JEE Mains Maths · Application of Derivatives

Local Maxima and Minima

Finding and counting the local maxima and minima of a function from the sign changes of its derivative, including at corners and cusps where the derivative does not exist.

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Why this matters

Nineteen PYQs, thirteen of them multiple choice, and three from 2026. Twelve find or count local extrema from where f′ changes sign — often f is defined by an integral, so f′ comes straight from the integrand; seven have corners or cusps, from a modulus, a piecewise rule or a fractional power, where f′ does not exist. Two ideas cover the page.

Concept 1 of 2: Sign changes of the derivative

A local maximum is where

f'

changes from

+

-

, a local minimum where it changes from

-

+

. A factor

(x-a)^n

f'

changes sign at

a

only when

n

is odd. When

f(x)=\int_0^{g(x)}h(t)\,dt

, the derivative is

h(g(x))\,g'(x)

, so the extrema come from the zeros of

h

g(x)

and of

g'

Definition

Maximum: $f'$ goes $+\to-$ ; minimum: $-\to+$ .
$(x-a)^n$ in $f'$ : a sign change only for odd $n$ .
$\frac{d}{dx}\int_a^{g(x)}h(t)\,dt=h(g(x))\,g'(x)$ .
Second-derivative test: $f'(a)=0$ , $f''(a)<0$ gives a maximum.

Derivative of an integral

\frac{d}{dx}\int_a^{g(x)}h(t)\,dt=h\big(g(x)\big)\,g'(x)

Worked example

Find the local extrema of

f(x)=\int_0^x t(t-2)^2\,dt

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q133Moderate

Example 1 · Application of Derivatives · Local Maxima and Minima

Let

f(x) =\int_{0}^{x^{2}} \frac{t^{2}- 8t + 15}{e^{t}}dt,x \in R

. Then the numbers of local maximum and local minimum points of

f

, respectively, are :

Even powers do not change sign

A zero of

f'

from a factor like

(t-4)^6

is a critical point but not an extremum. Count only the zeros where the sign actually flips.

Drill 11 more on sign changes of the derivative

Concept 2 of 2: Corners, cusps and piecewise functions

A critical point is also a point where

f

is defined but

f'

is not: the corner of

|x|

, the cusp of

x^{2/3}

, the join of a piecewise rule. Decide such a point by the sign of

f'

on either side, or by comparing values. For

|g|

, each zero of

g

(where

g

changes sign) is a minimum with value 0.

Definition

Critical points: $f'=0$ or $f'$ undefined, with $f$ defined there.
$|g|$ : zeros of $g$ are minima; extrema of $g$ stay extrema of $|g|$ .
Piecewise: check the value at the join against both sides.
A corner is an extremum only if the function turns there.

Critical points

f'(c)=0\ \text{or}\ f'(c)\ \text{does not exist}\ (c\in\text{domain})

Worked example

Find the local extrema of

|x^2-4|

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q155Moderate

Example 2 · Application of Derivatives · Local Maxima and Minima

The function

f(x) = 2x + 3(x)^{\frac{2}{3}},x \in R

, has

A corner need not be an extremum

If the function falls on both sides of a corner, the corner is not an extremum. Check the direction on each side instead of counting every kink.

Drill 6 more on corners, cusps and piecewise functions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Sign changes of the derivative
Derivative of an integral
$\frac{d}{dx}\int_a^{g(x)}h(t)\,dt=h\big(g(x)\big)\,g'(x)$
Corners, cusps and piecewise functions
Critical points
$f'(c)=0\ \text{or}\ f'(c)\ \text{does not exist}\ (c\in\text{domain})$

Watch out for (2)

Even powers do not change sign
→ Sign changes of the derivative
A corner need not be an extremum
→ Corners, cusps and piecewise functions

Test yourself on Application of Derivatives

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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