Multivariable calculus

Critical Points of xy - x - y

Q: Can a point be a saddle when f_xx is zero?

Yes. f_{xx} measures curvature along the x direction only, and a saddle can be flat along the axes while still bending up one diagonal and down the other. That is exactly this surface: f_{xx} = f_{yy} = 0 and yet D = -1 certifies a saddle.

The only critical point of f(x,y) = xy - x - y is (1, 1). There f_xx = 0, f_yy = 0 and f_xy = 1, so the discriminant is D = (0)(0) - 1^2 = -1. A negative discriminant means the point is a saddle, and the vanishing pure second partials do not change that.

f(x,y) = xy - x - y

$\left(1, 1\right)$ saddle pointdiscriminant D = -1

Where both partials vanish

Differentiating in $x$ treats $y$ as a constant, so the $xy$ term contributes $y$ and the $-y$ term contributes nothing.

\frac{\partial f}{\partial x} = y - 1, \qquad \frac{\partial f}{\partial y} = x - 1

Notice the swap: the $x$ derivative controls $y$ and the $y$ derivative controls $x$ . Setting both to zero forces $y = 1$ and $x = 1$ , so $(1,1)$ is the only critical point, with $f(1,1) = -1$ .

A saddle with no curvature along the axes

Differentiate again. Since $\frac{\partial f}{\partial x} = y - 1$ has no $x$ in it, differentiating in $x$ a second time gives zero, and the same happens in $y$ .

f_{xx} = 0, \qquad f_{yy} = 0, \qquad f_{xy} = 1

D = (0)(0) - 1^{2} = -1 < 0

The discriminant is negative, so $(1,1)$ is a saddle. Along the horizontal and vertical lines through the point the surface is perfectly straight, and all of the curvature lives in the diagonal directions.

The mistake students make

Seeing $f_{xx} = f_{yy} = 0$ , students often write that the test fails. It does not. The test is inconclusive only when $D = 0$ , and here $D = -1$ , a perfectly clear answer. The rule is that $D$ decides first, and $f_{xx}$ is consulted only when $D > 0$ .

You can see the saddle directly by factoring.

f(x,y) = (x-1)(y-1) - 1

Move away from $(1,1)$ with both factors positive and $f$ rises above $-1$ ; move with one factor positive and the other negative and $f$ drops below $-1$ . Rising in some directions and falling in others is what a saddle is.

Frequently asked questions

Can a point be a saddle when f_xx is zero?

Yes. $f_{xx}$ measures curvature along the $x$ direction only, and a saddle can be flat along the axes while still bending up one diagonal and down the other. That is exactly this surface: $f_{xx} = f_{yy} = 0$ and yet $D = -1$ certifies a saddle.

Are there other critical points off to the side?

No. Both partials are linear and each pins down one variable exactly, so $y = 1$ and $x = 1$ is the whole solution set. One critical point, and it is a saddle.