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函数值域是函数概念中三要素之一,是高考中必考内容,具有较强的综合性,贯穿整个高中数学的始终.而在高考试卷中的形式可谓千变万化,但万变不离其宗,真正实现了常考常新的考试要求.所以,我们应该掌握一些简单函数的值域求解的基本方法.例1求下列函数的值域:(1)y=-x~2+2x-2,z∈[-1,2);(2)y=z+2(x-1)~(1/2);(3)y=z+3-3x~2.思路分析(1)利用配方法得y=-(z-1)~2-1.因为z∈[-1,2),所以,当z=1时,y_(max)=-1,当z=-1时,y_(min)=-5,值域为[-5,-1].
The function value range is one of the three elements of the concept of function, and it is a must-examination content in the college entrance examination. It has a strong comprehensiveness and runs through the entire high school mathematics. The form of the examination paper can be described as ever-changing, but it will never change. Really achieve the new test requirements for regular exams. Therefore, we should master the basic methods for value domain solution of simple functions. Example 1 Find the range of the following functions: (1) y=-x~2+2x-2, Z∈[-1,2);(2)y=z+2(x-1)~(1/2);(3)y=z+3-3x~2. Idea Analysis (1) Usage and Matching Method Get y=-(z-1)~2-1. Because z∈[-1,2), when z=1, y_(max)=-1, when z=-1, y_(min ) = -5, and the range is [-5,-1].