Second Derivative Test (2D)
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.7: “Maximum and Minimum Values” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-7-maximum-and-minimum-values |
| Textbook used in class | Stewart, Calculus, Section 14.7: “Maximum and Minimum Values” (Examples 3, 4) |
Quick Reference
Discriminant: $D(a,b) = f_{xx}(a,b)\,f_{yy}(a,b) - [f_{xy}(a,b)]^2$.
| $D$ | $f_{xx}$ | Conclusion |
|---|---|---|
| $D > 0$ | $< 0$ | Local maximum |
| $D > 0$ | $> 0$ | Local minimum |
| $D < 0$ | any | Saddle point |
| $D = 0$ | any | Test inconclusive |
Motivation
Once critical points are found, you need to classify them. In one variable, the second derivative test says: if $f''(a) > 0$ at a critical point, then $f$ has a local minimum there; if $f''(a) < 0$, a local maximum; if $f''(a) = 0$, inconclusive. In two variables, there are infinitely many directions to test, so the classification involves a $2 \times 2$ matrix of second partial derivatives called the Hessian. The discriminant $D$ is the determinant of this Hessian.
Intuitively, $D > 0$ means the surface curves the same way in all directions (bowl up or bowl down), while $D < 0$ means it curves up in some directions and down in others (saddle).
Key Concepts
1. The Discriminant
At a critical point $(a,b)$: $$D = \begin{vmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{vmatrix} = f_{xx}f_{yy} - (f_{xy})^2.$$
This is the determinant of the Hessian matrix. $D > 0$ means the Hessian is positive definite (local min) or negative definite (local max); $D < 0$ means indefinite (saddle).
2. The Role of $f_{xx}$
When $D > 0$, the sign of $f_{xx}$ distinguishes min from max:
- $f_{xx} > 0$: the surface is concave up in the $x$-direction, so the critical point is a local minimum.
- $f_{xx} < 0$: concave down, so a local maximum.
Note: when $D > 0$, the sign of $f_{yy}$ is necessarily the same as $f_{xx}$ (you can also use $f_{yy}$ to check).
Worked Example
Classify the critical points of $f(x,y) = x^4 + y^4 - 4xy + 1$ found in the previous skill.
Critical points: $(0,0)$, $(1,1)$, $(-1,-1)$.
Compute second partials: $$f_{xx} = 12x^2, \quad f_{yy} = 12y^2, \quad f_{xy} = -4.$$
At $(0,0)$: $D = (0)(0) - (-4)^2 = -16 < 0$. Saddle point.
At $(1,1)$: $D = (12)(12) - (-4)^2 = 144 - 16 = 128 > 0$ and $f_{xx}(1,1) = 12 > 0$. Local minimum.
At $(-1,-1)$: $D = (12)(12) - (-4)^2 = 128 > 0$ and $f_{xx}(-1,-1) = 12 > 0$. Local minimum.
So $(0,0)$ is a saddle point, and $(1,1)$, $(-1,-1)$ are local minima.
$D > 0$ means there is a local maximum. $D > 0$ only tells you the critical point is NOT a saddle. Whether it is a max or min depends on the sign of $f_{xx}$ (or $f_{yy}$). Check the sign of $f_{xx}$: positive means min, negative means max. Forgetting this step and concluding “local max because $D > 0$” is a frequent error.
Leveled Practice
Problem 1. Find and classify the critical points of $f(x,y) = x^2 + xy + y^2 - 3y$.
(Critical point found in the previous skill: $(-1, 2)$.)
Show answer
$f_{xx} = 2$, $f_{yy} = 2$, $f_{xy} = 1$.
$D = (2)(2) - (1)^2 = 3 > 0$.
$f_{xx} = 2 > 0$, so $(-1, 2)$ is a local minimum.
$f(-1,2) = 1 + (-2) + 4 - 6 = -3$.