f(x)=-x^3+m,
令x1＜x2
f(x2)-f(x1)
=[-x2^3+m] - [-x1^3+m]
= x1^3-x2^3
=(x1-x2)(x1^2+x2^2+x1x2)
=(x1-x2) * [(x1+1/2 x2)^2 + 3/4 x2^2]
∵(x1+1/2 x2)^2 + 3/4 x2^2＞0,x1-x2＜0
∴f(x2)-f(x1)＜0,f(x2)＜f(x1)
∴函數(shù)在R上是減函數(shù)
∵函數(shù)是奇函數(shù)
∴f(-x)=-f(x)
∴-（-x)^3+m=-(-x^3+m),x^3+m=x^3-m
∴m=0