4.2 Extreme Values. Preliminary Questions 1. What is the definition of a critical point?

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1 68 C HAPTER APPLICATIONS OF THE DERIVATIVE so Thus, Differentiating both sides of the equation C 7 D with respect to ields d d C 7 C 7 D 0; d d d d D 7 C 7 : d ˇ 7./ D d ˇ. ;/./ C 7. / D 5 ; and the linearization at P D. ; / is L./ D. C / D 5 5 C 6 5 : Finall, the equation 7:7 D corresponds to D :, soweestimatethesolutionofthisequationnear D is L. :/ D 6. :/ C 5 5 D :056: 7. Show that for an real number k,. C / k C k for small.estimate.:0/ 0:7 and.:0/ 0:. Let f./d. C / k.thenforsmall,wehave f. / L. / D f 0.0/. 0/ C f.0/d k. C 0/ k. 0/ C D C k Let k D 0:7 and D 0:0. ThenL.0:0/ D C.0:7/.0:0/ D :0. Let k D 0: and D 0:0. ThenL.0:0/ D C. 0:/.0:0/ D 0: Let f D f.5c h/ f.5/,wheref./d.verifdirectlthate Dj f f 0.5/hj satisfies (5) with K D. and Let f./d.then f D f.5c h/ f.5/d.5 C h/ 5 D h C 0h E Dj f f 0.5/hj Djh C 0h 0hj Dh D./h D Kh : 76. Let f D f.c h/ f./where f./d.showdirectlthate Dj f f 0./hj is equal to h =. C h/. Thenprove that E h if h. Hint: In this case, C h. Let f./d.then and f D f.c h/ f./d h D C h C h E Dj f f 0./hj Dˇ ˇ h C h C h ˇ D h C h : If h, then C h and Ch. Thus, E h for h.. Etreme Values Preliminar Questions. What is the definition of a critical point? Acriticalpointisavalueoftheindependentvariable in the domain of a function f at which either f 0./ D 0 or f 0./ does not eist.

2 SECTION. Etreme Values 69 InQuestionsand,choosethecorrectconclusion.. If f./is not continuous on Œ0;,then (a) f./has no etreme values on Œ0;. (b) f./might not have an etreme values on Œ0;. The correct response is (b): f./might not have an etreme values on Œ0;. AlthoughŒ0; is closed, because f is not continuous, the function is not guaranteed to have an etreme values on Œ0;.. If f./is continuous but has no critical points in Œ0;, then (a) f./has no min or ma on Œ0;. (b) Either f.0/or f./is the minimum value on Œ0;. The correct response is (b):eitherf.0/or f./is the minimum value on Œ0;.Rememberthatetremevaluesoccur either at critical points or endpoints. If a continuous function on a closed interval has no critical points, the etreme values must occur at the endpoints.. Fermat s Theorem does not claim that if f 0.c/ D 0, thenf.c/ is a local etreme value (this is false). What does Fermat s Theorem assert? Fermat s Theorem claims: If f.c/is a local etreme value, then either f 0.c/ D 0 or f 0.c/ does not eist. Eercises. The following questions refer to Figure. (a) How man critical points does f./have on Œ0; 8? (b) What is the maimum value of f./on Œ0; 8? (c) What are the local maimum values of f./? (d) Find a closed interval on which both the minimum and maimum values of f./occur at critical points. (e) Find an interval on which the minimum value occurs at an endpoint FIGURE 8 f() (a) f./has three critical points on the interval Œ0; 8 : at D, D 5 and D 7. Twoofthese, D and D 5, arewherethe derivative is zero and one, D 7, iswhere the derivative does not eist. (b) The maimum value of f./on Œ0; 8 is 6; the function takes this value at D 0. (c) f./achieves a local maimum of 5 at D 5. (d) Answers ma var. One eample is the interval Œ; 8. AnotherisŒ; 6. (e) Answers ma var. The easiest wa to ensure this is to choose an interval on which the graph takes no local minimum. One eample is Œ0;.. State whether f./d (Figure ) has a minimum or maimum value on the following intervals: (a).0; / (b).; / (c) Œ; FIGURE Graph of f./d.

3 70 C HAPTER APPLICATIONS OF THE DERIVATIVE f./ has no local minima or maima. Hence, f./ onl takes minimum and maimum values on an interval if it takes them at the endpoints. (a) f./takes no minimum or maimum value on this interval, since the interval does not contain its endpoints. (b) f./takes no minimum or maimum value on this interval, since the interval does not contain its endpoints. (c) The function is decreasing on the whole interval Œ;. Hence,f./takes on its maimum value of at D and f./takes on its minimum value of at D. In Eercises 0, find all critical points of the function.. f./d C Let f./d C. Thenf 0./ D D 0 implies that D is the lone critical point of f.. f./d 7 Let f./d 7. Thenf 0./ D 7,whichisneverzero,sof./has no critical points. 5. f./d 9 5 C Let f./d 9 5 C. Thenf 0./ D 9 5 D. C /. 6/ D 0 implies that D and D 6 are the critical points of f. 6. f.t/d 8t t points of f. 7. f./d Let f.t/ D 8t t.thenf 0.t/ D t t D t.t / D 0 implies that t D 0 and t D are the critical Let f./d.then f 0./ D C D D 0 implies that D is the onl critical point of f.thoughf 0./ does not eist at D 0, thisisnotacriticalpointoff because D 0 is not in the domain of f. 8. g.z/ D z z Let g.z/ D z z.z / D D z z.z / z z : Then g 0 z.z/ D.z.z / D z/.z z/ D 0 implies that z D = is the onl critical point of g.thoughg 0.z/ does not eist at either z D 0 or z D, neitherisacriticalpoint of g because neither is in the domain of g. 9. f./d C Let f./d C. Then f 0./ D 0. f./d C 8 Let f./d C 8.Then. D 0 implies that D are the critical points of f. C / implies that D 0 and D are the critical points of f.. f.t/d t p t C Let f.t/d t p t C. Then f 0./ D. C 8/./. /. /. C 8/ D. C 8/ D 0 f 0.t/ D p t C D 0 implies that t D is a critical point of f.becausef 0.t/ does not eist at t D, thisisanothercriticalpointoff.

4 SECTION. Etreme Values 7. f.t/d t p t C Let f.t/d t p t C. Then f 0.t/ D t.t C / = D.t C / = t.t C / = D 0 implies that there are no critical points of f since neither the numerator nor denominator equals 0 for an value of t.. f./d p Let f./d p. Then f 0./ D p p C D p : This derivative is 0 when D 0 and when D p=; the derivative does not eist when D. Allfiveofthesevaluesare critical points of f. f./d Cj C j Removing the absolute values, we see that Thus, ; < f./d C ; f 0 ; <./ D ; and we see that f 0.0/ is never equal to 0. However, f 0. =/ does not eist, so D = is the onl critical point of f. 5. g./ D sin Let g./ D sin.theng 0./ D sin cos D sin D 0 implies that D n is acriticalvalueofg for all integer values of n. 6. R./ D cos C sin Let R./ D cos C sin.then R 0./ D sin C sin cos D sin.cos / D 0 implies that D n, are critical points of R for all integer values of n. 7. f./d ln D C n and D 5 C n Let f./d ln.thenf 0./ D C ln D 0 implies that D e D e istheonlcriticalpointoff. 8. f./d e Let f./d e.thenf 0./ D. C /e D 0 implies that D istheonlcriticalpointoff. 9. f./d sin Let f./d sin. Then f 0./ D p D 0 p implies that D are the critical points of f. 0. f./d sec ln Let f./d sec ln.then f 0./ D p : This derivative is equal to zero when p D, or when D p. Moreover, the derivative does not eist at D 0 and at D. Amongthesenumbers, D and D p are the onl critical points of f. D p, D, and D 0 are not critical points of f because none are in the domain of f.

5 7 C HAPTER APPLICATIONS OF THE DERIVATIVE. Let f./d C. (a) Find the critical point c of f./and compute f.c/. (b) Compute the value of f./at the endpoints of the interval Œ0;. (c) Determine the min and ma of f./on Œ0;. (d) Find the etreme values of f./on Œ0;. Let f./d C. (a) Then f 0.c/ D c D 0 implies that c D is the sole critical point of f.wehavef./d. (b) f.0/d f./d. (c) Using the results from (a) and (b), we find the maimum value of f on Œ0; is and the minimum value is. (d) We have f./d. Hencethemaimumvalueoff on Œ0; is and the minimum value is.. Find the etreme values of f./d 9 C on Œ0; and Œ0;. Let f./ D 9 C. First,wefindthecriticalpoints.Settingf 0./ D 6 8 C D 0 ields C D 0 so that D or D. Net,wecompare:first,forŒ0; : -value Value of f (critical point) f./d 5 (critical point) f./d 0 (endpoint) f.0/d 0 min (endpoint) f./d 9 ma Then, for Œ0; : -value Value of f (critical point) f./d 5 ma (endpoint) f./d 0 (endpoint) f.0/d 0 min. Find the critical points of f./d sin C cos and determine the etreme values on 0;. Let f./d sin C cos.thenontheinterval 0;, we have f 0./ D cos sin D 0 at D, the onl critical point of f in this interval. Since f. / D p and f.0/d f. / D, the maimum value of f on 0; p is, while the minimum value is.. Compute the critical points of h.t/ D.t / =.CheckthatouranswerisconsistentwithFigure.Thenfindtheetreme values of h.t/ on Œ0; and Œ0;. h(t) t FIGURE Graph of h.t/ D.t / =. Let h.t/ D.t / =.Thenh 0 t.t/ D.t D 0 implies critical points at t D 0 and t D. Theseresultsare / = consistent with Figure which shows a horizontal tangent at t D 0 and vertical tangents at t D. Since h.0/ D and h./ D 0, themaimumvalueonœ0; is h./ D 0 and the minimum is h.0/ D. Similarl,the minimum on Œ0; is h.0/ D and the maimum is h./ D = :5. 5. Plot f./ D p on Œ0; and determine the maimum value graphicall. Then verif our answer using calculus.

6 SECTION. Etreme Values 7 The graph of D p over the interval Œ0; is shown below. From the graph, we see that at D,thefunction achieves its maimum value of To verif the information obtained from the plot, let f./ D p. Then f 0./ D =. Solvingf 0./ D 0 ields the critical points D 0 and D.Becausef.0/D f./d 0 and f./d, weseethatthemaimumvalueoff on Œ0; is. 6. Plot f./ D ln 5 sin on Œ0:; and approimate both the critical points and the etreme values. The graph of f./d ln 5 sin is shown below. From the graph, we see that critical points occur at approimatel D 0: and D :. Themaimumvalueofapproimatel :6 occurs at 0:; theminimumvalueofapproimatel :6 occurs at : Approimate the critical points of g./ D cos and estimate the maimum value of g./. g 0./ D p C cos,sog 0./ D 0 when 0:6585.Evaluatingg at the endpoints of its domain, D, and at the critical point 0:6585, wefindg. / D, g.0:6585/ 0:56096, andg./ D 0.Hence,themaimumvalue of g./ is approimatel 0: Approimate the critical points of g./ D 5e tan in ;. Let g./ D 5e tan. Theng 0./ D 5e sec.thederivativeisdefinedforall ; and is equal to 0for :9895 and 0:8780. Hence,thecriticalpointsofg are :9895 and 0:8780. In Eercises 9 58, find the min and ma of the function on the given interval b comparing values at the critical points and endpoints. 9. D C C 5, Œ ; Let f./ D C C 5. Thenf 0./ D C D 0 implies that D is the onl critical point of f.the minimum of f on the interval Œ ; is f. / D, whereasitsmaimumisf./d. (Note: f. / D 5.) 0. D C C 5, Œ0; Let f./ D C C 5. Thenf 0./ D C D 0 implies that D is the onl critical point of f.the minimum of f on the interval Œ0; is f.0/ D 5, whereasitsmaimumisf./ D. (Note: The critical point D is not on the interval Œ0;.). D 6t t, Œ0; 5 Let f.t/d 6t t.thenf 0.t/ D 6 t D 0 implies that t D is the onl critical point of f.theminimumoff on the interval Œ0; 5 is f.0/d 0, whereasthemaimumisf./d 9. (Note: f.5/d 5.). D 6t t, Œ; 6 Let f.t/d 6t t.thenf 0.t/ D 6 t D 0 implies that t D is the onl critical point of f.theminimumoff on the interval Œ; 6 is f.6/d 0, whereasthemaimumisf./d 8. (Note: The critical point t D is not on the interval Œ; 6.). D 6 C 8, Œ; 6 Let f./ D 6 C 8. Thenf 0./ D D. / D 0 implies that D 0 and D are the critical points of f.theminimumoff on the interval Œ; 6 is f./d, whereasthemaimumisf.6/d 8.(Note: f./d and the critical point D 0 is not in the interval Œ; 6.). D C, Œ ; Let f./d C. Thenf 0./ D C D. /. C / D 0 implies that D = and D are critical points of f.theminimumoff on the interval Œ ; is f. / D, whereasthemaimumisf./ D 0. (Note: f. / D and f.=/d 5=7.)

7 7 C HAPTER APPLICATIONS OF THE DERIVATIVE 5. D t C t, Œ; Let f.t/ D t C t.thenf 0.t/ D 6t C 6t D 6t.t C / D 0 implies that t D 0 and t D are the critical points of f.theminimumoff on the interval Œ; is f./ D 5, whereasthemaimumisf./ D 8. (Note: Neither critical points are in the interval Œ;.) 6. D C, Œ0; Let f./ D C. Thenf 0./ D C D. 7/. / D 0 implies that D and D 7 are the critical points of f.theminimumoff on the interval Œ0; is f.0/d 0,whereasitsmaimumisf./D 0.(Note: f./d and the critical point D 7 is not in the interval Œ0;.) 7. D z 5 80z, Œ ; Let f.z/ D z 5 80z. Thenf 0.z/ D 5z 80 D 5.z 6/ D 5.z C /.z C /.z / D 0 implies that z D are the critical points of f.theminimumvalueoff on the interval Œ ; is f./ D 8, whereasthemaimumis f. / D 8. (Note: f. / D and f./d.) 8. D 5 C 5, Œ ; Let f./ D 5 C 5.Thenf 0./ D 0 C 0 D 0. C / D 0 implies that D 0 and D are critical points of f.theminimumvalueoff on the interval Œ ; is f. / D, whereasthemaimumisf./d 8.(Note: f. / D and f.0/d 0.) 9. D C, Œ5; 6 Let f./d C. Then f 0./ D. /. C /. / D 8. / D 0 implies D p7 are critical points of f. D is not a critical point because D is not in the domain of f.ontheinterval Œ5; 6,theminimumoff is f.6/d 7 p7 D 8:5, whereasthemaimumoff is f.5/d 6.(Note: The critical points D are not in the interval Œ5; 6.) 0. D C, Œ; Let f./d C.Then f 0./ D. C /. /. C /. /. C /. C / D. C / D 0 implies that D and D are critical points. Neither D 0 nor D is a critical point because neither is in the domain of f.ontheintervalœ;,themaimumvalueisf./d 0 and the minimum value is f./d 9. (Note: The critical point D is not in the interval Œ;.). D C, Œ0; Let f./d C. Then f 0./ D. /. C / D. C /. C / D 0 implies that D and D are critical points of f. D is not a critical point because D is not in the domain of f.theminimumoff on the interval Œ0; is f./ D, whereasthemaimumisf.0/ D f./ D 0. (Note: The critical point D is not in the interval Œ0;.). D p C, Œ0; Let f./d p C. Then f 0./ D p C D 0 q q implies that D are critical points of f.ontheintervalœ0;, theminimumisf D p and the maimum is f./d p 5. (Note: The critical point D q isnotintheintervalœ0;.)

8 SECTION. Etreme Values 75. D. C / p C. /, Œ0; Let f./d. C / p C. /.Then f 0./ D q C. /. C /. C. / / =. /. / D p D 0 C. / implies that D is the critical point of f.ontheintervalœ0;, theminimumisf.0/ D p 6 :9 and the maimum is f./d p 5:66. (Note: f./d p 5:.). D p C, Œ0; Let f./d p C.Then f 0./ D p C D 0 implies that f has no critical points. The minimum value of f on the interval Œ0; is f./ D p, whereas the maimum is f.0/d. 5. D p C p, Œ0; Let f./d p C p. Then f 0./ D. C / =. C / = D C p C p p C D 0 p p implies that D 0 and D are the critical points of f.neither D nor D p isacriticalpointbecauseneitheris in the domain of f.ontheintervalœ0;, theminimumoff is f 0: and the maimum is f./ 0:76. (Note: f.0/d 0.) 6. D.t t / =, Œ ; Let s.t/ D.t t / =.Thens 0.t/ D.t t / =. t/ D 0 at t D, acriticalpointofs. Othercritical points of s are t D 0 and t D, wherethederivativeofs does not eist. Therefore, on the interval Œ ;, theminimumofs is s. / D s./ D = :6 and the maimum is s. / D. /= 0:6. (Note: s.0/ D s./ D 0.) 7. D sin cos, 0; Let f./ D sin cos D sin. Ontheinterval 0;, f 0./ D cos D 0 when D. The minimum of f on this interval is f.0/d f. / D 0, whereas the maimum is f. / D. 8. D C sin, Œ0; Let f./ D C sin. Thenf 0./ D C cos D 0 implies that D is the onl critical point on Œ0;. The minimum value of f on the interval Œ0; is f.0/d 0,whereasthemaimumisf./D. (Note: f./d.) 9. D p sec, 0; Let f./ D p sec. OntheintervalŒ0;, f 0./ D p sec tan D 0 at D. The minimum value of f on this interval is f.0/ D, whereasthemaimumvalueoverthisintervalisf. / D p. / 0:09. (Note: f. / D p :5909.) 50. D cos C sin, Œ0; Let f./d cos C sin.ontheintervalœ0;, f 0./ D sin C cos D 0 where sin D cos,whichisatthe two points D and 5. The minimum value on the interval is f.5 / D p, whereas the maimum value on the interval is f. / D p. (Note: f.0/d f./d.) 5. D sin, Œ0; Let g./ D sin. OntheintervalŒ0;, g 0./ D cos D 0 at D and D 5. The minimum of g on this interval is g. / D p :685 and the maimum is g. 5 / D 5 C p 6:968. (Note: g.0/ D 0 and g./ D 6:8.) 5. D sin cos, Œ0; Let f./d sin cos.then f 0./ D sin cos C 6 cos sin D 6 cos sin.sin C / D 0 ields D 0; =; ; 7=6; =; =6; as critical points of f.theminimumvalueoff on the interval Œ0; is f.=/ D, whereasthemaimumisf.=/d. (Note: f.0/d f./d f./d and f.7=6/d f.=6/d =.)

9 76 C HAPTER APPLICATIONS OF THE DERIVATIVE 5. D tan, Œ0; Let f./ D tan. ThenontheintervalŒ0;, f 0./ D sec D 0 at D. The minimum of f is f. / D 0: and the maimum is f.0/d 0.(Note: f./d tan 0:59.) 5. D e, Œ0; Let f./d e.then,ontheintervalœ0;, f 0./ D e C e D. /e D 0 at D. Theminimum of f on this interval is f.0/d 0 and the maimum is f./d e 0: (Note: f./d e 0:7067.) 55. D ln, Œ; Let f./d ln. Then, on the interval Œ;, f 0./ D ln D 0 at D e. Theminimumoff on this interval is f./d 0 and the maimum is f.e/d e 0: (Note: f./d ln 0:660.) 56. D e e, ; Let f./ D e e.then,ontheinterval ;, f 0./ D e e D e. e / D 0 at D ln.=/. The minimum of f on this interval is f./ D e e 0: and the maimum is f.ln.=// D :5. (Note: f. =/ D e = e :57.) 57. D 5 tan, Œ; 5 Let f./d 5 tan.then,ontheintervalœ; 5, f 0./ D 5 C D 0 at D. Theminimumoff on this interval is f.5/ D 5 tan 5 5 :86700 and the maimum is f./ D 5 tan :557. (Note: f./d 5 :9699.) 58. D ln, ; Let f./d ln.then,ontheinterval ;, f 0./ D D 0 at D. Theminimumoff on this interval is f./ D 8 ln 8:655 and the maimum is f.=/ D 8 C ln 6:7605. (Note: f./d 7 ln 0:605.) 59. Let f./d sin C sin. (a) Show that is a critical point if cos D cos. (b) Show, using a unit circle, that cos D cos if and onl if D C k for an integer k. (c) Show that cos D cos if and onl if D C k or D 6 C k. (d) Find the si critical points of f./on Œ0; and find the etreme values of f./on this interval. (e) Check our results against a graph of f./. f./ D sin C sin is differentiable at all, sothewatofindthecriticalpointsistofindallpointssuchthat f 0./ D 0. (a) f 0./ D cos C cos. Iff 0./ D 0, then cos D cos, socos D cos. (b) Given the point.cos ;sin / at angle on the unit circle, there are two points with coordinate cos. Thegraphicshows these two points, which are: The point.cos. C /;sin. C // on the opposite end of the unit circle. The point.cos. /;sin. // obtained b reflecting through the ais. If we include all angles representing these points on the circle, we find that cos D cos if and onl if D. C / C k or D. / C k for integers k.

10 SECTION. Etreme Values 77 (c) Using (b), we recognize that cos D cos if D C C k or D C k. Solvingfor, weobtain D C k or D 6 C k. (d) To find all, 0 <indicated b (c), we use the following table: k 0 5 C k 6 C k The critical points in the range Œ0; are 6,, 5 6, 7 6,, and 6. On this interval, the maimum value is f. 6 / D f. 7 6 / D p and the minimum value is f. 5 6 / D f. 6 / D p. (e) The graph of f./d sin C sin is shown here: 5 6 We can see that there are si flat points on the graph between 0 and, aspredicted.therearelocaletrema,andtwopointsat. ; 0/ and. ; 0/ where the graph has neither a local maimum nor a local minimum. 60. Find the critical points of f./ D cos C cos in Œ0;. Checkouransweragainstagraphoff./. f./ is differentiable for all, sowearelookingforpointswheref 0./ D 0 onl. Setting f 0./ D 6sin 6 sin, wegetsin D sin. Lookingataunitcircle,wefindtherelationshipbetweenangles and such that sin D sin.thistechniqueisalsousedineercise59. From the diagram, we see that sin D sin if is either (i.) the point antipodal to ( D C C k) or(ii.)thepoint obtained b reflecting through the horizontal ais ( D Ck). Since sin D sin,wegeteither D C C k or D C k. Solvingeachoftheseequationsfor ields D C k and D 5 k, respectivel. The values of between 0 and are 0; 5 ; 5 ; ; 6 5 ; 8 5 ; and. The graph is shown below. As predicted, it has horizontal tangent lines at 5 k and at D. Each of these points is a local etremum. 5 6 In Eercises 6 6, find the critical points and the etreme values on Œ0;. InEercises6and6,refertoFigure π π π π = + = cos FIGURE

11 78 C HAPTER APPLICATIONS OF THE DERIVATIVE 6. Dj j Let f./ Dj j. For<,wehavef 0./ D. For>,wehavef 0./ D. Nowas!, we f./ f./. / 0 f./ f./. / 0 have D! ; whereasas! C; we have D!. Therefore, f 0./ D f./ f./ lim does not eist and the lone critical point of f is D. Alternatel,weeaminethegraphoff./ Dj j! shown below. To find the etremum, we check the values of f./at the critical point and the endpoints. f.0/d, f./ D, andf./ D 0. f./takes its minimum value of 0 at D,anditsmaimumofat D 0 and at D. 6. D j 9j Let f./dj 9j Dj j.for<,wehavef 0./ D.For>,wehavef 0./ D.Nowas!, f./ f./. / 0 f./ f./ we have D! ; whereas as! C; we have D. / 0!. Therefore, f 0 f./ f././ D lim does not eist and the lone critical point of f is D. Alternatel, we eamine the graph of f./dj 9j shown below.! To find the etrema of f./on Œ0;,wetestthevaluesoff./at the critical point and the endpoints. f.0/d 9, f./d 0 and f./d, sof./takes its minimum value of 0 at D, anditsmaimumvalueof9at D D j C j 8 6 Let f./ Dj C j Dj. C 6/. /j. Fromthegraphoff in Figure, we see that f 0./ does not eist at D 6and at D, sothesearecriticalpointsoff.thereisalsoacriticalpointbetween D 6and D at which f 0./ D 0.For 6 <<, f./d C,sof 0./ D D 0 when D.OntheintervalŒ0; the minimum value of f is f./ D 0 and the maimum value is f./ D 0. (Note: f.0/ D and the critical points D 6and D are not in the interval.) 6. Djcos j Let f./ Djcos j. Therearetwotpesofcriticalpoints:pointsoftheformn where the derivative is zero and points of the form n C = where the derivative does not eist. Onl two of these, D and D are in the interval Œ0;. Now, f.0/ D f./ D, f./ Djcos j 0:656, andf. / D 0, so f./takes its maimum value of at D 0 and D and its minimum of 0 at D. InEercises65 68,verifRolle stheoremforthegiveninterval. 65. f./d C, ; Because f is continuous on Œ ;, differentiableon. ; / and f D C D 5 D C D f./; we ma conclude from Rolle s Theorem that there eists a c. ; / at which f 0.c/ D 0. Here,f 0./ D D, so we ma take c D. 66. f./d sin, ; Because f is continuous on Œ ;, differentiable on. ; / and p f D f D ; wemaconcludefromrolle stheoremthatthereeistsac. ; / atwhichf 0.c/ D 0.Here,f 0./ D cos,sowematake c D.

12 SECTION. Etreme Values f./d 8 5, Œ; 5 Because f is continuous on Œ; 5, differentiableon.; 5/ and f./ D f.5/ D, wemaconcludefromrolle s Theorem that there eists a c.; 5/ at which f 0.c/ D 0.Here, so we ma take c D f./d sin cos, ; f 0./ D.8 5/./ 8. 5/.8 5/ D.8 5/ ; Because f is continuous on Œ ;, differentiable on. ; / and f D f D 0; we ma conclude from Rolle s Theorem that there eists a c. ; / atwhichf 0.c/ D 0. Here, so we ma take c D. f 0./ D sin cos cos. sin / D sin ; 69. Prove that f./d 5 C C has precisel one real root. Let s first establish the f./d 5 C C has at least one root. Because f is a polnomial, it is continuous for all.moreover,f.0/d < 0 and f./d > 0. Therefore,btheIntermediateValueTheorem,thereeistsac.0; / such that f.c/d 0. Net, we prove that this is the onl root. We will use proof b contradiction. Suppose f./d 5 C C has two real roots, D a and D b.thenf.a/d f.b/d 0 and Rolle s Theorem guarantees that there eists a c.a; b/ at which f 0.c/ D 0. However, f 0./ D 5 C 6 C for all, sothereisnoc.a; b/ at which f 0.c/ D 0. Basedonthiscontradiction,we conclude that f./d 5 C C cannot have more than one real root. Finall, f must have precisel one real root. 70. Prove that f./d C C 6 has precisel one real root. First, note that f.0/ D 0, sof has at least one real root. We will proceed b contradiction to establish that D 0 is the onl real root. Suppose there eists another real root, sa D a. Becausethepolnomialf is continuous and differentiable for all real, itfollowsbrolle stheoremthatthereeistsarealnumberc between 0 and a such that f 0.c/ D 0. However, f 0./ D C 6 C 6 D. C / C for all. Thus,thereisnoc between 0 and a at which f 0.c/ D 0. Basedonthis contradiction, we conclude that f./ D C C 6 cannot have more than one real root. Finall, f must have precisel one real root. 7. Prove that f./d C 5 C has no root c satisfing c>0. Hint: Note that D 0 is a root and appl Rolle s Theorem. We will proceed b contradiction. Note that f.0/ D 0 and suppose that there eists a c>0such that f.c/ D 0. Then f.0/ D f.c/ D 0 and Rolle s Theorem guarantees that there eists a d.0; c/ such that f 0.d/ D 0. However,f 0./ D C 5 C >for all >0,sothereisnod.0; c/ such that f 0.d/ D 0. Basedonthiscontradiction,weconcludethat f./d C 5 C has no root c satisfing c>0. 7. Prove that c D is the largest root of f./d 8 8. First, note that f./ D 8./ 8 D D 0, soc D is a root of f.wewillproceedb contradiction to establish that c D is the largest real root. Suppose there eists real root, sa D a, wherea>.becausethe polnomial f is continuous and differentiable for all real,itfollowsbrolle stheoremthatthereeistsarealnumberc.; a/ such that f 0.c/ D 0. However,f 0./ D 6 D. / > 0 for all >.Thus,thereisnoc.; a/ at which f 0.c/ D 0. Basedonthiscontradiction,weconcludethatf./D 8 8 has no real root larger than. 7. The position of a mass oscillating at the end of a spring is s.t/ D A sin!t,wherea is the amplitude and! is the angular frequenc. Show that the speed jv.t/j is at a maimum when the acceleration a.t/ is zero and that ja.t/j is at a maimum when v.t/ is zero. Let s.t/ D A sin!t.then and Thus, the speed v.t/ D ds dt D A! cos!t a.t/ D dv dt D A! sin!t: jv.t/j DjA! cos!tj

13 80 C HAPTER APPLICATIONS OF THE DERIVATIVE is amaimumwhenj cos!tj D, whichispreciselwhensin!t D 0; thatis,thespeedjv.t/j is at a maimum when the acceleration a.t/ is zero. Similarl, ja.t/j DjA! sin!tj is a maimum when j sin!tjd,whichispreciselwhencos!t D 0; thatis,ja.t/j is at a maimum when v.t/ is zero. 7. The concentration C.t/ (in mg=cm )ofadruginapatient sbloodstreamaftert hours is C.t/ D 0:06t t C t C Find the maimum concentration in the time interval Œ0; 8 and the time at which it occurs. C 0.t/ D 0:06.t C t C /.0:06t.t C // t C t.t C t C / D 0:06.t D 0:06 C t C /.t C / : C 0.t/ eists for all t 0, sowearelookingforpointswherec 0.t/ D 0. C 0.t/ D 0 when t D. Usingacalculator,wefindthat C./ D 0:00 mg cm. On the other hand, C.0/ D 0 and C.8/ 0:00. Hence,themaimumconcentrationoccursatt D hours and is equal to :00 mg cm. 75. Antibiotic Levels A stud shows that the concentration C.t/ (in micrograms per milliliter) of antibiotic in a patient s blood serum after t hours is C.t/ D 0.e 0:t e bt /,whereb is a constant that depends on the particular combination of antibiotic agents used. Solve numericall for the value of b (to two decimal places) for which maimum concentration occurs at t D h. You ma assume that the maimum occurs at a critical point as suggested b Figure C (mcg/ml) FIGURE 5 Graph of C.t/ D 0.e 0:t e bt / with b chosen so that the maimum occurs at t D h. t (h) Answer is b D :86.ThemaofC.t/ occurs at t D ln.5b/=.b 0:/ so we solve ln.5b/=.b 0:/ D numericall. Let C.t/ D 0.e 0:t e bt /.ThenC 0.t/ D 0. 0:e 0:t C be bt / D 0 when t D Substituting t D and solving for b numericall ields b :86. ln 5b b 0: : 76. In the notation of Eercise 75, find the value of b (to two decimal places) for which the maimum value of C.t/ is equal to 00 mcg/ml. From the previous eercise, we know that C.t/ achieves its maimum when t D ln 5b b 0: : Substituting this epression into the formula for C.t/,settingtheresultingepressionequalto00andsolvingforb ields b : In 99, phsicist Alfred Betz argued that the maimum efficienc of a wind turbine is around 59%. If wind enters a turbine with speed v and eits with speed v,thenthepoweretractedisthedifferenceinkineticenergperunittime: P D mv mv watts where m is the mass of wind flowing through the rotor per unit time (Figure 6). Betz assumed that m D A.v C v /=, where is the densit of air and A is the area swept out b the rotor. Wind flowing undisturbed through the same area A would have mass per unit time Av and power P 0 D Av.ThefractionofpoweretractedbtheturbineisF D P=P 0. (a) Show that F depends onl on the ratio r D v =v and is equal to F.r/ D. r /. C r/,where0 r. (b) Show that the maimum value of F.r/,calledtheBetz Limit,is6=7 0:59.

14 SECTION. Etreme Values 8 (c) Eplain wh Betz s formula for F.r/ is not meaningful for r close to zero. Hint: How much wind would pass through the turbine if v were zero? Is this realistic? F v v r (A) Wind flowing through a turbine. (B) F is the fraction of energ etracted b the turbine as a function of r = v /v. FIGURE 6 (a) We note that F D P P 0 D A.v Cv /.v v / Av D v v v v C v v! D v v C v v D. r /. C r/: (b) Based on part (a), F 0.r/ D. r / r. C r/ D r r C : The roots of this quadratic are r D and r D. Now, F.0/ D, F./ D 0 and F D 8 9 D 6 7 0:59: Thus, the Betz Limit is 6=7 0:59. (c) If v were 0, then no air would be passing through the turbine, which is not realistic. 78. The Bohr radius a 0 of the hdrogen atom is the value of r that minimizes the energ E.r/ D mr e 0 r where, m, e,and 0 are phsical constants. Show that a 0 D 0 =.me /.Assumethattheminimumoccursatacriticalpoint, as suggested b Figure 7. E(r) (0 8 joules) r (0 0 meters) FIGURE 7

15 8 C HAPTER APPLICATIONS OF THE DERIVATIVE Let Then implies Thus, de dr E.r/ D mr e 0 r : D mr C e 0 r D 0 r D 0 me : a 0 D 0 me : 79. The response of a circuit or other oscillator sstem to an input of frequenc! ( omega ) is described b the function.!/ D q.! 0! / C D! Both! 0 (the natural frequenc of the sstem) and D (the damping factor) are positive constants. The graph of is called a resonance curve, andthepositivefrequenc! q r > 0,where takes its maimum value, if it eists, is called the resonant frequenc.showthat! r D! 0 D if 0<D<! 0 = p and that no resonant frequenc eists otherwise (Figure 8) r r (A) D = 0.0 (B) D = 0. (C) D = 0.75 (no resonance) FIGURE 8 Resonance curves with! 0 D. Let.!/ D..! 0! / C D! / =.Then 0.!/ D!..! 0! / D /..! 0! / C D! / = q and the non-negative critical points are! D 0 and! D! 0 D.Thelattercriticalpointispositiveifandonlif! 0 D > 0, andsincewearegivend>0,thisisequivalentto0<d<! 0 = p q. Define! r D! 0 D.Now,.0/ D =! 0 and.!/! 0 as!!.finall,.! r / D ; D q! 0 D which, for 0<D<! 0 = p, is larger than =! 0.Hence,thepoint! D q! 0 D,ifdefined,isalocalmaimum. 80. Bees build honecomb structures out of cells with a heagonal base and three rhombus-shaped faces on top, as in Figure 9. We can show that the surface area of this cell is A./ D 6hs C s. p csc cot / with h, s, and as indicated in the figure. Remarkabl, bees know which angle minimizes the surface area (and therefore requires the least amount of wa). (a) Show that 5:7 ı (assume h and s are constant). Hint: Find the critical point of A./ for 0<<=. (b) Confirm, b graphing f./ D p csc cot, thatthecriticalpointindeedminimizesthesurfacearea.

16 SECTION. Etreme Values 8 θ h FIGURE 9 A cell in a honecomb constructed b bees. s (a) Because h and s are constant relative to, wehavea 0./ D s. p csc cot C csc / D 0. Fromthis,weget p csc cot D csc,orcos D p, whence D cos p D 0:9557 radians = 5:76 ı. (b) The plot of p csc cot,where is given in degrees, is given below. We can see that the minimum occurs just below 55 ı. h Degrees 8. Find the maimum of D a b on Œ0; where 0<a<b.Inparticular,findthemaimumof D 5 0 on Œ0;. Let f./d a b.thenf 0./ D a a b b.sincea<b, f 0./ D a.a b b a / D 0 implies critical points D 0 and D. a b /=.b a/,whichisintheintervalœ0; as a<bimplies a b < and consequentl D. a b /=.b a/ <. Also, f.0/ D f./ D 0 and a<bimplies a > b on the interval Œ0;, whichgivesf./ > 0 and thus the maimum value of f on Œ0; is a =.b a/ a a=.b a/ a b=.b a/ f D : b b b Let f./d 5 0.Thenbpart(a),themaimumvalueoff on Œ0; is! =5 f D D D : InEercises8 8,plotthefunctionusingagraphingutilitandfinditscriticalpointsandetremevaluesonŒ 5; D C j j Let f./d Cj j.theplotfollows: We can see on the plot that the onl critical point of f./ lies at D. Withf. 5/ D 7, f./ D and f.5/d 5, it follows that the maimum value of f./on Œ 5; 5 is f./d and the minimum value is f. 5/ D 7.

17 8 C HAPTER APPLICATIONS OF THE DERIVATIVE 8. D C j j C C j j Let The plot follows: f./d C j j C C j j : We can see on the plot that the critical points of f./ lie at the cusps at D and D and at the location of the horizontal tangent line at D 5 7. With f. 5/ D 70, f./ D f./ D 5, f. 5 / D 5 and f.5/ D 0 7, it follows that the maimum value of f./on Π5; 5 is f./d f./d 5 7 and the minimum value is f. 5/ D D j j Cj j Let f./d.thecuspsofthegraphofafunctioncontainingjg./j are likel to lie where g./ D 0, j jcj j so we choose a plot range that includes D and D : As we can see from the graph, the function has cusps at D and sharp corners at D. Thecuspsat.; / and. ; / are the locations of the maimum and minimum values of f./,respectivel. 85. (a) Use implicit differentiation to find the critical points on the curve 7 D. C /. (b) Plot the curve and the horizontal tangent lines on the same set of aes. (a) Differentiating both sides of the equation 7 D. C / with respect to ields 5 D. C / C d : d Solving for d=d we obtain d d D 7. C /. C / D.9. C / /. C / : Thus, the derivative is zero when C D. Substitutingintotheequationforthecurve,thisields D, or D. There are therefore four points at which the derivative is zero:. ; p /;. ; p /;.; p /;.; p /: There are also critical points where the derivative does not eist. This occurs when D 0 and gives the following points with vertical tangents:.0; 0/;. p 7; 0/: (b) The curve 7 D. C / and its horizontal tangents are plotted below.

18 SECTION. Etreme Values Sketch the graph of a continuous function on.0; / with a minimum value but no maimum value. Here is the graph of a function f on.0; / with a minimum value [at D ]butnomaimumvalue[sincef./! as! 0C and as! ] Sketch the graph of a continuous function on.0; / having a local minimum but no absolute minimum. Here is the graph of a function f on.0; / with a local minimum value [between D and D ] butnoabsolute minimum [since f./! as! 0C] Sketch the graph of a function on Œ0; having (a) Two local maima and one local minimum. (b) An absolute minimum that occurs at an endpoint, and an absolute maimum that occurs at a critical point. Here is the graph of a function on Œ0; that (a) has two local maima and one local minimum and (b) has an absolute minimum that occurs at an endpoint (at D 0 or D ) andhasanabsolutemaimumthatoccursatacriticalpoint Sketch the graph of a function f./ on Œ0; with a discontinuit such that f./ has an absolute minimum but no absolute maimum. Here is the graph of a function f on Œ0; that (a) has a discontinuit [at D ] and(b)hasanabsoluteminimum [at D 0] butnoabsolutemaimum[sincef./!as! ] A rainbow is produced b light ras that enter a raindrop (assumed spherical) and eit after being reflected internall as in Figure 0. The angle between the incoming and reflected ras is D r i, wheretheangleofincidencei and refraction r are related b Snell s Law sin i D n sin r with n : (the inde of refraction for air and water). (a) Use Snell s Law to show that dr di D cos i n cos r. s n (b) Show that the maimum value ma of occurs when i satisfies cos i D. Hint: Show that d di D 0 ifcosi D n cos r. Then use Snell s Law to eliminate r.

19 86 C HAPTER APPLICATIONS OF THE DERIVATIVE (c) Show that ma 59:58 ı. Incoming light ra i r θ r r Water droplet r i Reflected ra FIGURE 0 (a) Differentiating Snell s Law with respect to i ields cos i D n cos r dr di (b) Differentiating the formula for with respect to i ields or dr di D cos i n cos r : d di D dr cos i D di n cos r b part (a). Thus, d di D 0 when cos i D n cos r: Squaring both sides of this last equation gives cos i D n cos r; while squaring both sides of Snell s Law gives Solving this equation for cos r gives sin i D n sin r or cos i D n. cos r/: cos r D cos i n I Combining these last two equations and solving for cos i ields s n cos i D : (c) With n D :, cos i D s.:/ D 0:506 and cos r D cos i D 0:76: : Thus, r D 0: ı, i D 59:58 ı and ma D r i D :5 ı :

20 SECTION. Etreme Values 87 Further Insights and Challenges 9. Show that the etreme values of f./d a sin C b cos are pa C b. If f./ D a sin C b cos, thenf 0./ D a cos b sin, sothatf 0./ D 0 implies a cos b sin D 0. This implies tan D a b. Then, Therefore sin D a b p and cos D p a C b a C b : a f./d a sin C b cos D a p a C b C b b p a C b D a C b p a C b D pa C b : 9. Show, b considering its minimum, that f./d C takes on onl positive values. More generall, find the conditions on r and s under which the quadratic function f./ D C r C s takes on onl positive values. Give eamples of r and s for which f takes on both positive and negative values. Observe that f./d C D. / C >0for all. Letf./D C r C s.completingthesquare,wenote that f./d. C r/ C s r > 0 for all provided that s> r. Let f./d C D. /. /.Thenf takes on both positive and negative values. Here, r D and s D. 9. Show that if the quadratic polnomial f./d C r C s takes on both positive and negative values, then its minimum value occurs at the midpoint between the two roots. Let f./d C r C s and suppose that f./takes on both positive and negative values. This will guarantee that f has two real roots. B the quadratic formula, the roots of f are Observe that the midpoint between these roots is r C p r s D r pr s : C r p! r s D r : Net, f 0./ D C r D 0 when D r and, because the graph of f./is an upward opening parabola, it follows that f. r / is aminimum.thus,f takes on its minimum value at the midpoint between the two roots. 9. Generalize Eercise 9: Show that if the horizontal line D c intersects the graph of f./ D C r C s at two points. ;f. // and. ;f. //,thenf./takes its minimum value at the midpoint M D C (Figure ). c f() = c M FIGURE Suppose that a horizontal line D c intersects the graph of a quadratic function f./d C r C s in two points. ;f. // and. ;f. //. Thenofcoursef. / D f. / D c. Letg./ D f./ c. Theng. / D g. / D 0. BEercise 9, g takes on its minimum value at D. C /.Hencesodoesf./D g./ C c. 95. Acubicpolnomialmahavealocalminandma,oritmahaveneither(Figure).Findconditionsonthecoefficientsa and b of f./d C a C b C c that ensure that f has neither a local min nor a local ma. Hint: Appl Eercise 9 to f 0./.

21 88 C HAPTER APPLICATIONS OF THE DERIVATIVE (A) (B) FIGURE Cubic polnomials Let f./ D C a C b C c. UsingEercise9,wehaveg./ D f 0./ D C a C b>0for all provided b> a, in which case f has no critical points and hence no local etrema. (Actuall b a will suffice, since in this case [as we ll see in a later section] f has an inflection point but no local etrema.) 96. Find the min and ma of where p; q > 0. f./d p. / q on Œ0;, Let f./d p. / q ;0,wherep and q are positive numbers. Then f 0./ D p q. / q. / C. / q p p D p. / q.p. / q/ D 0 at D 0; ; The minimum value of f on Œ0; is f.0/d f./d 0, whereasitsmaimumvalueis p p p q q f D p C q.p C q/ pcq : p p C q 97. Prove that if f is continuous and f.a/ and f.b/are local minima where a < b,then there eists a value c between a and b such that f.c/ is a local maimum. (Hint: Appl Theorem to the interval Œa; b.) Show that continuit is a necessar hpothesis b sketching the graph of a function (necessaril discontinuous) with two local minima but no local maimum. Let f./be a continuous function with f.a/and f.b/local minima on the interval Œa; b.btheorem, f./must take on both a minimum and a maimum on Œa; b. Sincelocalminimaoccuratf.a/ and f.b/,themaimummustoccuratsome other point in the interval, call it c, wheref.c/is a local maimum. The function graphed here is discontinuous at D The Mean Value Theorem and Monotonicit Preliminar Questions. For which value of m is the following statement correct? If f./ D and f./ D 9, andf./is differentiable, then f has a tangent line of slope m. The Mean Value Theorem guarantees that the function has a tangent line with slope equal to Hence, m D makes the statement correct. f./ f./ D 9 D :. Assume f is differentiable. Which of the following statements does not follow from the MVT? (a) If f has a secant line of slope 0, then f has a tangent line of slope 0. (b) If f.5/<f.9/,thenf 0.c/ > 0 for some c.5; 9/. (c) If f has a tangent line of slope 0, then f has a secant line of slope 0. (d) If f 0./ > 0 for all,theneversecantlinehaspositiveslope.