Let $I$ be an interval and $f\colon I \to \mathbb{R}$ a differentiable function. Suppose the following definitions:
For $x_0 \in I$ the point $(x_0,f(x_0))$ is called saddle point if $f'(x_0) = 0$ but $x_0$ is not a local extremum of $f$.
For $x_W \in I$ the point $(x_W,f(x_W))$ is called point of inflection if there is a neighborhood $U$ from $x_W$ in $I$ such that $f'$ is strictly monotonic increasing (resp. decreasing) for $x < x_W$ on $U$ and strictly monotinic decreasing (resp. increasing) for $x > x_W$ on $U$.
What is the logical relation between saddle points and points of inflection?
My first intuitive guess was that a point $(x,f(x))$ is a saddle point iff it is a point of inflection and $f'(x) = 0$. However the implication "$\implies$" seems to be wrong. Consider the following counterexample:
$$
f(x) =
\begin{cases} x^4 \cdot \sin\left(\frac{1}{x}\right) & x \neq 0 \\
0 & x = 0
\end{cases}
$$
Then $(0,0)$ is a saddle point but not a point of inflection because the derivativative oscillates on every neighborhood of $0$.
Is this correct so far? Is the other implication true? If so, how to prove it?
Best Answer
Your example does indeed show that a saddle point need not be an inflection point. (The function $x^2\sin(1/x)$ also works, but your example has the virtue of being continuously differentiable.)
In the other direction, if $(a,f(a))$ is a point of inflection and $f'(a) = 0,$ then $(a,f(a))$ is a saddle point. To see this, suppose WLOG that for some small $\delta > 0$ that $f'$ strictly increases in $[a-\delta,a]$ and strictly decreases in $[a,a+\delta].$ In $[a-\delta,a)$ we have $f'(x)<0,$ because these values must be less than $f'(0)=0.$ The same reasoning shows that $f'(x) < 0$ for $x\in (a,a+\delta].$ The mean value theorem then shows $f$ strictly decreases on both $[a-\delta,a]$ and $[a,a+\delta].$ Hence $f$ strictly decreases on $[a-\delta,a+\delta].$ It follows that $f(a)$ is neither a local max. nor min. for $f$ at $a.$