已知:函数f(x)=log2
(1)判断函数f(x)的奇偶性; (2)证明函数f(x)有性质:f(x)+f(y)=f(
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(1)由
>0得:-1<x<1,1-x 1+x
由f(-x)=log2
=log2(1+x 1-x
) -1=-f(x)1-x 1+x
故知f(x)为奇函数
(2)f(x)+f(y)=log2
+log21-x 1+x
=log21-y 1+y
•1-x 1+x
=log21-y 1+y 1+xy-(x+y) 1+xy+(x+y)
=log2
=f(1- x+y 1+xy 1+ x+y 1+xy
)得证.x+y 1+xy