Does negative third derivative imply negative first derivative?

Question

Does negative third derivative imply negative first derivative? For a system, the negative of the derivative of the Lyapunov function means the system is stable. How about the negative third derivative?

Answer 1

No, consider $f(x) = x-x^3$ at $x=0$.

$f'(0) = 1 > 0$, while $f'''(0) = -6 < 0$.

Perhaps you are assuming additional contraints?

Answer 2

No. Consider $f(x) = x^4 - 4x^3 + 5x^2 + 1$. This function is everywhere positive: $f > 0$.

Also, $f' > 0$ on $(0,\infty)$ and $f''' < 0$ on $(-\infty, 1)$.

This means that on the interval $(0,1)$ we have both $f''' < 0$ and $f' > 0$.