Application of Derivatives — Class 12 Maths Chapter 6

Section 01

Basic Concepts

Understanding applications of derivatives

6.1 Introduction

In Chapter 5, we learnt how to find derivatives of composite functions, inverse trigonometric functions, implicit functions, exponential functions and logarithmic functions. In this chapter, we study applications of the derivative in various disciplines — engineering, science, social science, and many other fields.

Key Applications

We will learn how the derivative can be used to:

Determine rate of change of quantities
Find intervals of increasing / decreasing functions
Find turning points — local maxima and minima
Find absolute maximum / minimum values on closed intervals

6.2 Rate of Change of Quantities

Definition — Rate of Change

If a quantity $y$ varies with another quantity $x$, satisfying $y = f(x)$, then:

$$\frac{dy}{dx} \text{ (or } f'(x)\text{)}$$

represents the rate of change of $y$ with respect to $x$.

Further, if two variables $x$ and $y$ are varying with respect to another variable $t$, i.e., $x = f(t)$ and $y = g(t)$, then by Chain Rule:

$$\frac{dy}{dx} = \frac{dy}{dt} \Big/ \frac{dx}{dt}, \quad \text{if } \frac{dx}{dt} \neq 0$$

Note: $\frac{dy}{dx}$ is positive if $y$ increases as $x$ increases, and negative if $y$ decreases as $x$ increases.

6.3 Increasing and Decreasing Functions

Definition

Let $I$ be an interval in the domain of a real valued function $f$. Then:

(i) Increasing on I: $x_1 < x_2$ in I $\Rightarrow f(x_1) \leq f(x_2)$
(ii) Decreasing on I: $x_1 < x_2$ in I $\Rightarrow f(x_1) \geq f(x_2)$
(iii) Constant on I: $f(x) = c$ for all $x \in I$

Key Theorem

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then:

$f'(x) > 0$ for all $x \in (a,b)$ $\Rightarrow$ $f$ is increasing on $[a,b]$
$f'(x) < 0$ for all $x \in (a,b)$ $\Rightarrow$ $f$ is decreasing on $[a,b]$
$f'(x) = 0$ for all $x \in (a,b)$ $\Rightarrow$ $f$ is constant on $[a,b]$

6.4 Maxima and Minima

Definition — Local Maxima / Minima

Let $f$ be a real valued function and $c$ an interior point in its domain.

Local Maxima: There exists $h > 0$ such that $f(c) \geq f(x)$ for all $x \in (c-h, c+h)$
Local Minima: There exists $h > 0$ such that $f(c) \leq f(x)$ for all $x \in (c-h, c+h)$

Key Theorem

If $f$ has a local maxima or local minima at $x = c$, then either $f'(c) = 0$ or $f$ is not differentiable at $c$.

Critical Point: A point $c$ in the domain of $f$ where either $f'(c) = 0$ or $f$ is not differentiable.

6.4.1 Absolute Maxima and Minima

Definition — Absolute Extrema

Let $f$ be a function defined on a closed interval $[a,b]$.

Absolute Maximum: $f(c) \geq f(x)$ for all $x \in [a,b]$
Absolute Minimum: $f(c) \leq f(x)$ for all $x \in [a,b]$

Theorem

Every continuous function on a closed interval $[a,b]$ has an absolute maximum and an absolute minimum value.

Section 02

🎨 Visual Graphs

See applications of derivatives in action

V1 Rate of Change — Tangent Slope at a Point

dy/dx = slope of tangent = instantaneous rate of change

The derivative at a point equals the slope of the tangent line, which represents the instantaneous rate of change of the function at that point.

V2 Increasing & Decreasing Functions — $f(x) = x^2$

f'(x) < 0 → decreasing ; f'(x) > 0 → increasing

The sign of $f'(x)$ tells us whether the function is increasing or decreasing on an interval.

V3 Local Maximum — First Derivative Test

f'(x) changes from + to − at c → Local Maximum

At a local maximum, the function changes from increasing to decreasing — so $f'(x)$ changes from positive to negative.

V4 Local Minimum — First Derivative Test

f'(x) changes from − to + at c → Local Minimum

At a local minimum, the function changes from decreasing to increasing — so $f'(x)$ changes from negative to positive.

V5 Point of Inflection — $f(x) = x^3$

f'(c) = 0 but neither max nor min

A point of inflection is where the curve changes its concavity. Here $f'(x)$ does not change sign — so it's neither a maximum nor a minimum.

V6 Absolute Maximum & Minimum on $[a,b]$

Check all critical points + endpoints

On a closed interval, absolute maximum and minimum can occur at critical points inside the interval or at the endpoints.

V7 Second Derivative Test — Concavity

f''(c) < 0 → Max ; f''(c) > 0 → Min

The sign of the second derivative tells us the concavity of the curve, which helps identify maxima and minima.

V8 Applied Optimization — Box from Tin Sheet

Maximise volume by choosing optimal square cut

Optimization problems ask us to maximise or minimise a quantity (like volume) by choosing a variable (like the side of the cut square).

Section 03

⚡ Rate of Change — Deep Dive

Related rates and chain rule applications

R1 Method for Rate of Change Problems

Step-by-step procedure:

Identify the quantities that are changing with time
Write a relation between them (e.g., $A = \pi r^2$, $V = x^3$)
Differentiate both sides with respect to time $t$ using Chain Rule
Substitute the given values and solve for the required rate

Point to Remember

If a quantity is increasing, its rate is positive
If a quantity is decreasing, its rate is negative
Always write the relation before differentiating
Don't forget to use Chain Rule for composite quantities

R2 Common Rate Relations

Quantity	Formula	Rate Relation
Circle Area	$A = \pi r^2$	$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$
Circle Circumference	$C = 2\pi r$	$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$
Sphere Volume	$V = \frac{4}{3}\pi r^3$	$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
Sphere Surface	$S = 4\pi r^2$	$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$
Cube Volume	$V = x^3$	$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$
Cube Surface	$S = 6x^2$	$\frac{dS}{dt} = 12x \frac{dx}{dt}$
Cone Volume	$V = \frac{1}{3}\pi r^2 h$	$\frac{dV}{dt} = \frac{\pi}{3}\left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right)$
Rectangle Area	$A = xy$	$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}$

R3 Marginal Cost and Revenue

Definition

Marginal Cost (MC): The instantaneous rate of change of total cost with respect to output.

$$\text{MC} = \frac{dC}{dx}$$

Definition

Marginal Revenue (MR): The instantaneous rate of change of total revenue with respect to the number of items sold.

$$\text{MR} = \frac{dR}{dx}$$

Section 04

📈 Increasing & Decreasing Functions

Monotonicity and sign of derivative

M1 Method for Finding Intervals

Step-by-step procedure:

Find $f'(x)$
Solve $f'(x) = 0$ to get critical points
Plot these points on the real number line — they divide the line into disjoint intervals
Check the sign of $f'(x)$ in each interval (using test points)
If $f'(x) > 0$ → increasing; if $f'(x) < 0$ → decreasing

M2 Sign Scheme Table

Example: $f(x) = 4x^3 - 6x^2 - 72x + 30$

$f'(x) = 12x^2 - 12x - 72 = 12(x^2 - x - 6) = 12(x - 3)(x + 2)$

Critical points: $x = -2, 3$

Interval	Sign of $f'(x)$	Nature of $f$
$(-\infty, -2)$	$(-)(-) > 0$	Increasing
$(-2, 3)$	$(-)(+) < 0$	Decreasing
$(3, \infty)$	$(+)(+) > 0$	Increasing

M3 Standard Results

$\sin x$ is increasing on $(0, \pi/2)$ and decreasing on $(\pi/2, \pi)$
$\cos x$ is decreasing on $(0, \pi)$ and increasing on $(\pi, 2\pi)$
$e^x$ is increasing on $\mathbb{R}$
$\log x$ is increasing on $(0, \infty)$
$x^2$ is decreasing on $(-\infty, 0)$ and increasing on $(0, \infty)$
$x^3$ is increasing on $\mathbb{R}$

Section 05

🎯 Maxima & Minima — Deep Dive

First and second derivative tests

MM1 First Derivative Test

Let $f$ be continuous at a critical point $c$.

Test

If $f'(x)$ changes from positive to negative at $c$ → Local Maximum
If $f'(x)$ changes from negative to positive at $c$ → Local Minimum
If $f'(x)$ does not change sign at $c$ → Point of Inflection

MM2 Second Derivative Test

Let $f$ be twice differentiable at critical point $c$ with $f'(c) = 0$.

Test

If $f''(c) < 0$ → Local Maximum
If $f''(c) > 0$ → Local Minimum
If $f''(c) = 0$ → Test fails — go back to first derivative test

MM3 Important Observation

Common Mistake

$f'(c) = 0$ does NOT guarantee a local maximum or minimum.

Example: $f(x) = x^3$ has $f'(0) = 0$, but $x = 0$ is neither a maximum nor a minimum — it's a point of inflection.

Section 06

📊 Absolute Maxima & Minima

On closed intervals

A1 Working Rule

Step 1: Find all critical points of $f$ in $[a,b]$ (where $f'(x) = 0$ or $f$ is not differentiable).

Step 2: Take the endpoints $a$ and $b$.

Step 3: Evaluate $f$ at all these points.

Step 4: The maximum value is the absolute maximum and the minimum value is the absolute minimum.

A2 Example — $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1,5]$

$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$

Critical points: $x = 2, 3$

Point	Value of $f$
$x = 1$ (endpoint)	$f(1) = 24$
$x = 2$ (critical)	$f(2) = 29$
$x = 3$ (critical)	$f(3) = 28$
$x = 5$ (endpoint)	$f(5) = 56$

Absolute Maximum: 56 at $x = 5$

Absolute Minimum: 24 at $x = 1$

Quick Reference

📐 All Important Formulas

Memorize these for board exams

F1 Rate of Change

$$\frac{dy}{dx} = \frac{dy}{dt} \Big/ \frac{dx}{dt}, \quad \frac{dx}{dt} \neq 0$$

F2 Monotonicity Conditions

$f'(x) > 0 \Rightarrow$ Increasing

$f'(x) < 0 \Rightarrow$ Decreasing

$f'(x) = 0 \Rightarrow$ Constant

F3 First Derivative Test

$f'(x)$ changes $+ \to -$ at $c$ $\Rightarrow$ Local Max

$f'(x)$ changes $- \to +$ at $c$ $\Rightarrow$ Local Min

$f'(x)$ no sign change at $c$ $\Rightarrow$ Point of Inflection

F4 Second Derivative Test

$f'(c) = 0$ and $f''(c) < 0$ $\Rightarrow$ Local Max

$f'(c) = 0$ and $f''(c) > 0$ $\Rightarrow$ Local Min

$f'(c) = 0$ and $f''(c) = 0$ $\Rightarrow$ Test Fails

F5 Absolute Extrema on $[a,b]$

Evaluate $f$ at all critical points + endpoints

Highest value = Absolute Max

Lowest value = Absolute Min

F6 Marginal Cost & Revenue

$$\text{MC} = \frac{dC}{dx}, \quad \text{MR} = \frac{dR}{dx}$$

F7 Common Rate Relations

$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$ (Circle)

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ (Sphere)

$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$ (Cube)

Shortcuts

💡 Tricks & Shortcuts

Speed techniques for board exams

Trick 1 — Rate of Change Setup

"Write relation, then differentiate"

Identify what is given and what is asked
Write the formula connecting the quantities
Differentiate w.r.t. time using Chain Rule
Substitute known values and solve

Trick 2 — Sign Scheme for Monotonicity

"Factorise f'(x), mark roots, test intervals"

Solve $f'(x) = 0$ to get critical points
Mark them on number line
Pick a test point in each interval
Determine sign → increasing or decreasing

Trick 3 — f'(c) = 0 is NOT enough

"Zero derivative ≠ max/min always"

$f(x) = x^3$ has $f'(0) = 0$ but neither max nor min
Always check sign change or second derivative
Point of inflection is a valid outcome

Trick 4 — Second Derivative Test Fails

"f''(c) = 0 → go back to first derivative test"

Second derivative test is inconclusive when $f''(c) = 0$
Use sign scheme of $f'(x)$ instead
This often happens for polynomial functions

Trick 5 — Closed Interval — Always Check Endpoints

"Endpoints can be absolute extrema"

Absolute max/min on $[a,b]$ may occur at $a$ or $b$
Never forget to evaluate $f(a)$ and $f(b)$
Compare all values — don't assume

Trick 6 — Word Problems — Identify the Function

"What needs to be maximised/minimised?"

Area, volume, profit, cost, distance — identify the quantity
Write it as a function of one variable
Differentiate and set to zero
Verify max/min using second derivative or sign test

Trick 7 — Marginal Cost/Revenue Shortcut

"MC = dC/dx, MR = dR/dx"

Just differentiate the cost/revenue function
Substitute the given output level
Units matter — write them correctly

Trick 8 — Common Exam Traps

Watch out for:

Sign errors in rate problems (increasing vs decreasing)
Forgetting Chain Rule when differentiating w.r.t. time
Assuming $f'(c) = 0$ means max or min
Not checking endpoints for absolute extrema
Confusing local max with absolute max

Trick 9 — Point of Inflection

"f''(c) = 0 and concavity changes"

At point of inflection, $f''(c) = 0$ (usually)
Concavity changes from up to down or vice versa
$f'(x)$ does not change sign
Example: $y = x^3$ at $x = 0$

Trick 10 — Sanity Check Answer

Always verify:

Is the answer in the correct units?
Does the sign make sense? (positive = increasing)
For max/min problems — did you check the second derivative?
For absolute extrema — did you check endpoints?

Practice

📝 Solved Examples with Step Marking

Click "Show Answer" to reveal step-by-step solutions with marks

📋 Total: 2 Marks · Board Pattern

Q1. Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 5$ cm.

Step 1 — Write formula0.5 Mark

$A = \pi r^2$

Step 2 — Differentiate1 Mark

$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r$

Step 3 — Substitute r = 50.5 Mark

$\frac{dA}{dr}\Big|_{r=5} = 2\pi(5) = 10\pi$ cm²/cm

Rate of change of area = $10\pi$ cm²/cm

📋 Total: 3 Marks

Q2. The volume of a cube is increasing at a rate of 9 cm³/s. How fast is the surface area increasing when the edge is 10 cm?

Step 1 — Write relations0.5 Mark

$V = x^3$, $S = 6x^2$

Step 2 — Differentiate V w.r.t. t1 Mark

$\frac{dV}{dt} = 3x^2 \frac{dx}{dt} = 9$

$\Rightarrow \frac{dx}{dt} = \frac{3}{x^2}$

Step 3 — Differentiate S w.r.t. t1 Mark

$\frac{dS}{dt} = 12x \cdot \frac{dx}{dt} = 12x \cdot \frac{3}{x^2} = \frac{36}{x}$

Step 4 — Substitute x = 100.5 Mark

$\frac{dS}{dt}\Big|_{x=10} = \frac{36}{10} = 3.6$ cm²/s

Surface area is increasing at 3.6 cm²/s

📋 Total: 2 Marks

Q3. Show that $f(x) = x^3 - 3x^2 + 4x$ is increasing on $\mathbb{R}$.

Step 1 — Find f'(x)0.5 Mark

$f'(x) = 3x^2 - 6x + 4$

Step 2 — Complete the square1 Mark

$f'(x) = 3(x^2 - 2x + 1) + 1 = 3(x-1)^2 + 1$

Since $(x-1)^2 \geq 0$ for all $x$, we have $f'(x) \geq 1 > 0$.

Step 3 — Conclusion0.5 Mark

Since $f'(x) > 0$ for all $x \in \mathbb{R}$, $f$ is increasing on $\mathbb{R}$. ✓

$f$ is increasing on $\mathbb{R}$

📋 Total: 3 Marks

Q4. Find the intervals in which $f(x) = 4x^3 - 6x^2 - 72x + 30$ is (a) increasing (b) decreasing.

Step 1 — Find f'(x)0.5 Mark

$f'(x) = 12x^2 - 12x - 72 = 12(x^2 - x - 6) = 12(x-3)(x+2)$

Step 2 — Critical points0.5 Mark

$f'(x) = 0 \Rightarrow x = -2, 3$

Step 3 — Sign scheme1.5 Marks

Intervals: $(-\infty,-2)$, $(-2,3)$, $(3,\infty)$

In $(-\infty,-2)$: $f'(x) > 0$ → Increasing

In $(-2,3)$: $f'(x) < 0$ → Decreasing

In $(3,\infty)$: $f'(x) > 0$ → Increasing

Step 4 — Answer0.5 Mark

(a) Increasing on $(-\infty,-2) \cup (3,\infty)$

(b) Decreasing on $(-2,3)$

Increasing: $(-\infty,-2) \cup (3,\infty)$ ; Decreasing: $(-2,3)$

📋 Total: 3 Marks

Q5. Find all points of local maxima and local minima of $f(x) = x^3 - 3x + 3$.

Step 1 — Find f'(x)0.5 Mark

$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$

Step 2 — Critical points0.5 Mark

$f'(x) = 0 \Rightarrow x = 1, -1$

Step 3 — First derivative test1 Mark

At $x = -1$: $f'(x)$ changes from $+$ to $-$ → Local Maximum

$f(-1) = (-1)^3 - 3(-1) + 3 = -1 + 3 + 3 = 5$

Step 4 — Second point1 Mark

At $x = 1$: $f'(x)$ changes from $-$ to $+$ → Local Minimum

$f(1) = 1 - 3 + 3 = 1$

Local Max at $x = -1$ (value 5) ; Local Min at $x = 1$ (value 1)

📋 Total: 3 Marks

Q6. Find two positive numbers whose sum is 15 and the sum of whose squares is minimum.

Step 1 — Set up variables0.5 Mark

Let one number be $x$. Then the other is $15 - x$.

Let $S(x) = x^2 + (15-x)^2$

Step 2 — Simplify and differentiate1 Mark

$S(x) = x^2 + 225 - 30x + x^2 = 2x^2 - 30x + 225$

$S'(x) = 4x - 30$

Step 3 — Solve S'(x) = 00.5 Mark

$4x - 30 = 0 \Rightarrow x = 15/2$

$S''(x) = 4 > 0 \Rightarrow$ minimum at $x = 15/2$

Step 4 — Answer1 Mark

Numbers are $15/2$ and $15/2$ (both 7.5)

The two numbers are 7.5 and 7.5

📋 Total: 3 Marks

Q7. Find the absolute maximum and minimum values of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1,5]$.

Step 1 — Find f'(x)0.5 Mark

$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$

Critical points: $x = 2, 3$

Step 2 — Evaluate f at critical points + endpoints1.5 Marks

$f(1) = 2 - 15 + 36 + 1 = 24$

$f(2) = 16 - 60 + 72 + 1 = 29$

$f(3) = 54 - 135 + 108 + 1 = 28$

$f(5) = 250 - 375 + 180 + 1 = 56$

Step 3 — Compare0.5 Mark

Absolute Max = 56 at $x = 5$

Absolute Min = 24 at $x = 1$

Step 4 — Conclusion0.5 Mark

Absolute maximum = 56, Absolute minimum = 24

Abs Max = 56 at x=5 ; Abs Min = 24 at x=1

Exam Focus

📖 Previous Year Questions

Frequently asked questions in board exams

2024

Q. Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 5$ cm.

Step 1 — Formula0.5 Mark

$A = \pi r^2$

Step 2 — Differentiate1 Mark

$\frac{dA}{dr} = 2\pi r$

Step 3 — Substitute0.5 Mark

At $r = 5$: $\frac{dA}{dr} = 10\pi$ cm²/cm

$10\pi$ cm²/cm

2024

Q. The volume of a cube is increasing at 9 cm³/s. Find the rate of increase of surface area when edge is 10 cm.

Step 1 — Relations0.5 Mark

$V = x^3$, $S = 6x^2$

Step 2 — Differentiate1 Mark

$\frac{dV}{dt} = 3x^2 \frac{dx}{dt} = 9 \Rightarrow \frac{dx}{dt} = \frac{3}{x^2}$

Step 3 — Surface area rate1 Mark

$\frac{dS}{dt} = 12x \cdot \frac{3}{x^2} = \frac{36}{x}$

Step 4 — Substitute0.5 Mark

At $x = 10$: $\frac{dS}{dt} = 3.6$ cm²/s

3.6 cm²/s

2023

Q. Find the intervals in which $f(x) = x^2 - 4x + 6$ is (a) increasing (b) decreasing.

Step 1 — Find f'(x)0.5 Mark

$f'(x) = 2x - 4$

Step 2 — Critical point0.5 Mark

$f'(x) = 0 \Rightarrow x = 2$

Step 3 — Sign scheme1 Mark

For $x < 2$: $f'(x) < 0$ → Decreasing on $(-\infty, 2)$

For $x > 2$: $f'(x) > 0$ → Increasing on $(2, \infty)$

Increasing on $(2,\infty)$ ; Decreasing on $(-\infty,2)$

2023

Q. Find all points of local maxima and local minima of $f(x) = x^3 - 3x + 3$.

Step 1 — f'(x)0.5 Mark

$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$

Step 2 — Critical points0.5 Mark

$x = -1, 1$

Step 3 — First derivative test1 Mark

At $x = -1$: changes $+ \to -$ → Local Max, $f(-1) = 5$

At $x = 1$: changes $- \to +$ → Local Min, $f(1) = 1$

Step 4 — Conclusion1 Mark

Local Max at $x=-1$ (value 5) ; Local Min at $x=1$ (value 1)

Local Max: $(-1, 5)$ ; Local Min: $(1, 1)$

2022

Q. Find absolute maximum and minimum of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1,5]$.

Step 1 — Critical points1 Mark

$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$

Critical points: $x = 2, 3$

Step 2 — Evaluate f1 Mark

$f(1) = 24$, $f(2) = 29$, $f(3) = 28$, $f(5) = 56$

Step 3 — Compare1 Mark

Absolute Max = 56 at $x = 5$

Absolute Min = 24 at $x = 1$

Abs Max = 56 (at x=5) ; Abs Min = 24 (at x=1)

2022

Q. Show that $f(x) = x^3 - 3x^2 + 4x$ is increasing on $\mathbb{R}$.

Step 1 — Find f'(x)1 Mark

$f'(x) = 3x^2 - 6x + 4$

Step 2 — Simplify1 Mark

$f'(x) = 3(x-1)^2 + 1 > 0$ for all $x$

Step 3 — Conclusion1 Mark

Since $f'(x) > 0$ everywhere, $f$ is increasing on $\mathbb{R}$. ✓

$f$ is increasing on $\mathbb{R}$

2021

Q. Find two positive numbers whose sum is 15 and the sum of whose squares is minimum.

Step 1 — Set up1 Mark

Let numbers be $x$ and $15-x$. Then $S(x) = x^2 + (15-x)^2$

Step 2 — Differentiate1 Mark

$S'(x) = 4x - 30 = 0 \Rightarrow x = 15/2$

$S''(x) = 4 > 0$ → minimum

Step 3 — Answer1 Mark

Numbers are $15/2$ and $15/2$

The two numbers are 7.5 and 7.5

2020

Q. Find intervals of increase/decrease for $f(x) = 4x^3 - 6x^2 - 72x + 30$.

Step 1 — f'(x)1 Mark

$f'(x) = 12x^2 - 12x - 72 = 12(x-3)(x+2)$

Step 2 — Critical points0.5 Mark

$x = -2, 3$

Step 3 — Sign scheme1 Mark

Increasing: $(-\infty,-2) \cup (3,\infty)$

Decreasing: $(-2,3)$

Step 4 — Conclusion0.5 Mark

Answer as above

Increasing: $(-\infty,-2) \cup (3,\infty)$ ; Decreasing: $(-2,3)$

Basic Concepts

6.1 Introduction

6.2 Rate of Change of Quantities

6.3 Increasing and Decreasing Functions

6.4 Maxima and Minima

6.4.1 Absolute Maxima and Minima

🎨 Visual Graphs

V1 Rate of Change — Tangent Slope at a Point

V2 Increasing & Decreasing Functions — $f(x) = x^2$

V3 Local Maximum — First Derivative Test

V4 Local Minimum — First Derivative Test

V5 Point of Inflection — $f(x) = x^3$

V6 Absolute Maximum & Minimum on $[a,b]$

V7 Second Derivative Test — Concavity

V8 Applied Optimization — Box from Tin Sheet

⚡ Rate of Change — Deep Dive

R1 Method for Rate of Change Problems

R2 Common Rate Relations

R3 Marginal Cost and Revenue

📈 Increasing & Decreasing Functions

M1 Method for Finding Intervals

M2 Sign Scheme Table

M3 Standard Results

🎯 Maxima & Minima — Deep Dive

MM1 First Derivative Test

MM2 Second Derivative Test

MM3 Important Observation

📊 Absolute Maxima & Minima

A1 Working Rule

A2 Example — $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1,5]$

📐 All Important Formulas

F1 Rate of Change

F2 Monotonicity Conditions

F3 First Derivative Test

F4 Second Derivative Test

F5 Absolute Extrema on $[a,b]$

F6 Marginal Cost & Revenue

F7 Common Rate Relations

💡 Tricks & Shortcuts

📝 Solved Examples with Step Marking

✍️ MCQ Practice — 20 Questions

🎯 Assertion & Reasoning — 10 Questions

📖 Standard Options

📖 Previous Year Questions