In one variable, hills and valleys sit where the derivative vanishes. The same instinct
works in two variables — but with a twist. At a peak or a pit of a surface, the ground is
momentarily flat in every direction, so every
directional derivative
is zero. By the gradient identity
D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}, that forces the
whole gradient
to vanish:
\nabla f = (f_x,\, f_y) = (0,\, 0).
A point where \nabla f = \mathbf{0} is a
critical point — the multivariable cousin of a
critical point
in single-variable calculus. The twist is the new third option: besides a
local max or min,
the surface can do something a curve never can — rise one way and fall the other, forming a
saddle.
The second-derivative test
Finding \nabla f = \mathbf{0} locates the candidates;
classifying them needs the second derivatives. They assemble into the
Hessian, the matrix of second partials, and the test hinges on its
determinant, the discriminant:
D = \det \begin{pmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{pmatrix} = f_{xx}\, f_{yy} - f_{xy}^{\,2}.
At a critical point, the sign of D says whether the surface
curves the same way in every direction (a genuine extremum) or in
opposite ways along two directions (a saddle); then the sign of
f_{xx} tells max from min.
Let (a, b) be a critical point of a twice-differentiable
f (so \nabla f(a, b) = \mathbf{0}), and
write D = f_{xx} f_{yy} - f_{xy}^{\,2} evaluated at
(a, b). Then:
-
D > 0 and f_{xx} > 0 ⟹
local minimum (a bowl: concave up every way).
-
D > 0 and f_{xx} < 0 ⟹
local maximum (a dome: concave down every way).
-
D < 0 ⟹ saddle point (up one
way, down another — no extremum).
-
D = 0 ⟹ test inconclusive (the
second order isn't decisive; look closer).
A worked classification
Classify every critical point of
f(x, y) = x^3 - 3x + y^2.
Step 1 — the gradient. Differentiate in each variable:
f_x = 3x^2 - 3, \qquad f_y = 2y.
Step 2 — solve \nabla f = \mathbf{0}. Setting
both to zero: 3x^2 - 3 = 0 \Rightarrow x = \pm 1, and
2y = 0 \Rightarrow y = 0. Two critical points:
(1, 0) \quad\text{and}\quad (-1, 0).
Step 3 — the second partials. Differentiate again:
f_{xx} = 6x, \qquad f_{yy} = 2, \qquad f_{xy} = 0.
Step 4 — the discriminant.
D = f_{xx} f_{yy} - f_{xy}^{\,2} = (6x)(2) - 0^2 = 12x.
Step 5 — classify (1, 0). Here
D = 12(1) = 12 > 0 and
f_{xx} = 6(1) = 6 > 0 — so (1, 0) is a
local minimum.
Step 6 — classify (-1, 0). Here
D = 12(-1) = -12 < 0 — so (-1, 0) is a
saddle point, regardless of f_{xx}. Along the
x-axis the cubic x^3 - 3x peaks there,
while along y the term y^2 climbs — up
one way, down the other.
A saddle is the genuinely two-dimensional phenomenon: a mountain pass,
high relative to the valleys on either side yet low relative to the peaks fore and aft. A
one-variable function can't manage it — there's only one axis to disagree along. It is
exactly why \nabla f = \mathbf{0} is necessary but not
sufficient for an extremum.
The test above is local. To find the global maximum or minimum
of f on a closed, bounded region you must check two kinds of
candidate and compare their heights:
-
Interior critical points — solve
\nabla f = \mathbf{0} inside the region.
-
The boundary — the extreme value may sit on the edge, where
\nabla f need not vanish; restrict
f to the boundary curve and optimise there (often with
Lagrange multipliers).
A continuous function on a closed bounded region attains its global max and min
(the extreme value theorem), so the winner is simply the largest and smallest height among
all the candidates — interior and boundary alike.
Three kinds of flat point, one map
Here is the separable surface
f(x, y) = (x^3 - 3x) + (y^3 - 3y), whose contours are drawn as
faint rings. It has four critical points — a local min, a local max, and
two saddles. Pick a critical point with the buttons: the marker jumps to it, and the panel
reports its discriminant D = f_{xx} f_{yy} - f_{xy}^2 and the
verdict from the test. Notice the contours close into loops around the min and max,
but cross in an X at each saddle.