Let $f: \mathbb{R}^3 \to \mathbb{R}$, defined by $f(x,y,z)=x^2+4y^2+8z^2+4xy+6xz+12yz+8x+16y+24z$.
Find its local extremum points.
I find that the critic points are: $(-2a-4,a,0) , \forall a \in \mathbb{R}$.
Unfortunately, the Hessian matrix has a null determinant. I tried to use the second differential theorem and to look at the difference: $$ f(x,y,z)-f(-2a-4,a,0)=f(x,y,z)+16, $$ but, nothing.
