Is the local max point also a global max? Consider the following function $f$ of three variables, defined on $\mathbb{R^{3}}$: $$f(x,y,z) = 15x + xy - 4x^{2} - 2y^{2} - z^{2} + 2yz + 7$$
Is each of the critical points a local maximum? Is each of them a global maximum? Explain. 
I've found the critical points after taking the first-order derivatives $$f_{x} = 15 + y - 8x = 0,$$ $$f_{y} = x - 4y + 2z = 0,$$ $$f{z}=-2z + 2y = 0.$$
So there's only one critical point $\left ( 2,1,1 \right )$.
To figure out whether the critical point is a local maximum I've computed the partial derivatives and have obtained: 
$$H = \begin{bmatrix}
-f_{xx} &f_{xy}  &f_{xz} \\ 
 f_{yx}&f_{yy}  &f_{yz} \\ 
f_{zx} &f_{zy}  &zz 
\end{bmatrix}$$ 
$$ \Rightarrow  H = \begin{bmatrix}
-8 &1  &0 \\ 
 1&-4  &2 \\ 
0 &2  &-2 
\end{bmatrix}.$$ 
After using the Method of Leading principal minors, I've figured out that $H$ is negative definite so $\left ( 2,1,1 \right )$ is a local maximum. 
However, I'm a bit confused about whether this local maximum point is also a global maximum point. I know that for a global maximum point, the first-order condition must be satisfied and $H$ (the matrix of partial derivatives) must be negative definite for all $x, y, z \in \mathbb{R^{3}}$. 
So based on this, can I say that this local max point is also a global max because $H$ does not depend on the arguments?

When there is only one critical point and the critical point is a local maximum, that critical point is global maximum as well. 
