必须修读1—2-2-2对数函数和其性质应用.ppt

新疆奎屯市第一高级中学 特级教师王新敞 * 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. 目标与要求: Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 图 象 定义域 值 域 性 质 a1 0a1 必过 点： 在 R ＋上是 在 R＋ 上是 R ( 0 , + ∞ ) ( 1, 0 ) ,即 x = 1 时, y = 0 . 减函数 增函数 y x 0 x=1 (1．0) y x 0 x=1 (1．0) Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例1.求下列函数的定义域： (1) (2) (1)因为 ,即 ,所以函数 的定义域是 (2)因为 ,即 , 所以函数 的定义域是 解： Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 探究与深化 练习一：求下列函数的定义域 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 探究与深化 ) 1 ( 1 log 2 2 x y - = 解： 所以函数 的定义域是 1 log 2 2 x y - = Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 探究与深化 解： 所以函数 的定义域是 ) 1 ( log ) 2 ( 2 2 x y - = ) 1 ( log 2 2 x y - = Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 探究与深化 解： 所以函数 的定义域是 ) 1 ( log ) 3 ( ) 1 ( x y x + = - ) 1 ( log ) 1 ( x y x + = - Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例2.比较下列各组数中两个值的大小： (1) (2) (3) Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 解： 因为函数 在(0,+∞)上是增函数, 且3.4＜8.5,所以 (2)因为函数