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Jan 28, 2018 at 14:21 vote accept Tuki
Jan 28, 2018 at 14:19 comment added Mostafa Ayaz That's true in fact for showing a point to be local minima we must find a neighborhood where the value of the function in that minima is not greater than the value in whole points of the neghborhood.
Jan 28, 2018 at 14:14 comment added Tuki your using $\epsilon,\delta$ definition to prove that $(0,0)$ is local minima ? When you hold $x$ constant it is only necessary to show that @Mostafa Ayaz $\epsilon$ can get arbitrary small when $y\rightarrow -\infty$ which would mean that such values that are smaller than values in point $(0,0)$ exist $\rightarrow$ meaning point $(0,0)$ is only local minima ? Is this correct interpretation of your answer ? I am trying to understand this.
Jan 28, 2018 at 13:57 history answered Mostafa Ayaz CC BY-SA 3.0