1
令y=0
f(x)=f(x)+f(0)
f(0)=0
令y=-x
f(0)=f(x)+f(-x)
f(-x)=f(x),定義域為R
∴f(x)為奇函數(shù)
2.
f(3)=f(1)+f(2)=f(1)+f(1)+f(1)=3f(1)=9
∵f(x)為奇函數(shù)
∴f(-3)=-f(3)=-9第一題看不太懂 可以直接令y=0 令y=-x嗎可以直接令
對任意實數(shù)x y恒有f(x+y)=f(x)+f(y) (1) 求f(0)的值,并證明f(x)是奇函數(shù) (2) 若f(1)=3 ,求f(-3)的值
對任意實數(shù)x y恒有f(x+y)=f(x)+f(y) (1) 求f(0)的值,并證明f(x)是奇函數(shù) (2) 若f(1)=3 ,求f(-3)的值
數(shù)學(xué)人氣:703 ℃時間:2019-10-18 03:14:47
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