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采用换元法求函数的值域,其目的有两个,一是化简运算过程,避繁求简;二是转化函数的形式,化生为熟.本文就无理函数与部分三角函数值域的求解来说明其应用. 例1 求函数 的值域. 解令t-(1+x)～(1/(1+x))则x-t2-1(t≥0), 于是y=t2+t+1=(t+1/2)2+3/4,∵t≥0∴y≥1.∴函数的值域为[1,+∞). 说明1.通过换元把求无理函数的值域转化成求二次函数的值域,达到了化生为熟的目的;2.所换新元的范围由原函数的定义域及所换元的表达式来确定;3.此题还可利用函数的单调性求解.
The use of the equivalence method to find the range of functions has two purposes. One is to simplify the calculation process and avoid complications; the second is the form of the transformation function, and the metamorphosis is familiar. This paper presents irrational functions and some trigonometric function ranges. The solution to illustrate its application. Example 1 Find the value range of the function. Let t - (1 + x) ~ (1/ (1 + x)) then x-t2-1 (t ≥ 0), so y = t2 +t+1=(t+1/2)2+3/4, ∵t≥0∴y≥1. The range value of the ∴ function is [1,+∞). Note 1. The irrational function is obtained by changing the term. The range of values is converted into a range of quadratic functions, which achieves the purpose of becoming metastable. 2. The scope of the new element is determined by the domain of the original function and the expression of the substitution. 3. You can also use the monotonicity of the solution.