Let $a$ be a point in the domain of a continuous real valued function $f$. One says that $f$ has a flex point at $a$ if we can find a small interval $(a - \epsilon , a + \epsilon)$ in the domain of $f$ such that the following happens: (i) for all $x$ in the open interval $(a - \epsilon , a)$ the sign of $f ^ { \prime \prime } ( x )$ is constant and, (ii) for all $x$ in the open interval $(a , a + \epsilon)$ the sign of $f ^ { \prime \prime } ( x )$ is constant and opposite to the sign of $f ^ { \prime \prime } ( x )$ in $(a - \epsilon , a)$.
Statements
(29) If $f$ is a cubic polynomial with a local maximum at $x = p$ and a local minimum at $x = q$, then $f$ has a unique flex point at $x = \frac { p + q } { 2 }$. (30) If $f ^ { \prime \prime } ( a ) = 0$ then $f$ must have a flex point at $a$. (31) Let $f ( x ) = x ^ { 2 }$ for $x \geq 0$ and $f ( x ) = - x ^ { 2 }$ for $x < 0$. Then $f$ has no flex points. (32) If $f ^ { \prime }$ is monotonic on an open interval $I$, then $f$ cannot have a flex point in $I$.