How do you show that f has neither a local maximum nor a local minimum?

Question

Be $f : \mathbb R^2 \to \mathbb R$ defined by $f(x_1,x_2) = x_2x^4_1 −x_2$

How do you show that $f$ has for all $x \in \mathbb R^2$ with $Df (x) = 0$ neither a local maximum nor a local minimum?

I have already found the points which evaluate $Df (x) = 0$:

$(1,0)$ and $(-1,0)$.

The Hessian matrix is indefinite in both points. Is that an argument? In the lectures we only had the statement that: $f$ has a maximum / minimum in $a$, if $Df(a) = 0$ and $Hess f(a) \ge 0 (Hess f(a) \le 0)$.

$Df(x)$ means the derivative of $f$

Thanks in advance

Answer 1

The Hessian Matrix being indefinite is sufficient for both points to be saddle points (i.e. no local minima/maxima).

Furthermore, the statement in your lectures is wrong: It has to be: $$ x \text{ local minimum (maximum)} \implies Df(x) = 0 \text{ and }\mathrm{Hess}f(x)\geq0~(\mathrm{Hess}f(x)\leq 0) $$

"$\impliedby$" does not hold.