设f(x)=axx+bx+c
因为f(0)=0,所以c=0,f(x)=axx+bx
f(x+1)=f(x)+x+1
知f(1)=f(0)+0+1=1,f(2)=f(1)+1+1=3
所以有a+b=1,4a+2b=3,
得a=b=0.5所以f(x)=0.5xx+0.5x
g(x)=2f(-x)+x=xx
f(g(x))=f(xx)=0.5x^4+0.5x^2