For what values will f(x) be necessarily one-one? 
Let $g:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that
  $|g'(x)|\le M$ for all $x\in \mathbb{R}$. For what values of
  $\epsilon$ will the function $f(x)=x+\epsilon g(x)$ will be
  necessarily one-one?

Not getting any idea how to proceed, maybe Mean value theorem will work. 
 A: The necessary and sufficient condition is $M|\epsilon|<1$.


*

*If $M|\epsilon|<1$, then $|\epsilon g'(x)|\le M|\epsilon|<1$, so $f'(x)=1+\epsilon g'(x)>0$. Then $f$ is strictly increasing, and hence one-to-one.

*If $M|\epsilon|\ge 1$, consider the case $g(x)=-\epsilon^{-1}x$.

A: Yes, well Rolle's theorem is enough.
Let $a, b$ be such that $f(a)=f(b)$. There there is a $c\in [a, b]$ such that $f'(c)=1+\epsilon g'(c)=0$. Since $|g'(c)|\leq M$. Then if $|\epsilon|<\frac{1}{M}$ you are good.
A: The other answers are totally correct...
Next time, when you have no idea how to proceed, try a few simple cases! Let $M=1$ and let $g(x)=x$, or $g(x)=-x$, or $g(x)=\sin(x)$. Then try $\epsilon=0$, $\epsilon=1/2$, $\epsilon=1$, and $\epsilon=2$.
For each of these choices, write down $f(x)$. Graph it if you like. Is $f$ one-to-one? Now that you've looked at these cases, what patterns do you notice? Try $M=2$ or $M=1/2$. How do you need to adjust $g$ and $\epsilon$?
Keep going until you have a hypothesis. Then try to prove your hypothesis. You'll find it much easier to select the right tools when you know what the answer is!
