二次函数f(x)的二次项系数为负,f(x+2)=f(2-x)

二次函数f(x)的二次项系数为负,f(x+2)=f(2-x)

由题意可知，对称轴是2，开口向上。
2-(1/2)x^2<=2
2-(1/2)x^2在对称轴左侧或者对称轴上
分三种情况
1：都在对称轴左侧
2-(1/2)x^2>-x^2+6x-7
2:-x^2+6x-7在对称轴右侧
2-[2-(1/2)x^2]<[-x^2+6x-7]-2且-x^2+6x-7>2
3：最低点
2-(1/2)x^2=2
此时-x^2+6x-7＝-7
不等式仍然成立

Know from that intended, is the axial symmetry 2, opening up.
2 - (1 / 2) x ^ 2 <= 2
2 - (1 / 2) x ^ 2 symmetry axis in the left or symmetry axis
The three conditions
1: axial symmetry in the left
2 - (1 / 2) x ^ 2>-x ^ 2 +6 x-7
2:-x ^ 2 +6 x-7 in the right symmetry axis
2 - [2 - (1 / 2) x ^ 2] <[-x ^ 2 +6 x-7] -2 and - x ^ 2 +6 x-7> 2
3: the lowest point
2 - (1 / 2) x ^ 2 = 2
At this point - x ^ 2 +6 x-7 =- 7
Inequalities still set up

∵二次函数f(x)的二次项系数，则二次函数图像开口向上

又f（2-x）=f（2+x）

∴x=2是f（x）的对称轴

故对称轴x=2左侧单调递减，右侧单调递增

∵1-2x²＜2

1+2x-x²=2-（x-1）²≤2

即1-2x²和1+2x-x²都在对称轴左侧

∴f(1-2x^2)<f(1+2x-x^2),

即1-2x²＞1+2x-x²

即x²+2x＜0

∴x(x+2)＜0

∴-2＜x＜0