Show $f(x,y) = y^2 - x^2$ at $(0,0)$ has a critical point, but is not a max/min value

Question

So as always... I found the partial derivative with respect to $x$ and $y$ of $f(x,y)$ which gave me:

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$f_x=-2x$

$f_y=2y$

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So I wasn't too sure what to do next, but I set $f_x = 0$:

$0 = -2x$

$x=0$

And I got stuck again. How do I continue AND prove that $f(x,y)$ at $(0,0)$ has a critical point but is not max/min value? Thanks!

Answer 1

Intuitively:

If $x=0$, then $f(y)=y^2$ has a minimum point at $y=0$;

If $y=0$, then $f(x)=-x^2$ has a maximum point at $x=0$;

Thus, if $(x,y)=(0,0)$, then $f(x,y)=y^2-x^2$ has a saddle point.

Answer 2

You're on the right track: note that $f_x(0,0)=f_y(0,0)=0$, so $(0,0)$ is indeed a critical point. So it's enough to show that $(0,0)$ neither a max or a min. To show this, note that $f(0,0)=0$ but there are points $(x,y)$ arbitrarily close to the origin for which $f(x,y)>0$ or $f(x,y)<0$.

Answer 3

It's a saddle point. Look at the graph in the $y=x$ plane cross section. Not all maxima and minima occur along strictly the $x$ or $y$ direction in three dimensions.

See http://mathworld.wolfram.com/SecondDerivativeTest.html

For the derivation of the second derivative test discriminant see Kaplan's Advanced Calculus https://www.amazon.com/Advanced-Calculus-5th-Wilfred-Kaplan/dp/0201799375

Of course there are likely other sources for the derivation.