JEE Mains Maths · Application of Derivatives

Increasing and Decreasing Functions

Reading where a function rises or falls from the sign of its derivative, choosing a parameter so a function is monotonic, and using monotonicity to compare values.

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Why this matters

Twenty-seven PYQs, twenty-three of them multiple choice. Fourteen find where a function increases or decreases from the sign of f′; six choose a parameter so that a function is monotonic, or has no critical point; seven use the fact that f′ increases when f″ > 0, or that a monotonic function keeps the order of its inputs. Three ideas cover the page.

Concept 1 of 3: The sign of the derivative

A function increases where

f'>0

and decreases where

f'<0

. Factor

f'

, mark its zeros and the points where it or

f

is undefined, and read the sign on each interval. Zeros of

f'

at isolated points do not stop a function increasing:

x^3

increases everywhere.

Definition

$f'>0$ on an interval: $f$ strictly increasing there (isolated zeros allowed).
$f'<0$ : strictly decreasing.
Make a sign chart of $f'$ between its zeros and the gaps in the domain.
$(x^x)'=x^x(\ln x+1)$ .

Monotonicity test

f'(x)>0\ \text{on }I\ \Rightarrow\ f\ \text{increasing on }I

Worked example

Where does

f(x)=x^3-3x^2-9x

increase?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q166Moderate

Example 1 · Application of Derivatives · Increasing and Decreasing Functions

The function

f(x) =\frac{x}{x^{2}- 6x - 16}\mathbf{,x \in R - \{ - 2,8\}}

Each piece is not the union

\frac1x

decreases on

(-\infty,0)

and on

(0,\infty)

, but

\frac1{-1}<\frac11

, so it does not decrease on their union. Options often hinge on this.

Drill 13 more on the sign of the derivative

Concept 2 of 3: Parameters that make a function monotonic

'Increasing for all

x

' means

f'(x)\ge0

everywhere. When

f'

is a quadratic, that needs a positive leading coefficient and a discriminant

\le0

. On an interval, the least value of

f'

there must be

\ge0

. 'No critical point' means

f'

never vanishes: a constant plus a bounded term must never reach 0.

Definition

$ax^2+bx+c\ge0$ for all $x$ : $a>0$ and $b^2-4ac\le0$ .
Monotonic on $I$ : $\min_I f'\ge0$ (or $\max_I f'\le0$ ).
$f'=k+g$ with $|g|\le M$ : no zero when $|k|>M$ .

Quadratic never negative

ax^2+bx+c\ge0\ \forall x\ \Leftarrow\ a>0,\ b^2-4ac\le0

Worked example

For which

k

x^3+kx^2+3x

increasing for all

x

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q170Moderate

Example 2 · Application of Derivatives · Increasing and Decreasing Functions

Let

\lambda^{*}

be the largest value of

\lambda

for which the function

f_{\lambda}(x) = 4\lambda x^{3}- 36\lambda x^{2}+ 36x + 48

is increasing for all

x \in R

. Then

f_{\lambda}*(1) +f_{\lambda}*( - 1)

is equal to:

Include the boundary value

At the boundary value

f'

touches 0 at one point, and the function still increases. So 'increasing for all x' usually gives a closed condition like

\lambda\le\frac13

Drill 5 more on parameters that make a function monotonic

Concept 3 of 3: Increasing derivatives and order

f''>0

, then

f'

is increasing, so

f'(u)>f'(v)

exactly when

u>v

. That settles functions like

g(x)=f(x)+f(1-x)

g'(x)=f'(x)-f'(1-x)

is negative when

x<1-x

. Likewise, an increasing

f

keeps order —

f(u)>f(v)

exactly when

u>v

— which turns an inequality between values into one between inputs.

Definition

$f''>0$ : $f'$ increasing, so compare $f'(u)$ and $f'(v)$ by comparing $u$ and $v$ .
$f$ increasing: $f(u)>f(v)$ exactly when $u>v$ ; decreasing reverses it.
$\frac{\ln x}x$ decreases for $x>e$ , which orders powers like $a^b$ and $b^a$ .

Symmetric sum

g(x)=f(x)+f(c-x):\ g'(x)=f'(x)-f'(c-x)

Worked example

f''>0

(0,2)

and

g(x)=f(x)+f(2-x)

. Where does

g

decrease?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q159Moderate

Example 3 · Application of Derivatives · Increasing and Decreasing Functions

Let

g(x) = f(x) + f(1 - x)

and

f^{''}(x) > 0,x \in (0,1)

. If

g

is decreasing in the interval

(0,\alpha)

and increasing in the interval

(\alpha,1)

, then

\tan^{1}(2\alpha) +\tan^{- 1}\left( \frac{1}{\alpha} \right)+\tan^{- 1}\left( \frac{\alpha + 1}{\alpha} \right)

is equal to:

f″ > 0 is about f′, not f

f''>0

makes

f'

increase;

f

itself can still fall wherever

f'<0

. Keep track of which function the sign statement is about.

Drill 6 more on increasing derivatives and order

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 (3)

The sign of the derivative
Monotonicity test
$f'(x)>0\ \text{on }I\ \Rightarrow\ f\ \text{increasing on }I$
Parameters that make a function monotonic
Quadratic never negative
$ax^2+bx+c\ge0\ \forall x\ \Leftarrow\ a>0,\ b^2-4ac\le0$
Increasing derivatives and order
Symmetric sum
$g(x)=f(x)+f(c-x):\ g'(x)=f'(x)-f'(c-x)$

Watch out for (3)

Each piece is not the union
→ The sign of the derivative
Include the boundary value
→ Parameters that make a function monotonic
f″ > 0 is about f′, not f
→ Increasing derivatives and order

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