[2014创新设计高中数学苏教版第七章第4讲不等式的综合应用.ppt

考点梳理 1．不等式的证明 常用方法： (1)作差(作商)比较法． (2)放缩法，注意放缩的尺度要把握好． (3)综合法与分析法，即所谓的“由因导果”“执果索因”． (4)反证法，常见的否定性结论，“至多”“至少”等问题． (5)利用函数的单调性． 第4讲　不等式的综合应用 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2．不等式问题解法、证法的基本应用 (1)求函数的定义域、值域和最大值、最小值问题； (2)判断函数的单调性及其相应的单调区间； (3)利用不等式讨论方程的实根个数、分布范围和解含参数的方程； (4)将不等式同数学其他分支结合起来，解决一些有实际应用价值的综合题． 3．解不等式应用问题的几个主要步骤 (1)审题，必要时画出示意图； (2)建模，建立不等式模型，即根据题意找出常量与变量的不等关系；注意文字语言、符号语言、图形语言的转换； (3)求解，利用不等式的有关知识解题． Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 一个考情分析 对不等式性质的考查仍会与其他知识相结合来进行．对基本不等式的考查，不外乎大小判断，求最值，求取值范围等．不等式与函数、数列、三角的综合，以解答题呈现的可能性较大． 【助学·微博】 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 答案　5 考点自测 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2．已知两个正数x，y满足x＋4y＋5＝xy，则xy取最小值时，x，y的值分别为________． Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 3．已知过定点P(1,2)的直线在x轴与y轴正半轴的截距分别为a，b，则ab的最小值为________． 答案　8 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　x＋y－2＝0 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. (1)判断f(x)在(0，＋∞)上的增减性，并证明你的结论； (2)解关于x的不等式f(x)＞0； (3)若f(x)＋2x≥0在(0，＋∞)上恒成立，求a的取值范围． 考向一　不等式在方程及函数中的应用 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. Copyrigh