高中数学1—3—1函数的单调性与导数演示课件新人教a版选择修读.pptVIP

Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. [例5]　f′(x)是f(x)的导函数，f′(x)的图象如图所示，则f(x)的图象可能是 (　　) Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. [分析]　导函数值的正负决定了函数的增减，导函数值增减决定了函数值变化的快慢． 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 [解析]　由图可知，当bxa时，f′(x)0，故在[a，b]上，f(x)为增函数．且又由图知f′(x)在区间[a，b]上先增大后减小，即曲线上每一点处切线的斜率先增大再减小，故选D. [点评]　本题的关键是正确理解导函数与函数之间的关系． Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 如图所示，单位圆中弧AB的长为x，f(x)表示弧AB与弦AB所围成的弓形面积的2倍，则函数y＝f(x)的图象是(　　) [答案]　D Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 因为当0≤xπ时，f(x)＝x－sinx≤x，所以函数的图象在y＝x的下方；当π≤x≤2π时，f(x)＝x－sinx≥x，所以函数的图象在y＝x的上方．故选D. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 1．3　导数在研究函数中的应用 1．3.1　函数的单调性与导数 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 借助于函数的图象了解函数的单调性与导数的关系，能利用导数研究函数的单调性，会用导数法求函数的单调区间． Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 本节重点：利用导数研究函数的单调性． 本节难点：用导数求函数单调区间的步骤． Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 1．在利用导数讨论函数的单调区间时，首先要确定函数的定义域，解决问题的过程中，只能在定义域内通过讨论导数的符号，来判断函数的单调区间． 2．在对函数划分单调区间时，除了必须确定使导数等于零的点外，还要注意定义区间内的不连续点或不可导点． 3．注意在某一区间内f′(x)0(或f′(x)0)是函数f(x)在该区间上为增(或减)函数的充分条件．如f(x)＝x3是R上的可导函数，也是R上的单调递增函数，但当x＝0时，f′(x)＝0. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 4．由导数的几何意义可知，函数f(x)在x0的导数f′(x0)即f(x)的图象在点(x0，f(x0))的切线的斜率，在x＝x0处f′(x0)＞0，则切线的斜率k＝f′(x0)＞0，若在区间(a，b)内每一点(x0，