I have found the first order partial derivatives to be:

$$\frac{\partial f}{\partial x}=e^{-x^2-y^4}(y-2x^2y)$$ $$\frac{\partial f}{\partial y}=e^{-x^2-y^4}(x-4xy^4)$$

I understand critical points are when $\frac{\partial f}{\partial x}=0,\frac{\partial f}{\partial y}=0$.

The only one I have managed to find is (0,0) and I'm not even sure this is correct?

  • $\begingroup$ apologies, one should check what is written on the paper is the same as what is typed! $\endgroup$ – Maths student Jan 7 '14 at 18:48

The exponential $e^{-x^2-y^4}$ is never $0$, so equivalently we are solving the system $$y-2x^2y=0,\qquad x-4xy^4=0.$$

Suppose first that $y=0$. Then the first equation is satisfied. Substituting $y=0$ in the second equation, we get $x=0$. That gives the critical point $(0,0)$.

Suppose now that $y\ne 0$. Then $y-2x^2y=0$ precisely if $1-2x^2=0$, that is, $x=\pm \frac{1}{\sqrt{2}}$.

Since $x\ne 0$, the equation $x-4xy^4=0$ is satisfied precisely if $1-4y^4=0$, that is, if $y =\pm\frac{1}{\sqrt[4]{4}}=\pm\frac{1}{\sqrt{2}}$.

That gives four more critical points, $(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}})$.

  • $\begingroup$ The exponential term can be zero in the limit as either $x$ or $y$ tend toward $\to + \infty$, so you have a fixed point at $\infty$. $\endgroup$ – Bruce Dean Jan 7 '14 at 18:32
  • $\begingroup$ Also, what about the fixed points for complex values of $y$, i.e, $( \pm {\textstyle{1 \over {\sqrt 2 }}}, \pm {\textstyle{{\,{\rm{i}}\,} \over {\sqrt 2 }}})$. $\endgroup$ – Bruce Dean Jan 7 '14 at 18:36
  • $\begingroup$ @roybatty: I assumed that the question came from the usual first several variables course, with $x,y$ ranging over the reals. $\endgroup$ – André Nicolas Jan 7 '14 at 18:39
  • $\begingroup$ I don't think you should assume that. Where does he say that "I am taking a first several variables course, with x,y ranging over the reals" ? The question is asked with no stipulation on the number set. Even so, you still have not addressed the fixed point at real: $ + \infty $. $\endgroup$ – Bruce Dean Jan 7 '14 at 18:41
  • $\begingroup$ @roybatty: It doesn't, except indirectly via the tags. But after teaching the subject many times, one can usually recognize the implicit context. $\endgroup$ – André Nicolas Jan 7 '14 at 18:44

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