Monotonocity

Q: How do you find intervals of increase and decrease of a function?

Find the domain and f'(x), factorise f', mark its zeros and the points where it is undefined, and make a sign chart. Where f' > 0 the function increases; where f' < 0 it decreases. Include end points where f is continuous.

Q: Is x cubed strictly increasing even though its derivative is zero at 0?

Yes. f'(x) = 3x² is positive everywhere except at the single point x = 0. A derivative that vanishes only at isolated points does not stop a function from being strictly increasing; it would have to be zero on a whole interval.

Q: What does it mean for a function to be increasing at a point?

f is increasing at x = a if f(a - h) 0. For a differentiable function, f'(a) > 0 is enough; if f'(a) = 0, check the sign of f' on both sides of a.

Q: How is monotonicity used to prove inequalities?

To show f(x) > g(x) on an interval, let h = f - g, find a point where h = 0 (often an end point), and show h' has a fixed sign. If h' is unclear, use h'' to fix the sign of h' first.

MathsApplication Of DerivativesFor JEE aspirants

Monotonicity describes whether a function keeps rising or keeps falling as $x$ increases. The sign of $f^{'} (x)$ settles it: $f^{'} > 0$ means increasing, $f^{'} < 0$ means decreasing. With this one test we find intervals of increase and decrease, decide behaviour at a single point, find parameters that make a function monotonic, prove inequalities and compare numbers like $e^{π}$ and $π^{e}$ . Monotonicity, with concavity and points of inflection, is a regular JEE Main and JEE Advanced topic.

On this page1Definitions2Derivative test3Parameters4At a point5Inequalities6Concavity7Examples

Key Formulas - Quick Reference

Monotonically increasing (non-decreasing) on $S$ : $x_{1} < x_{2} \Rightarrow f (x_{1}) \leq f (x_{2})$ ; strictly increasing: $f (x_{1}) < f (x_{2})$
★ Must learn $f^{'} (x) > 0$ on an interval $\Rightarrow$ $f$ strictly increasing; $f^{'} (x) < 0$ $\Rightarrow$ strictly decreasing
★ Must learn $f^{'} (x) \geq 0$ with $f^{'} (x) = 0$ only at isolated points $\Rightarrow$ still strictly increasing (e.g. $x^{3}$ , $x - sin x$ )
At a point: $f^{'} (a) > 0$ $\Rightarrow$ increasing at $a$ ; $f^{'} (a) = 0$ $\Rightarrow$ look at the sign of $f^{'}$ on both sides
★ Must learnMonotonic for all $x$ : need $f^{'} (x) \geq 0$ (or $\leq 0$ ) for all $x$ ; for a quadratic $f^{'}$ : leading coefficient $> 0$ and $D \leq 0$
Inequality $f (x) > g (x)$ on $(a, b)$ : show $h = f - g$ is increasing and $h (a) \geq 0$
★ Must learn $f^{''} (x) > 0$ : concave up (tangents below, chords above); $f^{''} (x) < 0$ : concave down
Point of inflection: $f^{''}$ changes sign (usually $f^{''} (c) = 0$ , or $f^{''} (c)$ does not exist)

1. Monotonic Functions

Let $f$ be a real function with domain $D$ , and let $S \subseteq D$ . As $x$ moves to the right in $S$ , the values $f (x)$ may never go down, never go up, or do both.

Name	For all $x_{1} < x_{2}$ in $S$	Example
Monotonically increasing (non-decreasing)	$f (x_{1}) \leq f (x_{2})$	$[x]$ on $R$
Strictly increasing	$f (x_{1}) < f (x_{2})$	$x^{3}$ , $e^{x}$
Monotonically decreasing (non-increasing)	$f (x_{1}) \geq f (x_{2})$	$- [x]$
Strictly decreasing	$f (x_{1}) > f (x_{2})$	$e^{- x}$ , $cot x$ on $(0, π)$

Strictly increasing $\Rightarrow$ monotonically increasing, but not conversely. The same holds for decreasing.
A constant function is both monotonically increasing and monotonically decreasing.
$f$ is called increasing (decreasing) when it is increasing (decreasing) on its whole domain, and monotonic when it is either one.
If $f$ increases on one part of $S$ and decreases on another, it is non-monotonic on $S$ .

Figure 1: Two functions that never go down but are not strictly increasing. Left: rising pieces with flat parts on

(c, d)

and

(e, b)

, where

f^{'} (x) = 0

on whole intervals. Right:

f (x) = [x]

is constant on each

[n, n + 1)

. Both are monotonically increasing (non-decreasing), not strictly increasing.

Strictly increasing

$x_{1} < x_{2} \Rightarrow f (x_{1}) < f (x_{2})$ . The graph rises everywhere; no flat stretch.

Monotonically increasing

$x_{1} < x_{2} \Rightarrow f (x_{1}) \leq f (x_{2})$ . Flat stretches allowed, as in $[x]$ .

2. Derivative Test for Monotonicity

Let $I$ be an interval (open, closed or half-open) on which $f$ is differentiable.

If $f^{'} (x) > 0$ for all $x \in I$ , then $f$ is strictly increasing on $I$ .
If $f^{'} (x) < 0$ for all $x \in I$ , then $f$ is strictly decreasing on $I$ .

These follow from LMVT: $f (x_{2}) - f (x_{1}) = (x_{2} - x_{1}) f^{'} (c)$ has the sign of $f^{'}$ . Moreover, zeros of $f^{'}$ at isolated points do not spoil strictness: if $f^{'} (x) > 0$ on $I$ except at countably many points where $f^{'} (x) = 0$ , then $f$ is still strictly increasing on $I$ .

Figure 2:

f (x) = x^{3} - 3 x + 2

f^{'} (x) = 3 (x - 1) (x + 1)

. Short tangent segments tilt upward where

f^{'} > 0

(green bands:

x < - 1

and

x > 1

) and downward where

f^{'} < 0

(red band:

- 1 < x < 1

). So

f

increases on

(- \infty, - 1]

and

[1, \infty)

and decreases on

[- 1, 1]

What matters is whether $f^{'} = 0$ on a whole interval. For $x^{3}$ and $x - sin x$ the zeros are isolated; for the graph in Figure 1 (left) $f^{'} = 0$ on $(c, d)$ and $(e, b)$ , so that function is only non-decreasing.

Figure 3: Left:

f (x) = x^{3}

has

f^{'} (0) = 0

, yet it is strictly increasing on

R

. Right:

f (x) = x - sin x

has

f^{'} (x) = 1 - cos x = 0

x = 0, \pm 2 π, \pm 4 π, \dots

; these points are isolated, so

f

is still strictly increasing. It winds around the line

y = x

, touching it at

x = nπ

2.1 Finding intervals of increase and decrease

Find the domain of $f$ and compute $f^{'} (x)$ ; factorise it.
Mark the zeros of $f^{'}$ and the points where $f^{'}$ or $f$ is undefined. They split the domain into intervals.
Find the sign of $f^{'}$ on each interval (a sign chart, or "wavy curve"): signs alternate across simple factors and do not change across even powers.
$f^{'} > 0$ : increasing; $f^{'} < 0$ : decreasing. If $f$ is continuous at an end point, include it (closed brackets).

Figure 4:

f (x) = x^{2} (x - 2)^{2}

with the sign chart of

f^{'} (x) = 4 x (x - 1) (x - 2)

drawn under the same

x

-scale. Signs alternate

-, +, -, +

across

0, 1, 2

f

decreases on

(- \infty, 0]

and

[1, 2]

and increases on

[0, 1]

and

[2, \infty)

"Increasing on $(- \infty, - 2]$ and on $[0, \infty)$ " does not mean increasing on their union when the two pieces are separated by a break. For example $f (x) = - \frac{1}{x}$ increases on $(- \infty, 0)$ and on $(0, \infty)$ , yet $f (- 1) = 1 > f (1) = - 1$ . Write the intervals separately.

Exam Trick

Isolated zeros are allowed. In "find $a$ for which $f$ is increasing for all $x$ ", require $f^{'} (x) \geq 0$ , not $f^{'} (x) > 0$ : equality at isolated points is fine. Dropping the " $=$ " loses boundary values such as $b = 1$ in $f (x) = sin^{4} x + cos^{4} x + b x$ or $a = 3$ in $e^{2 x} - (a + 1) e^{x} + 2 x$ .

2.2 Parameters for monotonic functions

To make $f$ monotonic on $R$ we need $f^{'} (x) \geq 0$ for all $x$ (or $\leq 0$ for all $x$ ). Two standard routes:

Quadratic $f^{'}$ : $A x^{2} + B x + C \geq 0$ for all $x$ $⟺$ $A > 0$ , $D \leq 0$ (or $A = B = 0$ , $C \geq 0$ ).
Separate the parameter: write the condition as $k \leq g (x)$ for all $x$ , which means $k \leq min g$ (or $k \geq max g$ ).

Exam Trick

Separate, then use the range. For $f^{'} (x) = b - sin 4 x \geq 0$ for all $x$ : $b \geq max (sin 4 x) = 1$ . For $2 e^{2 x} - (a + 1) e^{x} + 2 \geq 0$ : divide by $e^{x}$ , so $a + 1 \leq 2 (e^{x} + e^{- x})$ for all $x$ , i.e. $a + 1 \leq 2 \times 2$ (minimum of $e^{x} + e^{- x}$ is $2$ ), giving $a \leq 3$ .

Quick Recall: tap to check

f (x) = x^{3}

strictly increasing even though

f^{'} (0) = 0

Yes.

f^{'} (x) = 3 x^{2} > 0

except at the single point

x = 0

Where is

x ln x

increasing?

f^{'} (x) = 1 + ln x \geq 0

for

x \geq \frac{1}{e}

: on

[\frac{1}{e}, \infty)

Condition for

A x^{2} + B x + C \geq 0

for all real

x

(with

A \neq = 0

A > 0

and

B^{2} - 4 A C \leq 0

Key idea

Factorise

f^{'}

, read its signs on a sign chart:

f^{'} > 0

increasing,

f^{'} < 0

decreasing; isolated zeros of

f^{'}

do not matter.

3. Monotonicity at a Point

$f$ is strictly increasing at $x = a$ if it is strictly increasing on some open interval containing $a$ ; in particular $f (a - h) < f (a) < f (a + h)$ for all small $h > 0$ . Similarly, $f$ is strictly decreasing at $a$ if $f (a - h) > f (a) > f (a + h)$ for small $h > 0$ .

The function need not be continuous at $a$ : only the order of the three values matters. If $a$ is an end point of the domain, use the one available side (a left end point: compare $f (a)$ with $f (a + h)$ ).

Figure 5: Monotonicity at a point compares

f (a)

with values just to the left and right. Panel 2:

f (a - h) > f (a) > f (a + h)

, so

f

is decreasing at

a

even with a jump. Panel 3:

f (a)

is bigger than both neighbours. Panel 4:

f (a)

is smaller than both. In the last two

f

is neither increasing nor decreasing at

a

3.1 Test for a differentiable function

At $x = a$	Conclusion
$f^{'} (a) > 0$	increasing at $a$
$f^{'} (a) < 0$	decreasing at $a$
$f^{'} (a) = 0$ , $f^{'} > 0$ on both sides	increasing at $a$ (e.g. $x^{3}$ at $0$ )
$f^{'} (a) = 0$ , $f^{'} < 0$ on both sides	decreasing at $a$
$f^{'} (a) = 0$ , $f^{'}$ changes sign	neither (a local maximum or minimum)

The test applies only when $f$ is continuous at $a$ ; for a jump, compare the values directly as in Figure 5. Try the four graphs below (answers in Solved Example 21).

Figure 6: Graphs for Solved Example 21: decide whether

f

is increasing, decreasing or neither at

x = a

by comparing

f (a)

with

f (a - h)

and

f (a + h)

for small

h > 0

(in (iii)

a

is a right end point, so only the left side is used).

Increasing on an interval

A property of a whole interval: every pair $x_{1} < x_{2}$ in it has $f (x_{1}) < f (x_{2})$ .

Increasing at a point

A local property: only values just left and right of $a$ are compared with $f (a)$ .

Key idea

At a point compare

f (a - h)

f (a)

f (a + h)

. If

f^{'} (a) = 0

, the sign of

f^{'}

on the two sides decides.

4. Using Monotonicity to Prove Inequalities

To compare $f (x)$ and $g (x)$ on an interval, study $h (x) = f (x) - g (x)$ :

Find the point of equality, usually an end point such as $x = 0$ , where $h = 0$ .
Show $h^{'} > 0$ (or $h^{'} < 0$ ) on the interval. If the sign of $h^{'}$ is not clear, differentiate again and use $h^{''}$ to settle the sign of $h^{'}$ .
Then $h (x) > h (0) = 0$ (or $< 0$ ) on the interval.

Figure 7: On

(0, \frac{π}{2})

the line

y = x

lies between

y = sin x

and

y = tan x

. All three start at the origin with slope

1

;

x - sin x

has derivative

1 - cos x > 0

and

x - tan x

has derivative

1 - sec^{2} x < 0

, so the gaps open up.

4.1 Comparing numbers

To decide which of two numbers is larger, write both as values of one function and use its monotonicity. Two functions do most of the work: $(1 + \frac{1}{x})^{x}$ and $x^{1/ x}$ .

Figure 8:

f (x) = (1 + \frac{1}{x})^{x}

is defined for

x < - 1

x > 0

and increases on each piece. It rises from

1

(at

0^{+}

) towards

e

on the right branch, and from

e

(as

x \to - \infty

) to

\infty

(as

x \to - 1^{-}

) on the left branch. Range:

(1, \infty) - {e}

Figure 9:

f (x) = x^{1/ x}

has

f^{'} (x) = x^{1/ x} \cdot \frac{1 - ln x}{x ^{2}}

: increasing on

(0, e]

, decreasing on

[e, \infty)

, with maximum

e^{1/ e} \approx 1.445

. Since

e < π

e^{1/ e} > π^{1/ π}

, so

e^{π} > π^{e}

; since

e < 100 < 101

10 0^{1/100} > 10 1^{1/101}

Exam Trick

$a^{b}$ versus $b^{a}$ . Compare $\frac{ln a}{a}$ and $\frac{ln b}{b}$ : the function $\frac{ln x}{x}$ increases on $(0, e]$ and decreases on $[e, \infty)$ . So for $e \leq a < b$ : $a^{b} > b^{a}$ . Examples: $e^{π} > π^{e}$ , $3^{4} > 4^{3}$ , $10 0^{101} > 10 1^{100}$ .

Quick Recall: tap to check

Which is larger,

e^{π}

π^{e}

e^{π}

, because

x^{1/ x}

(equivalently

\frac{ln x}{x}

) decreases for

x > e

To prove

sin x > x - \frac{x ^{3}}{6}

for

x > 0

, the sign of

h^{'} (x) = cos x - 1 + \frac{x ^{2}}{2}

is unclear. What next?

Differentiate again:

h^{''} (x) = x - sin x > 0

, so

h^{'}

increases from

h^{'} (0) = 0

What is

x \to 0 lim [\frac{tan ^{- 1} x}{x}]

0

, because

\frac{tan ^{- 1} x}{x}

tends to

1

from below.

Key idea

Inequalities:

h = f - g

, find where

h = 0

, then the sign of

h^{'}

(or

h^{''}

) carries it across the interval.

5. Concavity, Convexity and Points of Inflection

On an interval $(a, b)$ , the curve $y = f (x)$ is concave up if the tangent at every point lies below the curve, and concave down if every tangent lies above it. Concave up is also called convex (holds water, a cup); concave down is called concave (a cap). Some books use "concave" for concave up, so always say which way.

If $f^{''} (x) > 0$ for all $x \in (a, b)$ , the curve is concave up on $(a, b)$ : $f^{'}$ is increasing.
If $f^{''} (x) < 0$ for all $x \in (a, b)$ , the curve is concave down on $(a, b)$ : $f^{'}$ is decreasing.

Figure 10: Left:

y = e^{x}

has

f^{''} > 0

, so every tangent lies below the curve and every chord above it.

R

divides

P Q

in the ratio

1 : 2

, so

R

(height

\frac{2 e ^{x_{1}} + e ^{x_{2}}}{3}

) is above

S

(height

e^{(2 x_{1} + x_{2}) /3}

). Right:

y = sin x

(0, π)

has

f^{''} < 0

; tangents lie above and chords below.

5.1 Points of inflection

A point $(c, f (c))$ is a point of inflection if the curve is concave up on one side of $c$ and concave down on the other (within some $(c - δ, c + δ)$ ). At such a point the curve crosses its tangent.

If $f$ is continuous at $c$ and $f^{''}$ has opposite signs on the two sides of $c$ , then $(c, f (c))$ is a point of inflection.
If $f^{''} (c) = 0$ and $f^{'''} (c) \neq = 0$ , then $(c, f (c))$ is a point of inflection.
$f^{''} (c) = 0$ alone is not enough: $x^{4}$ has $f^{''} (0) = 0$ but is concave up on both sides. And $f^{''}$ may fail to exist at an inflection point: $x^{1/3}$ at $0$ .

Figure 11:

f (x) = 3 x^{4} - 4 x^{3}

f^{''} (x) = 12 x (3 x - 2)

changes sign at

x = 0

and

x = \frac{2}{3}

, so both are points of inflection; the curve crosses its tangent there. At

O

the tangent is horizontal (

f^{'} (0) = 0

) but there is no extremum. The minimum is at

(1, - 1)

and the roots are

0

and

\frac{4}{3}

5.2 Inequalities from concavity

On a concave-up curve every chord lies above the curve. A point dividing a chord $P Q$ in a given ratio is therefore above the curve point with the same $x$ : for $λ \in (0, 1)$ and $x_{1} \neq = x_{2}$ ,

f^{''} (x) > 0 \Rightarrow f (λ x_{1} + (1 - λ) x_{2}) < λ f (x_{1}) + (1 - λ) f (x_{2}),

with the inequality reversed when $f^{''} < 0$ .

JEE Advanced

Jensen's inequality. If $f^{''} > 0$ on an interval and $λ_{1}, \dots, λ_{n} > 0$ with $\sum λ_{i} = 1$ , then

f (\sum λ_{i} x_{i}) \leq \sum λ_{i} f (x_{i}),

with equality only when all $x_{i}$ are equal (reverse for $f^{''} < 0$ ). With $f = - ln x$ it gives AM $\geq$ GM; with $f = sin x$ on $(0, π)$ and $A + B + C = π$ it gives $sin A + sin B + sin C \leq 3 sin \frac{π}{3} = \frac{3 3}{2}$ . Centroid of points on the curve lies on the chord side: that is the whole proof.

Key idea

f^{''} > 0

: cup, tangents below, chords above. Inflection where

f^{''}

changes sign; then the curve crosses its tangent.

Figure 12: Two kinds of monotonicity questions. Intervals: factorise

f^{'}

and use a sign chart. Parameters: demand

f^{'} (x) \geq 0

(or

\leq 0

) for all

x

, keeping the equality, which is allowed at isolated points.

Figure 13: Mind map of monotonicity, inequalities and concavity.

6. Solved Examples

Solved Example 1

Let

f (x) = x^{3}

. Find the intervals of monotonicity.

Solution:

$f^{'} (x) = 3 x^{2} > 0$ for every $x$ except $x = 0$ , a single point. So $f$ is strictly increasing on $R$ (Figure 3).

Answer: strictly increasing on $R$ .

Solved Example 2

Let

f (x) = x - sin x

. Find the intervals of monotonicity.

Solution:

$f^{'} (x) = 1 - cos x \geq 0$ , with equality only at $x = 0, \pm 2 π, \pm 4 π, \dots$ . These points are isolated and do not form an interval, so $f$ is strictly increasing on $R$ (Figure 3, right).

Answer: strictly increasing on $R$ .

Solved Example 3

A function on

(a, b)

has the graph shown in Figure 1 (left): it rises, is flat on

(c, d)

, rises again and is flat on

(e, b)

. Is it increasing? Strictly increasing?

Solution:

$f^{'} (x) \geq 0$ on $(a, b)$ , but $f^{'} (x) = 0$ on the whole intervals $(c, d)$ and $(e, b)$ . There, $x_{1} < x_{2}$ gives $f (x_{1}) = f (x_{2})$ .

Answer: monotonically increasing (non-decreasing) on $(a, b)$ , but not strictly increasing.

Solved Example 4

Find the intervals in which

f (x) = x^{3} - 3 x + 2

is increasing.

Solution:

$f^{'} (x) = 3 (x^{2} - 1) = 3 (x - 1) (x + 1)$ .
$f^{'} (x) \geq 0$ $⟺$ $x \leq - 1$ or $x \geq 1$ (sign chart $+, -, +$ across $- 1, 1$ ).

Answer: increasing on $(- \infty, - 1]$ and on $[1, \infty)$ (Figure 2).

Solved Example 5

Find the intervals of monotonicity of (i)

f (x) = x^{2} (x - 2)^{2}

(ii)

f (x) = x ln x

(iii)

f (x) = sin x + cos x

x \in [0, 2 π]

Solution:

(i) $f^{'} (x) = 4 x (x - 1) (x - 2)$ ; signs $-, +, -, +$ across $0, 1, 2$ (Figure 4). Increasing on $[0, 1]$ and $[2, \infty)$ ; decreasing on $(- \infty, 0]$ and $[1, 2]$ .
(ii) Domain $x > 0$ . $f^{'} (x) = 1 + ln x \geq 0$ $⟺$ $x \geq \frac{1}{e}$ . Increasing on $[\frac{1}{e}, \infty)$ , decreasing on $(0, \frac{1}{e}]$ .
(iii) $f^{'} (x) = cos x - sin x \geq 0$ $⟺$ $cos x \geq sin x$ $⟺$ $x \in [0, \frac{π}{4}] \cup [\frac{5 π}{4}, 2 π]$ .

Answer (iii): increasing on $[0, \frac{π}{4}]$ and $[\frac{5 π}{4}, 2 π]$ ; decreasing on $[\frac{π}{4}, \frac{5 π}{4}]$ .

Solved Example 6

f (x) = [x]

is the greatest integer (step) function. Is it strictly increasing on

R

Solution:

No. For $x_{1} = 0.2$ , $x_{2} = 0.7$ : $x_{1} < x_{2}$ but $f (x_{1}) = f (x_{2}) = 0$ . Since $[x]$ never decreases, it is monotonically increasing (Figure 1, right).

Answer: non-decreasing, not strictly increasing.

Solved Example 7

f (x) = sin^{4} x + cos^{4} x + b x + c

, find the values of

b

and

c

for which

f

is monotonic for all

x \in R

Solution:

$f^{'} (x) = 4 sin^{3} x cos x - 4 cos^{3} x sin x + b = - 4 sin x cos x (cos^{2} x - sin^{2} x) + b = b - sin 4 x$ .
Increasing for all $x$ : $b \geq sin 4 x$ for all $x$ $⟺$ $b \geq 1$ .
Decreasing for all $x$ : $b \leq sin 4 x$ for all $x$ $⟺$ $b \leq - 1$ .

Answer: $b \in (- \infty, - 1] \cup [1, \infty)$ and $c \in R$ (at $b = \pm 1$ , $f^{'}$ vanishes only at isolated points).

Solved Example 8

Find the values of

a

for which

f (x) = e^{2 x} - (a + 1) e^{x} + 2 x

is monotonically increasing for all

x \in R

Solution:

$f^{'} (x) = 2 e^{2 x} - (a + 1) e^{x} + 2 \geq 0$ for all $x$ . Divide by $e^{x} > 0$ : $a + 1 \leq 2 (e^{x} + e^{- x})$ for all $x$ .
$e^{x} + e^{- x} \geq 2$ (equality at $x = 0$ ), so the right side has minimum $4$ : $a + 1 \leq 4$ .
Check by a second route: with $t = e^{x} > 0$ , $2 t^{2} - (a + 1) t + 2 \geq 0$ for all $t > 0$ holds if $D \leq 0$ , i.e. $a \in [- 5, 3]$ , or if both roots are $\leq 0$ , i.e. $a \leq - 5$ . The union is again $a \leq 3$ .

Answer: $a \in (- \infty, 3]$ .

Solved Example 9

Let

f (x) = x^{3} - 3 x + 2

. Examine the monotonicity of

f

x = 0, 1, 2

Solution:

$f^{'} (x) = 3 (x^{2} - 1)$ . $f^{'} (0) = - 3 < 0$ : decreasing at $x = 0$ .
$f^{'} (1) = 0$ . Just left of $1$ , $f^{'} < 0$ (e.g. $f^{'} (0.9) = - 0.57$ ); just right, $f^{'} > 0$ ( $f^{'} (1.1) = 0.63$ ). The sign changes, so $f$ has a local minimum at $1$ .
$f^{'} (2) = 9 > 0$ : increasing at $x = 2$ .

Answer: decreasing at $0$ , neither increasing nor decreasing at $1$ , increasing at $2$ .

Solved Example 10

For

x \in (0, \frac{π}{2})

, prove that

sin x < x < tan x

Solution:

$f (x) = x - sin x$ : $f^{'} (x) = 1 - cos x > 0$ on $(0, \frac{π}{2})$ , so $f (x) > f (0) = 0$ , i.e. $x > sin x$ .
$g (x) = x - tan x$ : $g^{'} (x) = 1 - sec^{2} x < 0$ there, so $g (x) < g (0) = 0$ , i.e. $x < tan x$ .

Answer: $sin x < x < tan x$ (Figure 7).

Solved Example 11

For

x \in (0, 1)

prove that

x - \frac{x ^{3}}{3} < tan^{- 1} x < x - \frac{x ^{3}}{6}

. Hence find

x \to 0 lim [\frac{tan ^{- 1} x}{x}]

, where

[\cdot]

is the greatest integer function.

Solution:

$f (x) = x - \frac{x ^{3}}{3} - tan^{- 1} x$ : $f^{'} (x) = 1 - x^{2} - \frac{1}{1 + x ^{2}} = - \frac{x ^{4}}{1 + x ^{2}} < 0$ , so $f (x) < f (0) = 0$ .
$g (x) = x - \frac{x ^{3}}{6} - tan^{- 1} x$ : $g^{'} (x) = 1 - \frac{x ^{2}}{2} - \frac{1}{1 + x ^{2}} = \frac{x ^{2} ( 1 - x ^{2} )}{2 ( 1 + x ^{2} )} > 0$ on $(0, 1)$ , so $g (x) > 0$ .
Dividing by $x > 0$ : $1 - \frac{x ^{2}}{3} < \frac{tan ^{- 1} x}{x} < 1 - \frac{x ^{2}}{6}$ . By the sandwich theorem the ratio tends to $1$ , but always from below (it is an even function, so the same holds for $x < 0$ ).

Answer: proved; the limit of $[\frac{tan ^{- 1} x}{x}]$ is $0$ .

Solved Example 12

For

x \in (0, \frac{π}{2})

, prove that

sin x > x - \frac{x ^{3}}{6}

Solution:

$f (x) = sin x - x + \frac{x ^{3}}{6}$ , $f^{'} (x) = cos x - 1 + \frac{x ^{2}}{2}$ : sign not obvious.
$f^{''} (x) = x - sin x > 0$ , so $f^{'}$ is increasing: $f^{'} (x) > f^{'} (0) = 0$ .
Hence $f$ is increasing: $f (x) > f (0) = 0$ .

Answer: $sin x > x - \frac{x ^{3}}{6}$ .

Solved Example 13

Which is greater on

(0, \frac{π}{2})

sin x tan x

x^{2}

? Hence evaluate

x \to 0 lim [\frac{sin x tan x}{x ^{2}}]

Solution:

$f (x) = sin x tan x - x^{2}$ ; $f^{'} (x) = sin x + sin x sec^{2} x - 2 x$ ; $f^{''} (x) = (cos x + sec x - 2) + 2 sec^{2} x sin x tan x$ .
$cos x + sec x - 2 = (cos x - sec x)^{2} \geq 0$ and $2 sec^{2} x sin x tan x > 0$ , so $f^{''} (x) > 0$ .
So $f^{'}$ increases: $f^{'} (x) > f^{'} (0) = 0$ ; then $f$ increases: $f (x) > f (0) = 0$ .

Answer: $sin x tan x > x^{2}$ . So $\frac{sin x tan x}{x ^{2}} > 1$ and tends to $1$ from above: the limit of the greatest integer is $1$ .

Solved Example 14

Prove that

f (x) = (1 + \frac{1}{x})^{x}

is increasing on its domain. Hence draw its graph and find its range.

Solution:

Domain: $1 + \frac{1}{x} = \frac{x + 1}{x} > 0$ , i.e. $x \in (- \infty, - 1) \cup (0, \infty)$ .
$f^{'} (x) = f (x) [ln (1 + \frac{1}{x}) - \frac{1}{x + 1}]$ . Since $f (x) > 0$ , the sign is that of $g (x) = ln (1 + \frac{1}{x}) - \frac{1}{x + 1}$ .
$g^{'} (x) = - \frac{1}{x ( x + 1 ) ^{2}}$ . For $x > 0$ : $g^{'} < 0$ and $g (x) > x \to \infty lim g (x) = 0$ . For $x < - 1$ : $g^{'} > 0$ and $g (x) > x \to - \infty lim g (x) = 0$ . So $f^{'} > 0$ on the domain.
Boundary values: $f \to e$ as $x \to \pm \infty$ , $f \to 1$ as $x \to 0^{+}$ , $f \to \infty$ as $x \to - 1^{-}$ .

Answer: increasing on each part of the domain; range $(1, \infty) - {e}$ (Figure 8).

Solved Example 15

Which is greater:

(100)^{1/100}

(101)^{1/101}

Solution:

Let $f (x) = x^{1/ x}$ . Then $f^{'} (x) = x^{1/ x} \cdot \frac{1 - ln x}{x ^{2}}$ .
$f^{'} (x) < 0$ for $x > e$ , so $f$ is decreasing on $[e, \infty)$ . Since $e < 100 < 101$ , $f (100) > f (101)$ .

Answer: $(100)^{1/100} > (101)^{1/101}$ (Figure 9).

Solved Example 16

Prove that for any two distinct numbers

x_{1}

and

x_{2}

\frac{2 e ^{x_{1}} + e ^{x_{2}}}{3} > e^{(2 x_{1} + x_{2}) /3}

Solution:

Take $P (x_{1}, e^{x_{1}})$ and $Q (x_{2}, e^{x_{2}})$ on $y = e^{x}$ . The point $R$ dividing $P Q$ in the ratio $1 : 2$ is $(\frac{2 x _{1} + x _{2}}{3}, \frac{2 e ^{x_{1}} + e ^{x_{2}}}{3})$ .
$S$ on the curve with the same $x$ -coordinate has height $e^{(2 x_{1} + x_{2}) /3}$ .
$y = e^{x}$ is concave up ( $y^{''} = e^{x} > 0$ ), so the chord $P Q$ lies above the curve: $R$ is above $S$ .

Answer: proved (Figure 10). The same result follows from AM $\geq$ GM applied to $e^{x_{1}}, e^{x_{1}}, e^{x_{2}}$ .

Solved Example 17

0 < x_{1} < x_{2} < x_{3} < π

, prove that

sin \frac{x _{1} + x _{2} + x _{3}}{3} > \frac{sin x _{1} + sin x _{2} + sin x _{3}}{3}

. Hence prove that if

A, B, C

are angles of a triangle, the maximum value of

sin A + sin B + sin C

\frac{3 3}{2}

Solution:

On $(0, π)$ , $y = sin x$ is concave down ( $y^{''} = - sin x < 0$ ). Take $A (x_{1}, sin x_{1})$ , $B (x_{2}, sin x_{2})$ , $C (x_{3}, sin x_{3})$ on the arc.
The centroid $G$ of triangle $A B C$ has $x$ -coordinate $\frac{x _{1} + x _{2} + x _{3}}{3}$ and $y$ -coordinate $\frac{sin x _{1} + sin x _{2} + sin x _{3}}{3}$ . It lies inside the triangle, which lies below the arc, so the point $F$ of the arc above $G$ is higher: this is the inequality.
For a triangle, $A + B + C = π$ : $\frac{sin A + sin B + sin C}{3} \leq sin \frac{π}{3} = \frac{3}{2}$ , with equality when $A = B = C$ .

Answer: maximum of $sin A + sin B + sin C$ is $\frac{3 3}{2}$ , for an equilateral triangle.

Solved Example 18

Find the points of inflection of

f (x) = sin^{2} x

x \in [0, 2 π]

Solution:

$f^{'} (x) = sin 2 x$ , $f^{''} (x) = 2 cos 2 x$ .
$f^{''} (x) = 0$ $⟺$ $2 x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2}$ $⟺$ $x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ .
$cos 2 x$ changes sign at each of these, so all four are points of inflection, each at height $sin^{2} x = \frac{1}{2}$ .

Answer: $x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ .

Solved Example 19

Find the points of inflection of

f (x) = 3 x^{4} - 4 x^{3}

and sketch the graph, showing concavity.

Solution:

$f^{'} (x) = 12 x^{3} - 12 x^{2} = 12 x^{2} (x - 1)$ ; $f^{''} (x) = 36 x^{2} - 24 x = 12 x (3 x - 2)$ .
$f^{''} (x) = 0$ at $x = 0, \frac{2}{3}$ , with signs $+, -, +$ : both are inflection points.
$f^{'}$ changes sign only at $x = 1$ (minimum $f (1) = - 1$ ); $f (0) = 0$ , $f (\frac{2}{3}) = - \frac{16}{27}$ ; roots $0$ and $\frac{4}{3}$ .

Answer: inflection points $(0, 0)$ and $(\frac{2}{3}, - \frac{16}{27})$ (Figure 11).

Solved Example 20

Find the values of

a

for which

f (x) = (a + 2) x^{3} - 3 a x^{2} + 9 a x - 1

is monotonically decreasing for all

x \in R

Solution:

$f^{'} (x) = 3 [(a + 2) x^{2} - 2 a x + 3 a] \leq 0$ for all $x$ .
$a = - 2$ gives $f^{'} (x) = 3 (4 x - 6)$ , which changes sign: rejected.
Otherwise need $a + 2 < 0$ and $\frac{D}{4} = a^{2} - 3 a (a + 2) = - 2 a (a + 3) \leq 0$ , i.e. $a \leq - 3$ or $a \geq 0$ .
Together with $a < - 2$ : $a \leq - 3$ .

Answer: $a \in (- \infty, - 3]$ .

Solved Example 21

For each graph in Figure 6, say whether

f

is increasing, decreasing or neither at

x = a

Solution:

(i) $f (a)$ is larger than the values on both sides: neither increasing nor decreasing.
(ii) $f (a - h) > f (a) > f (a + h)$ : decreasing at $a$ .
(iii) $a$ is a right end point and $f (a - h) > f (a)$ : decreasing at $a$ .
(iv) $f (a - h) < f (a) < f (a + h)$ : increasing at $a$ (the corner does not matter).

Answer: (i) neither (ii) decreasing (iii) decreasing (iv) increasing.

Solved Example 22

Let

f (x) = x

for

0 \leq x \leq 1

and

f (x) = [x]

for

1 \leq x \leq 2

. Comment on the monotonic behaviour of

f

x = 0, 1, 2

. Is

f

monotonically increasing on

[0, 2]

Solution:

$f = x$ on $[0, 1]$ , $f = 1$ on $[1, 2)$ , $f (2) = 2$ .
$x = 0$ (left end): $f (0 + h) = h > f (0)$ : increasing.
$x = 1$ : $f (1 - h) = 1 - h < f (1) = 1 = f (1 + h)$ : not strictly increasing and not decreasing, so neither.
$x = 2$ (right end): $f (2 - h) = 1 < f (2) = 2$ : increasing.
On $[0, 2]$ , $x_{1} < x_{2}$ always gives $f (x_{1}) \leq f (x_{2})$ .

Answer: increasing at $0$ and $2$ , neither at $1$ . On $[0, 2]$ it is monotonically increasing (non-decreasing) but not strictly increasing.

Solved Example 23

Which is greater:

\frac{1 + e ^{2}}{e}

\frac{1 + π ^{2}}{π}

Solution:

Both are values of $g (x) = x + \frac{1}{x}$ : $g (e)$ and $g (π)$ .
$g^{'} (x) = 1 - \frac{1}{x ^{2}} > 0$ for $x > 1$ , so $g$ is increasing there. Since $1 < e < π$ , $g (e) < g (π)$ .

Answer: $\frac{1 + π ^{2}}{π}$ is greater ( $\approx 3.460$ against $\approx 3.086$ ).

Solved Example 24

f

is monotonically decreasing,

f^{''} (x) > 0

and

f^{- 1}

exists, prove that

\frac{f ^{- 1} ( x _{1} ) + f ^{- 1} ( x _{2} )}{2} > f^{- 1} (\frac{x _{1} + x _{2}}{2})

for distinct

x_{1}, x_{2}

in the range of

f

Solution:

Let $g = f^{- 1}$ . Then $g^{'} (y) = \frac{1}{f ^{'} ( x )}$ with $x = g (y)$ , so $g^{'} < 0$ .
$g^{''} (y) = - \frac{f ^{''} ( x )}{( f ^{'} ( x ) ) ^{3}}$ . Here $f^{''} > 0$ and $(f^{'})^{3} < 0$ , so $g^{''} (y) > 0$ : $g$ is concave up.
On a concave-up curve the midpoint of a chord lies above the curve: $\frac{g ( x _{1} ) + g ( x _{2} )}{2} > g (\frac{x _{1} + x _{2}}{2})$ .

Answer: proved.

Solved Example 25

Find the intervals in which

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 15

is increasing or decreasing.

Solution:

$f^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$ .
Signs $+, -, +$ across $1, 2$ .

Answer: increasing on $(- \infty, 1]$ and $[2, \infty)$ ; decreasing on $[1, 2]$ .

Solved Example 26

The set of all values of

k

for which

f (x) = k x^{3} - 9 k x^{2} + 9 x + 3

is increasing on

R

is
(A)

[0, \frac{1}{3}]

(B)

(0, \frac{1}{3})

(C)

[0, \frac{1}{3})

(D)

(- \infty, \frac{1}{3}]

Solution:

Answer: (A). $f^{'} (x) = 3 k x^{2} - 18 k x + 9 \geq 0$ for all $x$ . For $k = 0$ , $f^{'} = 9 > 0$ : allowed. For $k \neq = 0$ : $k > 0$ and $D = 324 k^{2} - 108 k = 108 k (3 k - 1) \leq 0$ , so $0 < k \leq \frac{1}{3}$ (at $k = \frac{1}{3}$ , $f^{'} = (x - 3)^{2} \geq 0$ with an isolated zero). Hence $k \in [0, \frac{1}{3}]$ .

Solved Example 27

The function

f (x) = ln (1 + x) - \frac{2 x}{2 + x}

is increasing on
(A)

(- 1, \infty)

(B)

(- \infty, 0)

(C)

(- \infty, \infty)

(D)

(0, 1)

only

Solution:

Answer: (A). Domain $x > - 1$ . $f^{'} (x) = \frac{1}{1 + x} - \frac{4}{( 2 + x ) ^{2}} = \frac{x ^{2}}{( 1 + x ) ( 2 + x ) ^{2}} \geq 0$ on $(- 1, \infty)$ , zero only at $x = 0$ . So $f$ increases on the whole domain.

Solved Example 28

Find the intervals of concavity and the point of inflection of

f (x) = x^{3} - 6 x^{2} + 9 x + 1

Solution:

$f^{''} (x) = 6 x - 12 = 6 (x - 2)$ .
$f^{''} < 0$ on $(- \infty, 2)$ : concave down; $f^{''} > 0$ on $(2, \infty)$ : concave up.
$f (2) = 8 - 24 + 18 + 1 = 3$ .

Answer: concave down on $(- \infty, 2)$ , concave up on $(2, \infty)$ ; point of inflection $(2, 3)$ .

Practice Questions

Find the intervals of monotonicity of (i) $- x^{3} + 6 x^{2} - 9 x - 2$ (ii) $x + \frac{1}{x + 1}$ (iii) $x e^{x - x^{2}}$ (iv) $x - cos x$ .Answer: (i) increasing on $[1, 3]$ , decreasing on $(- \infty, 1]$ and $[3, \infty)$ (ii) increasing on $(- \infty, - 2]$ and $[0, \infty)$ , decreasing on $[- 2, - 1)$ and $(- 1, 0]$ (iii) increasing on $[- \frac{1}{2}, 1]$ , decreasing on $(- \infty, - \frac{1}{2}]$ and $[1, \infty)$ (iv) increasing on $R$
Let $f (x) = x - tan^{- 1} x$ . Prove that $f$ is monotonically increasing for $x \in R$ .Answer: $f^{'} (x) = \frac{x ^{2}}{1 + x ^{2}} \geq 0$ , zero only at $x = 0$
If $f (x) = 2 e^{x} - a e^{- x} + (2 a + 1) x - 3$ is increasing for all $x \in R$ , find the range of $a$ .Answer: $a \geq 0$
Let $f (x) = e^{2 x} - a e^{x} + 1$ . Prove that $f$ cannot be monotonically decreasing for all $x \in R$ , for any $a$ .Answer: $f^{'} (x) = e^{x} (2 e^{x} - a) > 0$ for large $x$
Let $f (x) = x^{3} - 3 x^{2} + 3 x + 4$ . Comment on the monotonic behaviour of $f$ at (i) $x = 0$ (ii) $x = 1$ .Answer: $f^{'} (x) = 3 (x - 1)^{2}$ : increasing at both points
Prove: (i) $x < - ln (1 - x)$ on $(0, 1)$ (ii) $x > tan^{- 1} x$ on $(0, \infty)$ (iii) $e^{x} > x + 1$ on $(0, \infty)$ (iv) $\frac{x}{1 + x} \leq ln (1 + x) \leq x$ on $(0, \infty)$ (v) $\frac{2}{π} < \frac{sin x}{x} < 1$ on $(0, \frac{π}{2})$ .Answer: in each case $h =$ difference, $h (0) = 0$ (for (v) use that $\frac{sin x}{x}$ decreases), and $h^{'}$ has a fixed sign
Using $f (x) = x^{1/ x}$ , identify which is larger: $e^{π}$ or $π^{e}$ .Answer: $e^{π}$ (Figure 9)
If $0 < x_{1} < x_{2} < x_{3} < π$ , prove that $sin \frac{2 x _{1} + x _{2} + x _{3}}{4} > \frac{2 sin x _{1} + sin x _{2} + sin x _{3}}{4}$ .Answer: weighted centroid of points on the concave-down arc of $sin x$ (Jensen)

Common Mistakes to Avoid

Watch out

Calling $x^{3}$ 'not strictly increasing' because $f^{'} (0) = 0$ . Isolated zeros of $f^{'}$ do not spoil strict monotonicity.
Merging intervals across a break: $- \frac{1}{x}$ increases on $(- \infty, 0)$ and on $(0, \infty)$ , but not on their union.
Ignoring the domain: $x ln x$ lives on $x > 0$ , and $x + \frac{1}{x + 1}$ excludes $x = - 1$ .
Changing sign across an even power in the sign chart: $(x - 1)^{2}$ does not change sign at $x = 1$ .
In 'monotonic for all $x$ ' questions, demanding $f^{'} > 0$ and losing the boundary values where $f^{'} = 0$ at isolated points.
Using the test $f^{'} (a) > 0$ at a point where $f$ is discontinuous; there, compare $f (a - h)$ , $f (a)$ , $f (a + h)$ directly.
Taking $f^{''} (c) = 0$ as proof of inflection ( $x^{4}$ at $0$ is not one), or missing inflection points where $f^{''}$ does not exist ( $x^{1/3}$ ).
Mixing up 'concave' and 'convex' between books. Decide by the sign of $f^{''}$ : $f^{''} > 0$ is a cup with chords above the curve.

Frequently Asked Questions

What is a monotonic function?

A function is monotonic on a set if it is either never decreasing or never increasing there. It is strictly increasing if $x_{1} < x_{2}$ always gives $f (x_{1}) < f (x_{2})$ , and monotonically increasing (non-decreasing) if it gives $f (x_{1}) \leq f (x_{2})$ .

How do you find intervals of increase and decrease of a function?

Find the domain and $f^{'} (x)$ , factorise $f^{'}$ , mark its zeros and the points where it is undefined, and make a sign chart. Where $f^{'} > 0$ the function increases; where $f^{'} < 0$ it decreases. Include end points where $f$ is continuous.

Is x cubed strictly increasing even though its derivative is zero at 0?

Yes. $f^{'} (x) = 3 x^{2}$ is positive everywhere except at the single point $x = 0$ . A derivative that vanishes only at isolated points does not stop a function from being strictly increasing; it would have to be zero on a whole interval.

What does it mean for a function to be increasing at a point?

$f$ is increasing at $x = a$ if $f (a - h) < f (a) < f (a + h)$ for all small $h > 0$ . For a differentiable function, $f^{'} (a) > 0$ is enough; if $f^{'} (a) = 0$ , check the sign of $f^{'}$ on both sides of $a$ .

How is monotonicity used to prove inequalities?

To show $f (x) > g (x)$ on an interval, let $h = f - g$ , find a point where $h = 0$ (often an end point), and show $h^{'}$ has a fixed sign. If $h^{'}$ is unclear, use $h^{''}$ to fix the sign of $h^{'}$ first.

What is a point of inflection?

It is a point where the curve changes from concave up to concave down or the reverse, so $f^{''}$ changes sign there. The curve crosses its tangent at such a point. $f^{''} (c) = 0$ alone is not enough, and $f^{''}$ may also fail to exist at an inflection point.

How is monotonicity asked in JEE Main?

JEE Main asks intervals of increase and decrease, values of a parameter for which a function is increasing on $R$ (usually through $D \leq 0$ for a quadratic derivative), comparisons like $e^{π}$ and $π^{e}$ , and simple inequalities proved with the derivative.

What monotonicity problems appear in JEE Advanced?

JEE Advanced combines monotonicity with inverse functions, composite functions and limits, asks inequality proofs that need the second derivative, uses concavity and Jensen-type arguments, and asks for the number of solutions of equations by comparing increasing and decreasing graphs.

Previous year questions on Monotonocity

5 questions from past papers, each with a step-by-step solution.

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