Investigating the bijectivity of $ 2 x + |\cos(x)| $. The question asks if the function
$$
f(x) = 2 x + |\cos(x)|
$$
if (one-one, onto), (many-one, onto) or (one-one, into). After a long process of plotting the graph, I managed to guess it’s one-one and onto. The textbook answer says it’s many-one and onto. I was unable to find any direct algebraic method (I tried to find the inverse function and find its domain) to prove my answer, so I can’t be sure. Any help in showing how it’s many-one is appreciated.
Edit: The function is defined from $ \Bbb{R} $ to $ \Bbb{R} $.
 A: You are right, the function is one to one. Note that $f(x)$ is continuous everywhere, and is differentiable except at odd multiples of $\pi/2$. Except at these points, we have $f'(x)=2\pm\sin x$. In particular, everywhere except at odd multiples of $\pi/2$, we have $f'(x)\ge 1$. So $f$ is monotone increasing, and therefore one to one.
We can make $f(x)$ arbitrarily large negative by choosing $x$ large enough negative, and arbitrarily large positive by making $x$ large enough positive. So by the Intermediate Value Theorem $f$ is onto.
A: Consider the function on the intervals $(n\pi,(n+1)\pi)$. It is easy to show that the function is strictly increasing on each interval.
Since it is continuous, if follows that it is strictly increasing, hence injective.
It is easy to see that $\lim_{x \to -\infty} f(x) = -\infty$,
$\lim_{x \to +\infty} f(x) = +\infty$
Since it is continuous, it follows that $f$ is onto.
A: I'm pretty sure your textbook is wrong. This is what Wolfram Alpha gives me. 
http://www.wolframalpha.com/input/?i=plot%20abs%28cos%28x%29%29%20%2B%202x
Looks like a pretty one-to-one function to me.
