f(x)在R上函数,且对于任意ab∈R。满足f(ab)=af(b)+bf(a)
发布网友
发布时间：2024-10-24 01:50
共1个回答
热心网友
时间：2024-10-28 14:41
(1)
f(ab)=af(b)+bf(a)
令a = 0,b = 0,则得:
f(0*0) = 0*f(0) + 0*f(0) = 0
即 f(0) = 0
令 a = 1,b = 1,得:
f(1) = 1*f(1) + 1*f(1)
则 f(1) = 0
(2)
令 a = -1,b = -1, 得:
f(1) = -1*f(-1) - 1*f(-1) = -2f(-1)
因为 f(1) = 0
所以 f(-1) = 0
f(-x) = f[(-1)*x] = -f(x) + xf(-1) = -f(x)
所以是奇函数
l
