证f(x+y)+f(x-y)=2f(x)f(y)为偶函数能否取-y?f(x+y)+f(x-y)=2f(x)f(y)取-y即把y换成-yf(x+y)+f(x-y)=2f(x)f(-y)比较两式得f(y)=f(-y)所以为偶函数?那为什么书上不这么证

f(x+y)=f(x)+f(y)+2xy f(xy)=f(x)+f(y),证明f(x/y)=f(x)-f(y) f(x+Y)+f(x-y)=2f(x)f(Y) 求其是偶函数 急 求证f（x+y）+f(x-y)=2f(x)f(y)是周期函数 f(x+y)+f(x-y)=2f(x)cosy,求证f(x)为周期函数 设f(x)=e的y次方,证明：（1）,f(x)f(y)=f(x+y) ,(2),f (x)/f(y)=f(x-y) 已知f(x)=3^x,求证：(1)f(x)·f(y)=f(x+y);(2)f(x)/f(y)=f(x-y). y=f(f(f(x))) 求导 f(x+y)+f(xy-1)=f(x)f(y)+2f(n)表达式 y=f(x), f[(x+y)/2] f(x+y)=f(x)-f(y),那么f(-x)=f(0)-f(-x)=-f(-x).2f(-x)=0? 导数：f(x+y)=f(x)f(y),且f'(o)=1,求f'(x)f(x+y)=f(x)f(y),且f'(o)=1,求f'(x)f(x+y)=f(x)+f(y)+2xy,且f'(o)存在,求f'(x) f(1+x)=af(x),且f'(0)=b,求f'(1) f(x+y)=f(x)*f(y)说明什么? f(xy,x-y)=x^2+y^2+xy;f(x,y)/∂x;f(x,y)/∂y 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 f(x+y)+f(x-y)=2f(x)f(y) 求证：f(0)=1 设f(x+y,x-y)=x^2-y^2,则f(x,y)=