微积分英文课件 -005.ppt

1、The best way to learn mathematics is to do mathematicsChapter 4 Applications of differentiation4.1 Maximum and minimum values4.2 The Mean value Theorem4.3 How Derivatives Affect the shape of a Graph4.4 Indeterminate Forms and L Hospitals Rule4.5 Summary of Curve Sketching4.6 Graphing with Calculus a

2、nd Calculators4.7 Optimization Problems4.8 Applications to business and Economics4.9 Newtons Method4.10 Antiderivatives A function f has an(or)at cif()for all xin D,where Dis the domain of f.absolute maxiumuglobal maximummaxThe number of()is called theof f on D.Similarly,f hasimum valueabsolute mini

3、manat cifD()()for all xefuinitioninmf cf xf cf cf xD and the number f(c)is called theof f on D.The maximum and minimum values of fare calledminmumvalueextreme valuethes of f.4.1 Maximum and minimum valuesThe maximum and minimum values of f are called the extreme values of f.cd(c,f(c)(d,f(d)A functio

4、n f has amaximum(or)at cif()when x is near c.This meanthat()for all x in some openinterval containing c.SimilaLocalrelrly,f hasative maximumlocal minmuDefinaat c if()()when x is near cti nm.i of cf xf cf xf cf xExample:4118163)(234xxxxxfxxfcos)(2)(xxf3)(xxfExample:Example:Example:int,maxmum()minimum

5、(),.If f is continuouson a closederval a b then f attainsan absolutevalue f c and anabsolutevalue f d at somenumbers cThe ExtremeTheoand d in ambreWe have seen that some functions have extreme values,whereas others do not.The following theoremgives conditions under which a function is guaranteed to

6、possess extreme values.Example)1,0(,xxyThis function has no maximum or minimum xoy1121,31,110,1)(xxxxxxfxoy1122机动 目录 上页 下页 返回 结束 ExampleThis function has no maximum or minimum Caution:nThe conditions cannot be weakened.Example22341()122223xxxf xxxxx Caution:The conditions is sufficient,but not neces

7、sary.This function has minimum value f(-1)=-4This function has maximum value f(-4)nProblem:nThe Extreme Value Theorem says that a continuous function on a closed interval has a maximum value and minimum value.nBut it does not tell us how to find these extreme values.nWe start by looking for local ex

8、treme values.If f has a local maximumor minmum at c,and if f(c)exiFermats Tsts,thenheof=e0r m(c)cProof:Without loss of generality,we consider local Maximum:(),()(),chcf chf c()()0,f chf chWe haveif 0,h()()0,f chf ch0,h if 0()()lim0hf chf ch()fc()fc()()()0.fcfcfc0()()lim0hf chf ch()fcThereforeBecause

9、 exists,ThusnCaution:n1)The condition is just sufficient.We cannot expect to locate extreme values simply by setting and solving for x.For example:2)There may be an extreme value even when does not exist.For example:0)(xf0)(3xatxxf()fc0)(xatxxfnFermats Theorem does suggest that we should at least st

10、art looking for extreme values of f at the numbers c where or does not exist.nDefinition:nA critical number of a function f is a number c in the domain of f such that either or does not exist.0)(cf0)(cf)(cf)(cf nConclusion:If f has a local maximum or minimum at c,then c is a critical number of f.nTo

11、 find an absolute maximum or minimum of a continuous function on a closed interval,we note that either it is local in which case it occur at a critical number by the above conclusion nor it occurs at an endpoint of the interval.nThe Closed Interval Method n To find the absolute maximum and minimum v

12、alues of a continous function f on a closed interval a,b:n1)Find the values of f at the critical numbers of f in(a,b).n2)Find the values of f at the endpoints of the interval.n3)The largest of the values from step 1)and step 2)is the absolute maximum value;the smallest of these values is the absolut

13、e minimum value.nExample Find the absolute maximum and minimum values of the functionnSolution:Since f is continuous on the given closed interval,we can use the Closed Interval Method:42113)(23xxxxf1233maximumminimum()(1)12Find the absoluteandvalueof the function f xxxx ，.解解答答23133(1)xyxx,10,13Thust

14、he critical numbers arexxand xTherefore,0y if130,xthat is,1,3x and dose not existywhen0,1.xor x解解答答x()fx134001/334/310232The values of at these critical numbers and at the endpoints of the interval areyComparing these numbers,we get 3max(2)2yf3min(1)4.yf kmp 4.2 The Mean Value TheoremnRolle s Theore

15、m:Let f be a function that satisfy the following three hypotheses:n1)f is continuous on the closed interval a,b.n2)f is differentiable on the open interval(a,b).n3)f(a)=f(b)nThen there is a number c in(a,b)such that 0)(cfbachaipisi:z nProof Because f(x)is continuous on a,b,by the Extreme Value Theor

16、em,we know that f(x)can take on its absolute maximum M and absolute minimum m there are two cases:nCase1:M=m means that f(x)is a constant function,then ,so any number c in(a,b)is OK!ab0)(xfn Case2:m0 for all x in(a,b),then f(x)is strictly increasing on a,b.(2)If f(x)0 for all x in(a,b),then f(x)is s

17、trictly decreasing on a,b.PROOF Choose any two points x1 and x2 in a,b with x10 for all x in I,then the graph of f(x)is concave upward on I.If f”(x)0 when x-1/3 and f”(x)0 when x-1/3.Intervalsign of f -+behavior of f(,1/3)1/3(1/3,)the sign of and the behavior of f are recorded below.f-12-8-4 4812-12

18、-8-44812 By Concavity Test,we get the following results:the graph of f(x)is concave downward on(-,-1/3)and is concave upward on(-1/3,+).So the point(-1/3,-88/27)is an inflection point.The graph of f is sketched in Figure 2.-12-8-4 4812-12-8-44812Figure 2The Second Derivative Test Suppose that f(x)is

19、 differentiable on an open interval containing c and that f(c)=0.1 If f”(c)0,then f(x)has a local minimum at c.Example 6 Discuss the curve y=f(x)=x4-4x3 with respect to local maxima and minima.Solution 3222()4124(3)()122412(2)fxxxxxfxxxx xTo find the critical numbers we set f(x)=0 and obtain x=0 and

20、 x=3.To use the Second Derivative Test we evaluate f”at these critical numbers:f”(0)=0 f”(3)=360 Since f(3)=0 and f”(3)0,f(3)=-27 is a local minimum.Since f”(0)=0,the Second Derivative Test gives no information about the critical number 0.But since f(x)0 for x0 and also for 0 x3,the First Derivative

21、 Test tells us that f does not have a local maximum or minimum at 0.Example 7 Sketch the graph of the function f(x)=x2/3(6-x)1/3.Solution 1/32/34/35/348()()(6)(6)Since()0 when4 and()does not exist when 0 or 6,the critical numbers are 0,4,and 6.xfxfxxxxxfxxfxxx Interval (-,0)(0,4)(4,6)(6,+)f +fdecrea

22、sing on(-,0)increasing on(0,4)decreasing on(4,6)decreasing on(6,+)By the First derivative Test,f(0)=0 is a local minimum and f(4)=25/3 is a local maximum.The sign of f does not change at 6,so there is no minimum or maximum there.Looking at the expression for f”(x),we have f”(x)0 for x0 and for 0 x0

23、for x6.Thus f is concave downward on(-,0)and(0,6)and concave upward on(6,+),and the only inflection point is(6,0).The graph is sketched in Figure 4.Figure 44.5 Summary of Curve sketchingGuidelines for Sketching a CurveA.Domain E.Intervals of Increase or DecreaseB.Intercepts F.Local Extreme ValuesC.S

24、ymmetry G.Concavity and Points of InflectionD.Asymptotes H.Sketch the CurveExample 1 Use the guidelines to sketch the curve222.1xyxExample 1 Use the guidelines to sketch the curve222.1xyxA.The domain isB.The x-and y-intercepts are both 0.C.Since f(-x)=f(x),the function f is even.D.Therefore,the line

25、 y=2 is a horizontal asymptote.(,1)(1,1)(1,).222lim21xxx2212lim1xxx 2212lim1xxx 2212lim1xxx 2212lim1xxx Therefore,the lines x=1 and x=-1 are vertical asymptotes.E.2222224(1)224()(1)(1)x xxxxfxxx Since f(x)=0,when x=0,we complete the following table.the only critical number is x=0.Interval (-,-1)(-1,

26、0)(0,1)(1,+)f +fincreasing on(-,-1)increasing on(-1,0)decreasing on(0,1)decreasing on(1,+)So f is increasing on(-,-1)and(-1,0)and decreasing on(0,1)and(1,).F.Since f changes from positive to negative at 0,f(0)is a local maximum.G.We have22224234(1)42(1)2124()(1)(1)xxxxxfxxxFigure 1 Interval f”+fCUCD

27、CU(,1)(1,1)(1,)Thus,the curve is concave upward on the intervals(-,-1)and(1,)and concave downward on(-1,1).It has no point of infl-ection.We sketch the curve in Figure 1.Slant(Oblique)AsymptotesThe line y=ax+b is a slant asymptote for a function f(x)if and only if exist.()limandlim()xxf xaf xaxbxExa

28、mple 2 Find the slant asymptotes,if any,of the function Solution The domain of f(x)is-x0 for all xc and f(x)c,then f(c)is the global maximum value of f.(b)If f(x)0 for all x0 for all xc,then f(c)is the global minimum value of f.Economics().()Suppose c x is thetotal cost that a companyincurs in produ

29、cing x units of a certaincommodity.The function c is called a cost functionc x is called the marginal cost3.3 Rates of Change in Natural and sciencesprdju:si kmditi ink:(1)().(1).Economists defined the marginal cost at aproduction level x to bec xc xwhich is the cost of producing one additionalunit

30、of the commodityor the cost of producingthe xst unit0(1)()(1)()1()()lim()hsincec xc xc xc xc xhc xc xhprudkn 2()()1000050.01()50.02500(500)50.02(500)$15/Suppose a company has estimated that cost in dollarsof producing x itemisC xxxThen the marginal cost function isC xxThe marginal cost at the produc

31、tion level ofitemisCite501(501)(500)$15.01/(500)(501)(500)mThe actual cost of producing thest itemisCCitemNoticethat CCC4.7 Applications to Business and EconomicsInflection point()yC x()c xx()C xThe cost function The marginal costThe average cost function()()C xc xx()C x()C xFigure 1We have 2()()()x

32、C xC xc xxNow c(x)=0 when xC(x)-C(x)=0 and this gives C(x)=C(x)/x=c(x)If the average cost is a minimum,then marginal cost=average costExample 1 A company estimates that the cost(in dollars)of producing x items is(a)Find the average cost and marginal cost of producing 1000 items,2000 items,and 3000 i

33、tems.2()260020.001.C xxx(b)At what production level will the average cost be lowest,and what is this minimum average cost?Solution (a)The average cost function isThe marginal cost function is()()/0.00122600/c xC xxxx()0.0022.C xxWe use these expressions to fill in the following table.x C(x)c(x)C(x)1

34、00056005.604.002000106005.306.003000176005.878.00(b)To minimize the average cost we must have()()0.00220.00122600/26000001612.451612C xc xxxxxTo see that this production level actually gives a minimum,we note that c”(x)=5200/x3 0,so c is concave upward on its entire domain.The minimum average cost i

35、s c(1612)=0.001(1612)+2+2600/1612=$5.22/itemLet p(x)be the price per unit that the company can charge if it sells x units.Then p is called the demand function(or price function)and we would expect it to be a decreasing function of x.If x units are sold and the price per unit is p(x),then the total r

36、evenue is R(x)=xp(x),and R(x)is called the revenue function.The derivative R(x)is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold.If x units are sold,then the total profit is P(x)=R(x)-C(x),and P is called the profit function.The mar

37、ginal profit function is P,the derivative of the profit function.In order to maximize profit we look for the critical numbers of P where the marginal profit is 0.But if P(x)=R(x)-C(x)=0,then R(x)=C(x).To ensure that this condition gives a maximum,we use the Second Derivative Test.Note that P”(x)=R”(

38、x)-C”(x)0,when R”(x)C”(x).Thus,the profit will be a maximum when R(x)=C(x)and R”(x)C”(x)inu Example 2 Determine the production level that will maximize the profit for a company with cost and demand functions C(x)=3800+5x-x2/1000 and p(x)=50-x/100Solution The revenue function is R(x)=xp(x)=50 x-x2/10

39、0,so the marginal revenue function is R(x)=50-x/50,and the marginal cost function is C(x)=5-x/500.Thus marginal revenue is equal to marginal cost when 50-x/50=5-x/500.Solving,we get x=2500.To check that this gives a maximum,we compute the second derivatives:R”(x)=-1/50 C”(x)=-1/500Thus,R”(x)0.Theref

40、ore,a production level of 2500 units will maximize profits.4.8 AntiderivativesDefinition A function F is called an antiderivative(or primitive function)of f(x)on an interval I if F(x)=f(x)for all x in I.For instance,let f(x)=x2.It is not difficult to see that the function F(x)=x3/3 is an antiderivat

41、ive of f(x)since F(x)=x2=f(x).However,the function G(x)=100+x3/3 also satisfies G(x)=x2=f(x),so G(x)is an antiderivative of f(x).Indeed any function of the form H(x)=x3/3+c,where c is a constant,is an antiderivative of f(x).Theorem If F is an antiderivative of f(x)on an interval I,then the most gene

42、ral antiderivative of f(x)on I is F(x)+cwhere c is an arbitrary constant.Going back to the function f(x)=x2,we see that the general antiderivative of f is x3/3+c.By assigning specific values to the constant c,we obtain a family of functions whose graphs are vertical translates of one another(see Fig

43、ure 1).Figure 1Example 1 Find the most general antiderivative of each of the following functions.()()sin()()1/()(),1na f xxb f xxc f xxn Solution (a)Since(-cosx)=sinx,we get an antidirivative of sinx is-cosx.So the most general antiderivative is-cosx+c.(b)Recall that We have the general antiderivati

44、ve of 1/x is(ln)1/(0).xxx 12ln0()lnln()0 xcif xF xxcxcif x(c)We use the Power Rule to discover an antiderivative of xn.In fact,then1n 1()1nnxxn Thus,the general antiderivative of f(x)=xn is1()1nxF xcnThis is valid for since then f(x)=xn is defined on an interval.If n is negative(but ),it is valid on

45、 any interval that does not contain 0.0n 1n If F=f,G=g,we list some particular antiderivatives in the following table.Function Particular antiderivative222()()()(1)1/cossinsecsectan1/11/(1)nxkf xf xg xxnxexxxxxxx 1()()()/(1)lnsincostansecarcsinarctannxkF xF xG xxnxexxxxxxExample 2 Solution The gener

46、al antiderivative of f(x)is2Findif()20/(1)and(0)2.xffxexf()20arctanTo determinewe use the fact that(0)2:(0)120arctan02Thus,we have3,so the particular solution is()20arctan3xxf xexccffccf xex The Geometry of AntiderivativesExample 3 The graph of a function f is given in Figure 2(a).Make a rough sketc

47、h of an antiderivative F,given that F(0)=2.Solution The slope of y=F(x)is f(x).We start at the point(0,2)and draw F as an initially decreasing function since f(x)is negative when 0 x1.Notice that f(1)=f(3)=0,so F has horizontal tangents when x=1 and x=3.Figure 2(a)(b)For 1x3,f(x)is negative and so F

48、 is decreasing on(3,).Since f(x)0 as x,the graph of F becomes flatter as x.Also notice that F”(x)=f(x)changes from positive to negative at x=2 and from negative to positive at x=4,so F has inflection points when x=2 and x=4.We use this information to sketch the graph of the antiderivative in Figure 2(b).