Question

已知函数f(x)=x^3+x（1）判断f(x)的奇偶性欲单调性并加以证明；（2）当x属于（-1,1）时，

有f（1-m）+f(1-m^2)<0，求m的取值范围。

Answer 1

f(-x)=-x^3-x=-(x^3+x)=-f(x)
所以是：奇函数！
f'(x)=3x^2+1>0
所以：f(x)是单调增函数！
（2）当x属于（-1,1）时，有f（1-m）+f(1-m^2)<0
则有：-1<1-m<1··········································1
-1<1-m^2<1········································2
f（1-m）+f(1-m^2)<0 则：f（1-m）<-f(1-m^2)
得：f(1-m)<f(m^2-1)
则有：1-m<m^2-1·······································3
联立1、2、3解得：1<m<√2