判断下列函数的奇偶性 

x＞0时,- x＜0f(-x) = (-x)[1+(-x)] = - x(1-x) = - f(x)x＜0时,- x＞0f(-x) = (-x)[1-(-x)] = - x(1+x) = - f(x)x=0时,讨论f(x)在x=0上的连续性 < x→ +0 > lim f(x) = lim x(1-x) = 0 ←（左逼近求极限）< x→ - 0 > lim f(x) = lim x(1+x) = 0 ←（右逼近求极限）∴< x→ +0 >lim f(x) = < x→ -0 > lim f(x) = f(0) = 0综上所述,原函数f(x)是定义在R上的奇函数。