《《高数双语》课件section 3_456.pptx》由会员分享，可在线阅读，更多相关《《高数双语》课件section 3_456.pptx（46页珍藏版）》请在三一文库上搜索。
1、Properties of FunctionsOverviewMonotonicity of FunctionsLocal Extreme Values of FunctionsGlobal Maxima and MinimaConvexity of FunctionsInflectionsGraphing Functions23Monotonicity of Functionsxyo()yf x abAB()0fx xyo()yf x()0fx abBA(1)The necessary and sufficient condition for the function f to bemono
2、tone increasing(decreasing)on I is(2)If 0f0()in I,then f is strictly monotone increasing,and differentiable in I.:fIR be continuous on I Let Then(decreasing)on I.(0f0)in I;Monotonicity of Functions4Proof(1)SufficiencyFor any 1212,(),x xI xx()f xis continuous on12,x xand is differentiable in12(,).x x
3、By the Lagrange theorem and the condition we have 21210()()()()0(),f xf xfxx where12(,).x x Hence,the function f to be monotone increasing(decreasing)on I.the interval Monotonicity of Functions5()()0(0).f xxf xx0()()()limxf xxf xfxx Proof(continued)(1)NecessitySuppose that f is monotone increasing(d
4、ecreasing)on I.For any xbelonging to the interior I,xxI then0(0).So,xtaking such that6Monotonicity of Functions()0fx ()0fx Since the roots of the equation()0fx are It easy to see that(,1)x if(3,)x and;Discuss the monotonicity of the function 32()395f xxxx.Solution2()3693(1)(3)fxxxxx,11x 23x and.(1,3
5、)x if.7Monotonicity of Functions()f xThen,we have the function is strictly monotone increasing in the interval Finish.Solution(continued)(,1)(3,)and;(1,3)and is strictly monotone decreasing in the interval.Mm Discuss the monotonicity of the function 32()395f xxxx.8Monotonicity of Functions,where2()(
6、1)(1)01.xf xx exxif2(1)(1)0,01.xx exxwhere 2()(12)1,01.xfxx ex Prove that21(01).1xxexxProofIn order to obtain the given inequatity,we need only prove that()fxif2()40,01.xfxxex LetthenBut the sign of is still not clear on 0,1).Let us repeat theprocedure to consider the derivative of function ().fxIt
7、is easy to obtainMonotonicity of Functions9if2()40,01.xfxxex()fxif()(0)0,01,f xfxif()(0)0,01.fxfxif2(1)(1)0,01.xx exxProof(continued)By Theorem 2.7.1,is strictly monotone decreasing on 0,1).Thus Again from Theorem 2.7.1,we know that or This is the desired conclusion.Finish.Prove that21(01).1xxexx10E
8、xtreme Values of Functionsoxyab()yf x 1x2x3x4x5x6x0 :,fIR0.xI Suppose that If there is a such that0()()f xf x 0()()f xf x (),then we say that the function f0.xhas a maximum(minimum)at Maximum and minimum values are given a joint name extreme value极值极值,is called a maximal(minimal)0(,),xU xI 0 xand th
9、e point point or extreme point极值点极值点.11Extreme Values of Functions()0fx 0 xx()0fx 0 xx(1)If for and for,then f has a maximum()0fx 0 xx()0fx 0 xx(2)If for and for,then f has a minimum 0 x then the point is not an extreme point.0,xLet the function f be differentiable in a neighborhood of a point 0()U
10、x0()0.fx,and 0()f x0 x at the point;()fx 0 x(3)If has the same sign on both the left and right sides of the point,0()f x0 x at the point;12Extreme Values of Functionsxyoxyo0 x0 x xyoxyo0 x0 x 13Extreme Values of Functions Find the extreme values of the function 323.()6f xxx2334()(6)xfxxx We need thr
11、ee steps to discuss this problem.,it is obvious that the stationary point is 0 x 6x and.Since Solution(1)Find the stationary points and points where the function is not differentiable.non-differentiability are and the points of4x ,()fx (2)In order to determine the sign of in the neighborhoods of the
12、se points,(,)we partition the interval of definition of the given function points and make a table as following by those (,0)(6,)()fx (0,4)()f xZ(4,6)0460 Minimal PointMaximal PointNon-extreme Pointx14Solution(continued)(,0)(6,)()fx (0,4)()f xZ(4,6)0460 Minimal PointMaximal PointNon-extreme PointxFi
13、nish.From the table we may see that the maximum of the function is(3)Determine the extreme values.3(4)2 4f,(0)0f and the minimum is .Extreme Values of Functions Find the extreme values of the function 323.()6f xxx15Extreme Values of Functions-10-5510-4-2246233()6f xxx Find the extreme values of the
14、function 323.()6f xxx16Extreme Values of Functionsf has a second derivative at a point 0()0fx and,fThen the function has a maximum(minimum)if 0()0fx 0()0fx ().Suppose that the function0 x,0()0fx .0()f x0 x at the point 220000001()()()()()()()2!f xf xfxxxfxxxo xx0()0fx 0 xSince f has a second derivat
15、ive at point 0()U x,in the neighborhood the Taylor formula of the function f with the second order and Peano remainder0().xU x where Since,we have22000001()()()()(),()2!f xf xfxxxo xxxU x.Proofis the following17Extreme Values of Functions0()xU x 0()f x0()()f xf x We can see from this that the sign o
16、f,the first term of the last expression.0()0fx ,or 0()()f xf x,then The minimum property may be proved similarly.Proof (continued)2200001()()()()()2!f xf xfxxxo xx is determined byHence,when0()xU x,f is a maximum of the function.Finish.0()()0f xf x,f has a second derivative at a point 0()0fx and,fTh
17、en the function has a maximum(minimum)if 0()0fx 0()0fx ().Suppose that the function0 x,0()0fx .0()f x0 x at the point Extreme Values of Functions1820,4f 5204f ()f x0,2 ()cossin0fxxx()f x()sincosfxxx we need only consider in a periodic interval.By the last theorem,we may find the extremeLet(2)Find th
18、e second derivative of and consider its sign at these Find the extreme value of the function()sincos.f xxxSolution()f xSince is a periodic function,()f x ,(1)Find the stationary points of the value by the following steps:0,2 x .5,44x.We have the stationary points .stationary points(3)Find the maximu
19、m and minimum of()f x.19Extreme Values of Functions-10-5510-1-0.50.510,2 24f ()f xThen the maximum of the function in the interval is;and the minimum is 524f .Solution(continued)Finish.Find the extreme value of the function()sincos.f xxx200()0fx f(1)()0000()()()0,()0nnfxfxfxfx L0()f x()0()0nfx()0()0
20、.nfx Note If we have 0 x at a stationary point Suppose that the function is n-times differentiable andThen0 x must be an extreme point,and()f xis a maximum of if and a minimum if(2)when n is odd,0 x is not an extreme point.Extreme Values of Functions,then we have to utilizethe higher derivatives of
21、the function.n is even,(1)when21()f x()f x,a bIf a function is continuous on a closed interval properties of continuous functions,must have a global maximumit must be a local maximum value or a local minimum value;if we want to find the global maximum value or global minimum value,()f a()f b and.,th
22、en by the,a band a global minimum on0 x.And if any of them,say(,)a b,is in,possibility is that0 x,a b may be one of the end points of.Therefore,we need to find all stationary points,compare all the function values at these points and values(1)(2)(3)Global maxima and minimaanothernon-differentiable p
23、oints and then22Global maxima and minimaAs we had seen in last lecture,by the definition of maximum and minimal value of a function,they are only local values.But for many problems,we need to find the largest value or smallest value in the fixed interval,and these value is referred as global maximum
24、 and global minimum.()f x0 x()f x,a bIf a function is continuous on a closed interval properties of continuous functions,must have a global maximum.And if any of them,say it must be a local maximum value or a local minimum value;anotherif we want to find the global maximum value or global minimum va
25、lue,()f a()f b and.,then by the,a band a global minimum on(,)a b,is in,possibility is that0 x,a b may be one of the end points of.Therefore,we need to find all stationary points,non-differentiable points and thencompare all the function values at these points and values(1)(2)(3)Global maxima and min
26、ima23Note It is worthwhile to indicate that for some special cases,finding global maxima or minima may be simplified.For instance,if f(x)is monotone increasing(decreasing)on the interval a,b,then the global maximum and minimum must be attained at the end points b(or a)or a(or b,respectively;if f(x)i
27、s continuous on a,b and has only one extreme point 0(,),xa b then x0 must be theglobal maximum(minimum)point provided it is a local maximum(minimum)point.Global maxima and minima24Note To solve a practical problem,we may should establish the objective function firstly and then find the global maximu
28、m or minimum by finding all the stationary points,non-differentiable points and comparing the function values and the function values at the end of the given interval.Global maxima and minima2522,(0,)SRHRH R 20VR H (1)Establish the objective function.SolutionThe least amount of material means the sm
29、allest surface area.Suppose that the surface area of the container is S,the height is H,and the radius of the bottom is R.ThenBy the last equation,we have 02VHR ,and then the objective function is202(0)VSRRR A cylindrical container with volume V0 and without cover is to be made of a sheet of iron.Ho
30、w should we design it if we wish to use the least amount of material?Global maxima and minima26202(0)VSRRR 0dSdR 2023420,Vd SdRR 03VR Solution(continued)(2)Find the global minimum.Let;we obtain the stationary point Since then we know that it is the global minimum point.We have 0222VdSRdRR .03VR .is
31、the minimum.(0,)Since this point is unique in,A cylindrical container with volume V0 and without cover is to be made of a sheet of iron.How should we design it if we wish to use the least amount of material?27Global maxima and minimaSolution(continued)03VR While,we haveTherefore,the amount of materi
32、al is minimized provided the height H and03VHR.Finish.the radius of the bottom R are equal.A cylindrical container with volume V0 and without cover is to be made of a sheet of iron.How should we design it if we wish to use the least amount of material?28Global maxima and minima It is easy to see tha
33、t the closest point from us to the enemy is the best position for us to shoot the enemy.Suppose that the enemys car go straight to the north from point A with the velocity of 1km/min and the width of the river is 0.5 kilometres.A tank of our army go along the river side and direct to the east from p
34、oint B with the velocity of 2km/min.(See the right figure)The question is where is the best position to shoot enemy?29Global maxima and minima0.5kilometres4kilometresB A)(ts(1)Find the relation between the position of our tank and the enemys car.SolutionSuppose that t is the time when our tank begin
35、 to chase the enemy from B and the distance between the enemy and usThen 22()(0.5)(42)s ttt(2)Find the global minimal value of()s t.2257.5()(0.5)(42)ts ttt ()0s t 1.5.t .Let,we find the stationary point It is easy to see that this point is the point of minimum value.after we began to chase the enemy
36、,is the best time we shoot the enemy.Finish.()s t.isTherefore,1.5 minuets,30Convexity of functionsxyo()yf x xyo()yf x Concave FunctionConvex Functionof being convex up or convex down for the graph of a function is called the convexityConvexity is another important property of functions.In generally,
37、a function,whose the graph is convex down,is called a concave function,while a function whose graph is convex up,is called convex function.The propertyof the function.31Convexity of functionsBy the graph of a function which is convex down on the interval ,a b.It is easy to see that for any two point
38、s 1x,2x(12xx)on ,a b,the graph of()f x on the interval 12,xx is always below the chord AB.abx1x2xABOxy()yf x 32Convexity of functionsAnd then,we have the following definition.0,1 1212(1)()(1)()fxxf xf xf12,x xI 12xx Definition(convex function)12,x xIIf and the inequalityholds,then(0,1)if,and,we have
39、then If the inequality of these two f is called convex function or strictly convex function,respectively.abx1x2xABOxy()yf x:fIRLet.I is called a concave function on;1212(1)()(1)()fxxf xf x,I f is called a strictly concave function on.inequality is reversed,then IonIt is not very easy to judge the co
40、nvexity of a given function directly from the definition.33Convexity of functionsTheorem Assume that f is twice differentiable in the interval I,then(1)fis a strictly concave(concave)function if()0(0)fx,xI ;(2)fis a strictly convex(convex)function if()0(0)fx,xI .xyo()yf x xyo()yf x abAB()is increasi
41、ngfx abBA0y ()is decreasingfx 0y This conclusion are easily understood geometrically.34Convexity of functions12xx 1212(1)()(1)()fxxf xf x12xx 12(1)xxxWe prove only the case of strict convexity;the other proofs are similar.()0fx ,It is easy to prove that(0,1),12,x xI,we have the inequalitySuppose tha
42、t;note that.It is enough to proveAdding two terms and then using the mean value theorem we haveProofSuppose that xI .12()(1)()()0,(0,1)f xf xf x .Theorem Assume that f is twice differentiable in the interval I,then(1)fis a strictly concave(concave)function if()0(0)fx,xI ;(2)fis a strictly convex(con
43、vex)function if()0(0)fx,xI .Convexity of functions3512()()()(1)()()f xf xf xf xf x111212(1)(1)(),xxxxxxxProof(continued)12()(1)()()f xf xf x121()()(1)()()f xf xf xf x1122()()(1)()()fxxfxx12()(1)()()0,(0,1)f xf xf x?221212(1)().xxxxxxx where 11xx,whileSubstituting these into the above equality and us
44、ing the mean value theorem,22xx,Theorem Assume that f is twice differentiable in the interval I,then(1)fis a strictly concave(concave)function if()0(0)fx,xI ;(2)fis a strictly convex(convex)function if()0(0)fx,xI .Convexity of functions361212(1)()()()xxf 0 Proof(continued)12()(1)()()f xf xf x1122()(
45、)(1)()()fxxfxx112212(1)()()(1)()()fxxfxxwhere Finish.12.The conclusion is proved.Theorem Assume that f is twice differentiable in the interval I,then(1)fis a strictly concave(concave)function if()0(0)fx,xI ;(2)fis a strictly convex(convex)function if()0(0)fx,xI .37Convexity of functions()(0,1);f xxx
46、 ()ln(0).f xx x(1)(2)Study the convexity of the following functions()lnf xx (1)Since()f xx so the power function(2)Since so the logarithm function Finish.Solution(0,)is strictly convex in the interval.21()0,0fxxx ,2()(1)0,(0,)fxxx ,(0,)is strictly concave in.Convexity of functions3812()()2f xf x()f
47、x If is continuous and strictly convex in an interval I,then then it must be the global minimum points in I.()f xhas at most one global minimum point,and if there exists a unique localminimum point in I12()()()f xf xmf x122xxf 122xxI 1x2x and,so that()f xBy the strict convexity of we haveBecause,a c
48、ontradiction appears.ProofSuppose that there exist two minimum points xI .m.Convexity of functions3900()(1)()f xf x01()(1)()f xf xProof(continued)0 xI 10 xx 10()()f xf x 01()(1)f xfxxSecondly,let Iit is the global minimum point in.Suppose that there exists 1xI,such that.By the definition of convex f
49、unction we have be a local minimum point;0,1 ,we will prove that0()f x.()f x If is continuous and strictly convex in an interval I,then then it must be the global minimum points in I.()f xhas at most one global minimum point,and if there exists a unique localminimum point in I40Convexity of function
50、s0,10()()f xf x is arbitrary on x,and 0 x can be taken arbitrarily close to In this case,contradicts assumption that 0 x is a localNote that()f xI is continuous in and Proof(continued)by taking close enough to 1.Finish.minimum.So,must be the global minimum points in I.0 x()f x If is continuous and s