1
$\begingroup$

I want to find the critical points of the function $$f(x,y)=x^4+y^4-y^2.$$ So, I found the partial derivatives and I set them to zero and solved for $x$ and $y$. I ended up with $2$ points: $(0,0)$ and $\left(0,\sqrt{\frac{1}{2}}\right)$. When I plugged them in to the Hessian matrix, I ended up with zero for the determinant for both points. So this means the test is inconclusive. But when I graph it there is a saddle point at $(0,0)$. So I don't understand how $(0,0)$ is a saddle point when the determinant of Hessian matrix is $0$?

$\endgroup$
  • 1
    $\begingroup$ The function $f$ is even in $y$, so if $(a, b)$ is a critical point, so is $(a, -b)$. $\endgroup$ – Travis Oct 25 '15 at 22:00
2
$\begingroup$

To say that the test is inconclusive when the determinant $f_{xx} f_{yy} - f_{xy}^2$ is zero at a point is to say just that: The test doesn't tell us anything, so if we want to determine the type of the critical point, we must do a little more. (Indeed, the functions $(x, y) \mapsto x^4 + y^4$, $(x, y) \mapsto x^4 - y^4$, and $(x, y) \mapsto -x^4 - y^4$ all have a critical point at $(0, 0)$ with zero Hessian determinant, but the three functions respectively have a minimum, a saddle point, and a maximum there.)

First, note that something like this issue already occurs for single-variable functions. Checking shows that $g(x) := x^4$ has a critical point at $x = 0$, and computing the second derivative there gives $g''(0) = 0$, so we cannot conclude whether $g$ has a minimum, a maximum, or neither, at that point. We can still determine the type of critical point, however, by observing that $g(x) > 0$ for any $x \neq 0$, and so the critical point must be an (isolated) minimum.

In the case of our two-variable function $f(x, y)$, we can proceed as follows: Computing gives that $$f(x, 0) = x^4, $$ which we've already said has an isolated minimum at $x = 0$. On the other hand, $$f(0, y) = y^4 - y^2 $$ Applying the (single-variable) Second Derivative Test gives $\frac{d^2}{dy^2} [f(0, y)] = -2$, so $f(0, y)$ has an isolated maximum at $y = 0$. So, $f(x, y)$ takes on both positive and negative values in any open set containing $(0, 0)$, so it must be a saddle point.

$\endgroup$
  • $\begingroup$ You're welcome, I hope you found it useful. $\endgroup$ – Travis Oct 25 '15 at 22:29
  • $\begingroup$ By using the same method, does $f(x,y)$ have a relative minimum at $(0,\frac1{\sqrt{2}})$ and a saddle point at $(0,-\frac1{\sqrt{2}})$? $\endgroup$ – vidyarthi Feb 6 at 11:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.