Let $f:\mathbb{R}^{3} \to \mathbb{R}$ be the function defined by: $$f(x,y,z)=x^{4}+x^{3}y+y^{3}z+z.$$ Find and classify the critical points of $f$.
I found the partial derivatives and thus the critical points. The difficulty lies on identifying them. one is $(0,-1,0)$ which I believe it is a saddle point and the other is $(\frac{3}{4},-1,\frac{-9}{64})$. How do I prove what it is with interval theory?
Best Answer
The Hessian matrix of $f$ at $(x,y,z)$ is$$\begin{bmatrix}12 x^2+6 y x & 3 x^2 & 0 \\ 3 x^2 & 6 y z & 3 y^2 \\ 0 & 3 y^2 & 0\end{bmatrix}.$$Therefore, the Hessian matrix at $\left(\frac34,-1,-\frac9{64}\right)$ is$$\begin{bmatrix}\frac{9}{4} & \frac{27}{16} & 0 \\ \frac{27}{16} & \frac{27}{32} & 3 \\ 0 & 3 & 0\end{bmatrix},$$whose determinant is $-\frac{81}4<0$. So, the product of the eigenvalues is smaller than $0$. But their sum is $\frac{99}{32}>0$. So, there is at least an eigenvalue greater than $0$ and there is at least one eigenvalue smaller than $0$, and therefore $\left(\frac34,-1,-\frac9{64}\right)$ is a saddle point.
The point $(0,-1,0)$ is also a saddle point, and this is much easier to justify. Just note that $f(x,-1,z)=x^4-x^3=x^3(x-1)$, which is greater than $0$ if $x<0$ and which is smaller than $0$ if $x\in(0,1)$.