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函数的值域是函数的三要素之一,也是三要素中的难点和重点,和函数的最值有着密切的联系,因此,如何求它就显得特别重要,本文介绍了求函数值域常用的几种方法及其具体的应用.一、利用已知的函数模型1.观察法.“直线类,反比例函数类”用此方法.2.配方法.利用的是二次函数的模型,采用配方与函数的图象相结合的方式求值域.适合的题型是二次型函数y=Af~2(x)+Bf(x)+C,这种方法要注意的是其结构是同一个函数中具备一个函数和这个函数的平方的关系,如:x与x~(1/2),e~(2x)与e~x等.例1求y=(-x~2-6x-5)~(1/2)的值域.解:设μ=-x~2-6x-5,则μ≥4;μ=-x~2-6x-5=-(x+3)~2+4≤4;又μ≥0,所以0≤μ≤4.μ~(1/2)∈[0,2],所以值域
The range of the function is one of the three elements of the function. It is also the difficulty and focus of the three elements, which is closely related to the value of the function. Therefore, how to find it is particularly important. This article introduces the commonly used Several methods and their specific applications. First, the use of known function model 1. Observation. “Straight class, inverse function class ” using this method .2 with the method. The use of the quadratic function model, The formula of the formula is used to find the value range of the image, and the suitable type is the quadratic function y = Af ~ 2 (x) + Bf (x) + C. The same function with a function and the square of this function, such as: x and x ~ (1/2), e ~ (2x) and e ~ x, etc. Example 1 seeking y = (- x ~ 2-6x -5) ~ (1/2) Solution: Let μ = -x ~ 2-6x-5, then μ≥4; μ = -x ~ 2-6x-5 = - (x + 3) ~ 2 + 4≤4; and μ≥0, so 0≤μ≤4.μ ~ (1/2) ∈ [0,2], so the range