f(x1)+f(x2)=2f((x1+x1)/2)f((x1-x2)/2)恒成立,求证f(x)是偶函数

问题描述:

f(x1)+f(x2)=2f((x1+x1)/2)f((x1-x2)/2)恒成立,求证f(x)是偶函数
f(x)恒不为0

令x₂=x₁2f(x₁) = 2 f(x₁) f(0)令x₂= -x₁f(x₁) +f(-x₁) = 2 f(0) f(x₁) 故 2f(x₁) = f(x₁) + f(-x₁)即 f(-x₁) = f(x₁)∴f(x)...