73 .8. NEWTON S METHOD 09 Newton s Method does not alwas guarantee that the values of n+ = n f(n) f ( n) will approimate a zero of f() to an accurac desired. For instance, let f() = + whose graph is given in Figure. The tangent line at (, ) has -intercept (0, 0), and the tangent line at (0, ) has -intercept (, 0). Appling Newton s Method with initial estimate = will generate the following values: = = 5 = and = = 6 = 0. A Sufficient Condition for Convergence The net theorem will give a condition when Newton s Method is guaranteed to work. That is, the values of n s generated b the method will approimate a given zero c of a function to an desired precision. In such a case, we write lim n = c. n Also, we sa n converges to c as n. The proof of the theorem relies on the Mean Value Theorem. Before stating the theorem, if a<b, we observe that the interval [a, b] is the middle third of [a b, b a]. Theorem. Let f(c) =0andsuppose f()f () (f ()) α< (7) for all in (c β,c + β) for some positive constants α, β. If lies in [c β,c + β] and n+ = n f(n) f ( n),then lim n = c. n Proof Let N() = f(). Observe N(c) =c and f () N () = (f ()) f()f () (f ()) = f()f () (f ()). In particular, N () α for all in (c β,c + β). Appling the Mean Value Theorem, we find c = N( ) N(c) α c. Since 0 <α<and c β, it follows that c <β.thus, lies in (c β,c+β). Similarl, b the Mean Value Theorem we obtain c = N( ) N(c) α c α c. In general, for an integer n wehave n+ c α n c and n+ lies in (c β,c + β). Let ε>0 be a given degree of accurac. Then n+ c <εwhenever n is sufficientl large. Hence, lim n = c. n

74 0 CHAPTER. APPLICATIONS OF DERIVATIVES Eample Convergent Approimations Using Newton s Method Let f() = and =. Prove the values of n generated b Newton s Method satisf lim n n = Solution The graph of f is seen in Figure a, and is the real zero of f(). The first two derivatives are f () = and f () =6. Then inequalit (7) in Theorem. reduces to f()f () [f ()] = ( )(6) 9 = The graph of g() = and g( 6 )=. 5 is shown in Figure b. We find g() = 5, g( ) <g(), Figure a: f() =. Figure b: g() =. From Figure b, we see that g() < 5 for all in [, ]. Let β = > 0. Then g() < 5 for all in ( β, +β). Thus, we can appl Theorem. with =. Hence, the values of n generated b Newton s Method satisf lim n n =. Tr This Let f() = 60 and = 8. Prove the values of n generated b Newton s Method satisf lim n n = 60. Eample Convergent Approimations Using Newton s Method Let f() =cos() and =. Iff(c) = 0, prove the values of n generated b Newton s Method satisf lim n = c. Thenfind to the nearest thousandth. n Solution The graph of f is seen in Figure 5, and its zero c is near. The first two derivatives are f () = sin() and f () = cos. Then inequalit (7) in Theorem. reduces to g() = f()f () [f ()] = (cos() )cos ( + sin )

75 In Figure 5, we find g() for all in [, ]..8. NEWTON S METHOD f g Figure 5: f() =cos() and g() = f()f () (f ()). Then Theorem. applies with = or for an in [, ]. Thus, the values of n generated b Newton s Method will satisf lim n = c. Finall,since =,wefind n = f() cos() = f ( ) sin() 0.79 Tr This As in Eample, butf() =sin +. As done earlier, let =. An equation f() =0issaidtobesolvable b radicals if its solutions can be determined in a finite number of steps using addition, subtraction, multiplication, division, and the radicals n.foreample,a + b + c =0with a = 0 is solvable b radicals for the solutions are = b ± b ac. a Moreover, rd and th degree polnomials are solvable b radicals as proved b Geronimo Cardano (50-576). For instance, the real solution to + p + q = 0 where p>0is = q + p 7 q q + p 7 + q The verification of this cubic solution is an eercise at the end of this section. However, the zeros of 5th and higher degree polnomials are not solvable b radicals. The latter fact was established b Evariste Galois (8-8). Newton s Method is a robust numerical algorithm for solving equations whether solvable b radicals or not..8 Check-It Out Use Newton s Method to find n given n,,andf() =0.. n =, =, + =0. n =, =, =0 True or False. If false, eplain or show an eample that shows it is false.. In Newton s Method, the (n +)stestimateis n+ = n f(n) f ( n). If f() =m + b, m = 0,and is an real number, then the second value generated b Newton s Method satisfies f( )=0.. If p() is a rd degree polnomial and =0,thenthevalues n generated b Newton s Method converge to a zero of p().. Let be a first estimate of the zero of f() = a. Then the second estimate of the zero of f() bnewton s Method is = ( a ).

76 CHAPTER. APPLICATIONS OF DERIVATIVES 5. Let = be a first estimate of the zero of f() = a. B Newton s Method the second estimate of the zero is = a. 6. Let f(c) = 0 and suppose f () iscontinuous.if f(c)f (c) (f (c)) <, then there eists satisfing lim n = c. n Eercises for Section.8 In Eercises -8, appl Newton s method using the given to find the net two estimates and that solves the indicated equation f() =0.Roundtheanswerstothenearestthousandth.. =, 5=0. =, 0 = 0. =, =0. =, 5 =0 5. =, cos(π) =0 6. =, sin() +=0 7. =, sin()+cos()+=0 8. =, cos(π) π =0 In Eercises 9-, eplain wh Newton s Method fails to find a zero of f() given the initial estimate. 9. =, f() = 5 0. =,f() =. =,f() =. =, f() = +9. =5,f() =. =.5, f() = if 0 if <0 In Eercises 5-8, ou are given an equation f() =0and an initial estimate of a zero of f(). ApplNewton s Method to find n+ that satisfies n+ n < Round n+ to four decimal places. 5. =, 5=0 6. =, = 0 7. =, sin( 6 ) =0 8. =, tan( ) =0 In Eercises 9-, given show that the values n generated b Newton s Method converges to a zero of the given function f. Thenapproimate to the nearest thousandth. Appl Theorem. and see Eamples and. 9. =, f() = 0 0. =, f() = =0.5, f() =cos(π ). =7, f() = sin() Theor and Proofs. Let f() = f(n) + +andlet be an real number. Prove lim f(n) = 0 where n+ = n. n f ( n). Cardano s Formulas Let + p + q = 0 where p>0. a) Verif the identit: (a b) +ab(a b) (a b )=0 b) If a = q + p 7 q and b = q + p 7 + q, prove ab = p and a b = q c) Using the notation in part b), show = a b is the onl real solution of + p + q =0. 5. Let f() =a + b + c, a>0, and b ac 0. If = b f(n) and n+ = n a f ( n), prove lim f(n) =0. n 6. Let f(c) = 0 where a<c<b. Suppose for some α we have f()f () (f ()) α< for all in (a b, b a). If lies in [a, b] and n+ = n f(n) f ( n), prove lim n = c. n

77 .9. LINEAR APPROXIMATIONS AND DIFFERENTIALS.9 Linear Approimations and Differentials Linear Approimations Differentials Approimating a Functional Value Propagated Error Linear Aproimations If f is differentiable at c, the tangent line to the graph of f at (c, f(c)) is L() =f (c)( c)+f(c). (8) The linear function in (8) iscalledthelinear approimation, linearization, ortangent line approimation of f at c. Clearl, L() f(c) as c. Also, f() f(c) as c b continuit. Combining, we obtain whenever is near c. Eample Finding a Linear Approimation f() f (c)( c)+f(c). (9) Find the tangent line approimation of f() =sin at (0, 0). Then approimate the values of sin(0.) and sin(0.05). Solution Note, f () =cos and f (0) = cos 0 =. Using (8) withc =0,thetangent line to f() =sin at (0, 0) is given b L() = f (0)( 0) + f(0) L() =. = ( 0) + 0 Thus, the tangent line approimation to f() =sin at (0, 0) is sin provided is near f sin 0.5 Hence, sin(0.) 0. and sin(0.05) These are good approimations for sin(0.) and sin(0.05) b using a calculator. Figure The tangent line = approimates f() = sin near =0. Tr This Appl (8) to find the linearization of f() =tan at (0, 0). Then appl (9) to approimate the value of tan( π )usingc =0and = π. Differentials Definition The Differential of a Function Let = f() be differentiable on an open interval containing c. a) The differential d of is an independent variable. b) The differential d of is a function of and d, anddefinedb d = f ()d.

78 CHAPTER. APPLICATIONS OF DERIVATIVES Also, the independent variable d is denoted b and is called the change in. In the definition of f () insection. we encountered the epression = f( + ) f() We call the change in. If = c, then the linearization in (9) ma be rewritten as follows: f() f (c)( c)+f(c) for near c f(c + ) f (c) + f(c) for 0 f(c + ) f(c) f (c) Since c is an arbitrar number, we ma replace c b : f( + ) f() f () for 0 (0) d In Figure, we see a (dotted) tangent line at to the graph of f. Tangent line at f d Figure : d if 0. The quantit represents a change in the functional values of f. While d represents a change in the functional values of the tangent line. Eample Comparing With d Tangent line at f If = +, evaluated and if =and =0.. Solution Then we obtain and Let f() =(+) /.BtheChainRulewefind f () = d = f ()d = +. d + = f( + ) f() Figure d is the change in the values of the tangent line, and is the change in the values of = f(). = Substitute =andd = =0.. Then we find and d = 0. =0.05 = Finall, note d since d =0. is approimatel zero. In Figure, we see the tangent line at =.

79 .9. LINEAR APPROXIMATIONS AND DIFFERENTIALS 5 Tr This If =,evaluated and given =and =0.. Approimating a Functional Value Differentials ma be used to approimate the values of a function. We rewrite (0) as follows: f( + ) f()+f () () where 0. The above approimation shows how to estimate f( + ) if we know the values of f(), f (), and. Eample Approimating a Square Root Approimate 0 b appling (). Solution In (), let f() =, =9,and =. Then 0 = f(0) = f(9) + f (9) = + 6 () for f () = Thus, =.75 The latter decimal is obtained b using a calculator. Tr This Estimate 5 b appling () with =6and =. Eample Approimating a Functional Value Approimate tan 7 b appling (). Solution We recall tan(5 π )=tan =and π = = π Let = π, d = π,andf() =tan. Thenf () =sec. 90 Appling () wefind π f + π π f + f π π π = tan +sec π π 90 π = + 90 π f + π 90 Since tan(7 )=f π + π 90,weobtain = + π 5 tan(7 ) + π f tan Π Figure The linear approimation at π underestimates f() =tan since the tangent line lies below the graph of f locall. The latter decimal is obtained b using a calculator.

80 6 CHAPTER. APPLICATIONS OF DERIVATIVES Tr This Estimate the value of sin 9 b appling () with = π 6 π and =. 80 Propagated Error Man scientific eperiments involve approimations of position, rate, etc. Suppose is an approimation, and + is the true value of a quantit. We sa is the error in the approimation. Let = f() be a function of. The propagated error is defined as f = f( + ) f(), therelative error in f() is given b f f(), andthepercentage relative error in f() is f 00%. f() Eample 5 Approimating a Propagated Error The approimate radius of a sphere is 6 in. with an error of at most ±0.05 in. of the true radius. Use differentials to approimate the propagated error and the percentage relative error in calculating the surface area of the sphere. Solution The surface area of a sphere of radius is f() =π. If the true radius is 6 +, then the propagated error is f = f(6 + ) f(6) Figure 5 A sphere with a true radius of 6 +. where Note f () =8π. Appling the differential in (0), we obtain f(6 + ) f(6) f (6) = 8π 8π(0.05) 7.5 sq.in. Thus, the propagated error is f 7.5 sq. in. Finall, the percentage relative error is f 8π(0.05) f(6) 00 00%.7%. π Tr This 5 A rectangular bo has a square base with an eact area of 6 sq. in. The height of the bo is 6 in. with an error of at most ±0. in. Find the propagated error and percentage relative error in the volume of the bo.