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1、微专题：构造函数法解选填压轴题高考中要取得高分，关键在于选准选好的解题方法，才能省时省力又有效果。近几年各地高考数学试卷中，许多方面尤其涉及函数题目，采用构造函数法解答是一个不错的选择。所谓构造函数法 是指通过一定方式，设计并构造一个与有待解答问题相关函数，并对其进行观察分析，借助函数本身性质如单调性或利用运算结果，解决原问题方法，简而言之就是构造函数解答问题。怎样合理的构造函数就是问题的关键，这里我们来一起探讨一下这方面问题。几种导数的常见构造：1对于xgxf，构造xgxfxh若遇到0aaxf，则可构axxfxh2对于0 xgxf，构造xgxfxh3对于()()0fxf x，构造xfexhx
2、4对于()()fxf x或()()0fxf x，构造()()xf xh xe5对于0 xfxxf，构造xxfxh6对于0 xfxxf，构造xxfxh一、构造函数法比较大小例1已知函数()yf x的图象关于y 轴对称,且当(,0),()()0 xf xxfx成立，0.20.22(2)af，log 3(log 3)bf，33log 9(log 9)cf,则,a b c的大小关系是 ().A abc.B acb.C cba.D bac【解析】因为函数()yfx关于y轴对称,所以函数()yxf x为奇函数.因为()()()xfxf xxfx,所以当(,0)x时,()()()0 xf xf xxfx,函
3、数()yxf x单调递减,当(0,)x时,函数()yxfx单调递减.因为0.2122,0131og,3192og,所以0.23013219ogog,所以bac,选 D.变式:已知定义域为R的奇函数()fx的导函数为()fx，当0 x时，()()0f xfxx，若111(),2(2),ln(ln 2)222afbfcf，则下列关于,a b c的大小关系正确的是（D）.A abc.B acb.C cba.D bac例 2已知()f x为R上的可导函数，且xR，均有()()f xfx，则有A2016(2016)(0)eff，2016(2016)(0)fefB2016(2016)(0)eff，2016
4、(2016)(0)fefC2016(2016)(0)eff，2016(2016)(0)fefD2016(2016)(0)eff，2016(2016)(0)fef【解析】构造函数()(),xf xg xe则2()()()()()()()xxxxfx eef xfxf xgxee，因为,xR均有()()f xfx，并且0 xe，所以()0gx，故函数()()xf xg xe在 R 上单调递减，所以(2016)(0)(2016)(0)gggg，即20162016(2016)(2016)(0)(0)ffffee，也就是20162016(2016)(0)(2016)(0)efffef，故选 D变式:已知
5、函数()f x为定义在R上的可导函数，且()()f xfx对于任意xR恒成立，e为自然对数的底数，则（C）2016.(1)(0)(2016)(0)A fe ffef、2016.(1)(0)(2016)(0)B fe ffef、2016.(1)(0)(2016)(0)C fe ffef、2016.(1)(0)(2016)(0)D fe ffef、例 3在数列na中，1()n 1,()nnanN则数列na中的最大项为()A2B33C55D不存在【解析】由已知12a，323a，4342a，545a易得12234,.aa aaa.猜想当2n时，na是递减数列又由11nnan知ln(1)ln1nnan，
6、令ln()xf xx，则221ln1ln()xxxxfxxx当3x时，ln1x，则1ln0 x，即()0fx()f x在3,内为单调递减函数，2n时，lnna是递减数列，即na是递减数列又12aa，数列na中的最大项为323a故选 B练习1 已知函数)(xfy对任意的)22(，x满足()cos()sin0fxxfxx,则（）A)4(2)0(ff B.)3(2)0(ff C.)4()3(2ff D.)4()3(2ff提示：构造函数()()cosf xg xx，选 D文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10
7、G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J
8、6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9
9、Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U
10、10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L
11、8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1
12、N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y
13、1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4二、构造函数法解恒成立问题例 1若函数y=)(xf在R上可导且满足不等式()()0 xfxf x恒成立，对任意正数a、b，若ab，则必有（）A()()af bbf aB()()bf aaf bC()()af abf bD()()bf baf a【解析】由已知()()0 xfxf x构造函数)()(xxfxF，则()Fx()()0 xfxf x，从而)(xF在R上为增函数。ab()()F aF
14、b即()()af abf b，故选 C。例 2已知)(xf是定义在（0，+）上的非负可导函数，且满足)()(xfxf x0，对任意正数a、b，若ab，则必有（）A()()af bbf aB()()bf aaf bC()()af abf bD()()bf baf a【解析】xxfxF)()(，0)()()(2xxfxxfxF，故xxfxF)()(在（0，+）上是减函数，由ba，有bbfaaf)()(，即()()af bbf a。故选 A。变 式1.设()()f xg x、是R上 的 可 导 函 数，()()fxg x、分 别 为()()f xg x、的 导 函 数，且 满 足()()()()0f
15、x g xf x gx，则当axb时，有（C）.()()()()A f x g bf b g x.()()()(B fx g afa g x.()()()()C f x g xf b g b.()()()(D fx g xfb g a变式 2.设函数bxaxgxfbaxgxf则当且上均可导在),()(,)(),(时，有（C）A)()(xgxfB)()(xgxfC)()()()(afxgagxf D)()()()(bfxgbgxf例 3设函数()f x在 R 上的导函数为()fx，且22()()f xxfxx，下面不等式恒成立的是（）A0)(xfB0)(xfCxxf)(Dxxf)(【解析】由已知
16、，首先令0 x得0)(xf，排除 B，D令2()()g xx f x，则()2()()g xxf xxfx，当0 x时，有2()2()()()0gxf xxfxxgxx，所以函数()g x单调递增，所以当0 x时，()(0)0g xg，从而0)(xf文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q
17、5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U1
18、0G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8
19、J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N
20、9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1
21、U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4
22、L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4当0 x时，有
23、2()2()()()0gxf xxfxxgxx，所以函数()g x单调递减，所以当0 x时，()(0)0g xg，从而0)(xf综上0)(xf故选 A例 4 如果22(1)(1)1xxyy，那么下面的不等式恒成立的是（）A0 xyB0 xyC0 xyD0 xy【解析】构造函数2()lg(1)()f xxxxR，易证()f x在 R 上是奇函数且单调递增22(1)(1)1xxyy2()()l g(1)fxfyxx+2lg(1)yy=22lg(1)(1)xxyy=lg1=0()()fxfy即：()()f xfy又()f x是增函数xy即0 xy。故选 B练习 1.已知yxyx)5.0(log)()
24、5.0(log31313131，则实数yx,的关系是（D ）A.0yx B.0yx C.0yx D.0 xy【解析】构造函数133()(log 2)xf xx，()fx是增函数，又()()f xfy，0 xy，故选 D练习 2.已知函数)(xfy是R上的可导函数，当0 x时，有0)()(xxfxf,则函数xxxfxF1)()(的零点个数是(B )A.0 B.1 C.2 D.3【解析】由xxxfxF1)()(，得1()xf xx，构造函数g()()xxfx，则g()()()xf xx f x，当0 x时，有0)()(xxfxf,当0 x时，()()0 xfxf xx即当0 x时，g()()()0
25、 xf xx f x，此时函数g()x单调递增，此时g()g(0)0 x，当0 x时，g()()()0 xf xx f x，此时函数g()x单调递减，此时g()g(0)0 x，作出函数g()x和函数1yx的图象，（直线只代表单调性和取值范围），由图象可知函数xxxfxF1)()(的零点个数为1 个故选B三、构造函数法解不等式例 1.函数 f(x)的定义域为R，f(1)2，对任意x R，()2fx，则 f(x)2x4 的解集为()A(1,1)B(1，)C(，1)D(，)文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U
26、10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L
27、8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1
28、N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y
29、1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C
30、4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:C
31、Y1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ
32、1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4【解析】构造函数G(x)f(x)2x4，所以()()2G xfx，由于对任意xR，()2fx，所以()()2G xfx0 恒成立，所以G(x)f(x)2x4 是 R 上的增函数，又由于 G(1)f(1)2(1)40，所以 G(x)f(x)2x40，即 f(x)2x 4的解集为(1，)，故选 B.变式 1.已知函数)(Rxxf满足1)1(f，且21)(xf，则212)(xxf的解集为（）A.
33、11xxB.1xxC.11xxx或D.1xx【解析】构造新函数1()()()22xF xfx,则11(1)(1)()1 1022Ff，1()()2Fxfx，对任意xR，有1()()02Fxfx，即函数()F x在 R上单调递减，所以()0F x的解集为(1,)，即212)(xxf的解集为(1,)，选 D.变式2.定义在R上的函数()f x，其导函数()fx满足()1fx，且23f，则关于x的不等式1fxx的解集为(,2)变式 3.已知函数()f x为定义在R上的可导函数，且()()f xfx对于任意xR恒成立，且(1)fe，则()1xf xe的解集为(,1)变式 4.函数)(xf的定义域是R，
34、2)0(f，对任意Rx，()()1f xfx，则不等式1)(xxexfe的解集为（A ）A.0 xx B.0 xx C.1x-1xx或 D.101xxx或例 2 设()f x是定义在R 上的奇函数，且(2)0f，当0 x时，有2()()0 xfxf xx恒成立，则不等式2()0 x f x的解集是解：因为当x0 时，有2()()0 xfxf xx恒成立，即()f xx0 恒成立，所以()f xx在(0,)内单调递减因为(2)0f，所以在（0，2）内恒有()0f x；在(2,)内恒有()0f x又因为()f x是定义在R 上的奇函数，所以在(,2)内恒有()0f x；在(2,0)内恒有()0f
35、x又不等式2()0 x f x的解集，即不等式()0f x的解集所以答案为(,2)（0，2）变式 1.已知定义在)0,(上的可导函数，其导函数为()fx，且有22()()f xxfxx，则不等式文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6
36、C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q
37、5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U1
38、0G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8
39、J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N
40、9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1
41、U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C40)2(4)2014()2014(2fxfx的解集为（C）A)2012
42、,(B.)02012(，C.)2016,(D.)02016(，变式 2.函 数()f x的定义域为R，(2)2016f，对任意xR，都有()2fxx成立，则不等式2()2012f xx的解集为（C）A.(2,2)B.(2,)C.(,2)D.(,)变式3.设)(xfy是定义在R上的函数，其导函数为()fx，若()()1f xfx，(0)2017f,则不等式()2016xxf x ee的解集为（D ）A.(2016,)B.(,0)(2016,)C.),0()0,(D.),0(变 式4.函 数)(xf是 定 义 在R上 的 偶 函 数,0)2(f,且0 x时,()()0f xxfx,则 不 等 式0
43、)(xxf的解集是 _ 2,02,)_（提示：构造的()()g xxf x为奇函数，(0)0f）例 4设()()fxg x、是R上的可导函数，()()()()0fx g xf x g x，(3)0g，则不等式()()0f x g x的解集为(3,)变式 1设()()f xg x、分别是定义在R上的奇函数、偶函数，当0 x时，()()()()0fx g xf x gx，(3)0g，则不等式()()0fx g x的解集为(,3)(0,3.变式 2已知R上的函数()()fxg x、满足()()xf xag x，且()()()()f x g xf x g x，若(1)(1)5(1)(1)2ffgg，则
44、关于x的不等式log1ax的解集为1(0)2，.变式 3.设奇函数)(xf定义在),0()0,(上，其导函数为()fx，且02f，当x0时，sincos0fxxfxx，则关于x的不等式fx2sin6fx的解集为 _(,0)(,)66.（提示：构造的()()sinf xg xx为偶函数）四、构造函数法求值例 1设()fx是R上的可导函数，且()()fxf x，(0)1f，21(2)fe.则(1)f的值为.提示：由()()fxf x得()()0fxf x，所以()()0 xxe fxe f x，即()0 xe f x，文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C
45、4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5
46、K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10
47、G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J
48、6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9
49、Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U
50、10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L8J6C4文档编码:CY1N9Q5K3D7 HQ1Y1U10G1I6 ZN7C4L