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 $0$?

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

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.

  • $\begingroup$ You're welcome, I hope you found it useful. $\endgroup$ – Travis Willse 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

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