求函数f(x)=(4-x2)(x2+mx+n)的图像关于直线x=1对称,则f(x)的最大值是

y=f(x)的图像关于直线x=1对称,
f(1+x)=f(1-x),
[4-(1+x)^2][(1+x)^2+m(1+x)+n]=[4-(1-x)^2][(1-x)^2+m(1-x)+n],
[3-x^2-2x][x^2+(2+m)x+m+n+1]=[3-x^2+2x][x^2-(2+m)x+m+n+1],
(2+m)(3-x^2)=2[x^2+m+n+1],
(4+m)x^2=m-2n+4,
4+m=0,m-2n+4=0,
解得m=-4,n=0,
∴f(x)=(4-x^2)(x^2-4x)[注]
=-x^4+4x^3+4x^2-16x,
f'(x)=-4x^3+12x^2+8x-16=-4(x^3-3x^2-2x+4)=-4(x-1)(x-1+√5)(x-1-√5),
用序轴标根法,x