微积分第四章课件.ppt

1、1 Chapter 4 Applications of Differentiation4.1 Maximum and Minimum valuesDefinition 1 A function f has an absolute maximum(or global maximum)at c if for all x in D,where D is the domain of f.The number f(c)is called the maximum value of f on D.f has an absolute minimum(or global minimum)at c if for

2、all x in D,where D is the domain of f.The number f(c)is called the minimum value of f on D.The maximum and minimum values of f are called the extreme values of f.()()f cf x()()f cf x2Definition 2 Let c be a number in the domain of a function f.1 f(c)is a local(or relative)maximum value of f if there

3、 is an open interval I containing c such that f(c)f(x)for all x in I.2 f(c)is a local(or relative)minimum value of f if there is an open interval I containing c such that f(c)f(x)for all x in I.3 In Figure 1,the function takes on a local minimum at c.f(b)is a local maximum value of f(x);f(d)is a max

4、imum value of f(x)on a,e and is also a local maximum value of f(x).y a 0 b c d e x Figure 1 f(a)is a minimum value of f(x)on a,e,but it is not a local minimum value of f(x)because there is no open interval I contained in a,e such that f(a)is the least value of f(x)on I.4Example 1 If f(x)=x2,then f(x

5、)f(0)for all x.Therefore,f(0)=0 is the absolute(and local)minimum value of f.However,this function has no maximum value.Example 2 The graph of the function is shown in Figure 2.We can see that f(1)=5 is a local maximum,whereas the absolute maximum is f(-1)=37.Also,f(0)=0 is a local minimum andf(3)=-

6、27 is both a local and an absolute minimum.432()3161814f xxxxx Figure 25Theorem 3 (Fermats Theorem)If f(x)has a local extreme value(that is,maximum or minimum)at c and if f(c)exists,then f(c)=0.Note We cant expect to locate extreme values simply by setting f(x)=0 and solving for x.dcFigure 36Proof F

7、or the sake of definiteness,suppose that f has a local maximum at c.Then,(,),()(),cxccf cxf c 0()()()limxf cxf cfcx()fc()fc00)0(x)0(x()0fc7Example 3 If f(x)=x3,then f(x)=3x2,so f(0)=0.But f has no maximum or minimum at 0.Example 4 The function has its(local and absolute)minimum value at 0,but that v

8、alue cant be found by setting f(x)=0 because f(0)does not exist.Definition 4 A critical number(or point)of a function f is a number c in the domain of f such that either f(c)=0 or f(c)does not exist.()f xx8Example 5 Find the critical numbers of Solution We have 3/5()(4).f xxx2/53/52/53128()(4)(1)55x

9、fxxxxx3Therefore,()0 if,and()does not exist23when0.Thus,the critical numbers areand 0.2fxxfxx9The Closed Interval Method To find the absolute maximum and minimum values of a continuous function f on a closed interval a,b:1.Find the values of f at the critical numbers of f in(a,b)2.Find the values of

10、 f at the endpoints of the interval a,b3.The largest of the values from Steps 1 and 2 is the absolute maximum value;the smallest of these values is the absolute minimum value.10Example 6 Find the absolute maximum and minimum values of the function Solution Since f is continuous on-1/2,4,We get 321()

11、314.2f xxxx2()363(2)fxxxx xSince()exists for all,the only critical numbers ofoccurwhen()0,that is,0 or 2.Notice that each of these critical numbers lies in the interval(-1/2,4).fxxffxxx11(0)1(2)3()(4)1728ffff Comparing these four numbers,we see that the absolute maximum value is f(4)=17 and the abso

12、lute minimum value is f(2)=-3.11 4.2 The Mean Value Theorem Rolles Theorem Let f be a function satisfying the following three hypotheses:1 f is continuous on a closed interval a,b;2 f is differentiable on the open interval(a,b);3 f(a)=f(b).Then there is a number c in(a,b)such that f(c)=0.PROOF If f(

13、x)is constantly k on a,b,the result is obvious.If f(x)is not constantly k on a,b,f(x)must take on a maximum value at some point d of a,b and a minimum value at some point c of a,b.Since f(a)=f(b),c or d cannot be a and cannot be b.12This means that c or d must lie in the open interval(a,b)and theref

14、ore f(c)or f(d)exists.Assume that f(x)takes on maximum value at d and d(a,b).We have f(d)=f-(d)=f+(d).By the definition of the derivative,We conclude therefore that f(d)=0.),()()()()(lim;),()()()()(limdxdfxfdfdxdfxfdxdfxfdfdxdfxfdxdx since,0 since,013xyoab)(xfy cFigure 1Example 1 Lets apply Rolles T

15、heorem to the position function s=f(t)of a moving object.If the object is in the same place at two different instants t=a and t=bthen f(a)=f(b).Rolles Theorem says that there is some instant of t=c between a and b when f(c)=0;that is,the velocity is 0.(In particular,you can see that this is true whe

16、n a ball is thrown directly upward.)14Example 2 Prove that the polynomial function p(x)=2x3+5x 1 has exactly one real zero.Proof Obviously,p(x)is continuous on 0,1 and p(0)=-10,we know that p(x)has at least one real zero.Suppose that p(x)has more than one real zero.In particular,suppose that p(a)=p(

17、b)=0,where a and b are real numbers and ab.15Without loss of generality,assume that a 0 for all x,(since x2 is nonnegative)so p(x)0 for all x.Thus,we have a contradiction and we can conclude that p(x)has exactly one real zero.16(The Mean Value TheoremLagranges Theorem)Let f be a function satisfying

18、the following hypotheses:1 f is continuous on a closed interval a,b;2 f is differentiable on the open interval(a,b);Then there is a number c in(a,b)such that f(c)=(f(b)-f(a)/(b-a),or,f(b)-f(a)=f(c)(b-a).xyoab)(xfy cFigure 217PROOF OF THE MEAN-VALUE THEOREM We can create a function g(x)that satisfies

19、 the conditions of Rolles Theorem.It is not hard to see that is exactly such a function.If f is differentiable on(a,b)and continuous on a,b,then so is g.As you can check,g(a)and g(b)are both 0.By Rolles Theorem,there is at least one number c in(a,b)for which g(c)=0.So,we obtain)()()()()()(afaxabafbf

20、xfxg.)()()(abafbfcf18Example 3 Lets consider f(x)=x3-x,a=0,b=2.since f is a polynomial,it is continuous and differentiable for all x,so it is certainly continuous on 0,2 and differentiable on(0,2).Therefore,by the mean value Theorem,there is a number c in(0,2)such that f(2)-f(0)=f(c)(2-0)so this equ

21、ation becomes 6=(3c2-1)2=6c2-2 which gives But c must lie in(0,2),so2/3.c 2/3.c Figure 319Example 4 Suppose that f(0)=-3 and f(x)5 for all values of x.How large can f(2)possibly be?Solution Since f is differentiable(and therefore continuous)everywhere,we can apply the Mean Value Theorem on the inter

22、val 0,2.There is a number c in(0,2)such that f(2)-f(0)=f(c)(2-0)so f(2)=f(0)+2f(c)=-3+2f(c)which gives f(2)-3+10=7.The large possible value for f(2)is 7.20Example 5 Given that for all real number x,show that for all real numbers x1 and x2.Solution Let since f(x)is continuous on x1,x2 and differentia

23、ble on(x1,x2),the function f(x)satisfies the hypotheses of Lagranges Theorem and there exists at least a number c in(x1,x2)such that With we have ()1fx1212()()f xf xxx12,xx2121()()()()f xf xfc xx()1fx212121()()()()f xf xfc xxxx21Example 6 Prove thatProof Let Since f(t)is continuous on 0,x and differ

24、entiable on(0,x),the function f(t)satisfies the hypotheses of Lagranges Theorem and there exists at least a number c in(0,x)such that or,equivalently,And we have ln(1),0.1xxxxx()ln(1),0,.f tttx()(0)()(0)f xffc x1ln(1).1xxc111,(0,),1110cxxcln(1),0.1xxxxx22Example 7 Suppose f(x)is continuous on 0,1 an

25、d differentiable on(0,1).Prove that there is at least a number c in(0,1)such that f(1)=2cf(c)+c2f(c).Solution We define a new function g(x)as follows:g(x)=x2f(x)for every x in 0,1.Since f(x)is continuous on 0,1 and differentiable on(0,1),the same is true for the function g(x).So,the function g(x)sat

26、isfies the hypotheses of Lagranges Theorem and there exists at least a number c in(0,1)such that g(1)g(0)=g(c)(1 0)or,equivalently,12f(1)02f(0)=2cf(c)+c2f(c),i.e.,f(1)=2cf(c)+c2f(c).23Corollary 1 If f(x)=0 for all x in an interval(a,b),then f(x)is constant on(a,b).Corollary 2 If f(x)=g(x)for all x i

27、n an interval(a,b),then f(x)=g(x)+c,where c is a constant.12121212Let and be any two numbers in(,)with.Since is differentiable on(,),it must be differentiable on(,)and continuous on ,.By the Mean Value Theorem,there exists anumbersucxxa bxxfa bx xx xcProof12h thatandxcx242121212112()()()()Since()0 f

28、or all,we have()0,and so this eqution becomes()()0 or()()Therefore,has thesame value at any twonumbersandin(,).This means thatis constant on(,).f xf xf c xxfxxf cf xf xf xf xfxxa bfa b25Example 8 Proof thatarctanarccot/2.xx22Let()arctanarccot.We get11()011for all values of.Therefore,(),a constant.To

29、 determine the value of,we put1.Then(1)arctan1arccot1/2Thus,arctanarccot/2.f xxxfxxxxf xccxcfxxSolution264.3 How Derivatives Affect the Shape of a GraphIncreasing/Decreasing Test Suppose f(x)is continuous on a,b and differentiable on(a,b).(1)If f(x)0 for all x in(a,b),then f(x)is strictly increasing

30、 on a,b.(2)If f(x)0 for all x in(a,b),then f(x)is strictly decreasing on a,b.PROOF Choose any two points x1 and x2 in a,b with x10 and the intervals on which f(x)0 or (3x+5)(x-1)0,we get x1.30Thus the function f(x)is increasing on(-,-5/3 and on 1,+).It is easy to check that f(x)0 if 5/3 x0 for all x

31、 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.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 F

32、igure 2.-12-8-4 4812-12-8-44812Figure 236Definition A point(c,f(c)on the graph of f(x)is called an inflection point if f(x)is continuous at c and the graph changes from concave upward to concave downward or from concave downward to concave upward at the point.The Second Derivative Test Suppose that

33、f(x)is 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.37Example 6 Discuss the curve y=f(x)=x4-4x3 with respect to concavity,points of inflection,and local maxima and minima.Use this information to sketch the curve.Solution 3222()4124(3)(

34、)122412(2)fxxxxxfxxxx xTo find the critical numbers we set f(x)=0 and obtain x=0 and x=3.To use the Second Derivative Test we evaluate f”at these critical numbers:f”(0)=0 f”(3)=36038 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 abo

35、ut the critical number 0.But since f(x)0 for x0 and also for 0 x3,the First Derivative Test tells us that f does not have a local maximum or minimum at 0.Since f”(x)=0 when x=0 or 2,we divide the real line into intervals with these numbers as endpoints and complete the following chart.39The points(0

36、,0)and(2,-16)are the inflection points.Using the local minimum,the intervals of concavity,and the inflection points,we sketch the curve in Figure 3.IntervalSign of f”+-+Concavityupwarddown ward upward(,0)(0,2)(2,)Figure 340Example 7 Sketch the graph of the function f(x)=x2/3(6-x)1/3.Solution 1/32/34

37、/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 +fdecreasing on(-,0)increasing on(0,4)decreasing on(4,6)decreasing on(6,+)41By the First derivative Test,f(0)=0 is a local minimum and f(4)=25/3 is a loc

38、al 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 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 Figu

39、re 4.Figure 4424.4 Indeterminate Forms and LHospitals RuleSuppose f and g are differentiable and g(x)0 near a(except possibly at a).Suppose thatlim()andlim()or thatlim()andlim()()()Thenlimlim()()if the limit on the right side exists(or is or-)0.0 xaxaxaxaxaxaf xg xf xg xf xfxg xg x43Example 1 FindSo

40、lution Since Remark It is important to understand that LHOSPITALS RULE does not apply to quotients in general;you should verify first that the numerator and denominator both tend to zero.limxxx12000000lim(21)0 and lim0,applying LHospitals Rule,we have21(21)2 ln 2limlimlimln 2.1xxxxxxxxxxxx44For exam

41、ple,but a blind application of LHospitals rule would lead toThis is incorrect.Example 2 FindSOLUTION 200000()()011limlim21lim.220 xxxxxxexexxe00lim0,cos1xxxx001limlim1.cossin1xxxxxx201lim.xxexx45Note LHospitals Rule is also valid for one sided limits and for limits at positive infinity or negative i

42、nfinite;that is,“xa”can be replaced by any of the following symbols:“xa+”,“xa-”,“x+”,“x-”.Example 3 Calculate SOLUTION23ln()limand()lim.xxxexabxx2()()limlimlim22()xxxxxxeeeaxx 332/3()ln1/3()limlimlim013xxxxxbxxx46 The Forms 0,00,0,1 and-(1)If and ,then the form is called an indeterminate form of typ

43、e 0.We can deal with it by rewriting the product f(x)g(x)as a quotient and then applying LHospitals Rule to the resulting indeterminate form of type 0/0 or/.0)(limxfax )(limxgax)()(limxgxfax,)()()()()()()()(xfxgxgxfxgxfxgxf1or 00147 (2)If and ,then the form is called an indeterminate form of type-.T

44、o find out the limit,we convert the difference f(x)-g(x)into a quotient or a product and then apply LHospitals Rule.(3)(limxfax )(limxgax)()(limxgxfax lim()ln()()00lim()(0,1)=xag xf xg xxaf xe48Example 4 FindSolution 0lim.xxx012000limlnln00lnlimlim1/lim0limlim1xxxxxxxxxxxxxxxxxeeeeeeThe graph of the

45、 function is shown in Figure 5.,0,xyxxFigure 549Example 5 FindSolution (1)The given limit is an indeterminate form of type 0.We write .lim)()ln(lim)(,lnlim)(1111303 and 1121xxxxxxxxxx.limlimlnlimlnlim033130403030 xxxxxxxxxxx.limlimlnlnlim)(lnlnlim)ln(lim)(21111111111212110011xxxxxxxxxxxxxxxxxxxxxx50

46、1111lnlimlim1111(3)The given limit is of type 1.We getlim.xxxxxxxxeee111111orlimlim1(1).xxxxxxeQ:sin1 coslimlimdoes not exist.1xxxxxx)sin1(limxxx1514.5 Summary of Curve sketchingGuidelines for Sketching CurveA.Domain E.Intervals of Increase or DecreaseB.Intercepts F.Local Extreme ValuesC.Symmetry G.

47、Concavity and Points of InflectionD.Asymptotes H.Sketch the Curve52Example 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 y=2 is a horizontal asymptote.(,1)(1,1)(1,).222lim21xxx2212lim1

48、xxx 2212lim1xxx 532212lim1xxx 2212lim1xxx Therefore,the lines x=1 and x=-1 are vertical asymptotes.E.Since f(x)0 when x0(x-1)and f(x)0,f is increasing on(-,-1)and(-1,0)and decreasing on(0,1)and(1,).F.The only critical number is x=0.Since f changes from positive to negative at 0,f(0)is a local maximu

49、m.2222224(1)224()(1)(1)x xxxxfxxx 54G.We haveThus,the curve is concave upward on the intervals(-,-1)and(1,)and concave downward on(-1,1).It has no point of inflection.We sketch the curve in Figure 1.22224234(1)42(1)2124()(1)(1)xxxxxfxxx2()0101.fxxx Figure 155Slant(Oblique)AsymptotesThe line y=ax+b i

50、s a slant asymptote for a function f(x)if and only if exist.()limandlim()xxf xaf xaxbxExample 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 mi