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I need help.. Question

An examination of the function $f:\mathbb{R}^2 \to \mathbb{R}$, $f(x,y) = (y-3 x^2)(y-x^2)$ will give an idea of the difficulty of finding conditions that guarantee that a critical point is a relative extremum Show that (a) the origin is a critical point of $f$ : I solved easily
(b) f has a relative minimum at $(0,0)$ on every straight line through $(0,0)$; that is, if $g(t)=(at, bt)$, then $f o g : \mathbb{R} \to \mathbb{R}$ has a relative minimum at $0$, for every choice of a and b;

(c) the origin is not a relative minimum of $f$.

It is hard to me (b) and (c)..

ml0105
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user142512
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  • For part (b), the straight lines through the origin then have equations $ \ y = \frac{b}{a} x \ $ . What does the function $ \ f \ $ becomes in terms of $ \ x \ $ ? – colormegone Apr 20 '14 at 04:56

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To prove that the function is a local minimum along any straight line through the origin, compose $f$ with a general form of a path through the origin and get a function from $\Bbb{R}$ to $\Bbb{R}$ which you can then analyze using basic calculus.

To show that it is not a local minimum consider the value of the function along the path $g: t \mapsto (t,t^2)$.

The derivative of $f\circ g$ is $0$ everywhere, since the function is equal to $0$, and hence constant, along this path. This path is not a straight line so it doesn't contradict (b). You can find points on this path arbitrarily close to the origin (but not the origin) where the value is $0$ hence the origin is not a local minimum.

guest196883
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