Recipe

Classify a critical point with the second-derivative test

Use when: you have a critical point $(x_0,y_0)$ of $f(x,y)$ , meaning $f_x=f_y=0$ there, and need to know whether it is a local max, a local min, or a saddle

Goal: a definite classification of the point, or a clear statement that the test can't decide

Taught in The second derivative test

Quick recipe

Confirm the point is critical
Compute all three second partials
Compute the discriminant $D$
Read off the classification from $D$ and $f_{xx}$

1
Check that $f_x=0$ and $f_y=0$ at the point. If either partial is nonzero there, the test doesn't apply at all; the point isn't a critical point yet, and finding the actual critical points is the previous section's job, not this one.

Example

Take $f(x,y)=x^3-3x+y^2-4y$ . Its partials are $f_x=3x^2-3$ and $f_y=2y-4$ , both zero at $(1,2)$ and at $(-1,2)$ . Both points pass this check.

Why: the whole test is a local statement about how $f$ curves right around a point where the tangent plane is flat. Away from a critical point the tangent plane already tilts, so the surface is guaranteed to rise on one side and fall on the other.
2
Find $f_{xx}$ , $f_{yy}$ , and $f_{xy}$ , the mixed partial. Continuous mixed partials mean the order of differentiation doesn't matter, so there is only one number to find, not two.

Example

For $f(x,y)=x^3-3x+y^2-4y$ : $f_{xx}=6x$ , $f_{yy}=2$ , and $f_{xy}=0$ .

Why: these three numbers are exactly the entries of the Hessian matrix, and its determinant is what the next step needs.
3
Evaluate $D=f_{xx}f_{yy}-(f_{xy})^2$ at the critical point itself, not as a general formula in $x$ and $y$ .

Example

At $(1,2)$ : $D=(6\cdot1)(2)-0^2=12$ . At $(-1,2)$ : $D=(6\cdot(-1))(2)-0^2=-12$ .

Why: $D$ is the determinant of the Hessian, and its sign tells you whether $f_{xx}$ and $f_{yy}$ are forced to agree in sign or are free to disagree.
4
With $D$ in hand, the sign split settles the question.

Example

At $(1,2)$ : $D=12>0$ and $f_{xx}=6>0$ , so $(1,2)$ is a local minimum, with $f(1,2)=1-3+4-8=-6$ . At $(-1,2)$ : $D=-12<0$ , so $(-1,2)$ is a saddle point regardless of its $f_{xx}$ .
- $D>0$ and $f_{xx}>0$ — local minimum
- $D>0$ and $f_{xx}<0$ — local maximum
- $D<0$ — saddle point, regardless of $f_{xx}$ 's sign
- $D=0$ — the test is inconclusive; compare $f$ at nearby points directly instead
Why: when $D>0$ , $f_{xx}$ and $f_{yy}$ are forced to share a sign, so checking either one alone settles whether the surface curves up or down in every direction. When $D<0$ the surface curves oppositely along two different directions, no matter what $f_{xx}$ says.

Common mistakes

Plugging $x$ and $y$ into $D$ 's formula without first confirming the point is actually critical.
Checking only $D$ 's sign and forgetting the second check on $f_{xx}$ when $D>0$ .
Treating $D=0$ as a saddle instead of as "inconclusive."
Computing $f_{xy}$ from only one of the two mixed-partial orders and never double-checking it against the other.