Is it ture that if we compose the function with itself many times for a certain $x$ the results either a fixed point or $,\infty$ or $-\infty$. ?
For ploynomials I guess it is true I tried more than one example and it is true. I look on this geometrically by graphing the function and the line $y=x$ then choose initial value for $x$ then make a projection in the $y$ direction to the $f(x)$ then make a projection to the line $y=x$ in the $x$ direction and so on.
On
A counterexample- let's say that you are currently incredibly indecisive about either studying for an exam or going to the movies with friends. You decide, ultimately, to go visit your friends. So, you drive over, and get halfway there before realizing that, no, you do in fact need to study. So, you drive halfway home, and then change your mind. Halfway back again, you change your mind yet again, and so on.
We can describe this as a function that takes any value $x\in[0.5, 1)$ and maps it to $x/2$, and takes any value $y\in[0,0.5)$ and maps it to $y+(1-y)/2$. If we self-compose this function over and over infinitely, we'll actually find that it alternates between $1/3$ and $2/3$, regardless of what value we chose to start with in the interval of $[0,1)$. Hence, this does not reach infinity or negative infinity, and does not reach a singular fixed point.
There are some obvious counterexamples if you truly mean a fixed point. As noted above in the comments, let $f(x)=1-x$ and let $x_0=0$. Define $x_{n+1}=f(x_n)$. Then $x_{n}= n \pmod{2}$; i.e., $f(x)$ alternates between $0$ and $1$. Or let $f(x) = \frac x2$ and choose $x_0=1$. Then $x_n$ is never fixed (though $\lim x_n=0$.)
What I think you really mean is that the sequence $x_n$ (defined by $x_{n+1}=f(x_n)$) goes to $\pm \infty$ or it has a cluster point. And that's obviously true. Any infinite sequence of real numbers that's bounded is contained in a compact set, and that means that it has a cluster point.