TITLE: Prove that the function $f(x) = (x_2-x_1^2)^2 +x_1^5$ has only one stationary point which is not extreme
QUESTION [0 upvotes]: Prove that the function $$f(x) = (x_2-x_1^2)^2 +x_1^5$$ has only one
  stationary point which is not a local minimizer or local maximizer

$$\frac{\partial f}{\partial x_1} = 2(x_2-x_1^2)(-2x_1)+5x_1^4$$
$$\frac{\partial f}{\partial x_2} = 2(x_2-x_1^2) = 0 \implies x_2 = x_1^2$$
substituting in $\frac{\partial f}{\partial x_1}$:
$$\frac{\partial f}{\partial x_1} = 5x_1^4 = 0 \implies x_1 = 0 \implies x_2 = 0$$
So $(0,0)$ is the only stationary point. I'm not going to calculate the hessian because in these exercises it's always inconclusive. Let's analyze a neighborhood of $(0,0)$:
What if I try:
$$f(\sqrt{x_2},x_2) = \sqrt{x_2}^5$$
can I say that this changes sign since it's a odd degree polynomial with roots only on $0$? Therefore $(0,0)$ is not an extreme point?

REPLY [1 votes]: No, you cannot say that, since $\sqrt{x_2}^5$ is always non-negative. But you can say that $f(x_1,{x_1}^2)={x_1}^5$, which can take any real value.
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