Question

Amc "for each $x$" in [0,1], define f(x) = 2x for how many values of x in [0,1] if f 2005 (x) = 1/2?

Answer 1

`2^(2005)`

Explanation:

We are given
`f^([2005])(x) = \frac {1}{2}`.

Hence,
`f(f^([2004])(x))=(1)/(2)`.

Assume
`y=f^([2004])(x)`.

Hence,
`f(y)=(1)/(2)` or
`y=(1)/(4)` or
`y=(3)/(4)`.

So we can conclude that
`f^([2004])(x)=(1)/(4)` or
`f^([2004])(x)=(3)/(4)`.

So we can find
`f(f^([2003])(x))=(1)/(4)` orf(f^{[2003]}(x))=\frac{3}{4}.

Assume
`z=f^([2003])(x)`.

Hence,
`f(z)=(1)/(8)` 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
`2^(2005)`.