= Solution
{c}
Write $h=x-x_0$, $k=y-y_0$, and let $a=f_{xx}$, $b=f_{xy}$ and $c=f_{yy}$ be evaluated at the <critical point>. Assuming $f$ is twice continuously differentiable nearby, the <Taylor expansion> is
$$
f(x_0+h,y_0+k)-f(x_0,y_0)
=\frac12(ah^2+2bhk+ck^2)+o(h^2+k^2).
$$
The linear terms vanish because the <gradient> is zero. For $a>0$, <completing the square> gives
$$
\frac12(ah^2+2bhk+ck^2)=\frac1{2a}\left((ah+bk)^2+(ac-b^2)k^2\right).
$$
Consequently the requested constant is the <Hessian determinant>
$$
\boxed{D=f_{xx}f_{yy}-f_{xy}^2.}
$$
The displayed identity in the question is an identity for the quadratic part; for a general $f$, its right-hand side needs the remainder $o(h^2+k^2)$.
If $D>0$, the <quadratic form> is positive for every nonzero $(h,k)$. On the <unit circle> it has a positive minimum, so it dominates the remainder sufficiently near the <critical point>. Hence the point is a \b[strict <local minimum>].
If $D<0$, along $k=0$ the quadratic term is $ah^2/2>0$, whereas along $h=-bk/a$ it is $Dk^2/(2a)<0$. Along each line the remainder is of smaller order. Thus values both above and below $f(x_0,y_0)$ occur arbitrarily nearby, giving a \b[<saddle point>].
Back to article page
