Answer:
[tex]2^{2005}[/tex]
Step-by-step explanation:
We are given [tex]f^{[2005]}(x) = \frac {1}{2}[/tex].
Hence, [tex]f(f^{[2004]}(x))=\frac{1}{2}[/tex].
Assume [tex]y=f^{[2004]}(x)[/tex].
Hence, [tex]f(y)=\frac{1}{2}[/tex] or [tex]y=\frac{1}{4}[/tex] or [tex]y=\frac{3}{4}[/tex].
So we can conclude that [tex]f^{[2004]}(x)=\frac{1}{4}[/tex] or [tex]f^{[2004]}(x)=\frac{3}{4}[/tex].
So we can find [tex]f(f^{[2003]}(x))=\frac{1}{4}[/tex] orf(f^{[2003]}(x))=\frac{3}{4}.
Assume [tex]z=f^{[2003]}(x)[/tex].
Hence, [tex]f(z)=\frac{1}{8}[/tex] or f(z)=\frac{7}{8} or f(z)=\frac{5}{8} or f(z)=\frac{3}{8}.
Therefore, we double the number of possible solutions for each iteration.
Thus, the number of solutions is [tex]2^{2005}[/tex].
