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分类讨论题是当前试题中的热点,而求函数y=A_1(t)f~2(x)+A_2(t)f(x)+A_3(t)的最值(函数在I上有定义,A_1(t)≠0,t为参数)又是分类讨论中常见的类型。如1992年及1993年上海市普通高级中学会考试题的压轴题,他们的模式便是本文议论的问题。 1.求该类函数的最值,其属求一元函数最值的范畴。函数 y=A_1(t)f~2(x)+A_2(t)f(x)+A_3(t)在I上有定义,A_1(t)≠0。若令f(x)=z,由x∈I得到f(x)∈(?),这样,原函数可化为y=A_1(t)z~2+A_2(t)z+A_3(t)A_1(t)≠0,z∈(?)。即y关于一元z的二次函数。由于t是参数,因此在求该类函数的最值时,它的思考方法和运
The classification discussion topic is the hot spot in the current test, and the function y=A_1(t)f~2(x)+A_2(t)f(x)+A_3(t) is defined as the function (the function is defined on I, A_1 (t) ≠ 0, t is a parameter) is another common type of discussion. For example, in the 1992 and 1993 editions of the Shanghai Senior High School Exams, their model was the subject of this article. 1. Find the most value of this type of function, which belongs to the category of the most value of the one-way function. The function y=A_1(t)f~2(x)+A_2(t)f(x)+A_3(t) is defined on I and A_1(t)≠0. If f(x)=z, f(x)∈(?) is obtained from x∈I. Thus, the original function can be changed to y=A_1(t)z~2+A_2(t)z+A_3(t). A_1(t)≠0,z∈(?). That is, y is a quadratic function with respect to a monad z. Since t is a parameter, when thinking about the best value of this kind of function, it’s thinking method and operation