定义在R上的函数f(x)满足f(x+y)=f(x)+f(y)+2xy (x,y属于R),f(1)=2,则f(-2)等于

问题描述:

定义在R上的函数f(x)满足f(x+y)=f(x)+f(y)+2xy (x,y属于R),f(1)=2,则f(-2)等于

f(x+1)=f(x)+f(1)+2x=f(x)+2x+2
=> f(x)=f(x+1)-2x-2
=> f(-2) = f(-1) +2 = f(0) + 2 = 2

f(2)=f(1)+f(1)+2=6
f(1)=f(2)+f(-1)-4=2 ,f(-1)=0
f(-2)=f(-1)+f(-1)+2=2

f(0)=f(0+0)=f(0)+f(0)+0 =2f(0)
所以f(0)=0
f(2)=f(1+1)=f(1)+f(1)+2 = 2+2+2=6
而0=f(0)=f(2-2)=f(2)+f(-2)-8=6+f(-2)-8=f(-2)-2
所以f(-2)=2