= Solution
Solving $f_y=0$ gives $y=-1-\alpha x$. Substitution into $f_x=0$ gives $x\{3x+2(1-\alpha^2)\}=0$. Therefore $(0,-1)$ is always a <critical point>. When $\alpha\ne-1$ as well as the printed restriction $\alpha\ne1$, there is a second <critical point>
$$
\boxed{\left(\frac23(\alpha^2-1),\,
-1-\frac{2\alpha}{3}(\alpha^2-1)\right).}
$$
At $(0,-1)$, the <Hessian matrix> is $\begin{pmatrix}2&2\alpha\\2\alpha&2\end{pmatrix}$ with determinant $4(1-\alpha^2)$. The <second-derivative test> gives a strict <local minimum> when $|\alpha|<1$ and a <saddle point of a scalar function> when $|\alpha|>1$.
The allowed value $\alpha=-1$ requires higher-order terms. With $v=y+1+\alpha x$, <completing the square> gives the exact identity
$$
f(x,y)+1=x^3+(1-\alpha^2)x^2+v^2.
$$
The coordinate change is invertible, so the <Hessian at a critical point under a change of coordinates> preserves the classification. At $\alpha=-1$, this is a <cubic degenerate saddle>: along $v=0$, the difference from the critical value is $x^3$ and takes both signs. Thus, throughout the printed parameter range,
$$
\boxed{\begin{cases}
\text{strict local minimum},&-1<\alpha<1,\\
\text{saddle point},&\alpha\leq-1\ \text{or}\ \alpha>1.
\end{cases}}
$$
There is no <local maximum>. When the second point is distinct, it is a <local minimum> for $|\alpha|>1$ and a <saddle point of a scalar function> for $|\alpha|<1$: in the $(x,v)$ coordinates its <Hessian matrix> is $\operatorname{diag}(-2(1-\alpha^2),2)$. The printed instruction to split only at $\alpha=1$ misses the negative-parameter cases.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/mathematics-2014-critical-point-classification.png]
{title=Classification of a <critical point> for $\alpha=0,-1,-2$}
{description=Original sign diagrams in coordinates $x$ and $v=y+1+\alpha x$. Orange means $f-f(0,-1)>0$ and blue means $f-f(0,-1)<0$; black curves are zero contours. Stars mark the point being classified and open circles mark the other critical point. The three panels use $\alpha=0,-1,-2$.}
Back to article page
