Is there a continuous, determinate function that is not injective, constant, or an injective part followed by a constant part? Let $y=f(x)$ be a continuous function, and let $f(x)=f(x')$ with $x \neq x'$. Say that $x$ and $x'$ have the same sequel if $f(x+k)=f(x'+k)$ for all $k \ge 0$. Now, say that $f$ is determinate if $x$ and $x'$ have the same sequel whenever $f(x)=f(x')$. It's clear that a function is determinate if it is: a) injective, b) constant, or c) composed of an injective part followed by a constant part (e.g., $f(x)$ = $x$ when $x \le 0$ and $0$ when $ x \ge 0$). 
Question: Is there a determinate function that that is not one of the aforementioned three types? 
 A: I assume the final problem statement excludes injective functions, because these are automatically determinate.
But take a function $f:\Bbb R\to\Bbb R$ which is injective only for $x\le 0$ and constant for $x\ge 0$. This function is not constant everywhere but still determinate, because all possible $(x,x')$-pairs can only be found on the constant part and their sequels are only to the right of zero.

As it turns out, all counterexamples are of the form above. Any determinate function becomes constant as soon as it becomes non-injective. More precisely:

Theorem. If $f$ is determinate and $a\not=b$ with $f(a)=f(b)$. Then $f(a)=f(x)$ for all $x\ge a$.

Proof.
So assume $\alpha:=f(a)=f(b)$ for $a>b$. The proof goes in two steps. At first we show that $f$ is constant on $[a,b]$, and then that it is constant everywhere right of $b$.
Step 1. We show that $f^{-1}(\alpha)$ is dense in $[a,b]$. Assume it is not. Because $f^{-1}(\alpha)$ is closed, we can find an interval $[s,t]\subseteq[a,b]$ with $f(s)=f(t)=\alpha$ but $f(x)\not=\alpha$ for all $x\in(s,t)$. IVT shows that there are $s < s'< 1/2 < t' < t$ with $f(s')=f(t')$. Note that $s'+(t-t')\in(s,t)$. But then
$$f(s'+(t-t'))=f(t'+(t-t'))=f(t)=\alpha$$
in contradiction to the choice of $[s,t]$. Therefor $f^{-1}(\alpha)$ is dense in $[a,b]$ and because $f$ is continuous $f$ is constant on $[a,b]$.
Step 2. Assume $f$ is not constant everywhere right of $b$. Because $f^{-1}(\alpha)$ is closed, we can find a maximal interval $[a,c]$ on which $f$ is constant. Note that $a+\xi\in[a,c]$ for all $\xi\in[0,c-a]$. Hence
$$f(c+\xi)=f(a+\xi)=\alpha.$$
This contradicts the choice of $[a,c]$ and hence $f$ must be constant on $[a,\infty)$. $\;\square$
