Intermediate

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Monotonicity, Rolle's Theorem & Mean Value Theorem

Master Monotonicity Rolle's Theorem and Mean Value Theorem for JEE Main JEE Advanced and Board Exams. Covers increasing decreasing functions sign chart method critical points concavity inflection points Rolle's theorem LMVT with applications and solved examples.

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One of the most powerful uses of the derivative is to determine where a function is increasing or decreasing — and to guarantee the existence of special points where the derivative takes a particular value. For JEE Main, monotonicity questions appear every year. The theorems of Rolle and Lagrange formalise existence theorems. Note that as per the updated 2026 syllabus, Rolle's Theorem and LMVT have been removed from JEE Main and are now strictly JEE Advanced topics.

1. Increasing and Decreasing Functions

A function $f$ is said to be:

Strictly Increasing on interval $I$ if $x_{1} < x_{2} \Rightarrow f (x_{1}) < f (x_{2})$ for all $x_{1}, x_{2} \in I$ .
Strictly Decreasing on interval $I$ if $x_{1} < x_{2} \Rightarrow f (x_{1}) > f (x_{2})$ for all $x_{1}, x_{2} \in I$ .
Monotone on $I$ if it is either entirely increasing or entirely decreasing on $I$ .

Test Using Derivatives

Condition on $f^{'} (x)$ in $(a, b)$	Behaviour of $f$ on $(a, b)$
$f^{'} (x) > 0$ for all $x$	Strictly increasing
$f^{'} (x) < 0$ for all $x$	Strictly decreasing
$f^{'} (x) = 0$ for all $x$	Constant
$f^{'} (x) \geq 0$ on $(a, b)$ , zero only at isolated points	Still strictly increasing

Steps to Find Intervals of Increase/Decrease

Step 1: Find $f^{'} (x)$ .
Step 2: Solve $f^{'} (x) = 0$ and find where $f^{'} (x)$ is undefined — these are the critical points.
Step 3: Mark critical points on the number line — they partition it into intervals.
Step 4: Test the sign of $f^{'} (x)$ in each interval using a test point.
Step 5: $f^{'} (x) > 0$ → increasing; $f^{'} (x) < 0$ → decreasing.

Worked Example

Find intervals where $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 15$ is increasing/decreasing.

$f^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$
Critical points: $x = 1$ , $x = 2$ .
Sign analysis: $f^{'} (x) > 0$ for $x < 1$ or $x > 2$ ; $f^{'} (x) < 0$ for $1 < x < 2$ .
Increasing: $(- \infty, 1) \cup (2, \infty)$ ; Decreasing: $(1, 2)$ .

2. Critical Points, Stationary Points and Points of Inflection

Term	Definition	Note
Critical point	$c$ where $f^{'} (c) = 0$ or $f^{'} (c)$ does not exist	Every extremum is a critical point
Stationary point	$c$ where $f^{'} (c) = 0$ specifically	Subset of critical points
Point of inflection	$c$ where $f^{″} (c) = 0$ AND $f^{″}$ changes sign at $c$	$f^{″} (c) = 0$ alone is NOT sufficient

Concavity

Concave upward on $(a, b)$ : $f^{″} (x) > 0$ — curve bends upward like $\cup$ .
Concave downward on $(a, b)$ : $f^{″} (x) < 0$ — curve bends downward like $\cap$ .
Point of inflection: concavity changes — $f^{″}$ changes sign at $c$ .

Classic trap: $f (x) = x^{4}$ at $x = 0$ : $f^{″} (0) = 0$ but $f^{″}$ does NOT change sign → $x = 0$ is a minimum, NOT a point of inflection.

3. Rolle's Theorem JEE Advanced Only

Statement

If $f$ satisfies all three conditions:

$f$ is continuous on $[a, b]$ ,
$f$ is differentiable on $(a, b)$ , and
$f (a) = f (b)$ ,

then there exists at least one $c \in (a, b)$ such that $f^{'} (c) = 0$ .

Geometric Interpretation

If a smooth curve starts and ends at the same height, there must be at least one point between where the tangent is horizontal (parallel to the $x$ -axis).

Important Notes on Rolle's Theorem

All three conditions must be verified before applying the theorem.
If any condition fails, the theorem cannot be applied (and $f^{'} (c) = 0$ may or may not exist).
The theorem guarantees existence of $c$ but does NOT say it is unique.
The theorem says nothing about the value of $c$ — we must solve $f^{'} (x) = 0$ to find it.

Worked Example

Verify Rolle's Theorem for $f (x) = x^{2} - 4 x + 3$ on $[1, 3]$ and find $c$ .

$f$ is a polynomial → continuous on $[1, 3]$ , differentiable on $(1, 3)$ .
$f (1) = 1 - 4 + 3 = 0$ ; $f (3) = 9 - 12 + 3 = 0$ → $f (1) = f (3)$ ✓.
All conditions satisfied.
$f^{'} (x) = 2 x - 4 = 0$ → $x = 2$ . Since $2 \in (1, 3)$ : $c = 2$ ✓.

4. Lagrange's Mean Value Theorem (LMVT) JEE Advanced Only

Statement

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then there exists at least one $c \in (a, b)$ such that:

$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$

Geometric Interpretation

There is at least one point on the curve where the tangent is parallel to the chord joining $(a, f (a))$ and $(b, f (b))$ .

Relation to Rolle's Theorem

Rolle's Theorem is a special case of LMVT when $f (a) = f (b)$ — the chord is then horizontal, so the parallel tangent also has slope zero.

Theorem	Conditions	Conclusion
Rolle's Theorem	Continuous $[a, b]$ , differentiable $(a, b)$ , $f (a) = f (b)$	$\exists c \in (a, b)$ : $f^{'} (c) = 0$
LMVT	Continuous $[a, b]$ , differentiable $(a, b)$	$\exists c \in (a, b)$ : $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$

Worked Example — LMVT

Find $c$ in LMVT for $f (x) = x^{3} - x^{2} - x + 1$ on $[0, 2]$ .

$f (0) = 1$ , $f (2) = 8 - 4 - 2 + 1 = 3$ .
Slope of chord $= \frac{f (2) - f (0)}{2 - 0} = \frac{3 - 1}{2} = 1$ .
$f^{'} (x) = 3 x^{2} - 2 x - 1$ . Set $f^{'} (c) = 1$ :
$3 c^{2} - 2 c - 1 = 1$ → $3 c^{2} - 2 c - 2 = 0$ → $c = \frac{2 \pm \sqrt{4 + 24}}{6} = \frac{2 \pm \sqrt{28}}{6} = \frac{1 \pm \sqrt{7}}{3}$ .
Only $c = \frac{1 + \sqrt{7}}{3} \approx 1.22 \in (0, 2)$ is valid.

Applications of MVT

Proving inequalities: If $f^{'} (x) > 0$ on $(a, b)$ , then $f (b) > f (a)$ — used to prove e.g., $\ln (1 + x) < x$ for $x > 0$ .
Estimating function values: $f (b) - f (a) = f^{'} (c) (b - a)$ for some $c$ — bounds on $f^{'}$ give bounds on $f (b) - f (a)$ .
Uniqueness of roots: If $f^{'} (x) \neq 0$ on an interval, $f$ can have at most one root there (by Rolle).

Monotonicity & MVT — Quick Decision Framework

Question Type	Approach
Find intervals of increase/decrease	Find $f^{'} (x)$ , set = 0, sign chart, conclude
Verify Rolle's Theorem	Check 3 conditions in order: continuity, differentiability, $f (a) = f (b)$
Apply LMVT to find $c$	Compute $\frac{f (b) - f (a)}{b - a}$ , set $f^{'} (c)$ equal to it, solve for $c$ in $(a, b)$
Prove inequality using MVT	Apply LMVT, use known bounds on $f^{'} (c)$ to bound $f (b) - f (a)$

JEE/Board Power Tips — Avoid These Common Mistakes

Rolle's Theorem requires ALL three conditions. Missing even one means the theorem cannot be applied. Forgetting $f (a) = f (b)$ is the most common error — always check all three explicitly in board answers.
$f^{'} (x) \geq 0$ with equality at isolated points still means strictly increasing. For example, $f (x) = x^{3}$ has $f^{'} (0) = 0$ but is strictly increasing on $R$ . This subtlety is tested in JEE.
$f^{″} (c) = 0$ does NOT guarantee a point of inflection — $f^{″}$ must also change sign at $c$ . The function $f (x) = x^{4}$ has $f^{″} (0) = 0$ but no inflection at $x = 0$ (it is actually a minimum).
LMVT gives existence, not uniqueness. There could be more than one value of $c$ satisfying the conclusion. Only find those that lie strictly within $(a, b)$ .
Rolle's is a special case of LMVT. When $f (a) = f (b)$ , the chord slope is zero, so LMVT gives $f^{'} (c) = 0$ — exactly Rolle's conclusion. Never treat them as independent results.

Practice Questions (JEE / Board Level)

Q1: The function $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 15$ is increasing when:

A) $x < 1$ or $x > 2$
B) $1 < x < 2$
C) $x < - 1$ or $x > - 2$
D) $x > 0$

Answer: A) $x < 1$ or $x > 2$ .

Explanation:

Find the first derivative of the function:
$f^{'} (x) = 6 x^{2} - 18 x + 12$

Factor the quadratic expression:
$f^{'} (x) = 6 (x^{2} - 3 x + 2) = 6 (x - 1) (x - 2)$

For the function to be strictly increasing, $f^{'} (x) > 0$ . This occurs when the factors $(x - 1)$ and $(x - 2)$ have the same sign:
Both positive: $x > 1$ and $x > 2 ⟹ x > 2$
Both negative: $x < 1$ and $x < 2 ⟹ x < 1$

Therefore, the function is increasing on the intervals $(- \infty, 1) \cup (2, \infty)$ .

Q2: Verify Rolle's Theorem for $f (x) = x^{2} - 4 x + 3$ on $[1, 3]$ and find the value of $c$ :

A) $c = 1$
B) $c = 2$
C) $c = 3$
D) $c = 0$

Answer: B) $c = 2$ .

Explanation:

First, verify the conditions for Rolle's Theorem. Since $f (x)$ is a polynomial, it is continuous on $[1, 3]$ and differentiable on $(1, 3)$ .

Evaluate the endpoints:
$f (1) = 1^{2} - 4 (1) + 3 = 0$
$f (3) = 3^{2} - 4 (3) + 3 = 0$

Since $f (1) = f (3) = 0$ , all conditions are met. Rolle's Theorem guarantees at least one $c \in (1, 3)$ such that $f^{'} (c) = 0$ .

Find the derivative and solve for $c$ :
$f^{'} (x) = 2 x - 4$
$f^{'} (c) = 2 c - 4 = 0 ⟹ c = 2$ .

Since $2 \in (1, 3)$ , the value is $c = 2$ .

Q3: Find $c$ from Lagrange's Mean Value Theorem (LMVT) for $f (x) = x^{2}$ on $[2, 4]$ :

A) $c = 2$
B) $c = 3$
C) $c = 4$
D) $c = 2.5$

Answer: B) $c = 3$ .

Explanation:

According to LMVT, there exists a point $c \in (a, b)$ such that $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$ .

Calculate the slope of the secant line between $x = 2$ and $x = 4$ :
$f (2) = 4$
$f (4) = 16$
$Slope = \frac{16 - 4}{4 - 2} = \frac{12}{2} = 6$ .

Find the derivative of the function:
$f^{'} (x) = 2 x$

Set the derivative at $c$ equal to the secant slope:
$2 c = 6 ⟹ c = 3$ .

Since $3 \in (2, 4)$ , the value is $c = 3$ .

Q4: On which interval is $f (x) = x^{3} - 3 x^{2} + 3 x + 10$ strictly increasing?

A) $(- \infty, 1)$ only
B) $(1, \infty)$ only
C) $(- \infty, \infty)$
D) $(0, 2)$

Answer: C) $(- \infty, \infty)$ .

Explanation:

Find the first derivative:
$f^{'} (x) = 3 x^{2} - 6 x + 3$

Factor the derivative:
$f^{'} (x) = 3 (x^{2} - 2 x + 1) = 3 (x - 1)^{2}$

Since $(x - 1)^{2} \geq 0$ for all real numbers, $f^{'} (x) \geq 0$ for all $x$ . The derivative is zero only at the isolated point $x = 1$ .

Because $f^{'} (x)$ is non-negative everywhere and zero only at a discrete point, the function does not flatten out over any interval. Therefore, $f (x)$ is strictly increasing on the entire real line $(- \infty, \infty)$ .

Q5: For $f (x) = \sin x$ on $[0, π]$ , Rolle's Theorem gives $c =$ :

A) 0
B) $π / 4$
C) $π / 2$
D) $π$

Answer: C) $π / 2$ .

Explanation:

First, confirm the conditions: $\sin x$ is continuous on $[0, π]$ and differentiable on $(0, π)$ .

Check the endpoints:
$f (0) = \sin (0) = 0$
$f (π) = \sin (π) = 0$

Since $f (0) = f (π)$ , Rolle's Theorem applies. There must be a $c$ where $f^{'} (c) = 0$ .

Find the derivative:
$f^{'} (x) = \cos x$

Set to zero and solve within the interval:
$\cos c = 0 ⟹ c = π / 2$ .

Q6: The function $f (x) = x^{3} + a x + b$ has Rolle's theorem applicable on $[- 1, 1]$ and $c = 1 / \sqrt{3}$ . The value of $a$ is:

A) 1
B) -1
C) 3
D) -3

Answer: B) -1.

Explanation:

For Rolle's theorem to be applicable on $[- 1, 1]$ , the function values at the endpoints must be equal: $f (- 1) = f (1)$ .

Evaluate both endpoints:
$f (- 1) = (- 1)^{3} + a (- 1) + b = - 1 - a + b$
$f (1) = (1)^{3} + a (1) + b = 1 + a + b$

Set them equal to each other:
$- 1 - a + b = 1 + a + b$
$- 2 a = 2 ⟹ a = - 1$

We can verify this using the provided $c$ value. If $a = - 1$ , then $f^{'} (x) = 3 x^{2} - 1$ .
Setting $f^{'} (c) = 0 ⟹ 3 c^{2} - 1 = 0 ⟹ c^{2} = 1 / 3 ⟹ c = \pm 1 / \sqrt{3}$ . The given $c = 1 / \sqrt{3}$ falls perfectly within $(- 1, 1)$ .

Q7 (JEE type): If $f^{'} (x) > 0$ for all $x \in (a, b)$ and $f (x)$ is continuous on $[a, b]$ , which of the following must be true?

A) $f (x)$ has a maximum in $(a, b)$
B) $f (x)$ is constant on $(a, b)$
C) $f (a) < f (x) < f (b)$ for all $x \in (a, b)$
D) $f (x)$ has no zeros in $(a, b)$

Answer: C) $f (a) < f (x) < f (b)$ for all $x \in (a, b)$ .

Explanation:

If the first derivative $f^{'} (x) > 0$ for all points within the interval $(a, b)$ , it means the function is strictly increasing throughout the entire interval.

By the definition of a strictly increasing function, for any $x$ between $a$ and $b$ (i.e., $a < x < b$ ), the function values must strictly follow the same inequality: $f (a) < f (x) < f (b)$ .

Option A is incorrect because a strictly increasing function reaches its maximum at the endpoint $b$ , not anywhere inside the open interval. Option D is incorrect because an increasing function can absolutely cross the x-axis (having a zero) as it rises.

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DifficultyIntermediate

Est. Read Time25 minutes

CategoryApplication Of Derivatives

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Monotonicity, Rolle's Theorem & Mean Value Theorem

1. Increasing and Decreasing Functions

Test Using Derivatives

Steps to Find Intervals of Increase/Decrease

Worked Example

2. Critical Points, Stationary Points and Points of Inflection

Concavity

3. Rolle's Theorem JEE Advanced Only

Statement

Geometric Interpretation

Important Notes on Rolle's Theorem

Worked Example

4. Lagrange's Mean Value Theorem (LMVT) JEE Advanced Only

Statement

Geometric Interpretation

Relation to Rolle's Theorem

Worked Example — LMVT

Applications of MVT

Monotonicity & MVT — Quick Decision Framework

Practice Questions (JEE / Board Level)

Related Study Material

Practice Questions

Topic Information

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