Injective function $f(x) = x + \sin x$ How can I prove that $f(x) = x + \sin x$ is injective function on set $x \in [0,8]$?
I think that I should show that for any $x_1, x_2 \in [0,8]$ such that $x_1 \neq x_2$ we have $f(x_1) \neq f(x_2)$ id est $x_1 + \sin x_1 \neq x_2 + \sin x_2$. But I don't know what can I do next. 
Equivalently I can show that for any $x_1, x_2 \in [0,8]$ such that $f(x_1) = f(x_2)$ we have $x_1 = x_2$ but I have again problem. 

Hint: What is the sign of $f'(x)$. What does it tell you about $f(x)$?

 Since $f'(x)\ge0$, the function is non-decreasing. So if $f(a)=f(b)$ for some $a<b$, then $f(x)$ would have to be constant on the interval $[a,b]$. This would imply $f'(x)=0$ for each $x\in[a,b]$. But there is no non-trivial interval such that $f'(x)$ is zero for each point of the interval.

