分段复合函数求解「详细过程」
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解决时间 2021-03-26 03:46
- 提问者网友：回忆在搜索
- 2021-03-25 11:58
分段复合函数求解「详细过程」
最佳答案
- 五星知识达人网友：千夜
- 2021-03-25 13:15
|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
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
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
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
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
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