Given a function $f(x,y)=(sin(x)+2)y^2$ for $(x,y) ∈ R2$ where $x$ is periodic.
I am trying to find the values $(x,y)$ where the function is convex.
I need to use the Sylvester criterion to fin the values. I have tried using the Hessian matrix and find where the determinat is greater than 0 but I don't quite understand how to proceed from the matrix.
The Hessian Matrix I have acquired is:
H =\begin{pmatrix} -y^2sen(x) & 2cos(x)y\\ 2cos(x)y & 2(sin(x)+2) \end{pmatrix}
I have tried calculating the $|a_{11}|$ and $|H|$ determinants and equaling them to $>0$. One of the solutions found is: $x=2*\pi*n+\pi, y=0$ for $n ∈ Z$.
I don´t really know if this is or why would be a possible answer and if it were, how could I answer the original question with this information.
Thanks for your help.