Proving an unspecified function is invertible Given $g: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function with bounded derivative i.e. satisfying $|g'(x)|\leq K>0 ,  \forall x \in \mathbb{R}$, I am trying to show that for some constant $\epsilon>0$ small enough, the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x):=x+\epsilon g(x)$ is invertible.
My idea is to show that $f$ is bijective.  It occurred to me that  $f$ is injective since its derivative can be made to be strictly positive so that it is strictly increasing if $\epsilon<1/K \implies f'=1+\epsilon g'>1+\epsilon(-K)>1-1>0$.  However, how can I prove $f$ is surjective?
 A: You could restrict the codomain of $f$ to its range:
Then $f:\mathbb{R}\to f(\mathbb{R})$ is surjective.
Update: It turns out we can show $f(\mathbb{R})=\mathbb{R}$.
For $x>0$, 
Consider $f(x)-f(0)=x+\epsilon(g(x)-g(0))=x+\epsilon(xg'(c_x))$ for some $c_x\in (0,x)$ by Mean Value Theorem.
Then $f(x)=f(0)+x(1+\epsilon g'(c_x))\geq f(0)+x(1+\epsilon (-K))$ for all $x>0$.
Since $\epsilon<\frac 1K$, $(1+\epsilon(-K))$ is postive. Then letting $x\to +\infty$ will see that $f(x)\to +\infty$.
Analogously consider $x<0$, see that $f(x)\to -\infty$ as $x\to -\infty$.
A: By the given construction, $f$ is strictly increasing. It is also continuous: $g$ is differentiable, hence continuous, and $f$ is a linear combination of continuous functions.
If an injective $f$ is bounded for some $\epsilon$, $\lim_{x\to\pm\infty}(\epsilon g(x))'=-1$, and decreasing $\epsilon$ will increase this limit such that $\lim_{x\to\pm\infty}f'(x)>0$, i.e. $f$ diverges. With the monotonicity and continuity of $f$ shown above, $f$ becomes surjective. The injectivity of $f$ is preserved in the lowering of $\epsilon$, so it is always possible to choose $\epsilon$ so $f$ is bijective – and therefore invertible.
A: The key argument here is to show that $f(x)$ can't be "too far" from $x$, so the fact that $x$ is unbounded above and below implies the same for $f(x)$.
One such method is
$$ \begin{align}
|f(x) - x| = \epsilon |g(x)| 
= \epsilon \left| g(0) + \int_0^x g'(t) \, \mathrm{d}t \right|
\leq \epsilon \left( |g(0)| + |x| K \right)
\end{align}
$$
For $x>0$, we have
$$ x (1 - K\epsilon ) - \epsilon |g(0) | \leq f(x) $$
showing that $f(x)$ goes to $\infty$ as $x \to \infty$, assuming $K\epsilon < 1$.
Similarly, for $x < 0$, we have
$$ f(x) \leq x (1 - K \epsilon) + \epsilon |g(0)| $$
