Past exam of the mathematical methods II course of the University of Oxford / 2015 resit / 5

For a twice continuously differentiable function $F$, first check the stationary-point condition $F_x=F_y=0$. Write its Hessian matrix as $H=\left(\begin{array}{cc}A& B\\ B& C\end{array}\right)$, with $A=F_{xx}$, $B=F_{xy}$ and $C=F_{yy}$. The second-derivative test uses $D=AC-B^2$: if $D>0$ and $A>0$, the point is a strict local minimum; if $D>0$ and $A<0$, it is a strict local maximum; if $D<0$, it is a saddle point. If $D=0$, the quadratic test is inconclusive and higher-order terms or a direct argument are needed.

For $F=e^{y^2-x^2}$, the first derivatives vanish at the origin andThus $\boxed{(0,0)\text{ is a saddle point, with }F(0,0)=1}$. Directly, $F(x,0)=e^{-x^2}<1$ for nearby nonzero $x$, while $F(0,y)=e^{y^2}>1$ for nearby nonzero $y$. The presence of both increasing and decreasing directions confirms the Hessian matrix classification.

Let $G=x^3{e}^{-x^2-y^2}$ and $a=\sqrt{3/2}$. Its partial derivatives areso both vanish at $(a,0)$ and $(-a,0)$. Evaluating the second derivatives at a point $(x,0)$ with $x^2=3/2$ givesTherefore the Hessian determinant is $18x^2{e}^{-3}=27e^{-3}>0$. Its two diagonal entries are negative at $x=a$ and positive at $x=-a$. HenceThese are also global extrema: for positive $x$, the factor $e^{-y^2}\le 1$ maximizes the positive value at $y=0$, and differentiating $x^3{e}^{-x^2}$ fixes its positive maximum at $a$. Oddness in $x$ supplies the corresponding global minimum.