Second derivative test — Hessian determinant: Worked examples

The scenario matrix

Before any example, let's lay out the full space of things that can happen. Every worked example below is tagged with the cell it covers.

Cell	Situation	What makes it tricky	Example
A	$D > 0, f_{xx} > 0$	clean minimum	Ex 1
B	$D > 0, f_{xx} < 0$	clean maximum	Ex 2
C	$D < 0$	clean saddle	Ex 3
D	Several critical points in different quadrants	must classify each separately	Ex 4
E	$f_{xx} = 0$ at the point	can't complete the square in $u$ — pivot to $f_{y y}$	Ex 5
F	$D = 0$ (degenerate)	test fails — investigate by hand	Ex 6
G	$D = 0$ but it's actually a saddle	shows $D = 0$ ≠ "flat valley"	Ex 7
H	Real-world word problem (optimisation)	translate words → $f$ , mind units	Ex 8
I	Exam twist: parameter that flips the classification	classify as a function of a constant	Ex 9

Example 1 — Cell A: clean minimum

Worked example

$f (x, y) = x^{2} + x y + y^{2} + 3 x$ Forecast first: the $x^{2}$ and $y^{2}$ terms both curve up, so guess a minimum. Where, though? The $+ 3 x$ tilts things — the low point won't be at the origin.

Step 1 — Find critical points. Set both first partials to zero. $f_{x} = 2 x + y + 3 = 0, f_{y} = x + 2 y = 0$ Why this step? Every extremum or saddle lives where the gradient vanishes; the tilt term $3 x$ moves the solution off-origin. From $f_{y} = 0$ : $x = - 2 y$ . Sub into $f_{x} = 0$ : $2 (- 2 y) + y + 3 = 0 \Rightarrow - 3 y + 3 = 0 \Rightarrow y = 1$ , then $x = - 2$ . Critical point: $(- 2, 1)$ .

Step 2 — Second partials. $f_{xx} = 2, f_{y y} = 2, f_{x y} = 1$ Why this step? These are the coefficients of the quadratic form that decides the shape.

Step 3 — Discriminant. $D = (2) (2) - (1)^{2} = 3 > 0$ Why this step? $D > 0$ says "extremum, not saddle"; its sign is the first gate.

Step 4 — Pick min vs max. $f_{xx} = 2 > 0$ , so local minimum at $(- 2, 1)$ . Why this step? $D > 0$ forces $f_{xx}$ and $f_{y y}$ to share a sign; positive means the bowl opens upward.

Verify: value there is $f (- 2, 1) = 4 - 2 + 1 - 6 = - 3$ . Nudge: $f (- 2, 1.1) \approx 4 - 2.2 + 1.21 - 6 = - 2.99 > - 3$ . Higher on all sides ✓. Consistent with a minimum.

Example 2 — Cell B: clean maximum

Worked example

$f (x, y) = - x^{2} - y^{2} + 4 x + 6 y$ Forecast: both squares are negative (open downward) — this is a dome, expect a maximum.

Step 1 — Critical point. $f_{x} = - 2 x + 4 = 0 \Rightarrow x = 2, f_{y} = - 2 y + 6 = 0 \Rightarrow y = 3$ Critical point $(2, 3)$ . Why this step? The gradient is where the tangent plane goes flat — the only place a max/min/saddle can live.

Step 2 — Second partials. $f_{xx} = - 2, f_{y y} = - 2, f_{x y} = 0$ . Why this step? These three numbers are the entire input the test needs; they measure the curvature in each direction and the tilt between them.

Step 3 — Discriminant. $D = (- 2) (- 2) - 0 = 4 > 0$ . Why this step? $D > 0$ rules out a saddle, leaving only min or max.

Step 4 — Verdict. $f_{xx} = - 2 < 0$ → local maximum. Why $f_{xx}$ ? Same-sign forced by $D > 0$ ; negative means frowning (dome).

Verify: $f (2, 3) = - 4 - 9 + 8 + 18 = 13$ . Nudge $f (2.1, 3) = - 4.41 - 9 + 8.4 + 18 = 12.99 < 13$ ✓. Peak.

Example 3 — Cell C: clean saddle

Worked example

$f (x, y) = x y$ Forecast: along the line $y = x$ we get $x^{2}$ (up), along $y = - x$ we get $- x^{2}$ (down). Up one way, down another → saddle.

Step 1 — Critical point. $f_{x} = y = 0, f_{y} = x = 0 \Rightarrow (0, 0)$ . Why this step? Only there does the surface flatten — the prerequisite for the test.

Step 2 — Second partials. $f_{xx} = 0, f_{y y} = 0, f_{x y} = 1$ . Why this step? We need the curvature data; strikingly, both pure curvatures are zero here, so all the shape information sits in the cross-term $f_{x y}$ .

Step 3 — Discriminant. $D = (0) (0) - (1)^{2} = - 1 < 0$ . Why stop here? $D < 0$ alone declares a saddle — no need to check $f_{xx}$ .

Step 4 — Verdict. Saddle point at $(0, 0)$ . Why this step? It states the conclusion the sign of $D$ forces, closing the classification.

The figure below shows this saddle as a contour map: the blue curves are places where $f = x y$ is positive (rising away from the point), the pink curves where it is negative (falling), and the yellow dashed axes are the $f = 0$ lines. Look how both signs surround the single white critical point — that co-existence of "up" and "down" directions is the visual fingerprint of $D < 0$ .

Verify: $f (0.1, 0.1) = 0.01 > 0$ but $f (0.1, - 0.1) = - 0.01 < 0$ — both signs near the point ✓. That two-sign behaviour is exactly what $D < 0$ predicts.

Example 4 — Cell D: many critical points, all four quadrants

Worked example

$f (x, y) = x^{3} - 12 x + y^{3} - 3 y$ Forecast: cubics give two turning points each in $x$ and $y$ , so expect four critical points spread over the quadrants — a mix of verdicts.

Step 1 — Critical points. The variables separate: $f_{x} = 3 x^{2} - 12 = 0 \Rightarrow x = \pm 2, f_{y} = 3 y^{2} - 3 = 0 \Rightarrow y = \pm 1$ Why this step? Because $f$ splits into an $x$ -part and a $y$ -part, the gradient conditions decouple; every combination of $x = \pm 2$ with $y = \pm 1$ is a critical point. Four points: $(2, 1), (2, - 1), (- 2, 1), (- 2, - 1)$ — one in each quadrant.

Step 2 — Second partials. $f_{xx} = 6 x, f_{y y} = 6 y, f_{x y} = 0$ , so $D = 36 x y$ . Why this step? With $f_{x y} = 0$ , the discriminant is just the product of the two diagonal curvatures — sign of $x y$ decides everything.

Step 3 — Classify each. (Recall $D = 36 x y$ , and if $D > 0$ read $f_{xx} = 6 x$ .)

Point	$D = 36 x y$	$f_{xx} = 6 x$	Verdict
$(2, 1)$	$+ 72$	$+ 12$	min
$(2, - 1)$	$- 72$	—	saddle
$(- 2, 1)$	$- 72$	—	saddle
$(- 2, - 1)$	$+ 72$	$- 12$	max

Why this table? Every quadrant produces a different verdict — this is the case class where you must not classify once and reuse.

Verify: at $(2, 1)$ , $f = 8 - 24 + 1 - 3 = - 18$ ; at $(- 2, - 1)$ , $f = - 8 + 24 - 1 + 3 = 18$ . The min value $<$ the max value, and the two saddles at $(2, - 1), (- 2, 1)$ give $f = 8 - 24 - 1 + 3 = - 14$ and $- 8 + 24 + 1 - 3 = 14$ , sitting between — geometrically sensible ✓.

Example 5 — Cell E: the $f_{xx} = 0$ pivot

Worked example

$f (x, y) = y^{2} + 4 x y$ (careful — the derivation assumed $f_{xx} \neq = 0$ ) Forecast: there's no $x^{2}$ term, so $f_{xx} = 0$ . Recall from the definition box that the parent derivation set $u = x - a, v = y - b$ and completed the square in $u$ , a move that quietly divided by $f_{xx}$ . Does the test still deliver a verdict when $f_{xx} = 0$ ?

Step 1 — Critical point. $f_{x} = 4 y = 0 \Rightarrow y = 0, f_{y} = 2 y + 4 x = 0 \Rightarrow x = 0$ Critical point $(0, 0)$ . Why this step? The test only applies where the gradient vanishes; we must locate that point before touching second derivatives. (Here $(a, b) = (0, 0)$ , so the shift variables are simply $u = x, v = y$ .)

Step 2 — Second partials. $f_{xx} = 0, f_{y y} = 2, f_{x y} = 4$ . Why this step? These curvatures feed the discriminant; note the flagged feature — $f_{xx} = 0$ — which is exactly the degeneracy this example is built to test.

Step 3 — Discriminant. $D = (0) (2) - (4)^{2} = - 16 < 0$ . Why this rescues us: $D < 0$ gives saddle directly and never uses $f_{xx}$ . The "complete the square in $u$ " worry only mattered for the $D > 0$ min/max split, where dividing by $f_{xx}$ was needed. When $f_{xx} = 0$ but $D < 0$ , we simply complete the square in $v$ instead (dividing by $f_{y y} = 2 \neq = 0$ ) — the geometry, two opposite-sign curvatures, is unchanged.

Step 4 — Verdict. Saddle at $(0, 0)$ . Why this step? It records the conclusion $D < 0$ dictates, showing the degenerate pivot never blocked the answer.

Verify: along $y = 0$ : $f = 0$ flat; along $x = - y$ : $f = y^{2} - 4 y^{2} = - 3 y^{2} < 0$ ; along $x = y$ : $f = y^{2} + 4 y^{2} = 5 y^{2} > 0$ . Both signs appear ✓ — saddle confirmed.

Example 6 — Cell F: genuine $D = 0$ , inconclusive then resolved

Worked example

$f (x, y) = x^{4} + y^{4}$ Forecast: fourth powers are flatter than quadratics near $0$ . The test may go blind ( $D = 0$ ), even though intuitively this is a minimum.

Step 1 — Critical point. $f_{x} = 4 x^{3} = 0, f_{y} = 4 y^{3} = 0 \Rightarrow (0, 0)$ . Why this step? We must confirm the gradient is zero here before the second derivative test is even allowed to speak.

Step 2 — Second partials. $f_{xx} = 12 x^{2}, f_{y y} = 12 y^{2}, f_{x y} = 0$ . At $(0, 0)$ all are $0$ . Why this step? We evaluate curvature at the critical point; discovering every entry is zero is the warning sign that the quadratic approximation is flat.

Step 3 — Discriminant. $D = (0) (0) - 0 = 0$ → inconclusive. Why the test fails: the second-order Taylor information is entirely flat; the true shape lives in the fourth-order terms the Hessian cannot see.

Step 4 — Investigate by hand. For any $(x, y) \neq = (0, 0)$ , $f = x^{4} + y^{4} > 0 = f (0, 0)$ . So it's a strict local (and global) minimum. Why this step: with the test silent, we bypass the quadratic form and inspect the sign of $f - f (0, 0)$ directly — the definition of a minimum.

Verify: $f (0.1, 0) = 0.0001 > 0$ and $f (0, 0.1) = 0.0001 > 0$ ; nowhere near the origin is $f$ negative ✓. Minimum, despite $D = 0$ .

Example 7 — Cell G: $D = 0$ that is actually a saddle

Worked example

$f (x, y) = x^{4} - y^{4}$ Forecast: looks like Example 6 but with a sign flip. That flip should turn "min" into "saddle" — proving $D = 0$ carries no guaranteed verdict.

Step 1 — Critical point. $f_{x} = 4 x^{3}, f_{y} = - 4 y^{3} \Rightarrow (0, 0)$ . Why this step? Same rule as always — locate where the gradient vanishes before classifying.

Step 2 — Second partials at $(0, 0)$ . $f_{xx} = 12 x^{2} = 0, f_{y y} = - 12 y^{2} = 0, f_{x y} = 0$ . Why this step? We read the curvature at the point; all zero again means the Hessian sees a flat quadratic and will be uninformative.

Step 3 — Discriminant. $D = 0$ → inconclusive again. Why this step? Computing $D$ is still the first move; here it merely confirms the test cannot decide, forcing a by-hand look.

Step 4 — By hand. Along the $x$ -axis ( $y = 0$ ): $f = x^{4} > 0$ (curves up). Along the $y$ -axis ( $x = 0$ ): $f = - y^{4} < 0$ (curves down). Two signs → saddle. Why this matters: Examples 6 and 7 have the same $D = 0$ yet opposite geometry — hard proof that $D = 0$ demands hands-on investigation, never a lazy default.

Verify: $f (0.1, 0) = 0.0001 > 0$ but $f (0, 0.1) = - 0.0001 < 0$ ✓ — genuine saddle.

Example 8 — Cell H: real-world optimisation (mind the units)

Worked example A box factory

An open-top rectangular box must hold $V = 32 m^{3}$ . With base $x \times y$ (metres) and height $h$ , we have $h = 32/ (x y)$ . Material used (in $m^{2}$ ) is the base plus four walls: $S (x, y) = x y + 2 x h + 2 y h = x y + \frac{64}{y} + \frac{64}{x}$ Forecast: minimising surface should give a nearly square base — expect $x = y$ .

Step 1 — Critical point. $S_{x} = y - \frac{64}{x ^{2}} = 0, S_{y} = x - \frac{64}{y ^{2}} = 0$ Why this step? Least material sits at a constrained extremum, here turned into an unconstrained one by substituting $h$ . From the first: $y = 64/ x^{2}$ . Sub into second: $x = 64/ y^{2} = 64 x^{4} /6 4^{2} = x^{4} /64 \Rightarrow x^{3} = 64 \Rightarrow x = 4$ . Then $y = 64/16 = 4$ . Critical point $(4, 4)$ , height $h = 32/16 = 2 m$ .

Step 2 — Second partials. $S_{xx} = \frac{128}{x ^{3}}, S_{y y} = \frac{128}{y ^{3}}, S_{x y} = 1$ At $(4, 4)$ : $S_{xx} = 128/64 = 2, S_{y y} = 2, S_{x y} = 1$ . Why this step? These curvatures tell us whether the critical point is truly the cheapest design or a saddle in disguise — a genuine risk with rational functions like $64/ x$ .

Step 3 — Discriminant. $D = (2) (2) - 1^{2} = 3 > 0$ , $S_{xx} = 2 > 0$ → local minimum ✓ (least material). Why $f_{xx} > 0$ matters: confirms we found the cheapest box, not a costly saddle.

Verify (units + numbers): dimensions $4 \times 4 \times 2 m$ , volume $= 4 \cdot 4 \cdot 2 = 32 m^{3}$ ✓ (constraint met). Surface $S = 16 + 64/4 + 64/4 = 16 + 16 + 16 = 48 m^{2}$ . All terms are areas ( $m^{2}$ ) ✓ — dimensionally consistent.

Example 9 — Cell I: exam twist with a parameter

Worked example Classify

$(0, 0)$ for $f (x, y) = x^{2} + k x y + y^{2}$ as $k$ varies Forecast: small $k$ should stay a bowl; large $k$ should tip into a saddle. There must be a threshold value of $k$ where it flips.

Step 1 — Critical point. $f_{x} = 2 x + k y, f_{y} = k x + 2 y$ . At $(0, 0)$ both vanish for every $k$ , so $(0, 0)$ is always critical. Why this step? No linear terms — the origin is fixed as the critical point regardless of $k$ , so we can study one point while $k$ moves.

Step 2 — Second partials. $f_{xx} = 2, f_{y y} = 2, f_{x y} = k$ . Why this step? Here the tilt term $f_{x y} = k$ is the tunable knob; isolating it shows the parameter enters the test only through the cross-curvature.

Step 3 — Discriminant as a function of $k$ . $D (k) = (2) (2) - k^{2} = 4 - k^{2}$ Why this step? The whole verdict now rides on the sign of $4 - k^{2}$ .

Step 4 — Case split on $k$ .

$∣ k ∣ < 2$ → $D > 0$ and $f_{xx} = 2 > 0$ → minimum. Why: small tilt keeps both curvatures dominant and same-signed, so the bowl survives.
$∣ k ∣ > 2$ → $D < 0$ → saddle. Why: a large cross-term overpowers the diagonal curvatures, forcing opposite-sign directions.
$∣ k ∣ = 2$ → $D = 0$ → inconclusive, must check by hand. Two symmetric sub-cases:
- $k = + 2$ : $f = x^{2} + 2 x y + y^{2} = (x + y)^{2} \geq 0$ , a degenerate minimum, flat along the line $x = - y$ (where $x + y = 0$ ). Why flat there: on that line $f = 0 = f (0, 0)$ , a valley floor rather than a single low point.
- $k = - 2$ : $f = x^{2} - 2 x y + y^{2} = (x - y)^{2} \geq 0$ , also a degenerate minimum, flat along $x = y$ (where $x - y = 0$ ). Why the mirror: flipping the sign of $k$ mirrors the flat valley to the other diagonal.

The figure plots $D (k) = 4 - k^{2}$ against $k$ . The blue region ( $D > 0$ ) is where the origin is a minimum; the pink regions ( $D < 0$ ) are saddles; the yellow dots at $k = \pm 2$ mark the $D = 0$ thresholds where the verdict flips and the valley goes flat. Reading left to right you see the single parameter carry the point from saddle → min → saddle.

Verify: at $k = 1$ , $D = 3 > 0$ (min); at $k = 3$ , $D = 4 - 9 = - 5 < 0$ (saddle); at $k = 2$ , $D = 0$ and $f = (x + y)^{2} \geq 0$ flat along $x + y = 0$ ; at $k = - 2$ , $D = 0$ and $f = (x - y)^{2} \geq 0$ flat along $x - y = 0$ ✓. Threshold at $∣ k ∣ = 2$ confirmed.

Recall Which cell was hardest, and why?

Cells F and G — because $D = 0$ makes the Hessian matrix blind. Question ::: The Hessian only sees second-order curvature; when both eigenvalues are $0$ , the true shape hides in higher-order terms and you must inspect $f$ directly.

Active recall

What verdict does

D < 0

give, regardless of

f_{xx}

Saddle point.

For

f = x^{2} + x y + y^{2} + 3 x

, where is the critical point and its type?

(- 2, 1)

, local minimum (

D = 3 > 0

f_{xx} = 2 > 0

For

f = x^{3} - 12 x + y^{3} - 3 y

, classify

(2, - 1)

Saddle (

D = 36 x y = - 72 < 0

When

f_{xx} = 0

but

D < 0

, is the test valid?

Yes —

D < 0

gives a saddle and never uses

f_{xx}

Why can

D > 0

never coexist with

f_{xx} = 0

Then

D = - f_{x y}^{2} \leq 0

, contradicting

D > 0

For

f = x^{4} + y^{4}

at origin:

D

and true type?

D = 0

; genuine minimum (found by inspecting

f > 0

For

f = x^{4} - y^{4}

at origin:

D

and true type?

D = 0

; saddle (up along

x

, down along

y

Box of volume 32 minimising surface: dimensions?

4 \times 4 \times 2

m, surface

48 m^{2}

For

f = x^{2} + k x y + y^{2}

, threshold

∣ k ∣

flipping min→saddle?

∣ k ∣ = 2

(

D = 4 - k^{2}

Connections

Second derivative test — Hessian determinant — the parent rule these examples exercise.
Critical points and gradient — every example starts by solving $\nabla f = 0$ .
Taylor series multivariable — why $D = 0$ hides the answer in higher-order terms.
Hessian matrix — the object whose determinant is $D$ .
Quadratic forms and definiteness — the $∣ k ∣ < 2$ / $∣ k ∣ > 2$ split is definiteness in action.
Lagrange multipliers — the constrained cousin behind the box problem.

Worked examples — Second derivative test — Hessian determinant

The scenario matrix

Example 1 — Cell A: clean minimum

Example 2 — Cell B: clean maximum

Example 3 — Cell C: clean saddle

Example 4 — Cell D: many critical points, all four quadrants

Example 5 — Cell E: the $f_{xx} = 0$ pivot

Example 6 — Cell F: genuine $D = 0$ , inconclusive then resolved

Example 7 — Cell G: $D = 0$ that is actually a saddle

Example 8 — Cell H: real-world optimisation (mind the units)

Example 9 — Cell I: exam twist with a parameter

Active recall

Connections

Concept Map