# 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

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I believe you are asking whether $f\colon \mathbb{R} \mapsto \mathbb{R}$ is invertible. Take a look at the plot and you will see that the answer is no. – Fabian Jan 1 '13 at 22:55

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 $$f(2n \pi - \pi/2) = f(2 n \pi) = f(2n \pi + \pi/2)$$

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but for big big 'x' the dominant term is $f(x)=x$ and the inverse of this is just $g(x)$ , besides any function which can be drawn can be inverted, just reflect each point trough the line $y=x$ to get the Numerical inverse of the function – Jose Garcia Jan 1 '13 at 22:59
@JoseGarcia $f(x) \sim g(x)$ and $g(x)$ is invertible does not necessarily mean that $f(x)$ is also invertible. – user17762 Jan 1 '13 at 23:01

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 $f(g(x))=x$

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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.

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HOWEVER the sine function is periodic $f(x)=sin(x)$ and the inverse exists , $f^{-1}(x)= arcsin (x)$ and the same for the cosine function, anyway any function can be 'numerically' inverted by reflecting each point of the function $y=f(x)$ through the line $y=x$ am i right ? – Jose Garcia Jan 4 '13 at 13:44
Not really, any function is invertible on a suitably restricted domain. – copper.hat Jan 4 '13 at 18:05
you mean several branches ?? of the function like for example in the case of the arc sinus ?? :D – Jose Garcia Jan 4 '13 at 19:16

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 ($n \in Z$):

$$0.632351 + 2\pi n < x < 2.509244 + 2\pi n$$

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you might want to add a bit more detail to validate your claim, for example, that $f'(\pi/2) = -1$. Regards – Amzoti Jan 1 '13 at 23:17