Critical points — finding, classifying: Visual walkthrough

Step 1 — What "flat in every direction" really means

WHAT. We start at a special point $(a, b)$ where the ground is level. "Level" means: if I take a tiny step in any horizontal direction, the height does not start to rise or fall. The two most basic directions are along $x$ and along $y$ .

WHY. The slope along $x$ is written $f_{x}$ — read it as "how fast height changes if I nudge $x$ and freeze $y$ ." The slope along $y$ is $f_{y}$ . At a hilltop, valley, or pass, both of these are zero. This is what the parent calls $\nabla f = 0$ ; the symbol $\nabla f = (f_{x}, f_{y})$ is just the pair of these two slopes bundled together.

PICTURE. Look at the two coloured tangent lines. The blue line runs along the $x$ -direction, the orange along the $y$ -direction. Both are perfectly horizontal — that is our starting condition.

Figure — Critical points — finding, classifying

Step 2 — Zooming in: the height difference is a pure quadratic

WHAT. Take a tiny step $(h, k)$ away from the flat point: $h$ east, $k$ north. We ask how much the height changed, $Δ = f (a + h, b + k) - f (a, b)$ . We approximate this with a Taylor expansion.

WHY this tool. A Taylor expansion writes any smooth surface near a point as: (its height) + (linear slope terms) + (curvature terms) + (tinier stuff). We use it because it turns a scary curved surface into a polynomial we can analyse by hand. Here the linear slope terms are $f_{x} h + f_{y} k$ — and both slopes are zero (Step 1). So the first surviving piece is the curvature piece.

$Δ \approx = 0 at a flat point f_{x} h + f_{y} k + \frac{1}{2} (f_{xx} h^{2} + 2 f_{x y} hk + f_{y y} k^{2}) .$

Term by term in the surviving bracket:

$f_{xx}$ — curvature along $x$ (does the east-west slice smile or frown?), multiplied by $h^{2}$ .
$f_{y y}$ — curvature along $y$ (north-south slice), multiplied by $k^{2}$ .
$f_{x y}$ — the twist: how the east-west slope $f_{x}$ changes as you move north (in $y$ ). Multiplied by $2 hk$ .
$\frac{1}{2}$ — a bookkeeping constant from the Taylor formula; it scales the whole bracket but, being a fixed positive number, it never affects any sign.

PICTURE. The flat landscape is replaced, near the point, by a smooth "cup or cap or saddle" bowl whose only ingredients are these three second-derivative numbers.

Step 3 — Why we can't just look at $f_{xx}$ and $f_{y y}$

WHAT. The naive hope: "if the $x$ -slice curves up ( $f_{xx} > 0$ ) and the $y$ -slice curves up ( $f_{y y} > 0$ ), surely it's a bowl." We show this hope can fail.

WHY it fails. The twist term $2 f_{x y} hk$ can be so strongly negative along the diagonal direction (where $h$ and $k$ have opposite signs) that it drags $Q$ below zero — even when both pure curvatures are positive.

PICTURE. Look at the counterexample $f = x^{2} + y^{2} - 3 x y$ . Along the axes (blue and orange slices) the ground curves up. But along the diagonal $y = x$ (red slice) the twist wins and the ground curves down. Both axis-slices lie: it's a saddle.

Step 4 — The trick: complete the square

WHAT. We rewrite $Q$ so that the twist term $2 f_{x y} hk$ is absorbed into a perfect square. Assume for now $f_{xx} \neq = 0$ .

WHY complete the square. A perfect square $(\dots)^{2}$ can never be negative. If we can force $Q$ into the form (positive-or-negative number) $\times$ (square) + (another number) $\times$ (square), then the signs of the two multipliers tell us everything — with no direction-hunting. Completing the square is the standard algebraic move to turn $a h^{2} + bhk + c k^{2}$ into squares.

$f_{xx} h^{2} + 2 f_{x y} hk + f_{y y} k^{2} = f_{xx} (new variable u h + \frac{f _{x y}}{f _{xx}} k)^{2} + (f_{y y} - \frac{f _{x y}^{2}}{f _{xx}}) k^{2} .$

Reading it:

$f_{xx}$ — the coefficient in front of the first square $u^{2}$ .
$u = h + \frac{f _{x y}}{f _{xx}} k$ — a shifted, tilted direction; the square $u^{2} \geq 0$ always.
$(f_{y y} - \frac{f _{x y}^{2}}{f _{xx}})$ — the coefficient in front of the second square $k^{2}$ ; call it the "leftover".

PICTURE. Picture rotating/tilting our coordinate axes so the twist disappears. In the new tilted axes, the bowl has no cross-term — it's a clean sum of two independent curvatures, which we can read off directly.

Step 5 — The leftover coefficient IS the discriminant $D$

WHAT. Simplify that leftover coefficient over a common denominator: $f_{y y} - \frac{f _{x y}^{2}}{f _{xx}} = \frac{f _{xx} f _{y y} - f _{x y}^{2}}{f _{xx}} = \frac{D}{f _{xx}}, where D = f_{xx} f_{y y} - f_{x y}^{2} .$

WHY this matters. So the whole quadratic form is now $Q = f_{xx} u^{2} + \frac{D}{f _{xx}} k^{2} .$ Both $u^{2}$ and $k^{2}$ are $\geq 0$ . The only things that can flip signs are the two coefficients: $f_{xx}$ and $D / f_{xx}$ . The number $D$ has appeared — not by magic, but as the surviving coefficient after completing the square.

Naming $D$ properly. Collect the four second derivatives into a $2 \times 2$ grid called the Hessian matrix, written $H$ : $H = (f_{xx} f_{y x} f_{x y} f_{y y}) = Clairaut (f_{xx} f_{x y} f_{x y} f_{y y}) .$

Top-left $f_{xx}$ , bottom-right $f_{y y}$ — the two pure curvatures.
Off-diagonal $f_{x y} = f_{y x}$ — the twist, equal by Clairaut's theorem, which is why $H$ is symmetric.

The determinant of this $2 \times 2$ matrix (top-left times bottom-right, minus the two off-diagonals multiplied) is precisely our $D$ : $det H = f_{xx} f_{y y} - f_{x y} f_{y x} = f_{xx} f_{y y} - (f_{x y})^{2} = D .$

Term by term in $Q$ :

$f_{xx} u^{2}$ — curvature along the first tilted axis. Sign $=$ sign of $f_{xx}$ .
$\frac{D}{f _{xx}} k^{2}$ — curvature along the second tilted axis. Sign $=$ sign of $D / f_{xx}$ .

PICTURE. Two independent parabolic slices, one per tilted axis, each with its own upward or downward opening controlled by a single coefficient.

Step 6 — Read every sign: all cases

WHAT. We now sweep through every possible sign combination of the two coefficients $f_{xx}$ and $D / f_{xx}$ .

WHY. A judge-proof test must never leave the reader stranded. Here are all cases.

Case A — $D > 0$ and $f_{xx} > 0$ (bowl / minimum). Then $D / f_{xx} > 0$ too. Both coefficients positive $\Rightarrow Q = (+) u^{2} + (+) k^{2} > 0$ for every nonzero step. Every direction goes up ⇒ local minimum.

Case B — $D > 0$ and $f_{xx} < 0$ (dome / maximum). If $D > 0$ but $f_{xx} < 0$ , then $D / f_{xx} < 0$ as well. Both coefficients negative $\Rightarrow Q < 0$ everywhere ⇒ local maximum. (Note: $D > 0$ forces $f_{y y}$ to share $f_{xx}$ 's sign, since $f_{xx} f_{y y} = D + f_{x y}^{2} > 0$ . So we never need to inspect $f_{y y}$ separately.)

Case C — $D < 0$ (saddle). Then $f_{xx}$ and $D / f_{xx}$ have opposite signs. One tilted axis curves up, the other down $\Rightarrow Q > 0$ some directions, $Q < 0$ others ⇒ saddle, no matter what $f_{xx}$ is.

Case D (degenerate) — $D = 0$ . The leftover coefficient $D / f_{xx} = 0$ . Then $Q = f_{xx} u^{2}$ — it is flat (exactly zero) along the entire line $u = 0$ . The quadratic can no longer decide; higher-order terms take over. Inconclusive. (Example $f = x^{4} + y^{4}$ : here $Q \equiv 0$ , yet the quartic makes it a genuine minimum — only direct inspection reveals this.)

Case E (degenerate) — $f_{xx} = 0$ . Our completing-the-square assumed $f_{xx} \neq = 0$ , so handle this separately. If $f_{xx} = 0$ then $D = - f_{x y}^{2} \leq 0$ .

If $f_{x y} \neq = 0$ : $D < 0$ ⇒ saddle (the twist alone creates opposite curvatures).
If also $f_{x y} = 0$ : $D = 0$ ⇒ inconclusive; fall back to $f_{y y} k^{2}$ and higher terms.

So the $D$ -first rule survives even when $f_{xx} = 0$ : nothing is left uncovered.

PICTURE. Four miniature landscapes side by side — bowl, dome, saddle, and a flat trough for $D = 0$ — each tagged with its sign pattern.

Step 7 — A full run on a real example

WHAT. Apply the built machine to $f = x^{3} - 3 x + y^{2}$ , exactly as the parent's Example 2.

WHY. To watch $D$ and $f_{xx}$ change from point to point on one function.

Find flat points: $f_{x} = 3 x^{2} - 3 = 0 \Rightarrow x = \pm 1, f_{y} = 2 y = 0 \Rightarrow y = 0.$ Two flat points: $(1, 0)$ and $(- 1, 0)$ .

Second derivatives: $f_{xx} = 6 x$ , $f_{y y} = 2$ , $f_{x y} = 0$ , so $D = f_{xx} f_{y y} - f_{x y}^{2} = (6 x) (2) - 0 = 12 x .$

At $(1, 0)$ : $D = 12 > 0$ and $f_{xx} = 6 > 0$ → minimum (Case A).
At $(- 1, 0)$ : $D = - 12 < 0$ → saddle (Case C), regardless of $f_{xx}$ .

PICTURE. One landscape holding both flat points: a valley at $(1, 0)$ and a pass at $(- 1, 0)$ .

The one-picture summary

Everything compresses to one flow: complete the square → the leftover coefficient is $D$ → $D$ decides, then $f_{xx}$ describes.

Recall Feynman retelling — the whole walkthrough in plain words

Stand on a flat spot of a hilly field. Flat means every direction starts level, so slope alone can't tell you if you're in a bowl or on a saddle. So you feel the curving. There are three curving-numbers: how it bends east-west, how it bends north-south, and a sneaky "twist" number for how the two mix (there's only one twist number because Clairaut's theorem says the two mixed rates are equal). Straight up we can't compare them, because the twist confuses things. The clever move: we tilt our head (complete the square) until the twist vanishes — now the ground is just two clean parabolas at right angles. Two numbers remain: one is $f_{xx}$ , the other is $D / f_{xx}$ where $D = f_{xx} f_{y y} - f_{x y}^{2}$ . If both parabolas open up, it's a bowl (minimum). Both down, a dome (maximum). One up and one down, a saddle. And that happens exactly when $D < 0$ . If $D = 0$ , one parabola went flat and we simply can't tell from curving alone — we must look closer. So: $D$ decides whether, $f_{xx}$ describes which.

Connections

Critical points — finding, classifying — the parent; this page derives its central result.
Taylor Series in Several Variables — Step 2's quadratic approximation.
Quadratic Forms and Definiteness — $Q$ definite/indefinite ↔ min/max/saddle.
Hessian Matrix — $D = det H$ ; eigenvalues are the tilted-axis curvatures.
Gradient and Directional Derivatives — $\nabla f = 0$ marks the flat point of Step 1.
Single-variable Second Derivative Test — the one-variable ancestor of "check the curvature".