Question

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

Answer 1

f(x+y)=f(x)+f(y)+2xy
令 x=y=0
f(0)=f(0)+f(0)+2*0*0
=>f(0)=0
x=y=1
f(2)=f(1)+f(1)+2*1*1
=2+2+2
=6

y=-x
f(x-x)=f(x)+f(-x)+2x*(-x)
f(0)=f(x)+f(-x)-2x^2
f(-x)=2x^2-f(x)
x=2
f(-2)=2*2^2-f(2)
=8-6
=2

Answer 2

两边对x求导
f'(x+y)=f'(x)+0+2yx+y²
两边对y求导
f'(x+y)=0+f'(y)+2xy+x²
两式相减得f'(x)-f'(y)+y²-x²=0
令y=0
f'(x)-1-x²=0
得
f'(x)=x²+1
积分得f(x)=x³/3+x+C(C为常数)