= Solution
At $x=0$, the <gradient> from the preceding part vanishes and the <Hessian matrix> is zero. The second-order test is therefore inconclusive. Instead inspect the <eigenline>:
$$
h(tb)=(t^2b^TAb)(t\|b\|^2)=\tfrac12t^3.
$$
For arbitrarily small positive and negative $t$, these values have opposite signs. Thus \b[$0$ is neither a <local maximum> nor a <local minimum>]. In dimension at least two it is a degenerate <saddle point>; in one dimension it is a stationary <inflection point>. Looking only at the zero <Hessian matrix> would not determine this classification.
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