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Checking the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}, (x, y) \rightarrow (y-3x^2)(y-x^2)$ we can take an idea for the difficulty of finding conditions that ensure that a critical point is a relative extrema, when the theorem is not satisfied. Show taht

  1. $(0, 0)$ is a critical point of $f$.
  2. $f$ has a relative minimum at $(0, 0)$ over each line that passes through $(0, 0)$. That means that if $g(t)=(at, bt)$, then $f \circ g: \mathbb{R} \rightarrow \mathbb{R}$ has a relative minimum at $0$, for each choice of $a$ and $b$.
  3. $(0, 0)$ is not a relative minimum of $f$.

I have done the following:

  1. $\nabla f(x, y)=(0, 0) \Rightarrow (-8xy+12x, 2y-4x^2)=(0, 0) \Rightarrow x(-2y+3)=0 \text{ and } y=2x^2$ The point $(0, 0)$ satisfies these two relataions. S, $(0, 0)$ is a critical point of $f$.

    Is this correct??

  2. Could you give me some hints what I could do ??

  3. How could we show this??

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You're correct on the first one.

The second one is just plugging in $g(t)$ in $f$. So $f(at,bt)$ is what you're looking for. Differentiate this function with respect to $t$. You will see that $t = 0$ is a critical point and using the second derivative test (may need the first derivative test for the other cases), you'll get a local minimum. You probably will need to consider cases when $b \neq 0$ and $b = 0$.

Use the local extremum test for a function of two variables. You should be able to get what you're looking for for the third one.

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