判断函数f(x)=x3?3x2+1,x>0x3+3x2?1,x<0的奇偶性.
解:
当x>0时,-x<0,则f(-x)=-x3+3x2-1=-(x3-3x2+1)=-f(x);当x<0时,-x>0,则f(-x)=-x3-3x2+1=-(x3+3x2-1)=-f(x);综上,当x≠0时,总有f(-x)=-f(x),∴f(x)为奇函数.