Second derivative test — Hessian determinant: Question bank

Quick reminder of the machinery you are being tested on (all built from scratch in the parent note):

A critical point is where both first partials vanish: $f_{x} = f_{y} = 0$ .
The Hessian determinant (a single number computed at that point) is $D = f_{xx} f_{y y} - (f_{x y})^{2}$ .
$D > 0$ with $f_{xx} > 0$ → minimum; $D > 0$ with $f_{xx} < 0$ → maximum; $D < 0$ → saddle; $D = 0$ → the test says nothing.

True or false — justify

Every statement is either subtly right or subtly wrong. The reveal gives the reason, not just the verdict.

f_{xx} > 0

at a critical point, then the point is a local minimum.

False.

f_{xx} > 0

only says the surface curves up along the

x

-direction; if

D < 0

it still curves down along another direction and you have a saddle. You must check

D

first, and only then use the sign of

f_{xx}

D > 0

then

f_{xx}

and

f_{y y}

must have the same sign.

True. Since

D = f_{xx} f_{y y} - (f_{x y})^{2} > 0

, we get

f_{xx} f_{y y} > (f_{x y})^{2} \geq 0

, so the product

f_{xx} f_{y y}

is strictly positive — meaning both factors share a sign. That is exactly why checking

f_{xx}

alone is enough.

D < 0

the point could still be a local minimum.

False.

D < 0

forces the two curvature eigenvalues to have opposite signs, so

Q

takes both positive and negative values — that is the definition of a saddle. A minimum is impossible when

D < 0

D > 0

guarantees the point is a genuine extremum (min or max), not a saddle.

True.

D > 0

means both eigenvalues (curvatures along principal axes) share a sign, so the surface bends the same way in every direction — a bowl or a dome. The sign of

f_{xx}

then chooses which.

At a critical point where

f_{xx} = 0

and

f_{y y} = 0

, the point must be a saddle.

True (given a nonzero cross-term). Then

D = 0 \cdot 0 - (f_{x y})^{2} = - (f_{x y})^{2}

. If

f_{x y} \neq = 0

this is negative → saddle. Only if

f_{x y}

is also zero does

D = 0

and the test go silent.

D = 0

, there is no extremum at the point.

False.

D = 0

means inconclusive, not "none." For

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

the origin is a clear minimum yet

D = 0

there — the second-order Taylor terms are flat and you must look at higher order by hand.

The Hessian is always symmetric, so

f_{x y} = f_{y x}

automatically.

True whenever the mixed partials are continuous — this is Clairaut's theorem. For the smooth functions in this course it always holds, which is why

D

needs only one cross-term

(f_{x y})^{2}

You can apply the test at any point of the surface as long as you compute the Hessian correctly.

False. The whole derivation drops the linear Taylor terms because the gradient is zero there. At a non-critical point those linear terms dominate and

D

tells you nothing about shape.

D = f_{xx} f_{y y}

is a fine shortcut when the cross-term is small.

False. Dropping

(f_{x y})^{2}

throws away the tilt of the curvature axes. Even a small

f_{x y}

can flip

D

from positive to negative — the whole point of the minus sign is that coupling between directions matters.

Spot the error

Each item states a piece of (wrong) student reasoning. The reveal names the flaw.

"At

(0, 0)

f_{xx} = 2

and

f_{y y} = - 2

, and

D > 0

, so it's a min because

f_{xx} > 0

." Find the flaw.

The two premises contradict each other:

f_{xx} = 2

f_{y y} = - 2

give

f_{xx} f_{y y} = - 4 < 0

, so

D

cannot be positive. When

D > 0

the two second derivatives are forced to share a sign — different signs signal an arithmetic slip.

f_{x} = f_{y} = 0

has no solution, so there are no critical points and the function has no minimum." Find the flaw.

A minimum on the interior requires a critical point, but extrema can also occur on a boundary or at infinity (or the function may simply be unbounded). "No interior critical point" does not mean "no minimum anywhere."

D < 0

, so the point is inconclusive — I'll investigate by hand." Find the flaw.

D < 0

is fully decisive: it is a saddle. Only

D = 0

is inconclusive. The student confused the two failure-looking cases.

"I found

D > 0

, so the point is a maximum." Find the flaw.

D > 0

alone never distinguishes max from min — it only says "same-sign curvatures." You must still read the sign of

f_{xx}

(

> 0

min,

< 0

max) to finish.

"The Taylor expansion gave

f (a, b) + f_{x} u + f_{y} v + \frac{1}{2} Q

, and I kept the linear terms in

Q

." Find the flaw.

At a critical point

f_{x} = f_{y} = 0

, so the linear terms are identically zero and

Q

is purely quadratic. Keeping nonzero linear terms means you never actually landed on a critical point.

D

is the sum of the eigenvalues, so

D > 0

means positive curvature." Find the flaw.

D = det H = λ_{1} λ_{2}

is the product, not the sum. The product being positive means same sign; the trace

f_{xx} + f_{y y} = λ_{1} + λ_{2}

is the sum, and only the trace's sign then tells min vs max.

"For

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

, since

f_{x y} = 0

, the surface has no coupling and

D

must be zero." Find the flaw.

f_{x y} = 0

makes

D = f_{xx} f_{y y} - 0 = f_{xx} f_{y y}

, which for this bowl is

2 \cdot 2 = 4 > 0

— very much nonzero. A zero cross-term means the axes aren't tilted, not that

D

vanishes.

Why questions

Force yourself to give the reason, not the rule.

Why does the test only use second-order Taylor terms and ignore third order?

At a critical point the linear terms vanish, so the quadratic form is the leading behaviour — it dominates for small displacements. Higher-order terms are negligible unless the quadratic form is degenerate (

D = 0

), which is exactly when we fall back on them.

Why is completing the square the right tool to analyse the sign of

Q

A raw quadratic

f_{xx} u^{2} + 2 f_{x y} uv + f_{y y} v^{2}

mixes

u

and

v

, so its sign is not obvious. Completing the square rewrites it as (positive square) plus (coefficient)

\times v^{2}

, where each squared piece is

\geq 0

— now the sign hinges purely on the two coefficients

f_{xx}

and

D / f_{xx}

Why does the coefficient

D / f_{xx}

appear naturally in the completed square?

Completing the square in

u

leaves the leftover

v^{2}

-coefficient as

f_{y y} - \frac{f _{x y}^{2}}{f _{xx}}

, which over a common denominator is

\frac{f _{xx} f _{y y} - f _{x y}^{2}}{f _{xx}} = \frac{D}{f _{xx}}

. So

D

isn't invented — it falls out of the algebra as the second curvature.

Why does

D < 0

produce a saddle rather than a very flat extremum?

With

D < 0

, the two coefficients

f_{xx}

and

D / f_{xx}

have opposite signs. Pick

v = 0

to make

Q

one sign, then kill the bracket to make

Q

the other sign — the surface genuinely rises in one direction and falls in another. That is a pass, not a flat spot.

Why is

f_{xx}

(rather than

f_{y y}

) the conventional tiebreaker when

D > 0

When

D > 0

both agree in sign, so either would work;

f_{xx}

is just the standard choice. If they disagree, you've made an error — a built-in consistency check.

Why does connecting

D

to eigenvalues explain everything at once?

The eigenvalues of the symmetric Hessian are the curvatures along the principal axes. Their product is

D

: same sign (

D > 0

) is a bowl/dome, opposite sign (

D < 0

) is a saddle. The whole test is just "read the signs of two curvatures."

Why must the Hessian be symmetric for the

- (f_{x y})^{2}

form to be valid?

The single cross-term

(f_{x y})^{2}

assumes

f_{x y} = f_{y x}

; without symmetry you'd have two distinct off-diagonal entries and

det H = f_{xx} f_{y y} - f_{x y} f_{y x}

. Clairaut's theorem guarantees they're equal for continuous mixed partials, collapsing it to

- (f_{x y})^{2}

Edge cases

The scenarios the clean rules quietly assume away.

What does the test say for

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

at the origin?

All second partials are zero there, so

D = 0

— inconclusive. Yet the origin is clearly a minimum (every displacement raises

f

). This is the poster child for "

D = 0

needs a by-hand argument."

What does the test say for

f = x^{3}

(as a function of two variables,

y

absent) at the origin?

f_{xx} = 6 x = 0

and

f_{y y} = f_{x y} = 0

at the origin, so

D = 0

— inconclusive. Directly,

x^{3}

goes negative for

x < 0

and positive for

x > 0

, so it's neither a max nor a min: a degenerate inflection the test can't classify.

If both eigenvalues of the Hessian are zero, what is

D

and what can you conclude?

D = λ_{1} λ_{2} = 0

, so the test is silent. The quadratic form is identically zero and all shape information hides in higher-order terms — you must expand further.

What happens to the "complete the square" derivation if

f_{xx} = 0

but

f_{y y} \neq = 0

You can't divide by

f_{xx}

, so complete the square in

v

instead (using

f_{y y}

). The conclusion is identical — the roles of

u

and

v

are symmetric — because

D

itself is symmetric in the two variables.

Can a function have

D > 0

and

f_{xx} > 0

at a point that is only a local minimum, not global?

Yes. The test is purely local — it reads curvature in an infinitesimal neighbourhood. A function can have several such basins, each a local min, with one deeper than the rest; the Hessian sees only the immediate bowl.

At a critical point on a plateau where

f

is constant in one direction, what happens?

One curvature is zero, so that eigenvalue is

0

and

D = λ_{1} \cdot 0 = 0

— inconclusive. The surface is flat along that direction to second order, so second-order data can't classify it.

Is it possible for

D > 0

while

f_{xx}

and

f_{y y}

are both zero?

No. If both are zero then

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

, which can never be positive.

D > 0

requires both diagonal entries to be nonzero and of matching sign.

Recall Self-check: which trap did you fall for most?

The single most common trap ::: Concluding min/max from the sign of $f_{xx}$ before checking $D$ — always compute $D$ first. The most common "false failure" ::: Reading $D < 0$ as inconclusive; it is decisively a saddle. Only $D = 0$ is inconclusive.

Connections

Second derivative test — Hessian determinant — the parent note these traps target.
Critical points and gradient — every item assumes you've already solved $\nabla f = 0$ .
Taylor series multivariable — why only second-order terms appear.
Hessian matrix — symmetry, eigenvalues, the determinant.
Quadratic forms and definiteness — the sign-analysis behind min/max/saddle.
Clairaut's theorem — why $f_{x y} = f_{y x}$ .
Lagrange multipliers — the constrained cousin of this unconstrained test.