Critical points and saddle points: Worked examples

This page is a workout. We will not learn new theory — we take the parent tools (gradient, Hessian, eigenvalues) and drive them through every kind of situation that can appear: every sign pattern of the curvature, the awkward zero cases, the degenerate cases where the usual test breaks, a physical word problem, and an exam twist.

Before any symbol appears, here is the one picture that anchors everything.

Figure — Critical points and saddle points

The scenario matrix

Every critical-point problem is one of these cells. The examples below each carry the cell tag they resolve, so together they tile the whole table.

Cell	What makes it special	Curvature signature	Answer type
A	1-D, clean second derivative	$f^{''} > 0$ or $f^{''} < 0$	min / max
B	1-D, second derivative zero	$f^{''} = 0$ , use higher term	inflection / degenerate
C	2-D bowl	both $λ > 0$	local minimum
D	2-D dome	both $λ < 0$	local maximum
E	2-D mixed signs	$λ_{1} > 0, λ_{2} < 0$	saddle
F	2-D with a zero eigenvalue	one $λ = 0$	degenerate (flat valley)
G	Off-diagonal Hessian (cross term)	must diagonalise first	depends on eigenvalues
H	High-D "one bad direction"	$n - 1$ positive, $1$ negative	saddle (ML trap)
I	Word problem (physics / economics)	translate → classify	applied min/max
J	Exam twist: many critical points	classify each separately	mixed

Example A — one variable, textbook minimum (and its mirror maximum)

Worked example Cell A — clean

$f^{''} > 0$ Classify the critical points of $f (x) = x^{2} - 4 x + 7$ .

Forecast: Guess before reading — is this a min, a max, or something weirder?

Step 1. Find where the slope is flat. $f^{'} (x) = 2 x - 4 = 0 ⟹ x^{*} = 2$ Why this step? A critical point is by definition where the first derivative (the slope) is zero — the only place a min or max can hide for a smooth function.

Step 2. Read the curvature. $f^{''} (x) = 2 > 0$ Why this step? $f^{''}$ tells us how the slope is changing. Positive means the slope is increasing — going from downhill to uphill — which is the shape of the bottom of a bowl.

Step 3. Conclude: $x^{*} = 2$ is a local (and global) minimum, value $f (2) = 4 - 8 + 7 = 3$ .

Verify: Plug neighbours: $f (1) = 4$ , $f (3) = 4$ , both above $f (2) = 3$ . Lower in the middle ⇒ minimum confirmed. ✓

Example B — one variable, the test fails (two flavours)

Worked example Cell B —

$f^{''} = 0$ , even power → minimum Classify the critical point of $f (x) = x^{4}$ .

Forecast: The parent note showed $x^{3}$ gives an inflection when $f^{''} = 0$ . Does $x^{4}$ do the same, or something different?

Step 1. $f^{'} (x) = 4 x^{3} = 0 ⟹ x^{*} = 0$ . Why this step? Same rule: find the flat spot.

Step 2. $f^{''} (x) = 12 x^{2}$ , so $f^{''} (0) = 0$ . Why this step? When the curvature is exactly zero, the quadratic bowl/dome picture gives no information — the surface is flatter than any parabola near $0$ . We must look at the next non-zero term.

Step 3. Look at the shape directly: $f (0 + h) = h^{4} \geq 0$ for every $h$ , and $> 0$ unless $h = 0$ . Why this step? Because the second-derivative test is silent, we test the definition of a minimum directly: is $f$ higher everywhere nearby? Yes.

Conclusion: $x^{*} = 0$ is a local minimum even though $f^{''} (0) = 0$ . Rule: even lowest power → extremum; odd lowest power → inflection.

Verify: $f (- 0.1) = 1 0^{- 4}$ , $f (0) = 0$ , $f (0.1) = 1 0^{- 4}$ . Middle is lowest ⇒ minimum. ✓

Worked example Cell B —

$f^{''} = 0$ , odd power → inflection Classify the critical point of $f (x) = x^{3}$ (the parent note's function, worked in full here).

Forecast: Same $f^{''} (0) = 0$ as $x^{4}$ . Does the odd power change the verdict?

Step 1. $f^{'} (x) = 3 x^{2} = 0 ⟹ x^{*} = 0$ . Why this step? Locate the flat spot.

Step 2. $f^{''} (x) = 6 x$ , so $f^{''} (0) = 0$ — inconclusive again. Why this step? The quadratic term vanishes, so the bowl/dome test says nothing; go to the next term.

Step 3. Inspect the shape: $f (0 + h) = h^{3}$ , which is positive for $h > 0$ and negative for $h < 0$ . Why this step? Testing the definition directly: the function goes below $f (0) = 0$ on one side and above on the other — so it is neither a min nor a max.

Conclusion: $x^{*} = 0$ is an inflection point — the curve passes through the flat spot without pausing. This is the "odd lowest power" branch of the rule.

Verify: $f (- 0.1) = - 0.001 < 0 = f (0) < 0.001 = f (0.1)$ . Straddles $0$ ⇒ not an extremum. ✓

Example C, D, E — the 2-D trio in one shot

These three cells differ only in the sign pattern of the two eigenvalues. See the surfaces:

Worked example Cells C, D, E — bowl, dome, saddle

Classify $(0, 0)$ for three functions:

(C) $g (x, y) = x^{2} + 3 y^{2}$
(D) $h (x, y) = - x^{2} - 3 y^{2}$
(E) $p (x, y) = x^{2} - 3 y^{2}$

Forecast: All three have $\nabla = 0$ at the origin. Which is min, max, saddle?

Step 1. Confirm each is critical at the origin. $\nabla g = (2 x, 6 y)$ , $\nabla h = (- 2 x, - 6 y)$ , $\nabla p = (2 x, - 6 y)$ . All vanish at $(0, 0)$ . ✓ Why this step? Before classifying, we must be at a critical point — otherwise the Hessian test does not apply.

Step 2. Write each Hessian (these are already diagonal, so the diagonal entries are the eigenvalues). $H_{g} = (2006), H_{h} = (- 2 0 0 - 6), H_{p} = (20 0 - 6)$ Why this step? The eigenvalues are the curvatures along the principal axes. Diagonal Hessian ⇒ the axes are already the principal axes, so no extra work.

Step 3. Read the signs.

$g$ : $λ = 2, 6$ both $> 0$ → local minimum (bowl).
$h$ : $λ = - 2, - 6$ both $< 0$ → local maximum (dome).
$p$ : $λ = 2, - 6$ mixed → saddle (up along $x$ , down along $y$ ). Why this step? All-positive = curves up everywhere; all-negative = curves down everywhere; mixed = the defining fingerprint of a saddle.

Verify (saddle $p$ ): $p (0.1, 0) = + 0.01 > 0$ but $p (0, 0.1) = - 0.03 < 0$ . One direction up, one down ⇒ saddle, exactly as the mixed signs predicted. ✓

Example F — the degenerate flat valley (zero eigenvalue)

Worked example Cell F — one eigenvalue is exactly zero

Classify $(0, 0)$ for $f (x, y) = x^{2} + y^{4}$ .

Forecast: One direction is a clean parabola, the other is $x^{4}$ -flat. Min, saddle, or "can't tell"?

Step 1. $\nabla f = (2 x, 4 y^{3}) = (0, 0) ⟹ (0, 0)$ critical. ✓ Why this step? The Hessian classification test is only valid at a critical point. So before touching curvature we confirm the gradient truly vanishes here — both components are zero at the origin, so we are allowed to proceed.

Step 2. Hessian at the origin: $H = (20 0 12 y^{2})_{(0, 0)} = (2000)$ Eigenvalues $λ_{1} = 2 > 0$ , $λ_{2} = 0$ . Why this step? A zero eigenvalue means the quadratic test is silent along that direction — the surface is flatter than any parabola in $y$ . The table calls this degenerate.

Step 3. Resolve by hand along the flat direction: on the $y$ -axis ( $x = 0$ ) we have $f (0, y) = y^{4} \geq 0$ , rising away from $0$ . Why this step? Whenever an eigenvalue is $0$ , fall back to the definition and inspect that specific direction with higher-order terms — here $y^{4}$ still curves up.

Conclusion: Both directions rise ⇒ $(0, 0)$ is a local minimum, but a degenerate one — the Hessian alone could not certify it.

Verify: $f (0, 0) = 0$ ; $f (0.1, 0) = 0.01$ , $f (0, 0.1) = 0.0001$ . Both above $0$ ⇒ minimum. ✓ (If instead $f = x^{2} - y^{4}$ , the flat direction would fall ⇒ that one is a saddle.)

Example G — off-diagonal Hessian: you MUST diagonalise

Worked example Cell G — cross term hides the truth

Classify $(0, 0)$ for $f (x, y) = x^{2} + x y + y^{2}$ .

Forecast: The diagonal entries of the Hessian are both $+ 2$ . Does that alone make it a minimum? Careful.

Step 1. $\nabla f = (2 x + y, x + 2 y) = (0, 0)$ . Solving the pair gives $x = y = 0$ : critical at the origin. ✓ Why this step? Curvature only classifies a point once the gradient is zero there. We solve the two equations $2 x + y = 0$ and $x + 2 y = 0$ simultaneously; the only solution is the origin, so that is where we test the Hessian.

Step 2. Hessian: $H = (2112)$ Why this step? The off-diagonal $1$ 's mean the $x$ and $y$ axes are not the principal axes — you cannot just read the diagonal. You must find the true eigenvalues (see Eigenvalues and Eigenvectors).

Step 3. Eigenvalues from $det (H - λ I) = 0$ : $(2 - λ)^{2} - 1 = 0 ⟹ 2 - λ = \pm 1 ⟹ λ = 1, 3$ Why this step? Eigenvalues are the real curvatures along the rotated (principal) axes. Here both are positive.

Conclusion: $λ = 1, 3 > 0$ ⇒ local minimum (a tilted bowl). The tilt is why the raw axes looked ambiguous.

Verify: A symmetric $2 \times 2$ matrix is positive-definite iff its diagonal $> 0$ and determinant $> 0$ . Here $det = 2 \cdot 2 - 1 \cdot 1 = 3 > 0$ and entries $> 0$ ⇒ positive definite ⇒ minimum. ✓

Example H — high-dimensional "one bad direction" (the ML saddle)

Worked example Cell H —

$999$ good directions, $1$ bad A loss $L (w)$ has a critical point where the Hessian is diagonal with $999$ eigenvalues equal to $+ 5$ and exactly one equal to $- 0.2$ . Is $w^{*}$ a minimum? Which way should we escape?

Forecast: Only one of a thousand curvatures is negative. Surely it's basically a minimum?

Step 1. Apply the rule: a critical point is a saddle if even one eigenvalue is negative. Why this step? A single downhill direction breaks the "bowl" — you can always slide off. This is the parent note's key point about why saddles dominate in high dimensions.

Step 2. Identify the escape direction: the eigenvector $v$ for $λ_{m i n} = - 0.2$ . Why this step? Negative curvature along $v$ means $L$ decreases if you move along $\pm v$ — the guaranteed exit.

Step 3. Estimate the drop for a step of size $ϵ = 0.5$ along $v$ , using the quadratic model $L (w^{*} + ϵ v) \approx L (w^{*}) + \frac{1}{2} λ_{m i n} ϵ^{2}$ (see Taylor Series): $Δ L \approx \frac{1}{2} (- 0.2) (0.5)^{2} = - 0.025$ Why this step? We quantify that the escape genuinely lowers the loss — it is not just "flat".

Conclusion: Saddle, escape along $\pm v$ ; loss drops by about $0.025$ . Optimisers like Momentum and Adam or SGD noise supply exactly this perturbation.

Verify: $\frac{1}{2} (- 0.2) (0.25) = - 0.025$ . Negative ⇒ loss decreases ⇒ not a minimum. ✓

Example I — word problem: the cheapest can

Worked example Cell I — applied minimisation

A cylindrical can must hold volume $V = 500 cm^{3}$ . Material cost is proportional to surface area $A = 2 π r^{2} + 2 π r h$ . Find the radius $r$ that minimises material.

Forecast: Guess — will the optimal can be tall & thin, short & fat, or with height $=$ diameter?

Step 1. Eliminate $h$ using the constraint $π r^{2} h = 500 \Rightarrow h = \frac{500}{π r ^{2}}$ . Why this step? Reduce to one free variable so we can use the 1-D critical-point machinery.

Step 2. Substitute this $h$ into the area formula: $A (r) = 2 π r^{2} + 2 π r \cdot \frac{500}{π r ^{2}} = 2 π r^{2} + \frac{1000}{r}$ Why this step? Replacing $h$ turns the two-variable area into a single-variable function of $r$ alone; only then can we set an ordinary derivative to zero. The $2 π r \cdot \frac{500}{π r ^{2}}$ term simplifies because the $π$ cancels and one power of $r$ cancels, leaving $\frac{1000}{r}$ .

Step 3. Set derivative to zero (critical point): $A^{'} (r) = 4 π r - \frac{1000}{r ^{2}} = 0 ⟹ r^{3} = \frac{1000}{4 π} = \frac{250}{π}$ $r = (\frac{250}{π})^{1/3} \approx 4.301 cm$ Why this step? The cheapest can is a flat spot of the cost curve.

Step 4. Confirm it is a minimum, not a max: $A^{''} (r) = 4 π + \frac{2000}{r ^{3}} > 0 for all r > 0$ Why this step? Positive second derivative = bowl = genuine minimum (Cell A logic in an applied skin).

Conclusion: Optimal $r \approx 4.30 cm$ , and $h = \frac{500}{π r ^{2}} \approx 8.60 cm = 2 r$ — the classic result: height equals diameter.

Verify (units & value): $r$ in cm; $A^{''} > 0$ ⇒ min. Numerically $r^{3} = 250/ π \approx 79.577$ , $r \approx 4.301$ , $h \approx 8.603 \approx 2 r$ . ✓

Example J — exam twist: several critical points at once

Worked example Cell J — classify each critical point

Find and classify all critical points of $f (x, y) = x^{3} - 3 x + y^{2}$ .

Forecast: How many flat spots, and are they the same type?

Step 1. $\nabla f = (3 x^{2} - 3, 2 y) = (0, 0)$ . From $2 y = 0 \Rightarrow y = 0$ . From $3 x^{2} - 3 = 0 \Rightarrow x = \pm 1$ . Critical points: $(1, 0)$ and $(- 1, 0)$ . Why this step? Solve the gradient system component by component; a cubic in $x$ gives two roots, so expect two critical points.

Step 2. Hessian in general: $H (x, y) = (6 x 0 02)$ Why this step? It depends on $x$ , so the same function can give different classifications at different points — the heart of the twist.

Step 3. Evaluate at each point.

At $(1, 0)$ : $H = (6002)$ , eigenvalues $6, 2 > 0$ ⇒ local minimum.
At $(- 1, 0)$ : $H = (- 6 0 02)$ , eigenvalues $- 6, 2$ mixed ⇒ saddle. Why this step? Read the signs at each location separately — never assume one classification for the whole function.

Verify: $f (1, 0) = 1 - 3 = - 2$ ; nudge $f (1.1, 0) = 1.331 - 3.3 = - 1.969 > - 2$ ⇒ min at $(1, 0)$ . $f (- 1, 0) = - 1 + 3 = 2$ ; $f (- 1.1, 0) = - 1.331 + 3.3 = 1.969 < 2$ (down along $x$ ) while $f (- 1, 0.1) = 2.01 > 2$ (up along $y$ ) ⇒ saddle at $(- 1, 0)$ . ✓

Recall Quick self-test

Which cell does $f = x^{2} - y^{4}$ at the origin fall in, and why? ::: Cell F (degenerate): Hessian is $diag (2, 0)$ ; the zero eigenvalue direction $y$ gives $- y^{4}$ which falls, so it is a saddle — the Hessian couldn't decide, higher terms did. If a $1 0^{6}$ -parameter Hessian has one negative eigenvalue, minimum or saddle? ::: Saddle — a single negative eigenvalue means an escape direction exists (Cell H). Why can't you read the diagonal of $(2112)$ to classify? ::: Off-diagonal terms mean the coordinate axes aren't the principal axes; you must find eigenvalues ( $1$ and $3$ ) first (Cell G).

See also: Second-Order Optimization, Loss Landscape Visualization, Convex Optimization, Taylor Series, Eigenvalues and Eigenvectors.