confusion over statement: If $x' = f(x)$, $f$ is continuously differentiable, then all solutions are monotone or constant.

Question

Here's the statement:

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If $x' = f(x)$, an autonomous scalar dynamical system for $x\in\mathbb{R}$, where $f$ is continuously differentiable, then all solutions are either strictly monotone or constant.

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If we have $f(x)=cosx$ then $f$ is still $C^1$, it is continuously differentiable, and the solution would be $x(t) = sin(x)+K$ for $K\in\mathbb{R}$. Here, x(t) is not constant and it is not strictly monotone.

So, how does the statement hold? Isn't my example a counter-example to it? Why and how is the statement true?

Edit: I understand why my counter-example doesn't work. How can I go about proving the statement?

Answer 1

Here's a proof that illustrates a common technique for these kinds of arguments.

Suppose $x(t)$ is a solution to $x'=f(x)$.

Let's assume that $x'$ changes sign from negative to positive at $t=c$.

Then $x'$ would be increasing at $t=c$ which would mean $x''(c)>0$

However, with implicit differentiation, it follows that

$$x''(c)=f'(x(c))x'(c)=0$$ which is a contradiction. You can proceed with an identical argument for the case that $x'$ changes sign from positive to negative at $t=c$