分段复合函数求解「详细过程」
1个回答
展开全部
|x| >2
=> x> 2 or x<-2
|x|≤2
-2≤x≤2
case 1: x<-2
f(x) = 0
f(f(x)) =4-x^2 = 4
case 2: -2≤x≤-√2
f(x) = 4-x^2
0≤4-x^2≤2
f(f(x)) = 4- (4-x^2)^2 =16+8x^2-x^4
case 3: -√2<x<0
f(x) = 4-x^2
2<4-x^2<4
f(f(x)) =0
case 4: 0≤x≤√2
f(x) = 4-x^2
2≤4-x^2≤4
f(f(x)) =0
case 5: √2<x<2
f(x) = 4-x^2
0<4-x^2<2
f(f(x)) =4-(4-x^2)^2=16+8x^2-x^4
case 6 : x≥2
f(x) = 0
f(f(x)) =4
=> x> 2 or x<-2
|x|≤2
-2≤x≤2
case 1: x<-2
f(x) = 0
f(f(x)) =4-x^2 = 4
case 2: -2≤x≤-√2
f(x) = 4-x^2
0≤4-x^2≤2
f(f(x)) = 4- (4-x^2)^2 =16+8x^2-x^4
case 3: -√2<x<0
f(x) = 4-x^2
2<4-x^2<4
f(f(x)) =0
case 4: 0≤x≤√2
f(x) = 4-x^2
2≤4-x^2≤4
f(f(x)) =0
case 5: √2<x<2
f(x) = 4-x^2
0<4-x^2<2
f(f(x)) =4-(4-x^2)^2=16+8x^2-x^4
case 6 : x≥2
f(x) = 0
f(f(x)) =4
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询