1 |
2 |
(2)设x∈[0,1],则-x∈[-1,0],则f(−x)=
−x |
x2+1 |
因为函数f(x)为偶函数,所以有f(-x)=f(x)
既f(x)=
−x |
x2+1 |
所以f(x)=
|
(3)设0<x1<x2<1,则f(x2)−f(x1)=
−x2 |
x22+1 |
−x1 |
x12+1 |
(x2−x1)(x1x2−1) |
(x22+1)(x12+1) |
∵0<x1<x2<1
∴x2-x1>0,x1x2-1<0…(14分)
∴
(x2−x1)(x1x2−1) |
(1+ x12)(1+x22) |
∴f(x2)<f(x1)
∴f(x)在[0,1]为单调减函数…(16分)