Question

已知函数f(X)对一切实数XY都有F(x+Y)=F(x)+F(Y),求f(0)，并证明f(x)为奇

已知函数f(X)对一切实数XY都有F(x+Y)=F(x)+F(Y),求f(0)，并证明f(x)为奇函数

Answer 1

因为f(0) = f(0 + 0) = f(0) + f(0) = 2f(0)
所以f(0) = 0

因为0 = f(0) = f(x - x) = f(x) + f(-x)
所以f(x) = -f(-x)
所以f(x)是奇函数

追问

若f(1)=3，求f(-3)

麻烦了，谢谢

追答

f(3) = f(1 + 2) = f(1) + f(1 + 1) = f(1) + f(1) + f(1) = 3f(1) = 3 * 3 = 9
f(-3) = -f(3) = -9

f(3) = f(1 + 1 + 1) = 3f(1) = 3 * 3 = 9
f(-3) = -f(3) = -9