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)
二次函数f(x)的二次项系数为负,且f(x+2)=f(2-x),x∈R,问f(1-2x^2)与f(1+2x-x^2)满足什么关系时,有-2
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]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 -x ^ 2 +6 x-7
2:-x ^ 2 +6 x-7 in the right symmetry axis
2 - [2 - (1 / 2) x ^ 2]2
3:the lowest point
2 - (1 / 2) x ^ 2 = 2
At this point - x ^ 2 +6 x-7 =- 7
Inequalities still set up
由题意可知,对称轴是2,开口向上.
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]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 -x ^ 2 +6 x-7
2:-x ^ 2 +6 x-7 in the right symmetry axis
2 - [2 - (1 / 2) x ^ 2]2
3:the lowest point
2 - (1 / 2) x ^ 2 = 2
At this point - x ^ 2 +6 x-7 =- 7
Inequalities still set up