Past exam of the mathematical methods II course of the University of Oxford / 2016 resit / 12

For $f$ with continuous second partial derivatives near $(a,b)$, its multivariable Taylor expansion iswhere all displayed derivatives are evaluated at $(a,b)$. Continuity of the second derivatives justifies equality of the mixed partials and the stated remainder. A cubic-order remainder requires additional regularity.

The notation $(\partial f/\partial x{)}_y$ means the derivative with respect to $x$ while $y$ is held fixed:For a differentiable function, its first-order height change is $df=f_{x}dx+f_{y}dy$. If both coefficients vanish, the tangent plane is horizontal and every directional derivative vanishes: the surface is flat to first order at that point. It need not be constant in a neighbourhood; $f=x^2+y^2$ at the origin is a counterexample. Existence of coordinate partials alone, without differentiability, would not justify this tangent-plane conclusion.

At a stationary point, the linear terms vanish, so $\Delta f\simeq \frac{1}{2}(A{h}^2+2Bhk+C{k}^2)$, with $A=f_{xx}$, $B=f_{xy}$, $C=f_{yy}$. Let $D=AC-B^2$ be the Hessian determinant. For $A\ne 0$, completing the square givesThe second-derivative test therefore yieldsWhen $D<0$ the quadratic form is indefinite, including the case $A=0$, $B\ne 0$. When $D=0$, the quadratic approximation is degenerate and the test is inconclusive: higher-order terms may produce a minimum, maximum or saddle. For nondegenerate cases the quadratic term dominates the Taylor remainder sufficiently close to the point.

The gradient components are $f_x=2x(1+6x^2)$ and $f_y=2y(1-6y^2)$. Their real simultaneous zeros are $(0,0)$ and $(0,\pm 1/\sqrt{6})$. The Hessian matrix is diagonal:At the origin it is positive definite; at each other critical point it is $\mathrm{diag}(2,-4)$ and indefinite. HenceThere are no local maxima. The contours are symmetric under either coordinate reflection. Small positive levels have a closed component surrounding the minimum; these meet the saddle-level separatrices at level $1/12$. That critical contour has the useful exact form $3x^4+x^2=3(y^2-1/6{)}^2$. Other components extend toward large $|y|$, where the $-3y^4$ term makes the function unbounded below.