Consider the ODE $$y'=g(y)$$ Show that if $|g(y)|$ is increasing or decreasing in some interval then the solution in that interval is concave up or concave down, accordingly.
My approach:
Consider $y\in (a,b)$ implies $|g(y)|$ is increasing. Then $|y'|$ is also increasing for $y\in (a,b)$. $(a,b)$ must contain the point $y_0$ satisfying $g(y_0)=0=y'(x_0)$ for some point $x_0$ where the slope of the solution is zero. Since $|g(y)|$ is increasing on $(a,b)$, |y'| is also increasing on $(a,b)$, so that $|y''|$ is increasing on $(a,b)$ as well. Thus $|y''(x_0)|$ is increasing. However, $y''(x_0)$ may be greater than $0$ or it can be less than $0$.
So how can we say that the solution is necessarily concave up if $|g(y)|$ is increasing, but $g(y)$ is negative on $(a,b)$?
If there is $y_0 \in (a, b)$ such that $g(y_0) = 0$ then $\lvert g \rvert$ cannot be increasing or decreasing on $(a, b)$. So, either $g$ is positive and increasing on $(a, b)$ (which implies that the solution is increasing and concave up), or it is negative and decreasing (which implies that the solution is decreasing and concave down), or else $g$ is positive and decreasing, etc, etc.