Global inverse using Inverse Function Theorem I'm tasked with solving the following question: let $0 \le \alpha < 1$ be a constant and consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x - \alpha \sin x$.  Show that $f$ is one-to-one and onto and that its inverse function is smooth.  
So far I've tried using the Inverse Function Theorem for a single variable.  My version of the Theorem is: suppose $f : \mathbb{R} \to \mathbb{R}$ is continuously differentiable.  Suppose that for $a \in \mathbb{R}$ we have $f^\prime (a) \ne 0$ and that $f(a) = b$.  Then there exist open neighborhoods $U$ of $a$ and $W$ of $b$ such that (i) $f(U) = W$ and $f|_U$ is one-to-one; (ii) if $h$ is the inverse of $f|_U$ then $h$ is continuously differentiable; (iii) for each $x\in U$ we have $h^\prime (f(x)) = \frac{1}{f^\prime (x)}$.  
So to use the IFT we check the condition: it's clear that $f(x) = x - \alpha \sin x$ is continuously differentiable regardless of the value of $\alpha$.  For any fixed $a \in \mathbb{R}$ we have $f^\prime (a) = 1 - \alpha \cos a$.  Now, if $\alpha = 0$ then $f^\prime (a) = 1 \ne 0$ and the condition is automatically satisfied.  So suppose that $0 < \alpha < 1$.  The condition that $0 = f^\prime (a) = 1 - \alpha \cos a$ is equivalent to $\cos a = \frac{1}{\alpha}$, but $0 < \alpha <1$ implies $1 < \frac{1}{\alpha} < \infty$ so in this case there's no solution to $\cos a = \frac{1}{\alpha}$.  Thus in every possible case $f^\prime (a) \ne 0$.  Therefore the Inverse Function Theorem guarantees a local (smooth) inverse about every point.  To me it seems that the problem statement is asking about some global inverse however.  Is there some way we can glue together local inverses to obtain a global one?
 A: Thanks to the ordering of the real line, there is a global version of the inverse function theorem in the one dimensional case. In what follows, open intervals are allowed to be infinite in extent, such that the case $\mathbb{R} = (-\infty, +\infty)$ is also covered.

Global Inverse Function Theorem
Let $I$ be a non-empty open interval in $\mathbb{R}$, and suppose that $f \colon I \to \mathbb{R}$ is a continuously differentiable function such that $f'$ does not assume the value zero. Then:

*

*The derivative $f'$ has constant sign on $I$, and $f$ is strictly monotone.


*The image $f(I) = \{ f(x) : x \in I \}$ of $f$ is an open interval.
Furthermore, if $g \colon f(I) \to I$ is the inverse function of $f$ (which exists by 1.), then $g$ is continuously differentiable, has the same monotonicity as $f$, and $$g' = \frac{1}{f' \circ g}.$$

If we suppose that the open interval $I$ has lower endpoint $a$ and upper endpoint $b$ (both of which may be infinite), the strict monotonicity of $f$ along with the intermediate value theorem allows us to conclude that the image of $f$ is the open interval $f(I) = (m, M)$, where
$$m = \inf_{x \in I} f(x) \quad \text{and} \quad M = \sup_{x \in I} f(x).$$
If $f$ is strictly increasing, one has the additional formulas
$$m = \lim_{x \downarrow a} f(x) \quad \text{and} \quad M = \lim_{x \uparrow b} f(x),$$
and, if $f$ is strictly decreasing, the formulas read
$$m = \lim_{x \uparrow b} f(x) \quad \text{and} \quad M = \lim_{x \downarrow a} f(x).$$
Combining the above results should enable you to solve the problem.
A: Note that:
$$f'(x)=1-\alpha \cos(x)>0$$
so $f$ is always increasing, hence it is (globally) invertible. Then you can apply inverse function theorem for differentiability of the inverse.
