Map a closed function $f: (1,4) \rightarrow (1,4)$ without fixed point

Question

I need to find a continuous function $f (1,4) \rightarrow (1,4)$ that has no fixed points. I realise that the entire function then either lies above $y=x$ or below $y=x$ but I don't know how to get to an actual function so I need some help to get there.

Thanks in advance.

Answer 1

$f(x)=1+(x-1)^2/3.$ By analogy to $g(x)=x^2/3<x$ for $x\in (0,3).$

Answer 2

So you need that $g(x)=f(x)-x$ has no roots inside the interval. However you will find that $g$ needs to have roots at $1$ and $4$. Thus try something like $g(x)=c(x-1)(x-4)$.