已知f(x)=a*x^2+b*x+c,g(x)=c*x^2+b*x+a

50分不好拿啊

f(1) =a+b+c f(0) =c f(-1) =a-b+c
且绝对植小于等于1
所以，a= [f(1)+f(-1)-2*f(0)]/2
b= [f(1)-f(-1)]/2
c= f(0)
Fmax(x)=max|f(x)|*max|g(x)|
|f(x)| =|{[f(1)+f(-1)-2*f(0)]/2}*x^2
+{[f(1)-f(-1)]/2}*x+f(0)|
<=|f(1)| * |x^2+x|/2+|f(-1)| * |x^2-x|/2
+|f(0)|*|x^2-1|
<=|x^2+x|/2+|x^2-x|/2+|x^2-1|
0<=x<1,|f(x)|=-x^2+x+1<=1(x=0时取等号）
-1<x<0,|f(x)|=-x^2-x+1<1
所以，|f(x)|<=1
同理，|g(x)|<=1
所以,|F(x)|<=1

得1吧......

得1吧