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1、第8课时函数与方程,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,2014高考导航,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,本节目录,教材回顾夯实双基,考点探究讲练互动,名师讲坛精彩呈现,知能演练轻松闯关,Evalua
2、tion only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,基础梳理 1.函数的零点 (1)函数零点的定义 对于函数yf(x)(xD)，把使_成立的实数x叫做函数yf(x)(xD)的零点 (2)几个等价关系 方程f(x)0有实数根函数yf(x)的图象与_有交点函数yf(x)有_,f(x)0,x轴,零点,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Pro
3、file 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,思考探究 1是否任意函数都有零点？ 提示：并非任意函数都有零点，只有f(x)0有根的函数yf(x)才有零点,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,(3)函数零点的判定(零点存在性定理) 如果函数yf(x)在区间a，b上的图象是连续不断的一条曲线，并且有_，那么函数yf(x)在区间(a，b)内有零点，即存
4、在c(a，b)，使得_，这个c也就是f(x)0的根 思考探究 2在上面的条件下，(a，b)内的零点有几个？ 提示：在上面的条件下，(a，b)内的零点至少有一个c，还可能有其他零点，个数不确定,f(a)f(b)0,f(c)0,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,2.二次函数yax2bxc(a0)的图象与零点的关系,(x1，0),(x2，0),Evaluation only. Created with
5、Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,3.二分法的定义 对于在区间a，b上连续不断且_的函数yf(x)，通过不断地把函数f(x)的零点所在的区间一分为二，使区间的两个端点逐步逼近_，进而得到零点近似值的方法叫做二分法,f(a)f(b)0,零点,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pt
6、y Ltd.,课前热身 1.如图所示的函数图象与x轴均有交点，其中不能用二分法求图中交点横坐标的是() AB C D 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.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Ev
7、aluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,4.函数f(x)x3x的零点是_ 答案：1，0，1 5.若二次函数f(x)ax2bxc中，ac0，则其零点个数是_ 答案：2,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evalu
8、ation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,【解析】f(x)exx4，f(x)ex10， 函数f(x)在R上单调递增，对于A项，f(1)e1(1)45e10，f(0)30，f(1)f(0)0，A不正确，同理可验证B、D不正确对于C项，f(1)e14e3 0，f(2)e224e220，f(1)f(2)0，故选C. 【答案】C,Evaluation only. Created with Aspose.Slides for
9、 .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,【规律小结】判定函数零点个数的几种方法： (1)直接求零点：令f(x)0，如果能求出解，则有几个解就有几个零点； (2)零点存在性定理：利用定理不仅要求函数在区间a，b上是连续不断的曲线，且f(a)f(b)0，还必须结合函数的图象与性质(如单调性、奇偶性)才能确定函数有多少个零点； (3)利用图象交点的个数：画出两个函数的图象，看其交点的个数，其中交点的横坐标有几个不同的值，就有几个不同的零点,Evaluation only. Created with A
10、spose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,跟踪训练,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 2
11、004-2011 Aspose Pty Ltd.,【解析】由参考数据知f(1.437 5)0，f(1.406 25)0，f(1.437 5)f(1.406 25)0，且精确到0.1时，1.437 51.4，1.406 251.4，所以函数f(x)的一个零点的近似值是1.4，也就是方程x3x22x20的一个近似根 【答案】1.4,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,【规律小结】利用二分法求近似解需注意
12、的问题： (1)第一步中：区间长度尽量小；f(a)、f(b)的值比较容易计算且f(a)f(b)0； (2)根据函数的零点与相应方程根的关系，求函数的零点与相应方程的根是等价的 提醒：对于方程f(x)g(x)的根，可以构造函数F(x)f(x)g(x)，函数F(x)的零点即为方程f(x)g(x)的根,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,跟踪训练 2.(2013武汉模拟)若函数f(x)在(1，2)内有一个
13、零点，要使零点的近似值满足精确度为0.01，则对区间(1，2)至少二等分() A5次 B6次 C7次 D8次,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.,【答案】D,Evalu
14、ation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,【规律小结】已知函数有零点(方程有根)求参数值常用的方法和思路： (1)直接法：直接求解方程得到方程的根，再通过解不等式确定参数范围； (2)分离参数法：先将参数分离，转化成求函数值域问题加 以解决； (3)数形结合：先对解析式变形，在同一平面直角坐标系中，画出函数的图象，然后观察求解,Evaluation only. Created with Aspose.Slides
15、for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,跟踪训练 3.是否存在这样的实数a，使函数f(x)x2(3a2)xa1在区间1，3上恒有一个零点，且只有一个零点？若存在，求出a的范围；若不存在，说明理由,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 As
16、pose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,1.函数yf(x)的零点即方程f(x)0的实根，是数不是点 2.若函数yf(x)在闭区间a，b上的图象是连续不间断的，并且在区间端点的函数值符号相反，即f(a)f(b)0，满足这些条件一定有零点，不满足这些条件也不能说就没有零点如图，f(a)f(b)0，f(x)在区间(a，b)上照样存在零点，而且有两个所以说零点存在性定理的条件是充分条件,但并不必要,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,数学思想,Evaluation only. Created wi
