Show $x+\epsilon g(x)$ is 1-1 if $g'$ is bounded and $\epsilon$ is small enough. Problem: Suppose $g$ is a real function on $\mathbb{R}$ with bounded derivative (say $|g'|<M$). Fix $\epsilon>0$, and define $f(x)=x+\epsilon g(x)$. Prove that $f$ is one-to-one if $\epsilon$ is small enough. 
(A set of admissible values of $\epsilon$ can be determined which depends only on $M$.)
Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 3. 

Suppose not. Then for every $\epsilon>0$, there exists $a,b\in\mathbb{R}$, $a\neq b$, with $f(a)=f(b)$. Using the mean value theorem, we see there exists $x\in(a,b)$ with $f'(x)=0$. Note that $f$ is differentiable, as it is the sum of two differentiable functions. For such an $x$, we have 
$$f'(x)=1+\epsilon g'(x)=0 \Rightarrow g'(x)=-\frac{1}{\epsilon}.$$
Taking $\epsilon$ small enough, we can force $g'(x)$ to be arbitrarily large. This is a contradiction, as $g'$ is bounded.
