令其为0可得f3(x)=
1 |
2 |
1 |
2 |
解得f2(x)=
3 |
4 |
1 |
4 |
3 |
4 |
1 |
4 |
而f(f1(x))=|2f1(x)-1|,令其等于
3 |
4 |
1 |
4 |
可得f1(x)=
1 |
8 |
7 |
8 |
5 |
8 |
3 |
8 |
由f1(x)=f(x)=|2x-1|=
1 |
8 |
7 |
8 |
5 |
8 |
3 |
8 |
可解得x=
9 |
16 |
7 |
16 |
15 |
16 |
1 |
16 |
13 |
16 |
3 |
16 |
11 |
16 |
5 |
16 |
故可得函数y=f4(x)的零点个数为:8
故答案为8
1 |
2 |
1 |
2 |
3 |
4 |
1 |
4 |
3 |
4 |
1 |
4 |
3 |
4 |
1 |
4 |
1 |
8 |
7 |
8 |
5 |
8 |
3 |
8 |
1 |
8 |
7 |
8 |
5 |
8 |
3 |
8 |
9 |
16 |
7 |
16 |
15 |
16 |
1 |
16 |
13 |
16 |
3 |
16 |
11 |
16 |
5 |
16 |