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1、函数专题,义乌三中 刘石洋,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,1确定函数零点的三种常用方法 (1)解方程判定法若方程易解时用此法 (2)零点定理法根据连续函数yf(x)满足f(a)f(b)0，判断函数在区间(a，b)内存在零点 (3)数形结合法尤其是方程两端对应的函数类型不同时多用此法求解,Evaluation only. Created with Aspose.Slides for .NET 3
2、.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,2解函数应用题的四步曲 (1)阅读理解，审清题意：分析出已知什么，求什么，从中提炼出相应的数学问题； (2)数学建模：弄清题目中的已知条件和数量关系，建立函数关系式； (3)解函数模型：利用数学方法得出函数模型的数学结果； (4)实际问题作答：将数学问题的结果转译成实际问题作出解答,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004
3、-2011 Aspose Pty Ltd.,高考对本部分内容的考查多以选择题或填空题的形式出现，考查求函数零点的存在区间、确定零点的个数以及两函数图像交点的横坐标或确定有几个交点,考情分析,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyrig
4、ht 2004-2011 Aspose Pty Ltd.,答案B,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,类题通法,(1)解决这类问题的常用方法有解方程法、利用零点存在的判定和数形结合法，尤其是方程两端对应的函数类型不同的方程多以数形结合求解 (2)函数零点(即方程的根)的应用问题，即已知函数零点的存在情况求参数的值或取值范围问题，解决该类问题关键是用函数方程思想或数形结合思想，构建关于参数的方程或不等
5、式求解,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,冲关集训,D,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,2(2012唐山统考)设f(x)exx4，则函数f(x)的零点位 于区间() A(1,0) B(0,1) C(
6、1,2) D(2,3) 解析：选 f(x)exx4，f(x)ex10， 函数f(x)在R上单调递增又f(1)e1(1)4 5e10，故f(1)f(2)0.,C,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,C,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011
7、Aspose Pty Ltd.,考情分析,此类问题命题以函数的图像与性质为背景创设新情景，通常从定义的新运算、新概念或新性质入手，考查函数的图像与单调性、最值(值域)以及零点等函数性质，常与方程、不等式问题结合今后对新定义函数的考查是高考的一大热点,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 C
8、lient Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,思路点拨利用承托函数的定义，一一分析即可 解析对于，结合函数f(x)的图像分析可知，不存在函数g(x)使得f(x)g(x)对一切实数x都成立，即f(x)不存在承托函数；对于，注意到f(x)2x0，因此存在函数g(x)0，使得f(x)g(x)对一切实数x都成立，f(x)存在承托函数；对于，结合函数f(x)的图像分析可知，不存在函数g(x)使得f(x)g(x)对一切实数x都成立，即f(x)不存在承托函数；对于，注意到f(x)xsin xx1，因此存在函数g(x)x1，使得f(x)g
9、(x)对一切实数x都成立，f(x)存在承托函数综上所述，存在承托函数的f(x)的序号为. 答案,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,类题通法,解决与函数有关的新信息题的思路： 第一步准确理解新的运算、概念或性质 第二步根据新的定义，类比与函数有关的运算、性质等将其转化为熟悉的函数问题 第三步利用函数的相关知识求解问题,Evaluation only. Created with Aspose.Slid
10、es for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,冲关集训,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,B,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011
11、 Aspose Pty Ltd.,5(2012东城综合测试)函数f(x)的定义域为A，若x1，x2A且 f(x1)f(x2)时总有x1x2，则称f(x)为单函数例如：函数 f(x)2x1(xR)是单函数 给出下列命题： 函数f(x)x2(xR)是单函数； 指数函数f(x)2x(xR)是单函数； 若f(x)为单函数，x1，x2A且x1x2，则f(x1)f(x2)； 在定义域上具有单调性的函数一定是单函数 其中的真命题是_(写出所有真命题的编号) 解析：根据单函数的定义，可判断命题、是真命题， 是假命题；根据一个命题与其逆否命题等价可知，命题是 真命题,答案：,Evaluation only. C
12、reated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,考情分析,该类试题以实际生活为背景，通过巧妙设计和整合命制考题，试题常与函数解析式的求法、函数最值、不等式、导数、解析几何、空间几何体等知识交汇预测2013年的高考以求函数的最值为热点,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose
13、Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5
14、Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,类题通法,解
15、决函数实际应用题的关键有两点：一是认真读题，缜密审题，确切理解题意，明确问题的实际背景；然后进行科学地抽象概括，将实际问题归纳为相应的数学问题；二是要合理选取参变量，设定变量之后，就要寻找它们之间的内在联系，选用恰当的代数式表示问题中的关系，建立相应的函数模型，最终求解数学模型使实际问题获解,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,冲关集训,Evaluation only. Created with A
16、spose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,D,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evalua
