Prove the function has infinitely many local maxima and no local minima

Question

I need to find the local minima and maxima of f $$f(x,y)=(1+e^y)\cos{x}-ye^y.$$ I have found that $\nabla{f}(2k\pi,0)=0$ and $\nabla{f}\big((2k+1)\pi,-2\big)=0,\forall{k\in{\mathbb{Z}}}$. How can I identify if these points at which the gradient vanishes is a local minima or maxima?

Answer 1

Check Hessian: $$H=\begin{pmatrix}f_{xx}&f_{xy}\\ f_{yx}&f_{yy}\end{pmatrix}=\begin{pmatrix}-(1+e^y)\cos x&-e^y\sin x\\ -e^y\sin x&e^y\cos x-2e^y-ye^y\end{pmatrix}$$ At the critical point $(x,y)=(2k\pi,0)$: $$H=\begin{pmatrix}-2&0\\ 0&-1\end{pmatrix}\\ |H_1|=|-2|=-2<0\\ |H_2|=\begin{vmatrix}-2&0\\ 0&-1\end{vmatrix}=2>0 \Rightarrow \\ f(2k\pi,0)=2 \ (\text{local max})$$ At the critical point $(x,y)=((2k+1)\pi,0)$: $$H=\begin{pmatrix}1+e^{-2}&0\\ 0&-e^{-2}\end{pmatrix}\\ |H_2|=\begin{vmatrix}1+e^{-2}&0\\ 0&-e^{-2}\end{vmatrix}=-e^{-2}(1+e^{-2})<0 \Rightarrow \\ f((2k+1)\pi,-2)=e^{-2}-1 \ (\text{saddle})$$