一、函数性质法若题设中给出函数的单调性和奇偶性时,可以根据这些性质把抽象的函数关系式转化为具体的代数关系式,从而使问题得以解决。例1函数f(x)在(-1,1)上既是奇函数又是增函数,如果f(2-a)+f(4-a2)<0,求实数a的取值范围。分析:本题涉及不等式f(2-a)+f(4-a2)<0,只需将不等式进行移项得:f(2-a)<-f(4-a2),然后结合单调性,奇偶性以及定义 First, the method of functional nature If the given monotonicity and parity of the function, you can be based on these properties of the abstract functional relationship into a specific algebraic relationship, so that the problem can be solved. Example 1 The function f (x) is both an odd function and an increasing function on (-1,1), and f (2-a) + f (4-a2) <0. Analysis: The problem involved in the inequality f (2-a) + f (4-a2) <0, only need to shift the inequality obtained: f (2-a) <- f (4-a2), and then combined with monotonicity, Parity and definition