令x=0,y=0f(0)+f(0)=2f(0)*f(0) f(0)=1令x=0f(y)+f(-y)=2f(0)f(y)=2f(y)f(-y)=f(y)所以f为偶函数令y=1/2f(x+1/2)+f(x-1/2)=2f(x)f(1/2)=0-f(x+1/2)=f(x-1/2)=-f(x-3/2)所以2是周期 令x=1/2,y=1/2f (1)+f(0)=...
定义域在R上的函数满足f(x+y)+f(x-y)=2f(x)f(y) f(0)≠0 f(1/2)=0 求:f(x)为偶函
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