Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function with bounded derivative. Prove that there exists a constant $c>0$ such that the function $\phi\left(x\right)=x+cf\left(x\right)$ is a bijection with differentiable inverse.
I tried to do the next. Since $f$ has bounded derirvative there exists a constant $L>0$ such that $\mid f'\left(x\right)\mid\leq L$ for all $x\in\mathbb{R}$. And using the mean value theorem is easy ro see that $f$ is a Lipschitz function with constant $L$. So for all $x,y\in\mathbb{R}$ $\mid f\left(x\right)-f\left(y\right)\mid\leq L\mid x-y\mid$. Since $L>0$ I can take $k>0$ sucth that $kL<1$. Then I deifned $\phi$ as $\phi\left(x\right)=x+kf\left(x\right)$. For the injective part we suppose that there exist $x,y\in\mathbb{R}$, $x\neq y$ with $\phi\left(x\right)=\phi\left(y\right)$. So using that $f$ is Lipschitz and $kL<1$ we have that:
\begin{align*} x+kf\left(x\right)) & =y+kf\left(y\right)\\ x-y & =k\left(f\left(y\right)-f\left(x\right)\right)\\ \mid x-y\mid & =k\mid f\left(x\right)-f\left(y\right)\mid\\ & \leq kL\mid x-y\mid\\ & <\mid x-y\mid \end{align*}
That is $\mid x-y\mid<\mid x-y\mid$ wich is a contradiction and therefore $\phi$ is injective. But I couldn't prove that $\phi$ is surjective and its inverse is differentiable. I don't know if the $k$ that I chose is correct. Could you help me please? Thanks.
Note that $$\phi'(x)=1+kf'(x)\ge1-kL>0 $$ for all $x$. Hence $\phi$ stay above the line $f(0)+(1-kL)x$ for positive $x$ and below said line for negative $x$. This means that $\phi$ is not bounded from above or below, and by continuity (IVT), $\phi$ is onto.
Assume $y_n\to y$ where $y=\phi(x)$. Let $x_n=\phi^{-1}(y_n)$. Then $|x_n-x|\le\frac1{1-kL}|y_n-y|$ and hence $x_n\to x$. Hence $$ \lim_{n\to\infty}\frac{\phi^{-1}(y_n)-\phi^{–1}(y)}{y_n-y}=\lim_{n\to\infty}\frac{x_n-x}{\phi(x_n)-\phi(x)}=\frac1{\lim_{n\to\infty}\frac{\phi(x_n)-\phi(x)}{x_n-x}}=\frac1{\phi'(x)}$$ (note that the denominator is $\ne0$). We conclude that $\phi^{-1}$ is differentiable and $$\left(\phi^{-1}\right)'(y)=\frac1{\phi'(\phi^{–1}(y))}.$$