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Let $f,g: \mathbb{R}^2 \longrightarrow \mathbb{R}^2$ be defined by

\begin{align} f(x,y)&= x^4 +y^2 \\ g(x,y)&=x^4+y^2-10x^2y \end{align}

Is it possible for $f$ and $g$ to have local minima? Here I used the formula of $(rt - s^2=0)$ hence we cannot conclude whether it has points of local extremum. So what can I do to check whether it has local extrema?

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    What is $rt-s^2=0$? Are you familiar with analyzing gradients and Hessians?2017-02-01
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    r is the second order partial derivative of a function with respect to x and t is the same for y and s is second order partial derivative first with respect to x and then with y . I havent heard of hessians but i do nkow gradients.2017-02-01

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Observe for $f(x, y)$, we see that the minimal value is $0$ and $(0, 0)$ gives that value. Moreover, any deviation from $(0, 0)$ will give you a positive value. Hence $(0, 0)$ is a local minimum.

In the case of $g(x, y)$, we see that \begin{align} g(x, y) = x^4+y^2-10x^2y \end{align} which means \begin{align} \nabla g(x, y) = (x(4x^2-20y), 2y-10x^2)=(0, 0) \end{align} means \begin{align} x\cdot (4x^2-20y)=&\ 0\\ 2y-10x^2=&\ 0. \end{align} Hence $(0, 0)$ is the only critical point. However, $(0, 0)$ is only a saddle since if we move in the direction $(x, -x)$ the function value is positive for small $x>0$ and in the direction $(x, x)$ the function value is negative, again for small $x>0$.

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    But in the answer it is given that neither f or g has a local minimum . I did figure out for f that its minimum is at (0,0) but wasnt sure for g.if we all agree that f has minimium at (0,0) and that g only has a critical point then we may discard the answer.2017-02-01
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    @Khamba https://www.wolframalpha.com/input/?i=x%5E4%2By%5E22017-02-01