已知函数f(x)满足f(x+y)+f(x-y)=2f(x)×f(y),且f(0)≠0,证明f(x)是偶函数
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已知函数f(x)满足f(x+y)+f(x-y)=2f(x)×f(y),且f(0)≠0,证明f(x)是偶函数
令y=0
f(x+y)+f(x-y)=2f(x)×f(y)
=>f(x)+f(x)=2f(x)*f(0)
=>f(0)=1
令x=0
f(x+y)+f(x-y)=2f(x)×f(y)
=>f(y)+f(-y)=2f(0)*f(y)
=>f(-y)=f(y)
=>f(x)是偶函数
令x=y=0
故:f(0+0)+f(0-0)=2f(0)f(0)
即:2f(0)=2 f(0)f(0)
因为f(0)≠0
故:f(0)=1
令x=0
故:f(y)+f(-y)=2f(0)f(y)
故:f(y)+f(-y)=2f(y)
故:f(-y)=f(y)
因为y任意
故:f(-x)=f(x)
故:f(x)是偶函数(定义域R关于原点对称)
要证明f(x)是偶函数即证明f(x)=f(-x)。注意题中已知条件f(0)不等于0。证明:令x=y=0,代入已知关系式,得2f(0)=2f(0)^2,所以f(0)=1(因为f(0)不等于0)。然后令x=0,得f(y)+f(-y)=2f(0)f(y)=2f(y),所以f(-y)=f(y)亦即f(-x)=f(x),f(x)是偶函数得证。
要证明f(x)为偶函数,只需证明f(x)=f(-x)
此题我们假设x=1,y=0则
f(1)+f(1)=2f(1)* f(0)
所以f(0)=1
再假设x=0则
f(y)+f(-y)=2f(0)*f(y)
即
f(y)+f(-y)=2f(y)
所以
f(y)=f(-y)
可得
f(y)为偶函数
证明完毕
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