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Determine all local and global extrema of the function \begin{equation*}f:\mathbb{R}^2\rightarrow \mathbb{R} , \ \ f(x,y)=x^3+3xy^2-3x+1\end{equation*} Determine also the type of extrema : minimum, maximum, local, global.

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I have done the following :

First we calculate the gradient : \begin{equation*}\nabla f=\begin{pmatrix}f_x \\ f_y\end{pmatrix}=\begin{pmatrix}3x^2+3y^2-3 \\ 6xy\end{pmatrix}\end{equation*} Then we set the gradient equal to $0$ and we solve the syste that we get: \begin{equation*}\begin{pmatrix}3x^2+3y^2-3 \\ 6xy\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\Rightarrow \begin{cases}3x^2+3y^2-3=0 \\ 6xy=0\end{cases}\end{equation*} From the second equation we get $x=0$ or $y=0$. If $x=0$ we get from the first equation $3y^2-3=0\Rightarrow y^2=1 \Rightarrow y\pm 1$. If $y=0$ we get from the first equation $3x^2-3=0\Rightarrow x^2=1 \Rightarrow x\pm 1$. So we get the extrema $(0,-1)$, $(0,1)$, $(-1,0)$, $(1,0)$.

Now we determine the Hessian matrix : \begin{equation*}H_f(x,y)=\begin{pmatrix}f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{pmatrix}=\begin{pmatrix}6x & 6y \\ 6y & 6x\end{pmatrix}\end{equation*}

  • \begin{equation*}H_f(0,-1)=\begin{pmatrix}0 & -6 \\ -6 & 0\end{pmatrix}\end{equation*} The eigenvalues are $\lambda_1=-6$ and $\lambda_2=6$. The matrix is indefinite. So at $(0,1)$ the function has a saddle point.
  • \begin{equation*}H_f(0,1)=\begin{pmatrix}0 & 6 \\ 6 & 0\end{pmatrix}\end{equation*} The eigenvalues are $\lambda_1=-6$ and $\lambda_2=6$. The matrix is indefinite. So at $(0,1)$ the function has a saddle point.
  • \begin{equation*}H_f(-1,0)=\begin{pmatrix}-6 & 0 \\ 0 & -6\end{pmatrix}\end{equation*} The eigenvalues are $\lambda_1=-6$ and $\lambda_2=-6$. The matrix is negative definite. So at $(-1,0)$ the function has a local maximum.
  • \begin{equation*}H_f(1,0)=\begin{pmatrix}6 & 0 \\ 0 & 6\end{pmatrix}\end{equation*} The eigenvalues are $\lambda_1=6$ and $\lambda_2=6$. The matrix is positive definite. So at $(1,0)$ the function has a local minimum.

Do we have to check also some other points or can we say that at $(-1,0)$ the function has also a global maximum and at $(1,0)$ a global minimum?

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  • $\begingroup$ You can rewrite the function as, $f(x,y) = x (x^2 + 3y^2 - 3) + 1$ and that shows that as $x \to +\infty$, $f(x, y) \to +\infty$ and similarly when $x \to - \infty$, $f(x, y) \to -\infty$ $\endgroup$
    – Math Lover
    Sep 3, 2021 at 14:34
  • $\begingroup$ I see! Thank you very much!! :-) @MathLover $\endgroup$
    – Mary Star
    Sep 3, 2021 at 15:19

1 Answer 1

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That function neither has a global maximum nor a global minimum. Note that $f(x,0)=x^3-3x+1$ and that $\lim_{x\to\pm\infty}x^3-3x+1=\pm\infty$.

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  • $\begingroup$ Ahh ok! So only the local minimum at $(1,0)$ and the local maximum at $(-1,0)$, right? $\endgroup$
    – Mary Star
    Sep 3, 2021 at 14:26
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    $\begingroup$ Yes, those are just local extreme points. $\endgroup$
    – José Carlos Santos
    Sep 3, 2021 at 14:28
  • $\begingroup$ Ok! In general to check if a local extremum is also a global one, do we always have to set one variable equal to zero and check the behaviour of the function when the other variable goes to infinity? $\endgroup$
    – Mary Star
    Sep 3, 2021 at 14:30
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    $\begingroup$ I am not aware of a general method that works every single time. But when this method works, it's usually simple to apply. $\endgroup$
    – José Carlos Santos
    Sep 3, 2021 at 14:35
  • $\begingroup$ Ok! Thank you very much!! :-) $\endgroup$
    – Mary Star
    Sep 3, 2021 at 15:19

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