is this function invertible ?? given the function 
$$ f(x)= x+\cos(x)+\sin(\cos(x)) $$ (1)
is this invertible ?? i mean it exists another function $ g(x) $ so
$$ f(g(x))=x $$
my guess is that for big $ x \gg 1 $ the function 'x' is asymptotic to $ g(x) \sim x $
since for big 'x' the function $ f(x) \sim x $ so for big big x we have that our function is always invertible
also $ f(x) $ is approximately always increasing $ f'(x) \ge 0 $, which is a necessary condition to get a function to have an inverse
 A: The statement that $f'(x) \ge 0$ is false. $f'(x) = 1 - \sin(x)(\cos(\cos(x)) + 1)$. This is false for all $x$ such that ($n \in Z$):
$$ 0.632351 + 2\pi n < x < 2.509244 + 2\pi n$$
A: Look at the plot of $f(x) = x + \cos(x) + \sin(\cos(x))$ to conclude that it is not invertible.
We also have $$f'(x) = 1 - \sin(x) - \sin(x) \cos(\cos(x))$$
We have $$f'(n \pi) = 1, f'(2n \pi + \pi/2) = -1, f'(2n \pi - \pi/2) = 3$$ Hence no inverse exists since the function is not monotone.
EDIT
$f(x) \sim g(x)$ and $g(x)$ being invertible does not necessarily mean that $f(x)$ is also invertible.
To see this, let us consider the function
$$f(x) = x - \dfrac{\pi}2 \sin(x)$$ Clearly, $f(x) \sim x$ as $x \to \infty$. But $f(x)$ is not invertible since $$f(2n\pi) = 2 n \pi$$ $$f(2n \pi + \pi/2) = 2n \pi + \pi/2 - \pi/2 = 2n \pi$$  $$f(2n \pi - \pi/2) = 2n \pi - \pi/2 + \pi/2 = 2n \pi$$
Hence, we have $$f(2n \pi - \pi/2) = f(2 n \pi) = f(2n \pi + \pi/2)$$
A: The derivative is 
$$f'(x)=1-\sin(x)-\sin(x)\cos(\cos(x))=1-\sin(x)(1+\cos(\cos(x))).$$
At $x=\frac \pi2$, this becomes negative, whereas the overall trend of $f$ is increasing. Therefore $f$ is not invertible (not injective, not always increasing) in the form $g(f(x))=x$.
However, it is surjective and therefore allows a right inverse in the form $f(g(x))=x$
A: The derivative is $f'(x)= 1-\sin x (\cos(\cos x)+1)$ which is periodic and $f'(0) =1, f'(\frac{\pi}{2}) = -1$. Hence $f$ cannot be invertible, since if it were, $f'$ would be monotonic.
