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Let $$f(x,y) = -e^{x^4+y^2}+x^2+y^2+1$$

Find global/local maximum and minimum and upper/lower bound of the function.

I found that the lower bound is ${-\infty} $ and also that, since $$-e^{x^4+y^2}\le -1-x^4-y^2, \forall (x,y)\in \mathbb{R}^2$$we have $$f(x,y)\le -x^4+x^2\le\frac{1}{4},\forall(x,y)\in\mathbb{R}^2$$ so that function is upper bounded.

Then I calculated the critical points, which are $(0,0)$ and $(\pm\alpha,0)$, with $\alpha$ the positive solution of the equation $$e^{x^4}=\frac{1}{2x^2}$$ and by the second-derivative test is apparent that $(0,0)$ is a saddle point while both $(\pm\alpha,0)$ are local maximum.

How do I prove that both $(\pm\alpha,0)$ are global maximum?

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The function $f$ is even, so $f(\alpha, 0)=f(-\alpha,0)$. The global maximum of $f$ is in particular a local maximum, hence $(\pm \alpha,0)$ must be the points where $f$ attains a global maximum.

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  • $\begingroup$ It could be just a local maximum. Since the domain is not compact, you can not guarantee that there is a global maximum. I know it is because I have the solution of the problem, but I don't know how to prove it. $\endgroup$ – zinne98 Nov 24 '20 at 21:47
  • $\begingroup$ $f$ is continuous and we have $\lim_{|(x,y)|\to \infty} f(x,y)=-\infty$, so $f$ must attain a global maximum. This global maximum is also local, hence it is attained in $(\pm\alpha,0)$. $\endgroup$ – ym94 Nov 28 '20 at 14:32
  • $\begingroup$ Thank you very much, I didn't known this property. $\endgroup$ – zinne98 Nov 28 '20 at 20:30

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