Continuous functions with orbit of period $3$

Question

I would like to build some continuous functions $f : E \to \Bbb R$ (where $E \subset \Bbb R$ is an interval), such that $$\exists x \in E,\;\; [f(x)≠x≠f(f(x)),\;\; f^3(x):= f(f(f(x)))=x]$$

I tried $f(x)=ax+b$ and $x_0=1$ and I got the condition $a^3+a^2b+ab+b=1$ which leads to $b=1-a$, so that $f(1)=a\cdot 1 + (1-a)=1$ and $x_0=1$ is a fixed point instead of being of period $3$.

(My aim is to test Sharkvosky's theorem on some examples).

Thank you!

Answer 1

$1-2\left|x-\frac12\right|$ has the three-cycle $2/7,4/7,6/7$. This example comes from the paper "The Sharkovsky Theorem: A Natural Direct Proof", by Keith Burns and Boris Hasselblatt. A preprint is freely available online, and it's a great read.

Answer 2

Just pull some function values out of a hat -- for example $$ f(0) = 42 \qquad f(42)=117 \qquad f(117)=0 $$ and then do Lagrange interpolation (or for that matter linear interpolation, whatever floats your boat) between those points.

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Your attempt with a first-degree polynomial failed because a first-degree polynomial iterated three times is still a first-degree polynomial, and so $f^3(x)=x$ can have only one root. Since the root of $f(x)=x$ is also a root of $f^3(x)=x$, there won't be room for any other ones.

But as soon as you move to quadratic polynomials there will be lots of opportunities. (Lagrange interpolation between three given points always results in a polynomial of degree $\le 2$).