# Saddle points and the second derivative changing sign

Let $f$ be a some real-valued function differentiable at least three times (I want to say twice, or maybe twice with a continuous second derivative, but I'll play it safe).

A saddle point of $f$ is some value $c$ in its domain, not a local extremum, which satisfies $f'(c) = 0$.

Proposition: $c$ is a saddle point of $f$ if and only if $f'(c) = 0$, $f''(c) = 0$ and $\exists \epsilon \in \mathbb{R_+}$ such that $\forall y \in (c - \epsilon, c)$ and $\forall x \in (c, c+ \epsilon), f''(x) \cdot f''(y)≤0$.

In other words, the second derivative "changes sign" at $c$. Intuitively, I am hypothesizing that a point is a saddle point iff it's a critical point and an inflection point.

Does this hold?

• Consider the function $f(x) = \begin{cases} (x+1)^3, & x<-1 \\ 0, & -1\le x\le 1 \\ (x-1)^3, & x>1\end{cases}$. What type of point is $(0,0)$? – user137731 Jan 10 '16 at 5:21
• (I deleted the previous comment). But could you elaborate, perhaps? – MathematicsStudent1122 Jan 10 '16 at 5:29
• This function is at least twice differentiable (I think it's probably three times differentiable, but I'm too lazy to check now). The point $(0,0)$ is certainly not an extremum, thus by your definition it is a saddle point. But taking $\epsilon < 1$ we get $f''(x)\cdot f''(y)=0$. – user137731 Jan 10 '16 at 5:32
• @Bye_World Was my definition of a "saddle point" careless? I never considered constant functions. – MathematicsStudent1122 Jan 10 '16 at 5:34
• It's fine -- I believe that's the usual definition. But I just showed an example where you can have a saddle point that is not an inflection point. So your proposition is false. – user137731 Jan 10 '16 at 5:35

Consider the function $f(x):=x^n\sin{\displaystyle{1\over x}}$ $(x\ne0)$, $\ f(0):=0$, with $n\geq2$ to your liking.
Concerning you comment below: It's not clear what your exact definition of a saddle point is. In any case, if $f(0)=f'(0)=0$ then the following formula, obtained through partial integration, may be of interest to you: $$f(x)=\int_0^x(x-t)f''(t)\>dt\ .$$