设函数fx=4x^2-2(p-2)x-2p^2-p+1[-1,1]上小于等于零恒成立求p范围

2024-12-29 16:22:21
推荐回答(5个)
回答1:

f(x)=4x²-2(p-2)x-2p²-p+1
函数f(x)的对称轴是x=(p-2)/4
1、若p>2,则对称轴在直线x=0右侧,则f(x)的最大值是f(-1)=-2p²+p+1≤0,得:p≥1或 p≤-1/2,则:p>2;
2、若p≤2,则对称轴在直线x=0左侧,则f(x)的最大值是f(1)=-2p²-3p+9≤0,得:p≥3/2或p≤-3,则:p<-3
综合,有:p>2或p<-3

或者:
利用二次函数图像,根据f(1)≤0且f(-1)≤0也可以解出。

回答2:

f(1)=4-2(p-2)-2p^2-p+1≤0
f(-1)=4+2(p-2)-2p^2-p+1≤0
解这个不等式组就可以了
2p^2+3p-9≥0
2p^2-p-1≥0

(2p-3)(p+3)≥0,(2p+1)(p-1)≥0
p≥3/2或p≤-3,p≥1或p≤-1/2
所以最后的解是
p≥3/2或p≤-3

回答3:

1)对称轴 x=p/4-1/2 当x=p/4-1/2>=1时 p>=6
p不存在
(2)对称轴 x=p/4-1/2 当x=p/4-1/2=<-1时 P=<4
p(-3,3/2)
(3) 当对称轴在(-1,1)之内 -1

p不存在
综上 p(-3,3/2)

回答4:

fx=4x^2-2(p-2)x-2p^2-p+1
因式分解得
fx=(2x-2p+1)(2x+p+1)
可得,fx于X轴的两个焦点横坐标为 (2p-1)/2 -(p+1)/2
要求[-1,1]上小于等于零恒成立
(必须画图看曲线,此曲线为开口朝上的抛物线,看懂了就明白了)
则要求以上两个 坐标必须在 这个区间内
即: -1 < =(2p-1)/2<=1 -1 <=-(p+1)/2<=1
解出交集p范围 为[-1/2 ,1]

回答5:

p>=3/2或p<=-3

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