已知f(x)=|1-2x|,x属于【0,1】,问方程f(f(f(x)))=x/2有多少实根

Answer 1

1.x∈[0,1/2]时,
f(x)=|1-2x|=1-2x
f[f(x)]=|1-2f(x)|
=|1-2(1-2x)|
=|4x-1|
1-1.
x∈[0,1/4]时,
f[f(x)]=|4x-1|
=-4x+1
f{f[f(x)]}=|1-2f[f(x)]|
=|1-2(-4x+1)|
=|8x-1|
1-1-1.
x∈[0,1/8]时,
f{f[f(x)]}=|8x-1|
=-8x+1
=x/2
x=2/17;
1-1-2.
x∈[1/8,1/4]时,
f{f[f(x)]}=|8x-1|
=8x-1
=x/2
x=2/15;
1-2.
x∈[1/4,1/2]时,
f[f(x)]=|4x-1|
=4x-1
f{f[f(x)]}=|1-2f[f(x)]|
=|1-2(4x-1)|
=|-8x+3|
1-2-1.
x∈[1/4,3/8]时,
f{f[f(x)]}=|-8x+3|
=-8x+3
=x/2
x=3/17不∈[1/4,3/8],舍去；
1-2-2.
x∈[3/8,1/2]时,
f{f[f(x)]}=|-8x+3|
=8x-3
=x/2
x=2/5;
2.x∈[1/2,1]时,
f(x)=|1-2x|=2x-1
f[f(x)]=|1-2f(x)|
=|1-2(2x-1)|
=|-4x+3|
2-1.
x∈[1/2,3/4]时,
f[f(x)]=|-4x+3|
=-4x+3
f{f[f(x)]}=|1-2f[f(x)]|
=|1-2(-4x+3)|
=|8x-5|
2-1-1.
x∈[1/2,5/8]时,
f{f[f(x)]}=|8x-5|
=-8x+5
=x/2
x=10/17;
2-1-2.
x∈[5/8,1]时,
f{f[f(x)]}=|8x-5|
=8x-5
=x/2
x=2/3;
2-2.
x∈[3/4,1]时,
f[f(x)]=|4x+3|
=4x+3
f{f[f(x)]}=|1-2f[f(x)]|
=|1-2(4x+3)|
=|-8x-5|
=|8x+5|
=8x+5
=x/2
x=2/3不∈[3/4,1],舍去.
综上所述共有5个2/17、2/15、2/5、10/17、2/3.