datarekha

Maxima, Minima & the 2nd-Derivative Test

Once you have a critical point, the second derivative tells you its type: concave up is a valley, concave down is a peak. The single most-tested calculus idea in GATE DA.

8 min read Intermediate GATE DA Lesson 43 of 122

What you'll learn

  • The second-derivative test: at a critical point, `f'' > 0` is a local min, `f'' < 0` is a local max
  • `f''(x*) = 0` is inconclusive — it can be a min, a max, or neither
  • Local vs global: a local extremum need not be the highest or lowest value overall
  • Classifying `x³ − 3x` with `f'' = 6x` — the standard GATE drill, and a real 2024 question

Before you start

Last lesson left you with a list of suspects — the flat spots where f'(x) = 0 — and a closing question: must you really probe both sides of every one? There is a shortcut, and it is the single number that lesson hinted at. Just look at the curvature at the point.

That is all the second derivative is. f'' tells you whether the curve is bending up like a bowl or down like a dome. A bowl with a flat bottom is a minimum; a dome with a flat top is a maximum. One sign, one answer — and it is the calculus idea GATE asks most often of all.

The same curvature check scales up: in machine learning the multivariable version (the Hessian) is how Newton’s method and convexity tests decide whether a flat point is a usable minimum, so this sign-of-f'' instinct is one you will lean on for real models too.

The second-derivative test

At a critical point x* where f'(x*) = 0:

f''(x*) > 0   ⇒   concave up (bowl)   ⇒   LOCAL MINIMUM
f''(x*) < 0   ⇒   concave down (dome)  ⇒   LOCAL MAXIMUM
f''(x*) = 0   ⇒   INCONCLUSIVE  (could be min, max, or neither)

The first two lines are the workhorses. The third is the trap, and we return to it below.

local minf″ > 0concave up (bowl)local maxf″ < 0concave down (dome)
At a flat critical point, positive curvature cradles a minimum and negative curvature caps a maximum.

Take x³ − x as a first taste of the test in action. Its derivative f'(x) = 3x² − 1 is zero at x = ±1/√3 ≈ ±0.58, and the second derivative is f''(x) = 6x. Plug each critical point straight in:

f''(−0.58) = 6(−0.58) ≈ −3.5  < 0   ⇒  concave down, a dome  ⇒  LOCAL MAX on the left
f''(+0.58) = 6(+0.58) ≈ +3.5  > 0   ⇒  concave up,  a bowl  ⇒  LOCAL MIN on the right

One substitution each, and the curvature has spoken — the left critical point is the maximum, matching (one hopes) the guess you locked in.

Local vs global

Read the test’s verdict carefully: it certifies a local extremum — the highest or lowest point in a small neighbourhood. It says nothing about the function as a whole. A valley can sit above a deeper valley elsewhere; a hilltop can be dwarfed by a taller hill further along. A global (or absolute) extremum is the largest or smallest value over the entire domain. So f''(x*) > 0 guarantees a local min — never automatically a global one. Finding global extrema needs the closed-interval method of the next lesson.

How GATE asks this

The classic is a one-line MCQ/NAT: “if f'(x*) = 0 and f''(x*) > 0, then f has a ___ at x*” — where the tempting wrong answer is global minimum. Or it hands you a cubic and asks you to classify its critical points. The drill never varies: solve f'(x) = 0 for the candidates, then plug each into f'' and read the sign.

Worked example — a real GATE DA 2024 question

f is twice differentiable, f'(x*) = 0, and f''(x*) > 0. What can you conclude about f at x*?

This is GATE DA 2024. By the second-derivative test, a flat point with positive curvature is concave up — a bowl — so:

f'(x*) = 0   and   f''(x*) > 0    ⇒    f has a LOCAL MINIMUM at x*

The trap answer is global minimum. Positive curvature is a purely local statement: it describes the bowl right around x* and says nothing about how low f might dip elsewhere. The correct GATE answer is local minimum — and the word “local” is the whole point of the question.

Now apply the same test to classify a concrete cubic — the x³ − 3x from the last lesson, this time with the second derivative instead of a sign table:

f(x)  = x³ − 3x
f'(x) = 3x² − 3 = 0      →   x = −1  and  x = +1     (the critical points)
f''(x) = 6x

f''(−1) = 6(−1) = −6  < 0    ⇒  LOCAL MAXIMUM at x = −1   (value f(−1) = 2)
f''(+1) = 6(+1) = +6  > 0    ⇒  LOCAL MINIMUM at x = +1   (value f(1) = −2)

One substitution per critical point settles it — no sign table needed, and both verdicts agree with the first-derivative analysis from the previous lesson. That agreement is the reassurance: two different tests, one answer.

A question to carry forward

The second-derivative test certifies a local min — the bottom of one particular valley. But a real optimisation problem rarely wants a local best; it wants the lowest point anywhere on the domain. And a local min can sit above a deeper valley elsewhere, or be undercut by the very edge of the domain, where the curve might be lower still without the slope ever going flat. Here is the thread onward: when the domain is a closed interval [a, b], how do we find the true global maximum and minimum — and why do the two endpoints suddenly demand a seat at the table alongside the critical points?

In one breath

  • Second-derivative test at a critical point x* (f'(x*) = 0): f''(x*) > 0local min (bowl), f''(x*) < 0local max (dome), f''(x*) = 0inconclusive.
  • It is one substitution per critical point — faster than the first-derivative sign table, and it agrees with it.
  • Local ≠ global: f'' > 0 proves only a local min; the function may dip lower elsewhere. Global extrema wait for the next lesson.
  • f'' = 0 decides nothing: x⁴ (min), −x⁴ (max), (neither) all have f''(0) = 0 — fall back to the sign of f'.
  • It is the most-tested calculus idea in GATE DA (the 2024 “local vs global” trap), and the scalar shadow of the Hessian used in ML.

Practice

Quick check

0/5
Q1Recall: which statements about the second-derivative test are TRUE? (select all that apply)select all that apply
Q2Recall (GATE DA 2024): f is twice differentiable with f'(x*) = 0 and f''(x*) > 0. Then at x* the function f has a:
Q3Trace: for f(x) = x³ − 3x, the local MINIMUM occurs at x = 1. What is the local minimum VALUE f(1)?numerical answer — type a number
Q4Apply: f(x) = x³ − 6x² + 9x + 2. The critical point at x = 1 is a local maximum — give its function value f(1).numerical answer — type a number
Q5Create: when the second-derivative test gives f''(x*) = 0 at a critical point, which outcomes for x* are possible? (select all that apply)select all that apply

Sign in to track your progress

Completed lessons, your XP, level, and streak save to your account — it's free and takes a few seconds.

Related lessons

Explore further

Skip to content