Maximum and minimum restricted I'm having troble with this problem, I don´t know what to do, is for a final exam tomorrow. Let $f:R^2\to R$ given by $(y-x^2)(y-2x^2)$ a) Draw the sets $A=${$(x,y)$|$f(x,y)>0$}$\ and\ $$B=${$(x,y)$|$f(x,y)<0$} b) Shows that for all $\vec u$=$(u_{1},u_{2})$ with ||$\vec u$||=$1$, the function: $g(t)=f(t\vec u$) has a local minimum in $t=0$ c) Shows that notwithstanding the foregoing, $\vec 0$ is not a local minimum of $f$

This is a simple but useful example.

a) These sets are bounded by parabolas $y=x^2$ and $y=2x^2$, the set $A$ is pink the set $B$ is blue, the parabolas are not included into the sets. This picture is simple, but it is a key to the problem.

![enter image description here](

b) Fix an arbitrary $\vec u $ with $\|\vec u\|=1$. The set $\ell(\vec u)=\\{t\vec u: t\in\Bbb R\\}$ is a straight line going through $\vec 0$. Let $t_0\vec u$ be a second intersection point of the straight line $\ell(\vec u)$ with the parabola $y=2x^2$ (if the straight line $\ell(\vec u)$ is vertical, put $t_0=+\infty$). Then for $0<|t|<$ the point $t\vec u$ belongs to the set $A$, thus the function $g(t)=f(t\vec u$) has a local minimum in $t=0$.

c) You can see this at the picture. We have $f(\vec 0)=0$, but the set $B$ extends arbitrary close to $\vec 0$, so $\vec 0$ is not a point of a local minimum of the function $f$.