Prove the number of fixed points of $x_{n+1} = 3.9x_{n}(1-x_{n})$ is infinite as $n \to \infty$ 
Define $f\colon[0,1]\to[0,1]$ such that $f(x) = 3.9x(1-x)$. Consider the iterated function sequence $f_{n+1} = f(f_{n} (x))$. Prove that there is no $N \in \mathbb{N}$ such that the number of fixed points ($f_{n}(x) = x$) of $f_{n}$ is $\leq N$ for all $n \in \mathbb{N}$.
Hint: Use the intermediate value theorem.

I know the intermediate value theorme:

Let $f$ be a continuous real function on $[a,b]$. If $f(a) < f(b)$, and if $c$ is a number such that $f(a)<c<f(b)$, then there exists a point $x \in (a,b)$ such that $f(x) = c$.

I understand I have to show that the set $S_{n} = \{x \in [0,1] : f_{n} (x) = x\}$ grows in size for increasing $n$. I am unsure how to go about this problem using the intermediate value theorem.
 A: Consider the action of $f(f(x))$ on $[0.5,.975]$. One application of $f$ maps contains $[.1,.975]$ in the image. Upon the second application of $f$, the interval below $0.5$ gets has image containing $[.4,.975]$ while the portion above $.5$ Has image containing. $[.1,.975]$. Both of these contain $[.5,.975]$, so applying $f$ twice effectively maps that interval to two of those intervals.
Repeating this inductively $n$ times allows us to to say that $f_{2n}$ maps $2^n$ disjoint open intervals within $(.5,.975)$ to $(.5,.975)$. The image of $f$ contains that interval, so $f_{2n+1}$ also has $2^n$ such intervals. Within the closure of each interval, there’s some points that maps to .5 and 0.975. This, for all $m$, $f_m(x)-x$ has $2^{\lfloor m/2\rfloor}$ such intervals within each of whom maps to a positive value at the $.5$ number and a negative value at the $.975$ number. By the intermediate value theorem, each such pair have a zero between them and each zero is a fixed point, so the number of fixed points as $n$ goes to infinity is unbounded.
