How to construct a function which has 5-period point and has no 3-period point I just learnt Sarkoveskii theorem. I try to construct a continuous function $f:[0,1]\to [0,1]$ which has 5-period point but has no 3-period point. But I have no idea how to construct a such function.
 A: No idea how to construct this sort of function from first principle.  
The first example of such a function was given by Li and Yorke in 1975${}^{\color{blue}{[1]}}$. 

Let $F: [1,5] \to [1,5]$ be the piecewise linear function with values
  $$F(1) = 3, F(2) = 5, F(3) = 4, F(4) = 2, F(5) = 1$$ and linear
  interpolate between the integers.

This function $F(x)$ has a fixed point of period $5$ (i.e. $1 \to 3 \to 4 \to 2 \to 5 \to 1$) but no fixed point of period $3$. If one set $f(x) = \frac{F(1+4x)-1}{4}$, it will be a function you seek.
Look at Li and Yorke's paper for arguments why $F(x)$ doesn't has any fixed point of period $3$.
References


*

*$\color{blue}{[1]}$ - T.Y. Li, and J.A. Yorke, "Period Three Implies Chaos", American Mathematical Monthly 82, 985 (1975).

A: Consider the basic map that has a 3-period point:
$$ x \mapsto 2|x - \frac12| $$
Iterating this map three times you see a 3-period point $x$ solves
$$ |||8x - 4|-2|-1| = x $$
This equation has 8 solutions, corresponding to each of the possible signs in the absolute value. These include the fixed points $x = 1/3$ and $x = 1$, as well as the 2 cycles
$$ 1/7 \to 5/7 \to 3/7 \qquad \text{and} \qquad 
1/9 \to 7/9 \to 5/9 $$
Consider now instead the map
$$ f(x) = \max(\frac5{32}, 2|x - \frac12| ) $$
First one checks easily that it has a period 5 point
$$ 5/32 \to 11/16 \to 3/8 \to 1/4 \to 1/2 \to 5/32 $$
It suffices to argue that $f$ doesn't have a 3-period point. 
Claim 1 Any periodic orbit must only contain points that are greater than 5/32. 
Proof: $f(x) \geq 5/32$ for any input $x$. 
Claim 2 Restricted to any periodic orbit of period $\neq 5$, $f$ acts the same as the $x\mapsto 2|x-\frac12|$ map. 
Proof: Since the orbit has no points less than 5/32, we know that $f(x) \geq 5/32$ for any point $x$ in the orbit. If $f(x) > 5/32$ for all points, we are done. If $f(x) = 5/32$ for some $x$ in the orbit, we know this must be the period 5 orbit computed before. 
Now suppose for contradiction that $f$ has a 3-period point, by the two claims above $f$ restricted to the 3-cycle is equal to the map $x\mapsto 2|x-\frac12|$. But the 3-cycles of this latter map we have solved explicitly, and both solutions pass through a point less than 5/32 (both 1/7 and 1/9 are less than 5/32). Therefore the 3-cycle is impossible. 
