Question

关于抽象函数的题目已知f(x)是定义在R上的函数.(1)当f(-x)=-f(x)时,求f(0)的值；(2)当f(x+2)

Answer 1

(1)令x=0

那式变为f(0)=-f(0)

2f(0)=0

f(0)=0

(2)

令x=2

那么f(4)=-f(2)=-1

令x=4

f(6)=-f(4)=1

令x=0

那么f(2)=-f(0)=1

f(0)=-1

令x=-2

f(0)=-f(-2)=-1

f(-2)=1

令x=-4

f(-2)=-f(-4)=1

f(-4)=-1