Nature of stationary points

I have

$$f(x_1,x_2) = 2x^4_1 + 2x_1x_2 + 2x_1 + (1+x_2)^2$$

How can I determine the nature of the stationary points?

I know;

$$f_{x_1,x_1}(x) = 24x_1^2$$ $$f_{x_2,x_2}(x) = 2$$ $$f_{x_1,x_2}(x) = 2$$

This gives me a hessian:

$$\nabla^2 f(x) = \left[ \begin{array}{ c c } 24x_1^2 & 2 \\ 2 & 2 \end{array} \right]$$

By examining the leading principal minors; $$D_1 = \| (24x_1^2)\| \geq 0$$ and $$D_2 = \det\left|\array{24x_1^2&2\\2&2}\right| = 48x_1^2 -4$$

Since I have no condition on $x_1^2$ then $D_2$ can be positive, negative or even 0.

So then does this mean this matrix is indefinite, and hence all the stationary points here are saddle points?

Have I made a mistake in the Hessian? I feel the $24x_1^2$ term is out of place.

any help is VERY MUCH appreciated.

• The stationary points satisfy $\nabla f(x_1,x_2) = 0$; that should tell you what $x_1$ is at each. – Matthew Leingang Jan 22 '15 at 2:34
• I know what each of the stationary points are, but to determine the type of stationary point, I have to examine the hessian - is that not correct? The stationary points I worked out as $(0,-1), (\frac{1}{2}, \frac{-3}{2}), (\frac{-1}{2}, \frac{1}{2})$ It was my understanding to figure out the nature of the stationary points, one must examine the 2nd derivatives. – diabloescobar Jan 22 '15 at 2:50
• Yes, the $x_1$ in $D_2$ is the first coordinate of the stationary point. So you will get a number at each point. – Matthew Leingang Jan 22 '15 at 2:54

We have \begin{align*} f_{x_1} (x_1,x_2) &= 2x_1^4 + 2x_2 +2 \\ f_{x_2} (x_1,x_2) &= 2x_1 + 2(x_2 + 1) \\ \end{align*} So if $(x_1,x_2)$ is a critical point we have \begin{align} 8x_1^3 + 2x_2 +2 &= 0 \\ 2x_1 + 2x_2 + 2 &= 0 \end{align} Subtracting the two gives $$8x_1^3 - 2x_1 = 0 = 4x_1(4x_1^2-1)$$ therefore $x_1 = 0, \pm \frac{1}{2}$. The critical points are $(0,-1)$, $\left(\frac{1}{2},-\frac{3}{2}\right)$, and $\left(-\frac{1}{2},-\frac{1}{2}\right)$.
You have already computed that that $D_1 > 0$ if $x_1 \neq 0$, and in this case $D_2 = 48 \cdot \frac{1}{4} - 4 = 8$. So these two points are local minima. $D_2 < 0$ at $(0,0)$ so this critical point is a saddle point.