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\bf{Question}

\quad For each of the following functions $\ f:\br^2\to\br\ $ find
the points (if any) where $f(x_1,x_2)=0$ but where the $0$-contour
is
\smallskip

{\it (i)} {\it not} locally the graph of a function $x_2=g(x_1)$,

\smallskip

{\it (ii)} {\it not} locally the graph of a function $x_1=h(x_2)$,

\smallskip

{\it (iii)} both {\it(i)} and {\it(ii)}\ :

\medskip
\begin{center}
\begin{tabular}{lll}
$f(x_1,x_2)=\hspace{.5cm}$&(a)&$x_1^2-2x_2^2-4$\\
&(b)&$x_1^3-3x_1-x_2^2+2$\\&(c)&$(x_1+x_2)(x_1^2+x_2^2-1)$
\end{tabular}
\end{center}



\bf{Answer}

\begin{description}
\item{(i)}
$x^2-2y^2=4$

\begin{center}
\epsfig{file=342-1A-7.eps, width=50mm}

hyperbola $f(x,y)=0$.
\end{center}

This can be expressed (locally) as $y=g(x)$ (smooth function $g$)
everywhere except where the hyperbola crosses the $x$-axis -
namely where $\frac{\partial f}{\partial y}=0$. (There the
hyperbola bends back.)

\item{(ii)}
$$ \ $$
\begin{center}
$\begin{array}{cc} \epsfig{file=342-1A-8.eps, width=45mm} \ \ & \
\ \epsfig{file=342-1A-9.eps, width=45mm}
\end{array}$

\epsfig{file=342-1A-10.eps, width=45mm}

(Think of graph of $f$)
\end{center}

$\frac{\partial f}{\partial y} = -2y$ which $=0$ on the $x$-axis:
points $(-2,0)$ and $(1,0)$. At these points $f=0$ fails to have
(locally) the form $y=g(x)$, since at $(-2,0)$ it bends back,
while at $(1,0)$ it has two intersecting branches.

\item{(iii)}
$f(x,y)=0$ where $x+y=0$ or $x^2+y^2=1$.

\begin{center}
\epsfig{file=342-1A-11.eps, width=45mm}
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$\frac{\partial f}{\partial y}=2xy+x^2+3y^2-1$; this vanishes (on
the locus $f=0$) at $(\pm 1, 0)$ (where $f=0$ bends back), and at
$\pm(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ (where two branches
intersect). Elsewhere $y=g(x)$ locally.
\end{description}




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