JEE Mains Maths · Application of Derivatives

Functions Built from Their Extrema

Working backwards: finding a polynomial, or a parameter, from where its maxima and minima are and from a limit or a value it must take.

All Application of Derivatives notes JEE Mains Maths strategy

Why this matters

Fourteen PYQs, eleven of them multiple choice. Nine build a polynomial from the points where it has extrema together with a limit or a value; five find a parameter from where a cubic's maximum and minimum sit. Two ideas cover the page.

Concept 1 of 2: A polynomial from its extrema

An extremum at

a

gives

f'(a)=0

. When the extrema are all known, write

f'

first —

f'(x)=k(x-a)(x-b)

— and integrate. A limit like

\lim_{x\to0}\frac{f(x)}{x^3}=c

says the terms below

x^3

are missing and the

x^3

coefficient is

c

. Then use the remaining values to fix the constants.

Definition

Extremum at $a$ : $f'(a)=0$ .
Known extrema $a,b$ : $f'(x)=k(x-a)(x-b)$ .
$\lim_{x\to0}\frac{f(x)}{x^k}=c\neq0$ : no terms below $x^k$ , and the $x^k$ coefficient is $c$ .
An odd polynomial has $f(-x)=-f(x)$ .

Start from the derivative

f'(x)=k(x-a)(x-b)\ \Rightarrow\ f(x)=k\left(\tfrac{x^3}3-\tfrac{a+b}2x^2+abx\right)+C

Worked example

A cubic has extrema at

x=0

and

x=2

, with

f(0)=1

and

f(2)=-3

. Find it.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q68Moderate

Example 1 · Application of Derivatives · Functions Built from Their Extrema

Let

f(x)

be a polynomial of degree 5 and have extrema at

x= 1

and

x= - 1

. If

\lim_{x\rightarrow 0} \left( \frac{f(x)}{x^{3}} \right)= - 5

, then

f(2) - f( - 2)

is equal to :

Use every condition once

Count unknowns and conditions: each extremum gives one equation, the limit fixes several coefficients at once, and a value gives one more. A condition used twice leaves a constant undetermined.

Drill 8 more on a polynomial from its extrema

Concept 2 of 2: Where a cubic's maximum and minimum sit

For a cubic with positive leading coefficient,

f'

is an upward parabola:

f'

is positive, then negative, then positive. So the smaller root of

f'

is the maximum and the larger is the minimum; a negative leading coefficient swaps them. Conditions such as 'maximum at a negative

x

, minimum at a positive

x

' become conditions on the roots of

f'

, handled by their sum and product.

Definition

Leading coefficient $>0$ : smaller root of $f'$ = maximum, larger = minimum.
Two extrema exist when $f'$ has two distinct real roots.
Roots of opposite sign: product of the roots of $f'$ is negative.

Order of the extrema

f(x)=ax^3+\dots,\ a>0:\ x_{\max}<x_{\min}

Worked example

For

a>0

f(x)=x^3-3a^2x

, and the local maximum exceeds the local minimum by 32. Find

a

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q126Moderate

Example 2 · Application of Derivatives · Functions Built from Their Extrema

Let

a > 0

. If the function

f(x) = 6x^{3}- 45ax^{2}+ 108a^{2}x + 1

attains its local maximum and minimum values at the points

x_{1}

and

x_{2}

respectively such that

x_{1}x_{2}=

54 , then

a +x_{1}+x_{2}

is equal to :-

Check the sign of the leading term

With a negative leading coefficient, or a parameter that may be negative, the smaller root of

f'

is the minimum. Settle the sign before naming which root is which.

Drill 4 more on where a cubic's maximum and minimum sit

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)

A polynomial from its extrema
Start from the derivative
$f'(x)=k(x-a)(x-b)\ \Rightarrow\ f(x)=k\left(\tfrac{x^3}3-\tfrac{a+b}2x^2+abx\right)+C$
Where a cubic's maximum and minimum sit
Order of the extrema
$f(x)=ax^3+\dots,\ a>0:\ x_{\max}<x_{\min}$

Watch out for (2)

Use every condition once
→ A polynomial from its extrema
Check the sign of the leading term
→ Where a cubic's maximum and minimum sit

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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