Second partial derivative test

At a critical point P of a function $f(x, y)$, both first partial derivatives are zero and the tangent plane is horizontal. The second partial derivative test tells what kind of point it is from the second derivatives $A = f_{xx}$, $B = f_{xy}$ and $C = f_{yy}$, through the discriminant $D = AC - B^2$, the determinant of the Hessian matrix. If f has continuous second partial derivatives near P:

The page cuts a surface with a vertical slice through a critical point P, in any direction θ. The surface is a quadratic form whose second derivatives you set, or one of several other functions. How much the cut bends at P is the second derivative along that direction, $f_{uu}$. The compass around P is filled where the slices bend up and hatched where they bend down. The test is a statement about all these slices at once.

As θ turns through half a circle, the bending makes one full wave around its midline $m = \tfrac12(A + C)$ with amplitude s, drawn in the small plot. Its highest and lowest values, $\lambda_1 = m + s$ and $\lambda_2 = m - s$, come in two perpendicular directions, and they are the eigenvalues of the Hessian. Multiplying out gives $D = m^2 - s^2 = \lambda_1\lambda_2$. So $D > 0$ means the wave never reaches zero. Every slice bends the same way, and P is a minimum or a maximum. Any one slice then tells which, and $f_{xx}$ is simply the bending along x. $D < 0$ means the wave crosses zero: some slices bend up and others down, and P is a saddle point.

This is why $f_{xx} > 0$ and $f_{yy} > 0$ are not enough. The cross term $f_{xy}$ turns the peaks of the wave away from the axes and makes the wave taller without moving its midline. Once $|f_{xy}|$ is large enough, a diagonal slice bends down although both axis slices bend up, and the bowl has become a saddle.

When $D < 0$, the bending is zero in exactly two directions, where the wave crosses zero, and the compass draws them as lines across the disc. They are the tangents of the level curve through P, which crosses itself there. This is the X that a saddle point makes on a contour map.

When $D = 0$ the wave touches zero or lies on it, so in some direction the second-order term vanishes and the test cannot tell. $x^2 - y^4$ and $x^2 + y^4$ have the same second derivatives at the origin, yet along y the first falls like $-t^4$ and the second rises like $t^4$, so one is a saddle and the other a minimum. On the monkey saddle $x^3 - 3xy^2$ every second derivative is zero at the origin, and the surface rises in three directions and falls in three. “The test can’t tell” does not mean “no extremum”.

At a critical point the first-order terms vanish, so near P the function is close to its second-order Taylor polynomial, which is f(P) plus a quadratic form. Quadratic approximation draws it over the disc. When $D > 0$, close enough to P the quadratic term outweighs the remainder in every direction, so P is a minimum or maximum like the approximation. When $D < 0$, it still does along the directions where it bends up or down, so f rises along some and falls along others: a saddle.