Critical points — finding, classifying: Question bank

Before you start, recall the machinery from the parent note: a critical point is where $\nabla f = 0$ (both partials vanish) or a partial fails to exist; the discriminant is $D = f_{xx} f_{y y} - (f_{x y})^{2} = det H$ where $H$ is the Hessian Matrix; and the test reads $D > 0, f_{xx} > 0 \Rightarrow$ min, $D > 0, f_{xx} < 0 \Rightarrow$ max, $D < 0 \Rightarrow$ saddle, $D = 0 \Rightarrow$ inconclusive.

The three shapes you are classifying look like this — memorise the pictures before you tackle the traps:

True or false — justify

TF1. "If

f_{xx} > 0

and

f_{y y} > 0

at a critical point, it must be a local minimum."

False. You ignored the mixed term.

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

has

f_{xx} = f_{y y} = 2 > 0

but

D = 4 - 9 = - 5 < 0

— a saddle. Cross-curvature can dominate.

TF2. "If

D < 0

the point is always a saddle, no matter what

f_{xx}

is."

True.

D < 0

means the quadratic form is indefinite (up in some directions, down in others), which is exactly a saddle;

f_{xx}

is irrelevant here.

TF3. "

D = 0

means the point is definitely neither a max, min, nor saddle."

False.

D = 0

is inconclusive, not a verdict.

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

has

D = 0

at the origin yet is a genuine minimum; the quadratic term simply failed to decide.

TF4. "Every point where

\nabla f = 0

is a local max, min, or saddle."

False. These are the only candidates, but a monkey saddle or a degenerate flat region (

D = 0

cases) may be none of the three cleanly;

\nabla f = 0

is necessary, not sufficient.

TF5. "The second derivative test can be applied at a corner of

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

False. The test needs

f

twice differentiable. At the cusp the partials don't even exist, so the Hessian is undefined — classify by direct inspection instead.

TF6. "If

f_{xx} = 0

at a critical point, the completing-the-square derivation breaks and so does the test."

False. The derivation as written assumed

f_{xx} \neq = 0

, but the conclusion (

D = det H

and its sign rules) still holds — you can complete the square in

k

first, or read

D

directly.

TF7. "A function can have a critical point that is a local max in every straight-line direction yet not be a genuine local max."

True. This is the classic Peano surface phenomenon: restricting to lines can mislead because the surface can dip along a curved path. Straight-line tests are not enough; the Hessian (or a curve) is needed.

TF8. "Swapping

f_{x y}

for

f_{y x}

could change the value of

D

False. For

f

with continuous second partials,

f_{x y} = f_{y x}

(Clairaut/Schwarz), so

H

is symmetric and

D

is unambiguous.

TF9. "If both eigenvalues of the Hessian are positive, the point is a minimum."

True. Positive eigenvalues mean the quadratic form is positive-definite — up in all directions — which is

D > 0, f_{xx} > 0

in disguise.

TF10. "

D > 0

alone is enough to conclude 'extremum' without knowing which kind."

True.

D > 0

forces

f_{xx}

and

f_{y y}

to share a sign (since

f_{xx} f_{y y} = D + f_{x y}^{2} > 0

), so it's definitely a max or min — you just need

f_{xx}

's sign to say which.

Spot the error

SE1. "

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

: I found

x = \pm 1

, so both

(1, 0)

and

(- 1, 0)

are minima because

f_{y y} = 2 > 0

Error: classification used only

f_{y y}

. Here

D = 12 x

, so

(1, 0)

has

D > 0

(min) but

(- 1, 0)

has

D < 0

(saddle). Always compute

D

first.

SE2. "For

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

I get

D = (2) (- 2) = - 4

, and since

f_{xx} = 2 > 0

, it's a minimum."

Error: once

D < 0

you stop — it's a saddle. Reading

f_{xx}

after finding

D < 0

is meaningless; the min/max branch only runs when

D > 0

SE3. "

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

at the origin:

D = 0

, so the test is inconclusive and we cannot say anything about the point."

Error: inconclusive means the test can't decide, not that we can't. Direct inspection (

f \geq 0

, equal only at origin) proves it's a global minimum.

SE4. "I solved

f_{x} = 0

to get

x = 2

and separately

f_{y} = 0

to get

y = 3

, so the critical point is

(2, 3)

Error: the equations must be solved simultaneously. If

f_{x} = 0

has solutions depending on

y

, you can't set

x

and

y

independently; substitute one into the other.

SE5. "

f = e^{x}

: I set

f_{x} = e^{x} = 0

; no solution, so there are no critical points and the test doesn't apply — the function must be flat somewhere."

Error: correct that

f_{x} = e^{x} > 0

never vanishes (so no critical point), but wrong to conclude "flat somewhere." No critical points means no extrema at all; the function just keeps climbing.

SE6. "The Hessian

H = (2312)

gives

D = det H = 4 - 3 = 1 > 0

Error: a Hessian must be symmetric (

f_{x y} = f_{y x}

), so off-diagonals can't be

1

and

3

. If genuinely

f_{x y} = f_{y x} = c

, then

D = 4 - c^{2}

SE7. "Since

\nabla f = 0

is required for extrema, checking the sign of

\nabla f

tells us max vs min."

Error: at a critical point

\nabla f

is zero, so it has no sign to read. Classification is a job for the second derivatives, not the gradient.

Why questions

W1. "Why do first derivatives tell us nothing about the type of a critical point?"

Because they're all zero there — a flat tangent plane is common to maxes, mins, and saddles alike. Only curvature (second derivatives) distinguishes the shapes.

W2. "Why does the mixed partial

f_{x y}

appear in

D

at all?"

It measures how curvature in

x

couples to

y

— a tilt of the bowl. A big tilt can rotate a would-be bowl into a saddle, which is exactly the

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

term pulling

D

downward.

W3. "Why is

D = det H

and not, say, the trace of

H

The determinant is the product of eigenvalues, so its sign detects whether they share a sign (definite → extremum) or differ (indefinite → saddle). The trace (their sum) can't tell a saddle from an extremum. The figure below shows the eigenvectors as the surface's principal axes and how their curvature signs multiply into

D

W4. "Why do we complete the square in the Taylor derivation rather than just look at

f_{xx}

and

f_{y y}

Completing the square rewrites the quadratic as a sum of independent squares, exposing the sign of the form in every direction at once — the cross term is folded into one square, leaving

D / f_{xx}

as the honest second coefficient. Worked out algebraically and pictured below:

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

W5. "Why must extrema of smooth

f

occur only where

\nabla f = 0

If any directional slope were nonzero you could step slightly to go lower (or higher), contradicting the point being a local extreme. So every partial must vanish — Fermat's theorem in several variables.

W6. "Why does the 1-D test never mention anything like

D

In one dimension there's only one direction to curve, so

f^{''}

's sign is the whole story. The 2-D

D

exists precisely because two directions can disagree (up in

x

, down in

y

) — a possibility that simply can't arise on a line.

W7. "Why can the test be recast using eigenvalues of the Hessian?"

The quadratic form

\frac{1}{2} v^{⊤} H v

is positive-definite iff both eigenvalues

> 0

(min), negative-definite iff both

< 0

(max), indefinite iff they differ in sign (saddle) — and

D = det H = λ_{1} λ_{2}

encodes exactly this sign agreement.

Edge cases

E1. "What happens if

D > 0

but

f_{xx} = 0

— can that occur?"

No.

D > 0

means

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

, so

f_{xx}

cannot be zero. The 'read

f_{xx}

' step is always well-defined when

D > 0

E2. "At the origin of

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

, is it a max, min, or saddle?"

All second partials vanish there, so

D = 0

— inconclusive. Direct inspection: along the line

y = x

f = 2 x^{3}

changes sign through

0

, so it's neither max nor min (a degenerate saddle-like point). The contour plot below shows the sign-change lobes around the origin.

E3. "For the monkey saddle

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

at the origin, what does the test say?"

f_{xx} = 6 x

f_{y y} = - 6 x

f_{x y} = - 6 y

all vanish at the origin, giving

D = 0

— inconclusive. It's actually a saddle with three descending valleys, but the quadratic test can't see it; the cubic terms decide. The three-fold pattern is unmistakable in the figure:

E4. "Can a critical point exist where

f

is not differentiable, so the Hessian doesn't exist?"

Yes — the definition explicitly includes points where a partial fails to exist (corners, cusps like

x^{2} + y^{2}

0

). Such points are candidates but must be classified by definition, not by

D

E5. "If

f_{xx} < 0

and

f_{y y} < 0

but

D < 0

, what is it?"

A saddle. Once

D < 0

the shared negative sign of the pure curvatures is irrelevant — the cross-term has made the form indefinite anyway.

E6. "Is a constant function

f = 5

all critical points, and what type?"

Every point has

\nabla f = 0

, so yes — all are critical. But

H

is the zero matrix,

D = 0

everywhere: inconclusive by the test, and by inspection each point is simultaneously a (weak) max and min since

f

never changes.

E7. "What does

D < 0

mean geometrically for the two Hessian eigenvalues?"

They have opposite signs (

λ_{1} λ_{2} = D < 0

), so the surface curves up along one principal direction and down along the perpendicular one — the defining picture of a saddle.

Active Recall

The whole test compresses into one sign chart — commit this picture to memory:

Recall Two things

$D$ can never do when it's positive. It can never leave $f_{xx}$ ambiguous (since $f_{xx} \neq = 0$ ), and it can never indicate a saddle. $D > 0$ always means a genuine extremum.

Recall The single fastest way to disprove "positive

$f_{xx}$ and $f_{y y}$ ⟹ minimum." Cite $f = x^{2} + y^{2} - 3 x y$ : both pure curvatures positive, yet $D = 4 - 9 = - 5 < 0$ , a saddle.

Recall Why is

$D = 0$ not a classification but a confession? It confesses the quadratic approximation is too flat to decide; you must fall back on higher-order terms or direct inspection.

Connections

Critical points — finding, classifying — the parent note this bank drills.
Hessian Matrix — $D = det H$ ; eigenvalue view of every trap above.
Quadratic Forms and Definiteness — definite/indefinite ↔ the max/min/saddle verdicts.
Taylor Series in Several Variables — source of the completing-the-square argument.
Gradient and Directional Derivatives — why $\nabla f = 0$ is necessary.
Single-variable Second Derivative Test — the 1-D case with no $D$ .
Lagrange Multipliers — the constrained-extremum cousin.