Proving surjectivity of a strictly monotone function Let $f:\mathbb{R}\rightarrow \mathbb{R}$ a $C^1$ function with $|f'(x)|<M$ for all $x$. Then I want to show that there exists an $a>0$ such that $g:\mathbb{R}\rightarrow \mathbb{R}$, $g(x)=x+af(x)$ is a bijection with differentiable inverse. 
For $a<1/M$ I have shown that it is strictly monotone, which implies it's injective and that it has a differentiable inverse. How can I show that is surjective?
 A: By Mean value theorem the condition $|f'(x)|<M$ implies $\frac{f(x)}{x}$ is bounded. So you know that:
$$\lim_{x \to \infty}g(x)=\lim_{x \to \infty}x+af(x)=\lim_{x \to \infty}x+ax\frac{f(x)}{x}=\lim_{x \to \infty}x(a+\frac{f(x)}{x})=\mp \infty$$
and 
$$\lim_{x \to -\infty}g(x)=\pm \infty$$
But $g$ is contionous, so by Intermediate value theorem $g$ is surjective.
A: Another look at surjectivity: suppose $y>x$ is such that $g(x)=g(y)$. Then
$$
y+af(y)=x+af(x)\implies(y-x)=a(f(x)-f(y))=a(x-y)f'(z)
$$
for some $z\in(x,y)$. Taking absolute values of both sides:
$$
|y-x|=a|y-x||f'(z)|\implies 1=a|f'(z)|.
$$
But $a|f'(z)|<\frac{1}{M}M=1$ so you have the desired contradiction.
A: Choose a positive $a<{1\over M}$. Then $m:=1-aM>0$. Since
$$g'(x)=1+af'(x)>1-a M=m\qquad(-\infty<x<\infty)$$
the function  $g$ is monotonically increasing, hence injective on ${\mathbb R}$. Furthermore
$$g(x)=g(0)+\int_0^x g'(t)\>dt>g(0)+mx\to\infty\qquad(x\to\infty)\ ,$$
and similarly $\lim_{x\to-\infty}g(x)=-\infty$. It follows that $g$ maps ${\mathbb R}$ bijectively onto ${\mathbb R}$. As $g'(x)>0$ for all $x$ the inverse map $\>g^{-1}\!\!:\>{\mathbb R}\to{\mathbb R}$ is $C^1$ again.
