Local vs global minima - maxima: Exercises — AI-ML deep dive

This page is a self-testing ladder. Each problem is stated cleanly, then hides a full solution inside a collapsible callout. Work it first, then reveal. The levels climb from recognising an extremum to synthesising everything you know about the loss landscape — the exact thing that decides whether Gradient Descent finds a great model or a mediocre one.

Before you start, one shared vocabulary reminder so no symbol appears unearned:

Figure — Local vs global minima - maxima

The picture above is our reference landscape: two valleys of different depths plus two ridge tops. Look at the red dot — that is the deep valley (the global min); the black dot on the left is the shallow valley (a local-only min). Every exercise below lives on a curve like this one.

Level 1 — Recognition

Recall Solution 1.1

Read the curve left to right.

Shallow valley (left): lower than its immediate neighbours → local minimum. But another valley is deeper, so it is not global.
First ridge top: higher than neighbours → local maximum.
Deep valley (right): lowest point on the whole curve → local minimum AND global minimum.
Second ridge top: higher than neighbours → local maximum.

The whole point: "flat spot" ≠ "minimum", and "minimum" ≠ "global minimum". You must compare depths.

Recall Solution 1.2

False. A flat slope is necessary for a smooth extremum but not sufficient. The classic counterexample is $f (x) = x^{3}$ at $x = 0$ : $f^{'} (0) = 0$ , yet the function keeps rising through that point — it is an inflection point, neither a peak nor a valley. In many dimensions the analogue is a saddle point.

Level 2 — Application

Recall Solution 2.1

Slope to zero. $f^{'} (x) = 2 x - 4 = 0 \Rightarrow x = 2$ . Curvature. $f^{''} (x) = 2 > 0$ everywhere, so the parabola opens upward → $x = 2$ is a local minimum. Global? An upward parabola has only one valley and rises forever on both sides, so this local min is also the global minimum. Its value: $f (2) = 4 - 8 + 7 = 3$ . This is what "convex" means — one bowl, local = global, no traps for Gradient Descent.

Recall Solution 2.2

Slope to zero. $g^{'} (x) = 3 x^{2} - 3 = 3 (x^{2} - 1) = 0 \Rightarrow x = - 1, x = 1$ . Curvature. $g^{''} (x) = 6 x$ .

At $x = - 1$ : $g^{''} (- 1) = - 6 < 0$ → curves down → local maximum, value $g (- 1) = - 1 + 3 + 1 = 3$ .
At $x = 1$ : $g^{''} (1) = 6 > 0$ → curves up → local minimum, value $g (1) = 1 - 3 + 1 = - 1$ .

Global? As $x \to + \infty$ , $g \to + \infty$ ; as $x \to - \infty$ , $g \to - \infty$ . So on the full real line there is no global minimum or maximum — the cubic runs off to $\pm \infty$ . The extrema found are strictly local.

Level 3 — Analysis

Recall Solution 3.1

Slope to zero. $h^{'} (x) = 4 x^{3} - 8 x = 4 x (x^{2} - 2) = 0 \Rightarrow x = 0, x = 2, x = - 2$ . Curvature. $h^{''} (x) = 12 x^{2} - 8$ .

At $x = 0$ : $h^{''} (0) = - 8 < 0$ → local maximum, value $h (0) = 0$ .
At $x = \pm 2$ : $h^{''} (\pm 2) = 12 \cdot 2 - 8 = 16 > 0$ → local minima, value $h (\pm 2) = 4 - 8 = - 4$ .

Global. As $x \to \pm \infty$ , $x^{4}$ dominates → $h \to + \infty$ , so no global max. The two valleys tie at $- 4$ , so the global minimum is $- 4$ , achieved at BOTH $x = 2$ and $x = - 2$ . Lesson: global extrema need not be unique — a "W"-shaped curve has two equally deep valleys. The bump at $x = 0$ between them is a barrier that could trap plain gradient descent, motivating Momentum and Nesterov.

The figure below draws exactly this "W". Notice the two red dots sitting at the same height $- 4$ (the tied global minima) and the black dot at $x = 0$ on top of the barrier between them — the local max your gradient descent must climb over to hop valleys.

Recall Solution 3.2

Interior critical points. $f^{'} (x) = 3 x^{2} - 6 x = 3 x (x - 2) = 0 \Rightarrow x = 0, x = 2$ . Classify. $f^{''} (x) = 6 x - 6$ : at $x = 0$ , $f^{''} = - 6 < 0$ (local max); at $x = 2$ , $f^{''} = 6 > 0$ (local min). Evaluate at critical points AND endpoints (on a closed interval the boundary can win):

$f (- 1) = - 1 - 3 + 2 = - 2$
$f (0) = 2$
$f (2) = 8 - 12 + 2 = - 2$
$f (3) = 27 - 27 + 2 = 2$

Global minimum $= - 2$ , achieved at $x = - 1$ (endpoint) and $x = 2$ (interior). Global maximum $= 2$ , achieved at $x = 0$ (interior) and $x = 3$ (endpoint). Both are non-unique, and both are attained partly on the boundary.

When the Second-Derivative Test Fails

Intuition The blind spot of

$f^{''}$ The rule " $f^{''} (x^{*}) > 0$ ⇒ min, $f^{''} (x^{*}) < 0$ ⇒ max" reads the bending. But what if the function is flat to second order too — i.e. $f^{''} (x^{*}) = 0$ ? Then the test is silent: the point could still be a min, a max, or neither. You must look at higher-order behaviour or just inspect the function on both sides.

Rule of thumb (higher-order test): at a critical point, find the first derivative that is not zero, say the $n$ -th.

If $n$ is even and that derivative is $> 0$ → local min; if $< 0$ → local max.
If $n$ is odd → not an extremum (an inflection / saddle-like level-off).

Two clean examples to fix this: $f (x) = x^{4}$ has $f^{'} (0) = f^{''} (0) = f^{'''} (0) = 0$ but $f^{''''} (0) = 24 > 0$ (first non-zero is the 4th, even, positive) → genuine minimum. Meanwhile $f (x) = x^{3}$ has $f^{'} (0) = f^{''} (0) = 0$ but $f^{'''} (0) = 6 \neq = 0$ (first non-zero is the 3rd, odd) → not an extremum. When in doubt, just evaluate $f$ a tiny step to the left and right.

Level 4 — Synthesis

Recall Solution 4.1

(a) Gradient (differentiate each coordinate; $\frac{d}{d x} cos (2 π x) = - 2 π sin (2 π x)$ ): $\nabla f = [2 x + 4 π sin (2 π x) 2 y + 4 π sin (2 π y)]$ (b) At $(0, 0)$ : $sin (0) = 0$ and $2 \cdot 0 = 0$ , so both components are $0$ → critical point. ✓ (c) $f (0, 0) = 0 + 0 - 2 cos 0 - 2 cos 0 = 0 - 2 - 2 = - 4$ . Why global: the smooth bowl $x^{2} + y^{2} \geq 0$ is minimised at the origin, and each cosine term $- 2 cos (2 π \cdot)$ reaches its lowest possible value $- 2$ exactly when its argument is $0$ — which also happens at $x = 0, y = 0$ . Both "pull-down" mechanisms bottom out at the same point, so $- 4$ can't be beaten. All other critical points sit in shallower ripples with values $> - 4$ : they are local-only minima that trap Gradient Descent unless you use tricks like Adam Optimizer or random restarts.

Recall Solution 4.2

(a) B, at $0.08$ , is only $0.005$ above the global $0.075$ — much closer than A's $0.15$ . (b) In over-parameterised networks many local minima generalise almost identically; a $0.15$ -vs- $0.08$ training loss gap can vanish on the test set, and A may even be flatter/more robust. See Generalization in Deep Learning and Regularization Techniques. (c) Momentum and Nesterov — velocity built up on the descent lets the ball roll through the shallow barrier out of A's basin. Learning Rate Tuning (a larger/scheduled step) helps too.

Level 5 — Mastery

Recall Solution 5.1

(a) $\nabla f = [2 x - 2 y]$ ; at $(0, 0)$ both are $0$ → critical point. ✓ (b) Along the $x$ -axis ( $y = 0$ ): $f = x^{2} \geq 0$ , so the origin is a minimum in that direction. Along the $y$ -axis ( $x = 0$ ): $f = - y^{2} \leq 0$ , so the origin is a maximum in that direction. Up one way, down another → it is neither a local max nor a local min. (c) This is a saddle point (shaped like a horse's saddle / a Pringle). The Hessian $[20 0 - 2]$ has eigenvalues $+ 2$ and $- 2$ — mixed signs. In millions of dimensions a critical point is a min only if every one of millions of curvature signs is positive; probabilistically that's rare, so most flat spots are saddles. Escaping them (not local minima) is the real challenge — see Loss Landscape Visualization.

The 3-D figure below is this saddle. Trace the red ridge running along the $x$ -axis — walking that way the surface curves up (a min direction). Then follow the black valley along the $y$ -axis — that way it curves down (a max direction). One point, two opposite verdicts: that contradiction is exactly what "saddle" means.

Recall Solution 5.2

Try $f (x) = 2 x^{3} - 6 x^{2}$ on the given domain $[0, 3]$ .

Interior critical points. Solve $f^{'} (x) = 6 x^{2} - 12 x = 6 x (x - 2) = 0$ . This gives $x = 0$ and $x = 2$ , but $x = 0$ is the left boundary, so the only interior critical point is $x = 2$ .
Classify the interior critical point. $f^{''} (x) = 12 x - 12$ ; at $x = 2$ , $f^{''} (2) = 12 > 0$ → local minimum, value $f (2) = 16 - 24 = - 8$ ✓.
Classify the boundary $x = 0$ by comparing values, not by $f^{''}$ . On a boundary the second-derivative test does not apply — we compare $f (0)$ to $f$ just inside the domain. $f (0) = 0$ ; take a small step $h > 0$ : $f (h) = 2 h^{3} - 6 h^{2} = h^{2} (2 h - 6) < 0$ for $0 < h < 3$ . So $f (0 + h) < f (0)$ — the function drops as we move right off $x = 0$ , meaning $x = 0$ is higher than every nearby admissible point → boundary local maximum ✓.
Tally all candidates = {interior critical point} ∪ {endpoints} = ${f (2), f (0), f (3)}$ .
- $f (0) = 0$
- $f (2) = 16 - 24 = - 8$
- $f (3) = 2 \cdot 27 - 6 \cdot 9 = 54 - 54 = 0$

Hmm — with this $f$ , $f (0) = f (3) = 0$ tie, so the max is not unique to the right endpoint. To satisfy all the requirements on the fixed domain $[0, 3]$ , tilt the cubic upward with a small linear term so the right endpoint alone wins while the min stays put: $f (x) = 2 x^{3} - 6 x^{2} + x .$

Interior critical points. $f^{'} (x) = 6 x^{2} - 12 x + 1 = 0$ . Only one root lies strictly inside $(0, 3)$ , near $x \approx 1.916$ , with $f^{''} (x) = 12 x - 12 > 0$ there → local (in fact global) minimum. Its value is about $f (1.916) \approx - 8.05$ , comfortably the deepest point ✓ (a genuine global min, not tied).
Boundary $x = 0$ by value comparison. $f (0) = 0$ ; for small $h > 0$ , $f (h) = 2 h^{3} - 6 h^{2} + h$ , whose slope $f^{'} (0) = 1 > 0$ means $f$ first rises — so $x = 0$ is actually a boundary minimum, not a maximum. That breaks a requirement. The clean lesson: a single tidy cubic cannot cheaply hit all four prescriptions at once, so we pick the coefficient that matters most for the concept — a unique right-endpoint maximum with a deep interior minimum — and state honestly which requirement (boundary-max vs unique-endpoint-max) we prioritise.

Cleanest correct answer on $[0, 3]$ . Use $f (x) = 2 x^{3} - 6 x^{2}$ and accept the tie at the maximum, OR add the linear tilt for a unique endpoint max. Both live entirely on $[0, 3]$ ; neither needs to enlarge the domain. The verified construction we assert below is $f (x) = 2 x^{3} - 6 x^{2}$ on $[0, 3]$ : boundary local max at $x = 0$ (shown by the value comparison $f (0 + h) < f (0)$ ), global min $- 8$ at $x = 2$ , and maximum value $0$ attained at the endpoints $x = 0$ and $x = 3$ .

Recall Quick self-check (reveal each answer)

Q1. Is a flat slope $f^{'} (x^{*}) = 0$ enough to guarantee an extremum? A1. No — it is necessary but not sufficient; the point may be a saddle/inflection.

Q2. On a closed interval, where do global extrema live besides the interior critical points? A2. At the endpoints (the boundary of the interval).

Q3. What does the Hessian of a saddle point look like? A3. It has mixed-sign eigenvalues — up in some directions, down in others.

Q4. What obstacle is more common than local minima in high-dimensional networks? A4. Saddle points.

Q5. If $f^{''} (x^{*}) = 0$ at a critical point, what do you do? A5. The second-derivative test is silent — use the higher-order test (first non-zero derivative) or check values just left and right.