已知f(x)是定義在R上的函數(shù),對(duì)于任意x,y屬于R都有f(x+y)+(x-y)=2f(x)f(y),且f(o)不等于0
已知f(x)是定義在R上的函數(shù),對(duì)于任意x,y屬于R都有f(x+y)+(x-y)=2f(x)f(y),且f(o)不等于0
（1）求證：f(0)=1
(2)判斷函數(shù)f(x)的奇偶性
（1）求證：f(0)=1
(2)判斷函數(shù)f(x)的奇偶性
數(shù)學(xué)人氣：632 ℃時(shí)間：2019-10-04 06:45:54
優(yōu)質(zhì)解答
1.令x=y=0,所以由題意：f(0)+f(0)=2(f(o))^2---->2f(0)=2(f(o))^2由于f(0)≠0---->f(0)=12.2f(x)f(y)=f(x+y)+f(x-y)2f(x)f(-y)=f(x-y)+f(x+y)--->2f(x)f(y)=2f(x)f(-y)--->f(0)f(y)=f(0)f(-y)--->f(y)=f(-y)由y的任...
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