Question

f(x)在R上函数,且对于任意ab∈R。满足f(ab)=af(b)+bf(a)

f(x)在R上函数,且对于任意ab∈R。满足f(ab)=af(b)+bf(a)
(1) 求f(0),f(1)的值
(2) 判断f(x)的奇偶性

过程谢谢！

Answer 1

(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

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