# Finding global maximum of a 2-variable function

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?

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.
• $f$ 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 $(\pm\alpha,0)$. – ym94 Nov 28 '20 at 14:32