max
dy/dx
Chapter 6  ·  Class XII Mathematics

Turning Slopes Into Decisions

Application of Derivatives

From Rates to Optima — Calculus That Solves Real Problems

📖 Read Full Notes ⚡ Formula Capsules

Chapter Snapshot

10Concepts
11Formulae
6–8%Exam Weight
3–4Avg Q's
Moderate-HighDifficulty

Why This Chapter Matters for Entrance Exams

CBSEJEE MainJEE Advanced

Maxima-minima optimisation problems are a JEE Advanced staple, often combined with geometry or physics-style setups. CBSE Boards consistently include a 5-mark applied optimisation problem (e.g. minimum cost, maximum area) plus a tangent/normal question.

Key Concept Highlights

Rate of Change of Quantities
Increasing and Decreasing Functions
Tangents and Normals
Approximations (Differentials)
Maxima and Minima
Critical Points
First Derivative Test
Second Derivative Test
Absolute Maximum and Minimum on a Closed Interval
Applied Optimisation Problems

Important Formula Capsules

$f'(x) > 0 \Rightarrow f \text{ increasing};\quad f'(x) < 0 \Rightarrow f \text{ decreasing}$
$\text{Slope of tangent} = \left.\dfrac{dy}{dx}\right|_{(x_1,y_1)}$
$y - y_1 = m(x-x_1)\ \text{(tangent)}$
$y - y_1 = -\dfrac{1}{m}(x-x_1)\ \text{(normal)}$
$\Delta y \approx dy = f'(x)\,\Delta x$
$\text{Critical point: } f'(x) = 0 \text{ or undefined}$
$\text{2nd Derivative Test: } f''(c)<0 \Rightarrow \text{local max};\ f''(c)>0 \Rightarrow \text{local min}$

What You Will Learn

Navigate to Chapter Resources

📖 Full Notes Complete NCERT Coverage 🧠 MCQ Practice Topic-wise Questions 🎯 PYQ Bank JEE · CBSE · BITSAT 📘✔️ Exercises Step by step Solutions to textbook exercises ✔️❌ True / False Concept-based True & False questions

🏆 Exam Strategy & Preparation Tips

Practise setting up the objective function from a word problem — this translation step, not the calculus, is where most students lose marks. Master both derivative tests; CBSE frequently demands the second-derivative test explicitly. Pair with Ch 5 for a strong 2-week calculus block.

Chapter 6 · CBSE · Class XII
∿

Rate of Change of Quantities

Application of Derivatives NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Derivatives Rate of Change Increasing Functions Decreasing Functions Maxima and Minima Absolute Maximum Absolute Minimum Local Maxima Local Minima First Derivative Test Second Derivative Test Optimization Problems Related Rates Marginal Cost Marginal Revenue JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
19 sections
📖 Introduction
Rate of Change of Quantities
📘 Definition
Rate of Change
📘 Definition
Average Rate of Change
📘 Definition
Instantaneous Rate of Change
🔍 Geometrical Interpretation
Consider the curve
\[y=f(x)\]
The average rate of change between two points on the curve gives the slope of the secant line. As the second point approaches the first point, the secant line approaches the tangent line.

Consequently,
\[\frac{dy}{dx}\]
represents the slope of the tangent to the curve at the corresponding point.

Thus, rate of change and slope of a curve are two interpretations of the same derivative.
🔍 Interpretation
Physical Interpretation
Suppose the position \(s\) of a particle depends on time \(t\):
\[s=f(t)\]
Then its instantaneous velocity is
\[v=\frac{ds}{dt}\]
If velocity itself varies with time, then acceleration is the rate of change of velocity with respect to time:
\[a=\frac{dv}{dt}\]
Since
\[v=\frac{ds}{dt},\]
we obtain
\[\boxed{a=\frac{d^2s}{dt^2}}\]
This illustrates how successive derivatives describe successive rates of change.
🔢 Formula
Fundamental Formula
📌 Rate of Change with Respect to Time
📐 Derivation
Derivation of \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\)
Suppose both \(x\) and \(y\) are functions of a third variable \(t\):
\[x=x(t),\qquad y=y(t)\]
Differentiating \(y\) with respect to \(t\), using the chain rule,
\[\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\]
If
\[\frac{dx}{dt}\neq0,\]
dividing both sides by \(\frac{dx}{dt}\), we get
\[\boxed{\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}}\]
This result is particularly useful when \(x\) and \(y\) are given parametrically in terms of \(t\).
📎 Side Note
Related Rates
Problems in which two or more quantities vary with time and are connected by a mathematical relationship are called related-rate problems.

For example, if the radius \(r\) of a sphere changes with time, then its volume
\[V=\frac{4}{3}\pi r^3\]
also changes with time.

Differentiating with respect to \(t\),
\[\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\]
Therefore,
\[\boxed{\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}}\]
This equation connects the rate at which the radius changes with the rate at which the volume changes.
📌 Sign of the Rate of Change
📎 Units of Rate of Change
The units of a rate of change are obtained by dividing the units of the dependent variable by the unit of the independent variable.

For example, if distance is measured in metres and time in seconds,
\[\frac{ds}{dt}\]
has the unit
\[\mathrm{m\,s^{-1}}\]
Similarly, if area is measured in \(\mathrm{m^2}\) and time in seconds, then
\[\frac{dA}{dt}\]
has the unit
\[\mathrm{m^2\,s^{-1}}\]
✏️ Example
1
Example
If \(y=3x^2+5x-2\), find the rate of change of \(y\) with respect to \(x\) at \(x=2\).
The required rate is
\[\left.\frac{dy}{dx}\right|_{x=2}\]
  1. 1

    Differentiate the given function with respect to \(x\).

  2. 2

    Substitute \(x=2\).

  3. 3

    Interpret the resulting value as the instantaneous rate.

Given:
\[y=3x^2+5x-2\]
  1. Differentiate with respect to \(x\):
    \[\frac{dy}{dx}=6x+5\]
  2. At \(x=2\),
    \[\left.\frac{dy}{dx}\right|_{x=2}=6(2)+5=17\]
  3. Therefore, the rate of change of \(y\) with respect to \(x\) at \(x=2\) is
    \[\boxed{17}\]
2
Example
The radius of a spherical balloon is increasing at the rate of \(2\,\mathrm{cm/s}\). Find the rate at which its volume is increasing when the radius is \(5\,\mathrm{cm}\).
The volume of a sphere is
\[V=\frac{4}{3}\pi r^3\]
Since \(r\) changes with time, \(V\) also changes with time.
  1. 1

    Write the geometrical relation between \(V\) and \(r\).

  2. 2

    Differentiate both sides with respect to time \(t\).

  3. 3

    Substitute the given values of \(r\) and \(\frac{dr}{dt}\).

Given:
\[V=\frac{4}{3}\pi r^3\]
  1. Differentiate both sides with respect to \(t\):
    \[\frac{dV}{dt}=\frac{4}{3}\pi(3r^2)\frac{dr}{dt}\]
  2. Hence,
    \[\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\]
  3. Given
    \[r=5\text{ cm},\quad\frac{dr}{dt}=2\text{ cm/s}\]
  4. Therefore,
    \[\frac{dV}{dt}=4\pi(5)^2(2)\]
    \[\frac{dV}{dt}=4\pi(25)(2)=200\pi\]
  5. Therefore,
    \[\boxed{\frac{dV}{dt}=200\pi\ \mathrm{cm^3/s}}\]
3
Example
If
\[ x=t^2+1,\qquad y=t^3-2t, \]
find \(\frac{dy}{dx}\) at \(t=1\).
Since both \(x\) and \(y\) are expressed in terms of \(t\), use
\[ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}} {\dfrac{dx}{dt}}. \]
Given:
\[x=t^2+1,\quad y=t^3-2t.\]
  1. Differentiate both equations with respect to \(t\):
    \[\frac{dx}{dt}=2t\]
    and
    \[\frac{dy}{dt}=3t^2-2\]
  2. Therefore,
    \[\frac{dy}{dx}=\frac{3t^2-2}{2t}\]
  3. At \(t=1\),
    \[\left.\frac{dy}{dx}\right|_{t=1}=\frac{3(1)^2-2}{2(1)}=\frac{1}{2}\]
  4. Hence,
    \[\boxed{\frac{dy}{dx}=\frac12}\]
4
Example
The displacement of a particle is given by
\[ s=t^3-6t^2+9t+4. \]
Find its velocity at \(t=2\) seconds.
Given:
\[s=t^3-6t^2+9t+4\]
  1. Velocity is the rate of change of displacement with respect to time:
    \[v=\frac{ds}{dt}\]
  2. Therefore,
    \[v=3t^2-12t+9\]
  3. At \(t=2\),
    \[v=3(2)^2-12(2)+9\]
    \[=12-24+9=-3\]
  4. Hence,
    \[\boxed{v=-3\text{ units/s}}\]
    📝 The negative sign indicates that the particle's instantaneous velocity is directed opposite to the chosen positive direction.
5
Example
If
\[\frac{dy}{dx}=4x-6,\]
determine whether \(y\) is increasing or decreasing at \(x=1\).
  1. At \(x=1\),
    \[\frac{dy}{dx}=4(1)-6=-2\]
  2. Since
    \[\frac{dy}{dx}<0,\]
    the function is decreasing at \(x=1\).
  3. Therefore,
    \[\boxed{\text{\(y\) is decreasing at }x=1.}\]
🔗 Important Rate-of-Change Relations
The following relations occur frequently in NCERT, CBSE, JEE and other competitive examination problems.
  • If \(y=f(x)\),
    \[ \frac{dy}{dx}=f'(x). \]
  • If \(x=x(t)\) and \(y=y(t)\),
    \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt}\neq0. \]
  • If \(A=\pi r^2\),
    \[ \frac{dA}{dt} = 2\pi r\frac{dr}{dt}. \]
  • If \(V=\frac{4}{3}\pi r^3\),
    \[ \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}. \]
  • If \(C=2\pi r\),
    \[ \frac{dC}{dt} = 2\pi\frac{dr}{dt}. \]
  • If \(s=s(t)\),
    \[ v=\frac{ds}{dt}. \]
  • If \(v=v(t)\),
    \[ a=\frac{dv}{dt} = \frac{d^2s}{dt^2}. \]
🗒️ Significance for Board Examinations

The rate-of-change concept is important for CBSE board examinations because it connects the derivative with a practical interpretation rather than treating differentiation merely as an algebraic procedure. Questions may involve geometric quantities, motion, changing dimensions, or interpretation of the sign and magnitude of a derivative.

For board preparation, students should be able to:

  • identify the dependent and independent variables;
  • translate a word statement into a mathematical relation;
  • differentiate the relation with respect to the appropriate variable;
  • use the chain rule when quantities depend on time;
  • substitute the numerical values only after differentiation;
  • write appropriate units for the final rate;
  • interpret the sign of the derivative.
Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other entrance examinations, rate-of-change questions frequently test conceptual understanding of derivatives, chain rule, parametric differentiation, maxima and minima, tangent and normal, and related rates.

Competitive problems may combine rate of change with geometry, coordinate geometry, trigonometry, logarithmic differentiation, implicit differentiation or physical motion. The main skill is therefore not merely differentiation, but selecting the correct variable with respect to which differentiation must be performed.

⚡ Exam Tip
❌ Common Mistakes
  • Confusing average and instantaneous rate: The expression
    \[ \frac{\Delta y}{\Delta x} \]
    gives an average rate over an interval, whereas
    \[ \frac{dy}{dx} \]
    gives an instantaneous rate.
  • Ignoring the chain rule: If \(x\) depends on \(t\), simply differentiating \(y\) with respect to \(x\) may not answer a question asking for \(\frac{dy}{dt}\).
  • Substituting values too early: This can hide the functional relationship required for differentiation.
  • Forgetting units: A rate of change has units determined by the dependent and independent quantities.
  • Misinterpreting a negative derivative: A negative derivative indicates decreasing behaviour relative to the selected direction or independent variable.
  • Using \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\) when \(\frac{dx}{dt}=0\): The quotient form requires
    \[ \frac{dx}{dt}\neq0. \]
📋 Case Study
CBSE Case Study / HOTS

A spherical balloon is being inflated. Its radius \(r\) is increasing with time at a constant rate

\[ \frac{dr}{dt}=0.5\text{ cm/s}. \]
The volume of a sphere is
\[ V=\frac{4}{3}\pi r^3. \]
Determine the rate of increase of volume when \(r=10\) cm.

Question 1

Which variable is changing directly with time?

\[ \boxed{r} \]

Question 2

Which rate is given?

\[ \boxed{\frac{dr}{dt}=0.5\text{ cm/s}} \]

Question 3

Differentiate the volume relation with respect to \(t\).

\[ V=\frac{4}{3}\pi r^3 \]

\[ \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}. \]

Question 4

Find the rate of increase of volume when \(r=10\) cm.

\[ \frac{dV}{dt} = 4\pi(10)^2(0.5) \]

\[ \frac{dV}{dt} = 200\pi\text{ cm}^3/\text{s}. \]

Hence,

\[ \boxed{ \frac{dV}{dt}=200\pi\text{ cm}^3/\text{s} }. \]

Quick Revision

If \(y=f(x)\), then

\[ \boxed{\text{Rate of change of }y\text{ with respect to }x=\frac{dy}{dx}} \]

At \(x=x_0\),

\[ \boxed{ \text{Instantaneous rate}=f'(x_0) }. \]

Average rate of change:

\[ \boxed{ \frac{\Delta y}{\Delta x} = \frac{f(x_2)-f(x_1)}{x_2-x_1} }. \]

Parametric rate:

\[ \boxed{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt}\neq0 }. \]

Rate of change of area of a circle:

\[ \boxed{ \frac{dA}{dt} = 2\pi r\frac{dr}{dt} }. \]

Rate of change of volume of a sphere:

\[ \boxed{ \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} }. \]

Velocity:

\[ \boxed{ v=\frac{ds}{dt} }. \]

Acceleration:

\[ \boxed{ a=\frac{dv}{dt} = \frac{d^2s}{dt^2} }\]
.

🔑 Key Takeaway
∿

Example 1

📋
Table of Contents
6 sections
🗒️ Quetion
Find the rate of change of the area of a circle with respect to its radius \(r\) when \(r=5\) cm.
💡 Concept
🗺️ Roadmap

  1. Write the formula for the area of the circle.

  2. Differentiate the area with respect to \(r\).

  3. Substitute \(r=5\) cm into the derivative.

  4. Write the appropriate unit of the rate of change.

🧩 Solution
  1. The area of a circle of radius \(r\) is
    \[A=\pi r^2\]
  2. Differentiating both sides with respect to \(r\),
    \[\frac{dA}{dr}=\frac{d}{dr}\left(\pi r^2\right)\]
  3. Since \(\pi\) is a constant,
    \[\frac{dA}{dr}=\pi\frac{d}{dr}(r^2)\]
  4. Using
    \[\frac{d}{dr}(r^n)=nr^{n-1},\]
  5. we get
    \[\frac{dA}{dr}=\pi(2r)=2\pi r\]
  6. Hence, the rate of change of area with respect to radius is
    \[\boxed{\frac{dA}{dr}=2\pi r}\]
  7. At \(r=5\) cm,
    \[\left.\frac{dA}{dr}\right|_{r=5}=2\pi(5)=10\pi\]
  8. Therefore,
    \[\boxed{\left.\frac{dA}{dr}\right|_{r=5}=10\pi\ \mathrm{cm}}\]
🌟 Exam Insight

If the question instead stated that the radius is increasing at a particular rate, for example

\[ \frac{dr}{dt}=2\ \mathrm{cm/s}, \]
and asked for the rate of change of area with respect to time, then we would need
\[ \frac{dA}{dt}. \]

Using the chain rule,

\[ \frac{dA}{dt} = \frac{dA}{dr}\frac{dr}{dt}. \]

Since

\[ \frac{dA}{dr}=2\pi r, \]
we obtain
\[ \boxed{ \frac{dA}{dt} = 2\pi r\frac{dr}{dt} }. \]

This distinction between

\[ \frac{dA}{dr} \quad\text{and}\quad \frac{dA}{dt} \]
is a frequent source of errors in rate-of-change and related-rate problems.

🔑 Key Takeaway
∿

Example 2

📋
Table of Contents
9 sections
❓ Question
The volume of a cube is increasing at a rate of \(9\text{ cm}^3/\text{s}\). How fast is the surface area increasing when the length of an edge is \(10\) cm?
💡 Concept
🗒️ Geometrical Relations

For a cube with edge length \(x\), its volume is

\[ V=x^3 \]
and its total surface area is
\[ S=6x^2. \]

Since \(x\) changes with time, both \(V\) and \(S\) are functions of \(t\).

🗺️ Roadmap

  1. Write the formula for the volume of the cube.

  2. Differentiate the volume equation with respect to time.

  3. Use the given value of \(\frac{dV}{dt}\) to calculate \(\frac{dx}{dt}\).

  4. Write the surface-area formula.

  5. Differentiate it with respect to time.

  6. Substitute \(x=10\) cm and the previously calculated value of \(\frac{dx}{dt}\).

  7. State the answer with the correct units.

🧩 Solution
Find the Rate of Change of the Edge Length
  1. The volume of a cube with edge length \(x\) is
    \[V=x^3\]
  2. Since \(x\) varies with time, differentiate both sides with respect to \(t\):
    \[\frac{dV}{dt}=\frac{d}{dt}(x^3)\]
  3. Applying the chain rule,
    \[\frac{dV}{dt}=3x^2\frac{dx}{dt}\]
  4. We are given
    \[\frac{dV}{dt}=9\text{ cm}^3/\text{s}.\]
  5. Therefore,
    \[3x^2\frac{dx}{dt}=9\]
  6. When
    \[x=10\text{ cm},\]
  7. we get
    \[3(10)^2\frac{dx}{dt}=9\]
  8. Hence,
    \[300\frac{dx}{dt}=9\]
  9. Therefore,
    \[\frac{dx}{dt}=\frac{9}{300}=\frac{3}{100}\text{ cm/s}\]
  10. Thus,
    \[\boxed{\frac{dx}{dt}=\frac{3}{100}\text{ cm/s}}\]
Find the Rate of Change of Surface Area
  1. The total surface area of a cube is
    \[S=6x^2\]
  2. Differentiating both sides with respect to time \(t\),
    \[\frac{dS}{dt}=\frac{d}{dt}(6x^2)\]
  3. Applying the chain rule,
    \[\frac{dS}{dt}=12x\frac{dx}{dt}\]
  4. Substitute
    \[x=10\]
    and
    \[\frac{dx}{dt}=\frac{3}{100}\]
  5. Therefore,
    \[\frac{dS}{dt}=12(10)\left(\frac{3}{100}\right)\]
  6. Simplifying,
    \[\frac{dS}{dt}=\frac{360}{100}=3.6\]
  7. Hence,
    \[\boxed{\frac{dS}{dt}=3.6\text{ cm}^2/\text{s}}\]
Therefore, when the edge of the cube is \(10\) cm, its surface area is increasing at the rate of
\[ \boxed{3.6\text{ cm}^2/\text{s}}. \]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\frac{dV}{dt}=3x^2\): This omits the chain-rule factor \(\frac{dx}{dt}\). The correct expression is
    \[ \frac{dV}{dt}=3x^2\frac{dx}{dt}. \]
  • Differentiating \(S=6x^2\) as \(12x\): Because \(x\) changes with time, the correct derivative is
    \[ \frac{dS}{dt}=12x\frac{dx}{dt}. \]
  • Using \(9\text{ cm}^3/\text{s}\) directly as the answer: This is the rate of volume increase, not the rate of surface-area increase.
  • Forgetting the value of \(x\): The required rate depends on the instantaneous edge length, so \(x=10\) cm must be substituted.
  • Writing incorrect units: Since \(S\) is measured in \(\mathrm{cm}^2\), its rate with respect to time must be measured in \(\mathrm{cm}^2/\mathrm{s}\).
🌟 CBSE and Competitive Exam Significance

This example illustrates a standard related-rates application of derivatives. For CBSE Class 12 Mathematics, it develops the ability to translate a geometrical situation into a differentiable relation and apply the chain rule correctly.

For JEE Main, JEE Advanced and other competitive entrance examinations, the same principle can be extended to spheres, cones, cylinders, rectangles, triangles and composite geometrical figures. Problems may also combine related rates with optimisation, implicit differentiation, coordinate geometry or trigonometric relations.

🔑 Key Takeaway
∿

Example 3

📋
Table of Contents
10 sections
❓ Question
A stone is dropped into a quiet lake and waves move in circles at a speed of \(4\text{ cm/s}\). At the instant when the radius of the circular wave is \(10\) cm, how fast is the enclosed area increasing?
💡 Concept
🗺️ Roadmap

  1. Write the formula for the area of a circle.

  2. Differentiate the area with respect to time.

  3. Use the chain rule because \(r\) is changing with time.

  4. Substitute \(r=10\) cm and \(\frac{dr}{dt}=4\) cm/s.

  5. Interpret the result as the instantaneous rate at which the enclosed area is increasing.

🧩 Solution
  1. Write the Area Formula The area enclosed by a circular wave of radius \(r\) is
    \[A=\pi r^2\]
  2. Differentiate with Respect to Time Since \(r\) changes with time, differentiate both sides with respect to \(t\):
    \[\frac{dA}{dt}=\frac{d}{dt}(\pi r^2)\]
  3. Since \(\pi\) is constant,
    \[\frac{dA}{dt}=\pi\frac{d}{dt}(r^2)\]
  4. Applying the chain rule,
    \[\frac{d}{dt}(r^2)=2r\frac{dr}{dt}\]
  5. Therefore,
    \[\boxed{\frac{dA}{dt}=2\pi r\frac{dr}{dt}}\]
  6. Substitute the Given Values We have
    \[r=10\text{ cm}\]
    and
    \[\frac{dr}{dt}=4\text{ cm/s}\]
  7. Hence,
    \[\frac{dA}{dt}=2\pi(10)(4)\]
  8. Therefore,
    \[\frac{dA}{dt}=80\pi\]
  9. Thus, the enclosed area is increasing at the rate
    \[\boxed{80\pi\text{ cm}^2/\text{s}}\]
When the radius of the circular wave is \(10\) cm, the enclosed area is increasing at the instantaneous rate
\[\boxed{80\pi\text{ cm}^2/\text{s}}\]
🔍 Physical Interpretation

At \(r=10\) cm, the circumference of the wave is

\[ 2\pi r=20\pi\text{ cm}. \]

The boundary of the wave is expanding outward at

\[ 4\text{ cm/s}. \]

Therefore, the area swept out per second is

\[(20\pi)(4)=80\pi\text{ cm}^2/\text{s}\]

This gives an intuitive interpretation of the derivative:

\[\boxed{\frac{dA}{dt}=\text{circumference}\times\text{radial speed}}\]

👁️ Important Observation
💡 Sign of the Rate of Change
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\frac{dA}{dt}=2\pi r\): This is incomplete because it represents \(\frac{dA}{dr}\). The correct time rate is
    \[ \frac{dA}{dt}=2\pi r\frac{dr}{dt}. \]
  • Forgetting the given wave speed: The value \(4\) cm/s must be used as
    \[ \frac{dr}{dt}=4. \]
  • Using \(r=10\) before differentiating: The functional relation should first be differentiated and then the instantaneous value substituted.
  • Writing \(\mathrm{cm/s}\) instead of \(\mathrm{cm^2/s}\): The required quantity is an area rate, so its unit must be square centimetres per second.
  • Confusing rate of radius increase with rate of area increase:
    \[ \frac{dr}{dt}=4\text{ cm/s} \]
    is not the required answer. The required rate is
    \[ \frac{dA}{dt}=80\pi\text{ cm}^2/\text{s}. \]
🔑 Key Takeaway
∿

Example 4

📋
Table of Contents
9 sections
❓ Question
The length \(x\) of a rectangle is decreasing at the rate of \(3\text{ cm/min}\), and its width \(y\) is increasing at the rate of \(2\text{ cm/min}\). When \(x=10\) cm and \(y=6\) cm, find the rates of change of:

  1. the perimeter of the rectangle;
  2. the area of the rectangle.
💡 Concept
🗺️ Roadmap

  1. Write the formula for the perimeter and differentiate it with respect to time.

  2. Substitute the given rates of change of length and width.

  3. Write the formula for the area and differentiate it using the product rule.

  4. Substitute the instantaneous dimensions and their rates of change.

  5. Interpret the signs of the two final answers.

🧩 Solution
Given:
  • Length:
    \[ x=10\text{ cm} \]
  • Width:
    \[ y=6\text{ cm} \]
  • Rate of change of length:
    \[ \frac{dx}{dt}=-3\text{ cm/min} \]
  • Rate of change of width:
    \[ \frac{dy}{dt}=2\text{ cm/min} \]
Part (a)
Rate of Change of Perimeter
  1. The perimeter \(P\) of a rectangle is
    \[P=2(x+y)\]
  2. Since both \(x\) and \(y\) vary with time, differentiate both sides with respect to \(t\):
    \[\frac{dP}{dt}=\frac{d}{dt}\left[2(x+y)\right]\]
  3. Since \(2\) is constant,
    \[\frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)\]
  4. Substituting
    \[\frac{dx}{dt}=-3\]
    and
    \[\frac{dy}{dt}=2,\]
  5. we get
    \[\frac{dP}{dt}=2(-3+2)\]
  6. Therefore,
    \[\frac{dP}{dt}=2(-1)=-2\]
  7. Hence,
    \[\boxed{\frac{dP}{dt}=-2\text{ cm/min}}\]
  8. The negative sign means that the perimeter is <strong>decreasing</strong> at the rate of
    \[2\text{ cm/min}\]
  9. Equivalently, we may state:
    \[\boxed{\text{Perimeter decreases at }2\text{ cm/min}}\]
Part (b)
Rate of Change of Area
  1. The area \(A\) of a rectangle is
    \[A=xy\]
  2. Since both \(x\) and \(y\) are functions of time, differentiate with respect to \(t\). Using the product rule,
    \[\frac{dA}{dt}=\frac{d}{dt}(xy)\]
  3. Therefore,
    \[\boxed{\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}}\]
  4. Substitute
    \[x=10,\quad y=6,\]
    together with
    \[\frac{dx}{dt}=-3,\quad\frac{dy}{dt}=2\]
  5. Hence,
    \[\frac{dA}{dt}=10(2)+6(-3)\]
  6. Therefore,
    \[\frac{dA}{dt}=20-18=2\]
  7. Thus,
    \[\boxed{\frac{dA}{dt}=2\text{ cm}^2/\text{min}}\]
    📝 The positive sign means that the area is increasing at the rate of \(2\text{ cm}^2/\text{min}\).
Final Answer
  1. Rate of change of perimeter:
    \[ \boxed{ \frac{dP}{dt}=-2\text{ cm/min} } \]
    or the perimeter is decreasing at \(2\text{ cm/min}\).
  2. Rate of change of area:
    \[ \boxed{ \frac{dA}{dt}=2\text{ cm}^2/\text{min} } \]
    or the area is increasing at \(2\text{ cm}^2/\text{min}\).
🔍 Important Mathematical Insight

This example demonstrates that the signs of individual rates and the sign of the final rate must be interpreted carefully.

A quantity can increase even when one of the variables on which it depends is decreasing. For

\[ A=xy, \]
the net rate depends on the combined effect
\[ x\frac{dy}{dt}+y\frac{dx}{dt}. \]

Therefore, it is not correct to decide whether the area is increasing or decreasing merely by looking at the sign of \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\) individually.

⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\frac{dx}{dt}=3\): This ignores the word “decreasing”. The correct value is
    \[ \frac{dx}{dt}=-3. \]
  • Using \(\frac{dA}{dt}=xy\): This gives the area itself, not its rate of change.
  • Forgetting the product rule: Since
    \[ A=xy, \]
    the correct derivative is
    \[ \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}. \]
  • Assuming the area must decrease because the length decreases: The width is increasing simultaneously, so the net effect must be calculated.
  • Ignoring the units: Perimeter rate has units of length per time, while area rate has units of area per time.
🌟 Board and Competitive Exam Significance

This example combines three fundamental differentiation skills: related rates, linear differentiation and the product rule. It is especially useful for understanding how derivatives model simultaneous changes in multiple physical dimensions.

For CBSE board examinations, students should focus on correctly translating increasing and decreasing rates, differentiating the appropriate geometrical formula and interpreting the sign of the answer. For JEE Main, JEE Advanced and other competitive examinations, the same framework can be extended to more complex figures and problems involving several changing variables.

🔑 Key Takeaway
∿

Example 5

📋
Table of Contents
8 sections
❓ Question
The total cost \(C(x)\), in rupees, associated with the production of \(x\) units of an item is given by
\[ C(x)=0.005x^3-0.02x^2+30x+5000. \]
Find the marginal cost when \(3\) units are produced, where marginal cost means the instantaneous rate of change of total cost at any level of output.
💡 Concept
🗺️ Roadmap

  1. Identify the total cost function \(C(x)\).

  2. Differentiate \(C(x)\) with respect to \(x\).

  3. Obtain the marginal cost function \(MC(x)\).

  4. Substitute \(x=3\).

  5. Interpret the result economically.

🧩 Solution
Given:
\[C(x)=0.005x^3-0.02x^2+30x+5000\]
Differentiate the Total Cost Function
  1. Differentiating both sides with respect to \(x\),
    \[\frac{dC}{dx}=\frac{d}{dx}\left(0.005x^3-0.02x^2+30x+5000\right)\]
  2. Applying the power rule,
    \[\frac{d}{dx}(x^n)=nx^{n-1},\]
  3. we obtain
    \[\frac{dC}{dx}=0.005(3x^2)-0.02(2x)+30.\]
  4. Therefore,
    \[\frac{dC}{dx}=0.015x^2-0.04x+30\]
  5. Hence, the marginal cost function is
    \[\boxed{MC(x)=0.015x^2-0.04x+30}\]
Find the Marginal Cost at \(x=3\)
  1. Substituting
    \[x=3,\]
  2. we get
    \[MC(3)=0.015(3)^2-0.04(3)+30\]
  3. Since
    \[(3)^2=9,\]
  4. we have
    \[MC(3)=0.015(9)-0.12+30\]
  5. Therefore,
    \[MC(3)=0.135-0.12+30\]
  6. Hence,
    \[MC(3)=30.015\]
  7. Therefore,
    \[\boxed{MC(3)=\text{₹}30.015\text{ per unit}}\]
Final Answer

The marginal cost when \(3\) units are produced is
\[ \boxed{\text{₹}30.015\text{ per unit}}. \]
⚡ Exam Tip
❌ Common Mistakes
  • Forgetting the coefficient \(3\):
    \[ \frac{d}{dx}(0.005x^3)=0.015x^2. \]
  • Incorrectly differentiating the fixed cost:
    \[ \frac{d}{dx}(5000)=0. \]
  • Using \(C(3)\) instead of \(C'(3)\): The question asks for marginal cost, so the derivative of the cost function is required.
  • Confusing marginal cost with average cost:
    \[ MC=C'(x), \qquad AC=\frac{C(x)}{x}. \]
  • Ignoring units: Marginal cost is expressed as cost per unit of output.
🌟 Board and Competitive Exam Significance

This example demonstrates one of the most important real-world interpretations of derivatives: marginal analysis. For CBSE Class 12 Mathematics, it connects differentiation with economics and provides a direct application of the derivative as an instantaneous rate of change.

For competitive examinations, the same framework can be extended to marginal revenue, marginal profit, average cost, optimisation of production and relationships between total, average and marginal quantities.

🔑 Key Takeaway
∿

Example 6

📋
Table of Contents
4 sections
❓ Question
The total revenue in rupees received from the sale of \(x\) units of a product is given by
\[ R(x)=3x^2+36x+5. \]
Find the marginal revenue when \(x=5\), where marginal revenue means the rate of change of total revenue with respect to the number of items sold at an instant.
💡 Concept
🗺️ Roadmap

  1. Write the given total revenue function.

  2. Differentiate it with respect to \(x\).

  3. Obtain the marginal revenue function.

  4. Substitute \(x=5\).

  5. Interpret the resulting value as the instantaneous revenue rate.

🧩 Solution
Given:
\[R(x)=3x^2+36x+5\]
  1. Differentiate the Revenue Function Differentiating both sides of given function with respect to \(x\),
  2. Using the power rule,
  3. we obtain
  4. Therefore,
  5. Hence, the marginal revenue function is
  6. Find the Marginal Revenue at \(x=5\) Substituting \(x=5\;\) into the marginal revenue function,
  7. Therefore,
Final Answer
Hence, the marginal revenue when \(5\) units are sold is
\[ \boxed{ MR(5)=\text{₹}66\text{ per additional unit} }. \]
∿

Increasing and Decreasing Functions

📋
Table of Contents
19 sections
🗺️ Overview
The concepts of increasing and decreasing functions are fundamental applications of derivatives in NCERT Class 12 Mathematics Chapter 6, Application of Derivatives. They provide a systematic method for determining the intervals over which a function rises, falls, or remains constant.

The derivative plays a central role in this analysis. In general, a positive derivative indicates increasing behaviour, while a negative derivative indicates decreasing behaviour, subject to the differentiability conditions of the function.
📘 Definition
Decreasing Functions
Increasing Functions
📘 Definition
Let \(I\) be an interval contained in the domain of a real-valued function
\[ f:I\rightarrow\mathbb{R}. \]
The behaviour of \(f\) on \(I\) can be classified using comparisons between \(f(x_1)\) and \(f(x_2)\) whenever
\[ x_1 < x_2. \]
Strictly Increasing Function
A function \(f\) is said to be strictly increasing on an interval \(I\) if
\[x_1 < x_2,\qquad x_1,x_2\in I\]
implies
\[\boxed{f(x_1) < f(x_2)}\]
In simple terms, whenever \(x\) increases, the value of \(f(x)\) also increases.
Increasing Function
A function \(f\) is said to be increasing or non-decreasing on an interval \(I\) if
\[x_1 < x_2,\qquad x_1,x_2\in I\]
implies
\[\boxed{f(x_1)\leq f(x_2)} \]
Equality is permitted in this definition. Therefore, a function may remain constant over part of the interval and still be non-decreasing.
Strictly Decreasing Function
A function \(f\) is said to be strictly decreasing on an interval \(I\) if
\[x_1 < x_2,\qquad x_1,x_2\in I\]
implies
\[\boxed{f(x_1)>f(x_2)}\]
Thus, as \(x\) increases, the value of \(f(x)\) strictly decreases.
Decreasing Function
A function \(f\) is said to be decreasing or non-increasing on an interval \(I\) if
\[x_1 < x_2,\qquad x_1,x_2\in I\]
implies
\[\boxed{f(x_1)\geq f(x_2)}\]
Equality is permitted, so the function may remain constant over a portion of the interval.
Constant Function
A function \(f\) is said to be constant on an interval \(I\) if there exists a constant \(c\) such that
\[\boxed{f(x)=c,\qquad \forall x\in I}\]
A constant function has the same output for every input in the interval.

Its derivative, wherever it is defined, is
\[\boxed{f'(x)=0}\]
📝 Definition Summary
📜 Derivative Test for Increasing Functions
Conditions
📜 Conditions
One of the most important results in Application of Derivatives connects the sign of the first derivative with the monotonic behaviour of a function.

Let \(f\) be differentiable on an interval \(I\).

If
\[\boxed{f'(x)>0}\]
for every \(x\in I\), then \(f\) is strictly increasing on \(I\).

More generally, if
\[\boxed{f'(x)\geq0}\]
throughout \(I\), then \(f\) is non-decreasing on \(I\), under the standard differentiability assumptions.
📜 Derivative Test for Decreasing Functions
Conditions
📜 Conditions
If \(f\) is differentiable on an interval \(I\) and
\[\boxed{f'(x)<0}\]
for every \(x\in I\), then \(f\) is strictly decreasing on \(I\).

Similarly, if
\[\boxed{f'(x)\leq0}\]
throughout \(I\), then \(f\) is non-increasing on \(I\), under the standard differentiability assumptions.
🤔 Did You Know?
Why Does the Derivative Determine Increasing or Decreasing Behaviour?
The derivative
\[f'(x)\]
represents the instantaneous rate of change of \(f(x)\) with respect to \(x\).

If
\[f'(x)>0,\]
the tangent to the graph has positive slope. As \(x\) increases, the function therefore moves upward.

If
\[f'(x)<0,\]
the tangent has negative slope. As \(x\) increases, the function moves downward.

If
\[f'(x)=0,\]
the tangent is horizontal at that point. However, a zero derivative at one point does not by itself establish that the function is increasing, decreasing, maximum or minimum there.
📐 Derivation Using the Mean Value Theorem
The derivative test can be justified using the Mean Value Theorem (MVT). Suppose \(f\) is continuous on \([x_1,x_2]\) and differentiable on \((x_1,x_2)\), where
\[x_1 < x_2\]
By the Mean Value Theorem, there exists some \(c\in(x_1,x_2)\) such that
\[\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c)\]
Since
\[x_2-x_1>0,\]
the sign of
\[f(x_2)-f(x_1)\]
is determined by the sign of \(f'(c)\).

If
\[f'(x)>0\]
throughout the interval, then
\[f'(c)>0\]
Hence,
\[\frac{f(x_2)-f(x_1)}{x_2-x_1}>0\]
Since \(x_2-x_1>0\),
\[f(x_2)-f(x_1)>0\]
Therefore,
\[f(x_2)>f(x_1)\]
Hence \(f\) is strictly increasing on the interval.

Similarly, if
\[f'(x)<0\]
throughout the interval, then
\[f(x_2)-f(x_1)<0,\]
which gives
\[f(x_2) < f(x_1)\]
Hence, \(f\) is strictly decreasing.
🤔 How to Find Intervals of Increase and Decrease

To determine the intervals on which a differentiable function is increasing or decreasing, use the following systematic procedure.

  1. Find the first derivative \(f'(x)\).
  2. Solve
    \[ f'(x)=0 \]
    to obtain stationary or critical points where the derivative exists.
  3. Identify points where \(f'(x)\) does not exist but \(f(x)\) is defined.
  4. Use these points to divide the domain into suitable intervals.
  5. Determine the sign of \(f'(x)\) in each interval.
  6. Where \(f'(x)>0\), the function is increasing.
  7. Where \(f'(x)<0\), the function is decreasing.
  8. State the intervals carefully using the domain of the function.
🗒️ First Derivative Sign Chart
A sign chart is an efficient method for analysing monotonicity.

Suppose the critical points are
\[x=a,\quad x=b,\qquad a < b.\]
The real line is divided into
\[(-\infty,a),\quad (a,b),\quad (b,\infty)\]
Determine the sign of \(f'(x)\) in each interval:
Interval Sign of \(f'(x)\) Behaviour of \(f(x)\)
\((-\infty,a)\) \(+\) Increasing
\((a,b)\) \(-\) Decreasing
\((b,\infty)\) \(+\) Increasing
A change from \(+\) to \(-\) indicates a possible local maximum, while a change from \(-\) to \(+\) indicates a possible local minimum.
✏️ Example
1
Question
Find the intervals on which
\[f(x)=x^3-3x^2-9x+5\]
is increasing or decreasing.
  1. 1

    Find \(f'(x)\).

  2. 2

    Solve \(f'(x)=0\).

  3. 3

    Divide the real line into intervals using the critical points.

  4. 4

    Determine the sign of \(f'(x)\) in each interval.

  5. 5

    State the intervals of increase and decrease.

Given:
\[f(x)=x^3-3x^2-9x+5\]
  1. Differentiating
    \[f'(x)=3x^2-6x-9\]
  2. Factorising,
    \[f'(x)=3(x^2-2x-3)\]
  3. Therefore,
    \[f'(x)=3(x-3)(x+1)\]
  4. Critical points are obtained from
    \[f'(x)=0\]
  5. Hence,
    \[x=-1,\quad x=3\]
  6. These points divide the real line into
    \[(-\infty,-1),\quad(-1,3),\quad(3,\infty)\]
  7. For \(x<-1\), both \(x-3\) and \(x+1\) are negative, so
    \[f'(x)>0\]
  8. Hence, \(f\) is increasing on
    \[(-\infty,-1)\]
  9. For \(-1 < x <3\),
    \[x+1>0,\quad x-3 < 0,\]
    so
    \[f'(x) < 0\]
  10. Hence, \(f\) is decreasing on
    \[(-1,3)\]
  11. For \(x>3\), both factors are positive, so
    \[f'(x)>0\]
    Hence, \(f\) is increasing on
    \[(3,\infty)\]
  12. Therefore,
    \[\boxed{f(x)\text{ is increasing on }(-\infty,-1)\cup(3,\infty)}\]
    and
    \[\boxed{f(x)\text{ is decreasing on }(-1,3)}\]
Determine the intervals on which
\[f(x)=x^2-4x+7\]
is increasing and decreasing.
  1. Differentiate:
    \[f'(x)=2x-4\]
  2. Set
    \[f'(x)=0:\]
    \[2x-4=0\]
  3. Therefore,
    \[x=2\]
  4. For \(x < 2\),
    \[2x-4<0\]
  5. Hence, \(f\) is decreasing on
    \[(-\infty,2)\]
  6. For \(x>2\)
    \[2x-4>0\]
  7. Hence, \(f\) is increasing on
    \[(2,\infty)\]
  8. Therefore,
    \[ \boxed{f\text{ is decreasing on }(-\infty,2)}\]
    and
    \[\boxed{f\text{ is increasing on }(2,\infty)}\]
Constant Function: Consider
\[f(x)=7.\]
  1. For any \(x_1,x_2\),
    \[f(x_1)=7=f(x_2)\]
  2. Therefore, \(f\) is constant throughout its domain. Also
    \[f'(x)=0\]
  3. Thus, a constant function is both non-decreasing and non-increasing, but it is neither strictly increasing nor strictly decreasing.
📌 Important Distinction: \(f'(x)=0\)
📎 Critical Points and Monotonicity
A critical point is a point in the domain where either
\[f'(x)=0\]
or
\[f'(x)\]
does not exist.

Critical points are important because the sign of the derivative can change around them. Therefore, they are natural points at which to divide the domain while constructing a monotonicity table.

However, a critical point does not automatically represent a maximum or minimum. The sign of the derivative on either side must be examined.
🔗 Connection with Local Maxima and Minima
The first derivative test provides the following useful interpretation:
  • If \(f'(x)\) changes from positive to negative at \(x=a\), then \(f\) changes from increasing to decreasing, indicating a local maximum at \(x=a\).
  • If \(f'(x)\) changes from negative to positive at \(x=a\), then \(f\) changes from decreasing to increasing, indicating a local minimum at \(x=a\).
  • If the sign of \(f'(x)\) does not change around \(x=a\), then \(a\) is not a local maximum or minimum on the basis of the first derivative test.
🛠️ Applications of Increasing and Decreasing Functions
The analysis of monotonicity is useful in several areas of mathematics and applied sciences.
  • Determining intervals of increase and decrease of a function.
  • Locating local maxima and local minima.
  • Solving optimisation problems.
  • Analysing the behaviour of polynomial, rational, exponential and trigonometric functions.
  • Studying marginal cost, marginal revenue and marginal profit in economics.
  • Analysing velocity and acceleration in physical models.
  • Understanding the shape and behaviour of mathematical graphs.
⚡ Exam Tip
❌ Common Mistakes
  • Reversing the inequalities: For an increasing function,
    \[ x_1 < x_2\Rightarrow f(x_1) < f(x_2). \]
  • Confusing increasing with strictly increasing: In the non-decreasing definition, equality is allowed:
    \[ f(x_1)\leq f(x_2). \]
  • Confusing decreasing with strictly decreasing: For non-increasing functions,
    \[ f(x_1)\geq f(x_2), \]
    whereas strict decrease requires
    \[ f(x_1)>f(x_2). \]
  • Assuming \(f'(x)=0\) means maximum or minimum: A sign-change test is required.
  • Forgetting domain restrictions: Monotonicity intervals must lie within the domain of the function.
  • Using only one test point for the entire domain: The sign of \(f'(x)\) may change between critical points.
🌟 Significance for CBSE Board Examinations

Increasing and decreasing functions are central to questions involving the first derivative test, monotonicity, local extrema and optimisation in Class 12 Mathematics.

For board examinations, students should be comfortable with:

\[ f'(x)>0\Rightarrow\text{increasing} \]
and
\[ f'(x)<0\Rightarrow\text{decreasing}. \]

They should also be able to construct a sign table, identify critical points and state intervals correctly.

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other competitive examinations, monotonicity frequently appears as part of multi-concept problems involving inequalities, extrema, parameter-based functions, roots of equations and optimisation.

A strong understanding of derivative signs allows students to analyse the behaviour of a function without necessarily drawing its complete graph.

⚡ Quick Revision
For a differentiable function \(f\) on an interval \(I\):

\[\boxed{f'(x)>0\text{ on }I\Rightarrow f\text{ is strictly increasing on }I}\]
\[\boxed{f'(x)<0\text{ on }I\Rightarrow f\text{ is strictly decreasing on }I}\]
\[\boxed{f'(x)\geq0\text{ on }I\Rightarrow f\text{ is non-decreasing on }I}\]

\[\boxed{f'(x)\leq0\text{ on }I\Rightarrow f\text{ is non-increasing on }I }\]

If

\[f'(x)=0\]
throughout an interval, then
\[\boxed{f\text{ is constant on that interval}}\]
🔑 Key Takeaway
∿

Increasing or Decreasing at a Point

Application of Derivatives NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Derivatives Rate of Change Increasing Functions Decreasing Functions Maxima and Minima Absolute Maximum Absolute Minimum Local Maxima Local Minima First Derivative Test Second Derivative Test Optimization Problems Related Rates Marginal Cost Marginal Revenue JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
3 sections
📘 Definition
🔗 Connection with the Derivative
For a differentiable function, the sign of the derivative provides a convenient way of determining local increasing or decreasing behaviour.
,br> If there exists an open interval around \(x_0\) such that
\[ f'(x)>0 \]
throughout that interval, then \(f\) is strictly increasing in that neighbourhood and hence is increasing at \(x_0\).

Similarly, if
\[ f'(x) < 0 \]
throughout an open interval containing \(x_0\), then \(f\) is strictly decreasing in that neighbourhood and hence is decreasing at \(x_0\).

Thus, locally,
\[ \boxed{ f'(x)>0\Rightarrow f\text{ is increasing} } \]
and
\[ \boxed{ f'(x)<0\Rightarrow f\text{ is decreasing} }. \]
📌 Important Note
∿

Example 7

📋
Table of Contents
9 sections
❓ Question
Show that the function
\[ f(x)=7x-3 \]
is increasing on \(\mathbb{R}\).
💡 Concept
🗺️ Roadmap

  1. Choose arbitrary \(x_1,x_2\in\mathbb{R}\).

  2. Assume \(x_1

  3. Multiply the inequality by the positive number \(7\).

  4. Subtract \(3\) from both sides.

  5. Express the resulting inequality in terms of \(f(x_1)\) and \(f(x_2)\).

  6. Apply the definition of a strictly increasing function.

🧩 Solution
Solution Using the Definition
  1. Let
    \[x_1,x_2\in\mathbb{R}\]
    be any two real numbers such that
    \[x_1 < x_2\]
  2. Multiplying both sides by \(7\), which is positive, preserves the inequality:
    \[7x_1 < 7x_2\]
  3. Subtracting \(3\) from both sides gives
    \[7x_1-3 < 7x_2-3\]
  4. Since
    \[f(x)=7x-3\]
  5. we have
    \[f(x_1)=7x_1-3\]
    and
    \[f(x_2)=7x_2-3\]
  6. Therefore,
    \[f(x_1) < f(x_2)\]
  7. Hence, by the definition of a strictly increasing function,
    \[\boxed{f(x)=7x-3\text{ is strictly increasing on }\mathbb{R}.}\]
🔍 Graphical Interpretation
The function
\[f(x)=7x-3\]
is a straight line with slope
\[m=7.\]
Since its slope is positive, the graph rises from left to right. This is consistent with
\[f'(x)=7>0\]
Thus, the algebraic definition, derivative test and graphical interpretation all lead to the same conclusion.
❌ Common Mistakes
  • Writing \(f(x_1)>f(x_2)\) for an increasing function: The correct inequality is
    \[ f(x_1) whenever \(x_1 < x_2\).
  • Forgetting that \(x_1\) and \(x_2\) are arbitrary: The proof must work for every pair \(x_1,x_2\in\mathbb{R}\) satisfying \(x_1 < x_2\).
  • Ignoring the sign of the coefficient: Multiplication by a positive number preserves an inequality, whereas multiplication by a negative number reverses it.
  • Confusing increasing with decreasing: Always compare the corresponding function values after assuming \(x_1 < x_2\).
  • Stopping at \(7x_1 < 7x_2\): The proof should continue until the result is written as
    \[ f(x_1) < f(x_2). \]
⚡ Exam Tip
🌟 Board Examination Significance

This example is important because it demonstrates the formal definition of an increasing function. CBSE questions may ask students to prove monotonicity either directly from the definition or by using derivatives.

Students should be able to distinguish between:

\[ x_1 < x_2 \]
and
\[ f(x_1) < f(x_2). \]
The first compares inputs, while the second compares corresponding outputs.

Competitive Examination Significance

The derivative-based approach becomes particularly powerful for polynomial, rational, exponential and trigonometric functions. In competitive examinations, monotonicity is often combined with inequalities, equations, parameter-based functions, maxima-minima and graph analysis.

🔑 Key Takeaway
∿

Theorem 1: Derivative Test for Increasing and Decreasing Functions

📋
Table of Contents
10 sections
🧮 Theorem
🧮 Theorem
Statement
Let \(f\) be a real-valued function that is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\). Then:
  1. If
    \[ f'(x)>0 \]
    for every \(x\in(a,b)\), then \(f\) is strictly increasing on \([a,b]\).
  2. If
    \[ f'(x)<0 \]
    for every \(x\in(a,b)\), then \(f\) is strictly decreasing on \([a,b]\).
  3. If
    \[ f'(x)=0 \]
    for every \(x\in(a,b)\), then \(f\) is a constant function on \([a,b]\).
🤔 Why This Theorem Is Important
This theorem establishes the fundamental connection between the sign of the derivative and the monotonic behaviour of a function.

In practical problems, instead of comparing \(f(x_1)\) and \(f(x_2)\) directly for every possible pair of points, we can calculate \(f'(x)\) and study its sign.

The central idea can be summarized as:
  • Positive derivative means the function rises as \(x\) increases.
  • Negative derivative means the function falls as \(x\) increases.
  • Zero derivative throughout an interval means the function remains constant on that interval.
🧮 Theorem
🧮 Theorem
Part (a)
To Prove: We need to prove that \(f\) is strictly increasing on \([a,b]\).
  1. Choose Two Arbitrary Points
  2. Let
    \[x_1,x_2\in[a,b]\]
    be any two points such that
    \[x_1 < x_2\]
  3. Since \(x_1\) and \(x_2\) are arbitrary, proving
    \[f(x_1) < f(x_2)\]
    will establish that \(f\) is strictly increasing on \([a,b]\).
  4. Apply the Mean Value Theorem
  5. Since
    \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), it is also continuous on \([x_1,x_2]\) and differentiable on \((x_1,x_2)\).
  6. Therefore, by the Mean Value Theorem, there exists a point
    \[c\in(x_1,x_2)\]
    such that
    \[\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c)\]
  7. Multiplying both sides by \(x_2-x_1\), we obtain
    \[f(x_2)-f(x_1)=f'(c)(x_2-x_1)\]
  8. Determine the Sign of the Right-Hand Side
  9. Since
    \[c\in(x_1,x_2)\subset(a,b),\]
    the given condition \(f'(x)>0\) gives
    \[f'(c)>0\]
  10. Also, because
    \[x_1 < x_2,\]
    we have
    \[x_2-x_1>0\]
  11. Therefore,
    \[f'(c)(x_2-x_1)>0\]
  12. Determine the Sign of \(f(x_2)-f(x_1)\)
  13. From
    \[f(x_2)-f(x_1)=f'(c)(x_2-x_1),\]
    and the fact that the right-hand side is positive, we obtain
    \[f(x_2)-f(x_1)>0\]
  14. Hence,
    \[f(x_2)>f(x_1)\]
  15. Equivalently,
    \[f(x_1) < f(x_2)\]
  16. Apply the Definition of an Increasing Function
  17. We have shown that for arbitrary
    \[x_1,x_2\in[a,b],\quad x_1 < x_2,\]
    we have
    \[f(x_1) < f(x_2)\]
  18. Therefore, by the definition of a strictly increasing function,
    \[\boxed{f\text{ is strictly increasing on }[a,b].}\]
Part (b)
To Prove: \(f'(x)<0\) Implies \(f\) Is Decreasing
  1. Choose Two Arbitrary Points
  2. Let
    \[x_1,x_2\in[a,b]\]
    such that
    \[x_1 < x_2\]
  3. Apply the Mean Value Theorem
  4. Since \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), the Mean Value Theorem guarantees the existence of some
    \[c\in(x_1,x_2)\]
    such that
    \[\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c)\]
  5. Therefore,
    \[f(x_2)-f(x_1)=f'(c)(x_2-x_1)\]
  6. Determine the Sign
  7. Since
    \[c\in(a,b),\]
    and
    \[f'(x)<0\]
    throughout \((a,b)\), we have
    \[f'(c)<0\]
  8. Also,
    \[x_2-x_1>0\]
  9. Hence,
    \[f'(c)(x_2-x_1)<0\]
  10. Thus,
    \[f(x_2)-f(x_1) < 0\]
  11. Therefore,
    \[f(x_2) < f(x_1),\]
    or equivalently,
    \[f(x_1)>f(x_2)\]
  12. Conclusion
  13. We have shown that for every
    \[x_1,x_2\in[a,b]\]
    satisfying
    \[x_1 < x_2,\]
  14. we have
    \[f(x_1)>f(x_2)\]
  15. Hence, by the definition of a strictly decreasing function,
    \[\boxed{f\text{ is strictly decreasing on }[a,b].}\]
Part (c)
To Prove: \(f'(x)=0\) Implies \(f\) Is Constant
  1. Choose Two Arbitrary Points
  2. Let
    \[x_1,x_2\in[a,b]\]
    be any two points such that
    \[x_1 < x_2\]
  3. Apply the Mean Value Theorem
  4. By the Mean Value Theorem, there exists a point
    \[c\in(x_1,x_2)\]
    such that
    \[\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c)\]
  5. Hence,
    \[f(x_2)-f(x_1)=f'(c)(x_2-x_1)\]
  6. Use \(f'(c)=0\)
  7. Since
    \[c\in(a,b)\]
    and
    \[f'(x)=0\]
    for every \(x\in(a,b)\), we have
    \[f'(c)=0\]
  8. Therefore,
    \[f(x_2)-f(x_1)=0(x_2-x_1)=0\]
  9. Hence,
    \[f(x_2)-f(x_1)=0,\]
    which gives
    \[f(x_2)=f(x_1)\]
  10. Establish Constancy
  11. Since \(x_1\) and \(x_2\) were arbitrary points of \([a,b]\), the function takes the same value at every pair of points in the interval.

    Therefore, there exists a constant \(C\) such that
    \[f(x)=C\qquad\text{for every }x\in[a,b]\]
  12. Hence,
    \[\boxed{f\text{ is a constant function on }[a,b].}\]
📝 Complete Theorem at a Glance
🔍 Geometrical Interpretation

The derivative

\[ f'(x) \]
represents the slope of the tangent to the graph of \(y=f(x)\).

  • If
    \[ f'(x)>0, \]
    the tangent has positive slope, so the graph rises from left to right.
  • If
    \[ f'(x)<0, \]
    the tangent has negative slope, so the graph falls from left to right.
  • If
    \[ f'(x)=0 \]
    at every point of an interval, the tangent slope is zero throughout that interval and the graph is horizontal.
📌 Important Distinction
🗒️ Role Of The Mean Value Theorem

The Mean Value Theorem is the key mathematical tool behind this result. It connects the average rate of change between two points with the instantaneous rate of change at some intermediate point.

Specifically,

\[ \frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c) \]
for some
\[ c\in(x_1,x_2). \]

Since

\[ x_2-x_1>0, \]
the sign of
\[ f(x_2)-f(x_1) \]
is exactly the sign of
\[ f'(c). \]

This is the fundamental reason why the sign of the derivative determines whether a function is increasing, decreasing or constant.

🗒️ Exam Oriented Shortcut

For a differentiable function, the practical procedure is:

  1. Find \(f'(x)\).
  2. Determine where \(f'(x)>0\).
  3. Determine where \(f'(x)<0\).
  4. Use the resulting intervals to identify where \(f\) is increasing or decreasing.
  5. If \(f'(x)=0\) throughout an entire interval, conclude that \(f\) is constant there.

Thus, the central result to remember is

\[ \boxed{ \begin{aligned} f'(x)>0 &\Rightarrow f\text{ is strictly increasing},\\ f'(x)<0 &\Rightarrow f\text{ is strictly decreasing},\\ f'(x)=0 &\Rightarrow f\text{ is constant}, \end{aligned} } \]
provided the hypotheses of the theorem are satisfied.

❌ Common Mistakes
  • Applying the theorem without checking the required continuity and differentiability conditions.
  • Writing \(f'(x_0)=0\) and concluding that \(f\) is constant. The derivative must be zero at every point of the interval.
  • Forgetting that the Mean Value Theorem produces a point \(c\) strictly between \(x_1\) and \(x_2\).
  • Forgetting that
    \[ x_10. \]
  • Reversing the conclusion for a negative derivative. A negative derivative implies decreasing, not increasing.
  • Using \(f'(x)\geq0\) or \(f'(x)\leq0\) as though they were exactly the same hypotheses as the strict inequalities in this theorem. Those cases require the appropriate non-strict monotonicity result and should be handled carefully.
🔑 Key Takeaway
∿

Example 8

📋
Table of Contents
7 sections
❓ Question
Show that the function
\[ f(x)=x^3-3x^2+4x,\quad x\in\mathbb{R} \]
is increasing on \(\mathbb{R}\)
💡 Concept
🗺️ Roadmap

  1. Differentiate the given function.

  2. Factor or rewrite the derivative in a form whose sign is easy to determine.

  3. Show that the derivative is strictly positive for every \(x\in\mathbb{R}\).

  4. Apply the increasing-function theorem.

🧩 Solution
Given:
\[f(x)=x^3-3x^2+4x\]
  1. Find the First Derivative Differentiating term by term,
    \[f'(x)=3x^2-6x+4\]
  2. We now need to determine the sign of
    \[3x^2-6x+4\]
  3. Complete the Square Rewrite the derivative as
    \[f'(x)=3x^2-6x+4\]
  4. Taking \(3\) common from the first two terms,
    \[f'(x)=3(x^2-2x)+4\]
  5. Complete the square:
    \[x^2-2x=(x-1)^2-1\]
  6. Therefore,
    \[ \begin{aligned} f'(x) &=3\left[(x-1)^2-1\right]+4\\ &=3(x-1)^2-3+4\\ &=3(x-1)^2+1. \end{aligned} \]
  7. Determine the Sign of \(f'(x)\) For every real number \(x\),
    \[(x-1)^2\geq0\]
  8. Therefore,
    \[3(x-1)^2\geq0\]
  9. Adding \(1\) to both sides,
    \[3(x-1)^2+1\geq1\]
  10. Hence,
    \[f'(x)=3(x-1)^2+1\geq1>0\quad\forall x\in\mathbb{R}\]
  11. In particular,
    \[\boxed{f'(x)>0\quad\forall x\in\mathbb{R}.}\]
  12. Apply the Theorem Since \(f\) is a polynomial, it is continuous and differentiable for every
    \[x\in\mathbb{R}\]
  13. We have also established that
    \[f'(x)>0\quad\forall x\in\mathbb{R}\]
  14. Therefore, by the derivative test for increasing functions,
    \[\boxed{f(x)=x^3-3x^2+4x\text{ is strictly increasing on }\mathbb{R}}\]
❌ Common Mistakes
  • Writing the derivative incorrectly as
    \[ 3x^2-6x^2+4. \]
    The correct derivative is
    \[ \boxed{f'(x)=3x^2-6x+4}. \]
  • Forgetting the linear term:
    \[ \frac{d}{dx}(4x)=4. \]
  • Concluding that \(f'(x)\geq0\) without establishing whether equality can occur. Here,
    \[ f'(x)\geq1, \]
    so the derivative is strictly positive.
  • Checking the derivative at only a few values of \(x\). To prove increase on all of \(\mathbb{R}\), the inequality must hold for every real \(x\).
  • Forgetting that a polynomial is continuous and differentiable everywhere.
⚡ Exam Tip
🔑 Key Takeaway
∿

Example 9

Application of Derivatives NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Derivatives Rate of Change Increasing Functions Decreasing Functions Maxima and Minima Absolute Maximum Absolute Minimum Local Maxima Local Minima First Derivative Test Second Derivative Test Optimization Problems Related Rates Marginal Cost Marginal Revenue JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
8 sections
❓ Question
Prove that the function
\[f(x)=\cos x\]
is:
  1. decreasing in \((0,\pi)\),
  2. increasing in \((\pi,2\pi)\),
  3. neither increasing nor decreasing in \((0,2\pi)\).
💡 Concept
🗺️ Roadmap

  1. Find \(f'(x)\).

  2. Determine the sign of \(\sin x\) on \((0,\pi)\).

  3. Use \(f'(x)=-\sin x\) to determine the sign of the derivative.

  4. Repeat the analysis on \((\pi,2\pi)\).

  5. Combine the results to determine the behaviour on the entire interval \((0,2\pi)\).

🧩 Solution
Given:
\[f(x)=\cos x\]
Part (a): Prove That \(f(x)=\cos x\) Is Decreasing in \((0,\pi)\)
  1. Find the Derivative Differentiating with respect to \(x\),
    \[\boxed{f'(x)=-\sin x}\]
  2. For
    \[0 < x < \pi,\]
    the sine function is positive:
    \[\sin x>0\]
  3. Therefore,
    \[f'(x)=-\sin x < 0\qquad\text{for every }x\in(0,\pi)\]
  4. Since the derivative is strictly negative throughout \((0,\pi)\), by the derivative test,
    \[\boxed{f(x)=\cos x\text{ is strictly decreasing in }(0,\pi).}\]
Part (b): Prove That \(f(x)=\cos x\) Is Increasing in \((\pi,2\pi)\)
  1. for
    \[\pi < x < 2\pi,\]
    the sine function is negative:
    \[\sin x < 0.\]
  2. Therefore,
    \[f'(x)=-\sin x>0 \quad\text{for every }x\in(\pi,2\pi).\]
  3. Since
    \[f'(x)>0\]
    throughout the interval, the derivative test gives
    \[\boxed{f(x)=\cos x\text{ is strictly increasing in }(\pi,2\pi)}\]
Part (c): Prove That \(f(x)=\cos x\) Is Neither Increasing Nor Decreasing in \((0,2\pi)\)
  1. From parts (a) and (b), we know that:
    \[f(x)=\cos x\]
    is decreasing on
    \[(0,\pi)\]
    and increasing on
    \[(\pi,2\pi)\]
  2. Therefore, its monotonic behaviour changes at
    \[x=\pi\]
  3. We can also see this directly from its derivative:
    \[f'(x)=-\sin x\]
    On \((0,\pi)\),
    \[f'(x) < 0,\]
    whereas on \((\pi,2\pi)\),
    \[f'(x)>0\]
  4. Thus, \(f'(x)\) does not have one fixed sign throughout \((0,2\pi)\).
    Consequently,
    \[\boxed{f(x)=\cos x\text{ is neither increasing nor decreasing on }(0,2\pi)}\]
  5. Sign Chart for \(f'(x)\)
    Interval Sign of \(\sin x\) \(f'(x)=-\sin x\) Behaviour of \(f(x)=\cos x\)
    \((0,\pi)\) \(+\) \(-\) Decreasing
    \((\pi,2\pi)\) \(-\) \(+\) Increasing
🔍 Graphical Interpretation
The graph of
\[y=\cos x\]
falls continuously from \(x=0\) to \(x=\pi\), reaching its minimum value
\[-1\]
at
\[x=\pi.\]
It then rises from \(x=\pi\) to \(x=2\pi\), reaching the value
\[1\]
at
\[x=2\pi.\]
Thus, the change from decreasing to increasing occurs at
\[x=\pi.\]
Important Observation About \(x=\pi\)
At
\[x=\pi,\]
we have
\[f'(\pi)=-\sin\pi=0.\]
Moreover, the derivative changes sign from negative to positive:
\[f'(x)<0\quad\text{on }(0,\pi),\]
and
\[f'(x)>0\quad\text{on }(\pi,2\pi).\]
Therefore, \(x=\pi\) is a point where the function changes from decreasing to increasing. In fact,
\[f(\pi)=\cos\pi=-1,\]
so \(f\) has a local minimum at \(x=\pi\).
❌ Common Mistakes
  • Using endpoint derivative values: Calculating \(f'(0)\) and \(f'(\pi)\) does not prove the sign of \(f'\) inside \((0,\pi)\).
  • Forgetting the negative sign:
    \[ \frac{d}{dx}(\cos x)=-\sin x, \]
    not \(\sin x\).
  • Claiming that \(\cos x\) is decreasing on all of \((0,2\pi)\): It starts increasing after \(x=\pi\).
  • Using the sign of \(\sin x\) instead of the sign of \(f'(x)\): Since
    \[ f'(x)=-\sin x, \]
    the sign must be reversed.
  • Assuming that \(f'(\pi)=0\) alone proves a minimum: The sign change
    \[ -\to+ \]
    is what establishes the local minimum through the first derivative test.
⚡ Exam Tip
🔑 Key Takeaway
∿

Example 10

Application of Derivatives NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Derivatives Rate of Change Increasing Functions Decreasing Functions Maxima and Minima Absolute Maximum Absolute Minimum Local Maxima Local Minima First Derivative Test Second Derivative Test Optimization Problems Related Rates Marginal Cost Marginal Revenue JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
9 sections
❓ Question
Find the intervals in which the function
\[f(x)=x^2-4x+6\]
is:
  1. increasing
  2. decreasing
💡 Concept
🗺️ Roadmap

  1. Find \(f'(x)\).

  2. Set \(f'(x)>0\) to determine the interval of increase.

  3. Set \(f'(x)<0\) to determine the interval of decrease.

  4. Identify the critical point where \(f'(x)=0\).

  5. State the final intervals using interval notation.

🧩 Solution
Given:
\[f(x)=x^2-4x+6\]
Differentiating with respect to \(x\),
\[f'(x)=2x-4.\]
Factorizing,
\[f'(x)=2(x-2)\]
Part (a)
Interval in Which \(f\) Is Increasing
  1. A function is increasing wherever
    \[f'(x)>0\]
  2. Therefore,
    \[2x-4>0\]
  3. Adding \(4\) to both sides,
    \[2x>4\]
  4. Dividing by the positive number \(2\),
    \[x>2\]
  5. Hence,
    \[\boxed{x\in(2,\infty)}.\]
  6. Therefore,
    \[\boxed{f(x)\text{ is strictly increasing on }(2,\infty)}\]
Part (b)
Interval in Which \(f\) Is Decreasing
  1. A function is decreasing wherever
    \[f'(x) < 0\]
  2. Therefore,
    \[2x-4 < 0\]
  3. Adding \(4\) to both sides,
    2x < 4
  4. Dividing by the positive number \(2\),
    \[x < 2\]
  5. Hence,
    \[\boxed{x\in(-\infty,2)}\]
  6. Therefore,
    \[\boxed{f(x)\text{ is strictly decreasing on }(-\infty,2).}\]
📌
Note The critical point is obtained by solving
\[f'(x)=0.\]
Thus,
\[2x-4=0,\]
giving
\[x=2.\]
The point \(x=2\) separates the domain into two intervals:
\[(-\infty,2)\quad\text{and}\quad(2,\infty).\]
🗒️ Sign Analysis Of The Derivative
Interval Sign of \(x-2\) Sign of \(f'(x)=2(x-2)\) Behaviour of \(f\)
\((-\infty,2)\) Negative Negative Decreasing
\(x=2\) Zero Zero Critical point
\((2,\infty)\) Positive Positive Increasing
🔍 Graphical Interpretation
The function can be written as
\[f(x)=x^2-4x+6\]
Completing the square,
\[\begin{aligned}f(x)&=x^2-4x+4+2\\&=(x-2)^2+2\end{aligned}\]
This is an upward-opening parabola. Its vertex occurs at
\[x=2\]
At the vertex,
\[f(2)=(2-2)^2+2=2\]
Hence, the vertex is
\[\boxed{(2,2)}\]
The graph decreases as \(x\) approaches \(2\) from the left and increases as \(x\) moves to the right of \(2\). Therefore, \(x=2\) is the point where the function changes from decreasing to increasing.
❌ Common Mistakes
  • Writing
    \[ f'(x)=2x-4 \]
    and stopping without determining its sign.
  • Including \(x=2\) inside the increasing or decreasing intervals. At \(x=2\),
    \[ f'(2)=0, \]
    so the strict derivative tests \(f'(x)>0\) and \(f'(x)<0\) do not apply at that point.
  • Forgetting that dividing an inequality by a positive number preserves its direction.
  • Reversing the intervals and claiming that the function is increasing for \(x<2\). The correct behaviour is decreasing for \(x<2\) and increasing for \(x>2\).
  • Confusing the critical point \(x=2\) with the minimum value \(f(2)=2\). The former is an \(x\)-coordinate; the latter is the corresponding function value.
⚡ Exam Tip
🔑 Key Takeaway
∿

Example 11

📋
Table of Contents
8 sections
❓ Question
Find the intervals in which the function
\[f(x)=4x^3-6x^2-72x+30\]
is:
  1. increasing
  2. decreasing
💡 Concept
🗺️ Roadmap

  1. Find \(f'(x)\).

  2. Factorize \(f'(x)\).

  3. Find the critical points by solving \(f'(x)=0\).

  4. Divide the real line into intervals using the critical points.

  5. Determine the sign of \(f'(x)\) in each interval.

  6. Use the derivative test to state the increasing and decreasing intervals.

🧩 Solution
Given:
\[f(x)=4x^3-6x^2-72x+30\]
Part (a)
  1. Find the First Derivative Differentiating term by term,
    \[\begin{aligned}f'(x)&=12x^2-12x-72\end{aligned}\]
  2. Taking \(12\) common,
    \[f'(x)=12(x^2-x-6)\]
  3. Factorizing the quadratic,
    \[x^2-x-6=(x-3)(x+2)\]
  4. Therefore,
    \[\boxed{f'(x)=12(x-3)(x+2)}\]
  5. Critical points occur where
    \[f'(x)=0\]
  6. Therefore,
    \[12(x-3)(x+2)=0\]
  7. Since \(12\neq0\),
    \[(x-3)(x+2)=0\]
  8. Hence,
    \[x-3=0\]
    or
    \[x+2=0\]
  9. Thus,
    \[\boxed{x=3,\quad x=-2}\]
  10. These two critical points divide the real number line into three intervals:
    \[(-\infty,-2),\qquad(-2,3),\qquad(3,\infty)\]
Part (b)
Interval in Which \(f\) Is Increasing
  1. The function is increasing where
    \[f'(x)>0\]
  2. Since
    \[f'(x)=12(x-3)(x+2),\]
    and \(12>0\), we need
    \[(x-3)(x+2)>0\]
  3. ⭐ A product of two factors is positive when both factors have the same sign.
  4. Case 1: Both Factors Are Positive We require
    \[x-3>0\]
    and
    \[x+2>0\]
  5. These give
    \[x>3\]
    and
    \[x>-2\]
  6. Both conditions are satisfied when
    \[x>3\]
  7. Therefore,
    \[x\in(3,\infty)\]
  8. Case 2: Both Factors Are Negative We would require
    \[x-3<0\]
    and
    \[x+2<0\]
  9. These give
    \[x < 3\]
    and
    \[x < -2\]
  10. Both conditions are satisfied when
    \[x < -2\]
  11. However, this case also gives a positive product:
    \[(x-3)(x+2)>0\]
  12. Therefore, the complete solution is
    \[\boxed{x\in(-\infty,-2)\cup(3,\infty)}\]
  13. Hence,
    \[\boxed{f(x)\text{ is strictly increasing on }(-\infty,-2)\cup(3,\infty)}\]
Part (c)
Interval in Which \(f\) Is Decreasing
  1. The function is decreasing where
    \[f'(x) < 0\]
  2. Therefore,
    \[12(x-3)(x+2) < 0\]
  3. Since \(12>0\),
    \[(x-3)(x+2) < 0\]
  4. ⭐ A product of two factors is negative when the two factors have opposite signs.
  5. Sign Analysis Consider the three intervals determined by the critical points:
    \[(-\infty,-2),\quad(-2,3),\quad(3,\infty)\]
  6. For
    \[x\in(-\infty,-2),\]
    both
    \[x-3 < 0\]
    and
    \[x+2 < 0,\]
    so
    \[(x-3)(x+2)>0\]
  7. Hence,
    \[f'(x) > 0\]
  8. For
    \[x\in(-2,3),\]
  9. we have
    \[x-3 < 0\]
    but
    \[x+2>0\]
  10. Therefore,
    \[(x-3)(x+2)<0\]
  11. Hence,
    \[f'(x) < 0\]
  12. For
    \[x\in(3,\infty),\]
    both
    \[x-3>0\]
    and
    \[x+2>0,\]
    so
    \[(x-3)(x+2)>0\]
  13. Hence,
    \[f'(x)>0\]
  14. Therefore,
    \[\boxed{x\in(-2,3)}\]
  15. Hence,
    \[\boxed{f(x)\text{ is strictly decreasing on }(-2,3)}\]
  16. Sign Chart
    Interval \(x-3\) \(x+2\) \(f'(x)=12(x-3)(x+2)\) Behaviour
    \((-\infty,-2)\) \(-\) \(-\) \(+\) Increasing
    \(x=-2\) \(-\) \(0\) \(0\) Critical point
    \((-2,3)\) \(-\) \(+\) \(-\) Decreasing
    \(x=3\) \(0\) \(+\) \(0\) Critical point
    \((3,\infty)\) \(+\) \(+\) \(+\) Increasing
🗒️ Sign Line Method
The derivative can be written as
\[f'(x)=12(x+2)(x-3)\]
The critical points in ascending order are
\[-2,\quad 3\]
The sign pattern of the two linear factors is:
\[\begin{array}{c|ccc}x & (-\infty,-2) & (-2,3) & (3,\infty)\\ \hline x+2 & - & + & +\\ x-3 & - & - & +\\ \hline f'(x) & + & - & +\end{array}\]
Therefore, the derivative sign pattern is
\[\boxed{+\quad-\quad+}\]
Hence the function:
\[\boxed{\text{increases}\quad\to\quad\text{decreases}\quad\to\quad\text{increases}.}\]
❌ Common Mistakes
  • Forgetting to include both intervals when solving a quadratic inequality of the form
    \[ (x-a)(x-b)>0. \]
  • Assuming that \(x>3\) is the only solution of
    \[ (x-3)(x+2)>0. \]
  • Solving the two inequalities separately and taking their union incorrectly. For a positive product, both factors must have the same sign.
  • Forgetting the critical points
    \[ x=-2,\quad x=3. \]
  • Including critical points in intervals where the derivative is required to be strictly positive or strictly negative.
⚡ Exam Tip
🔑 Key Takeaway
∿

Example 12

📋
Table of Contents
9 sections
❓ Question
Find the intervals in which the function
\[ f(x)=\sin 3x, \qquad x\in\left[0,\frac{\pi}{2}\right] \]
is increasing and decreasing.
💡 Concept
🗺️ Roadmap

  1. Differentiate \(f(x)=\sin 3x\).

  2. Solve \(f'(x)=0\) within the given domain.

  3. Use the critical point to divide the domain into subintervals.

  4. Determine the sign of \(f'(x)\) on each subinterval.

  5. State the increasing and decreasing intervals.

🧩 Solution
Given:
\[f(x)=\sin 3x\]
  1. Find the First Derivative Using the chain rule,
    \[\frac{d}{dx}(\sin 3x)=\cos 3x\cdot\frac{d}{dx}(3x)\]
  2. Hence,
    \[\boxed{f'(x)=3\cos 3x}\]
  3. Find the Critical Points Critical points in the given interval occur where
    \[f'(x)=0\]
  4. Therefore,
    \[3\cos 3x=0\]
  5. Since \(3\neq0\),
    \[\cos 3x=0\]
  6. The general solution of \(\cos\theta=0\) is
    \[\theta=\frac{(2n+1)\pi}{2},\quad n\in\mathbb{Z}.\]
  7. Hence,
    \[3x=\frac{(2n+1)\pi}{2}\]
  8. Since
    \[x\in\left[0,\frac{\pi}{2}\right],\]
  9. multiplying the entire interval by \(3\) gives
    \[3x\in\left[0,\frac{3\pi}{2}\right].\]
  10. The values of \(3x\) in this interval for which \(\cos 3x=0\) are
    \[3x=\frac{\pi}{2}\]
    and
    \[3x=\frac{3\pi}{2}\]
  11. These correspond to
    \[x=\frac{\pi}{6}\]
    and
    \[x=\frac{\pi}{2}\]
  12. The point
    \[x=\frac{\pi}{2}\]
    is an endpoint of the given domain, not an interior point. Therefore, the only interior critical point is
    \[\boxed{x=\frac{\pi}{6}}\]
  13. Divide the Domain
    The interior critical point
    \[x=\frac{\pi}{6}\]
    divides the domain into
    \[\left(0,\frac{\pi}{6}\right)\]
    and
    \[\left(\frac{\pi}{6},\frac{\pi}{2}\right)\]
    📝 We now determine the sign of \(f'(x)\) on each interval.
Part (a): Increasing Interval
  1. We have
    \[f'(x)=3\cos 3x\]
  2. For
    \[0 < x < \frac{\pi}{6},\]
  3. multiplying by \(3\) gives
    \[0 < 3 x < \frac{\pi}{2}\]
  4. In the first quadrant,
    \[\cos 3x>0\]
  5. Therefore,
    \[f'(x)=3\cos 3x>0\]
  6. Hence, \(f\) is strictly increasing on
    \[\boxed{\left(0,\frac{\pi}{6}\right)}\]
  7. Since the original function is continuous on the closed interval and increasing up to the critical point, it is also appropriate in elementary monotonicity language to state that the function is increasing on
    \[\boxed{\left[0,\frac{\pi}{6}\right]}\]
    The strict derivative condition itself applies on the open interval.
Part (b): Decreasing Interval
  1. For
    \[\frac{\pi}{6} < x < \frac{\pi}{2},\]
  2. multiplying by \(3\) gives
    \[\frac{\pi}{2} < 3 x < \frac{3\pi}{2}\]
  3. In this range,
    \[\cos 3x < 0\]
  4. Therefore,
    \[f'(x)=3\cos 3x < 0\]
  5. Hence, \(f\) is strictly decreasing on
    \[\boxed{\left(\frac{\pi}{6},\frac{\pi}{2}\right)}.\]
🗒️ Sign Analysis Table
Interval Range of \(3x\) Sign of \(\cos 3x\) Sign of \(f'(x)\) Behaviour
\(\left(0,\frac{\pi}{6}\right)\) \(\left(0,\frac{\pi}{2}\right)\) Positive Positive Increasing
\(x=\frac{\pi}{6}\) \(3x=\frac{\pi}{2}\) Zero Zero Critical point
\(\left(\frac{\pi}{6},\frac{\pi}{2}\right)\) \(\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\) Negative Negative Decreasing
🔍 Graphical Interpretation
The function is
\[f(x)=\sin 3x\]
At
\[x=0,\]
we have
\[f(0)=0\]
At
\[x=\frac{\pi}{6},\]
we get
\[f\left(\frac{\pi}{6}\right)=\sin\left(3\cdot\frac{\pi}{6}\right)=\sin\frac{\pi}{2}=1\]
Thus, the function rises from \(0\) to \(1\) as \(x\) moves from \(0\) to \(\pi/6\). After
\[x=\frac{\pi}{6},\]
the function decreases. At the right endpoint,
\[f\left(\frac{\pi}{2}\right)=\sin\frac{3\pi}{2}=-1\]
Therefore, on the given domain, the function first increases and then decreases:
\[0\longrightarrow 1\longrightarrow-1\]
❌ Common Mistakes
  • Forgetting the chain-rule factor \(3\) in
    \[ \frac{d}{dx}(\sin 3x)=3\cos 3x. \]
  • Treating \(3x\) as \(x\) while solving
    \[ \cos 3x=0. \]
  • Forgetting that the domain restriction
    \[ x\in\left[0,\frac{\pi}{2}\right] \]
    must be applied before selecting the critical points.
  • Considering \(x=\pi/2\) an interior critical point. It is an endpoint of the specified domain.
  • Saying that \(\cos 3x<0\) throughout the entire interval without checking the range of \(3x\).
  • Confusing the critical point \(x=\pi/6\) with the maximum value \(1\).
⚡ Exam Tip
🔑 Key Takeaway
∿

Example 13

📋
Table of Contents
8 sections
❓ Question
Find the intervals in which the function
\[ f(x)=\sin x+\cos x, \qquad 0\leq x\leq2\pi \]
is increasing or decreasing.
💡 Concept
🗺️ Roadmap

  1. Differentiate the given function.

  2. Solve \(f'(x)=0\) within the given domain.

  3. Use the critical points to divide the domain into subintervals.

  4. Determine the sign of \(f'(x)\) in each subinterval.

  5. State the increasing and decreasing intervals.

🧩 Solution
Given:
\[f(x)=\sin x+\cos x\]
Part (a)
Find the First Derivative
  1. Differentiating term by term,
    \[\begin{aligned}f'(x)&=\frac{d}{dx}(\sin x)+\frac{d}{dx}(\cos x)\\&=\cos x-\sin x\end{aligned}\]
  2. Hence,
    \[\boxed{f'(x)=\cos x-\sin x}\]
  3. Find the Critical Points Critical points occur where \(f'(x)=0\). Therefore,
    \[\cos x-\sin x=0\]
  4. Hence,
    \[\cos x=\sin x\]
  5. Since
    \[\cos x\neq0\]
    at a solution of \(\sin x=\cos x\), we can divide by \(\cos x\):
    \[\tan x=1\]
  6. The general solution is
    \[x=\frac{\pi}{4}+n\pi,\quad n\in\mathbb{Z}.\]
  7. Since
    \[0\leq x\leq2\pi,\]
    the relevant solutions are
    \[\boxed{x=\frac{\pi}{4},\frac{5\pi}{4}}\]
  8. These points divide the domain into:
    \[\left(0,\frac{\pi}{4}\right),\quad\left(\frac{\pi}{4},\frac{5\pi}{4}\right),\quad\left(\frac{5\pi}{4},2\pi\right).\]
  9. Determine the Sign of \(f'(x)\) We have
    \[f'(x)=\cos x-\sin x\]
  10. A useful identity is
    \[\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)\]
  11. Therefore,
    \[f'(x)=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)\]
  12. ⭐ We can now determine its sign on each interval.
  13. Interval 1: \(0 < x < \frac{\pi}{4}\) Choose, for example,
    \[x=\frac{\pi}{8}\]
  14. Then
    \[f'\left(\frac{\pi}{8}\right)=\cos\frac{\pi}{8}-\sin\frac{\pi}{8}>0\]
  15. Therefore,
    \[f'(x)>0\quad\text{for }0 < x <\frac{\pi}{4}.\]
  16. Hence,
    \[\boxed{f\text{ is strictly increasing on }\left(0,\frac{\pi}{4}\right)}\]
  17. Interval 2: \(\frac{\pi}{4} < x < \frac{5\pi}{4}\) Choose a convenient test point:
    \[x=\frac{\pi}{2}\]
  18. Then
    \[f'\left(\frac{\pi}{2}\right)=\cos\frac{\pi}{2}-\sin\frac{\pi}{2}=0-1=-1<0\]
  19. Therefore,
    \[f'(x)<0\quad\text{for }\frac{\pi}{4} < x < \frac{5\pi}{4}]
  20. Hence,
    \[\boxed{f\text{ is strictly decreasing on }\left(\frac{\pi}{4},\frac{5\pi}{4}\right)}\]
  21. Interval 3: \(\frac{5\pi}{4} < x < 2 \pi\) Choose
    \[x=\frac{3\pi}{2}\]
  22. Then
    \[f'\left(\frac{3\pi}{2}\right)=\cos\frac{3\pi}{2}-\sin\frac{3\pi}{2}=0-(-1)=1>0\]
  23. Therefore,
    \[f'(x)>0\quad\text{for }\frac{5\pi}{4} < x < 2\pi\]
  24. Hence,
    \[\boxed{f\text{ is strictly increasing on }\left(\frac{5\pi}{4},2\pi\right)}\]
🗒️ Sign Chart
Interval Test Point Sign of \(f'(x)\) Behaviour of \(f\)
\(\left(0,\frac{\pi}{4}\right)\) \(\frac{\pi}{8}\) \(+\) Increasing
\(x=\frac{\pi}{4}\) Critical point \(0\) Stationary point
\(\left(\frac{\pi}{4},\frac{5\pi}{4}\right)\) \(\frac{\pi}{2}\) \(-\) Decreasing
\(x=\frac{5\pi}{4}\) Critical point \(0\) Stationary point
\(\left(\frac{5\pi}{4},2\pi\right)\) \(\frac{3\pi}{2}\) \(+\) Increasing
🗒️ Endpoint Consideration

The derivative test determines strict increase or decrease on open intervals. Since the given domain is the closed interval

\[ [0,2\pi], \]
the function is continuous at the endpoints.

Therefore, in the usual NCERT convention, the result may be stated as:

\[ \boxed{ f\text{ is increasing on } \left[0,\frac{\pi}{4}\right] \text{ and } \left[\frac{5\pi}{4},2\pi\right], } \]
and
\[ \boxed{ f\text{ is decreasing on } \left[\frac{\pi}{4},\frac{5\pi}{4}\right]. } \]

For strict derivative-sign intervals, the corresponding open intervals are

\[ \boxed{ \left(0,\frac{\pi}{4}\right) \quad\text{and}\quad \left(\frac{5\pi}{4},2\pi\right) } \]
for increasing behaviour, and
\[ \boxed{ \left(\frac{\pi}{4},\frac{5\pi}{4}\right) } \]
for decreasing behaviour.

❌ Common Mistakes
  • Using \(f'(x)=0\) as the condition for increasing. It is only the condition used to locate stationary critical points.
  • Forgetting that
    \[ \tan x=1 \]
    has the general solution
    \[ x=\frac{\pi}{4}+n\pi. \]
  • Finding only \(x=\pi/4\) and missing
    \[ x=\frac{5\pi}{4}. \]
  • Failing to check the sign of \(f'(x)\) between the critical points.
  • Confusing critical points with intervals of increase or decrease.
🔑 Key Takeaway
∿

Maxima and Minima

📋
Table of Contents
11 sections
📖 Introduction
📘 Definition
Maximum Minimum Extreme Values
🔍 Maximum and Minimum: Simple Interpretation
Concept Condition Meaning
Maximum value
\[ f(c)>f(x) \]
for all other \(x\in I\) 		\(f(c)\) is the largest value of \(f\) on \(I\)
Minimum value
\[ f(c) for all other \(x\in I\)
\(f(c)\) is the smallest value of \(f\) on \(I\)
Extreme value Maximum or minimum value Either the largest or smallest value
⚖️ Important Distinction: Point Versus Value
A common source of confusion is the difference between the point at which a maximum or minimum occurs and the corresponding function value.

Suppose
\[ f(x)=-(x-2)^2+5 \]
The maximum occurs at
\[ x=2 \]
Therefore, the point of maximum is
\[ \boxed{x=2} \]
The corresponding maximum value is
\[ f(2)=5 \]
Thus, the maximum value is
\[ \boxed{5} \]
The corresponding point on the graph is
\[ \boxed{(2,5)} \]
Therefore:
  • \(x=2\) is the point of maximum value in terms of the independent variable.
  • \(f(2)=5\) is the maximum value.
  • \((2,5)\) is the maximum point on the graph.
🔍 Graphical Interpretation
Geometrically, a maximum value corresponds to the highest function value attained by the graph within the specified interval.

Similarly, a minimum value corresponds to the lowest function value attained by the graph within that interval.

If \(c\) is an interior point at which a differentiable function has a local maximum or minimum, the tangent to the graph is horizontal, so typically
\[ f'(c)=0. \]
This observation leads to the important concept of a critical point, which is central to the study of maxima and minima.
📘 Local Maximum and Local Minimum
🗒️ Example of a Local Maximum and Local Minimum
Consider
\[f(x)=x^3-3x\]
Its derivative is
\['(x)=3x^2-3=3(x-1)(x+1)\]
Therefore, the critical points are
\[x=-1,\quad x=1\]
The derivative changes from positive to negative at \(x=-1\), so \(x=-1\) is a local maximum. The derivative changes from negative to positive at \(x=1\), so \(x=1\) is a local minimum.

Their corresponding values are
\[f(-1)=2\]
and
\[f(1)=-2\]
Thus,
\[\boxed{\text{Local maximum value}=2}\]
and
\[\boxed{\text{Local minimum value}=-2}\]
However, on the entire real line, this function has neither an absolute maximum nor an absolute minimum because
\[f(x)\to\infty\quad\text{as }x\to\infty\]
and
\[f(x)\to-\infty\quad\text{as }x\to-\infty\]
Maximum or Minimum Does Not Necessarily Occur Where \(f'(x)=0\)
For an interior point \(c\), if \(f\) is differentiable and has a local maximum or local minimum there, then
\[f'(c)=0\]
However, for a function defined on a closed interval, an absolute maximum or minimum can occur at an endpoint, where the usual interior derivative condition does not apply.

For example, consider
\[f(x)=x\]
on
\[[0,2]\]
We have
\[f'(x)=1,\]
so there is no point in \((0,2)\) where
\[f'(x)=0\]
Nevertheless,
\[f(0)=0\]
is the absolute minimum and
\[f(2)=2\]
is the absolute maximum on \([0,2]\).

This is why endpoint values must be checked when finding absolute maxima and minima on a closed interval.
⚖️ Extreme Point Versus Critical Point
These two terms should not be treated as synonyms.

A critical point is a point in the domain where
\[f'(c)=0\]
or \(f'(c)\) does not exist, provided \(c\) belongs to the domain.

An extreme point is a point where the function actually attains a maximum or minimum under the relevant definition.

Therefore, every differentiable interior extremum is a critical point, but every critical point need not be an extremum.

For example,
\[f(x)=x^3\]
has
\[f'(0)=0,\]
but \(x=0\) is not a local maximum or minimum because the function continues increasing through \(x=0\).
📌 Key Terminology
🌟 Exam-Oriented Understanding
In applications of derivatives, finding maxima and minima generally involves the following strategy:
  1. Find the domain or specified interval of the function.
  2. Calculate \(f'(x)\).
  3. Find critical points from \(f'(x)=0\) and points where \(f'(x)\) does not exist.
  4. Check whether the critical points lie in the relevant domain.
  5. For an absolute maximum or minimum on a closed interval, also evaluate the function at the endpoints.
  6. Compare all relevant function values.
  7. Identify the maximum and minimum values and their corresponding points.
🔑 Key Takeaway
∿

Example 14

📋
Table of Contents
7 sections
❓ Question
Find the maximum and minimum values, if any, of the function
\[f(x)=x^2,\qquad x\in\mathbb{R}\]
💡 Concept
🗺️ Roadmap

  1. Find the first derivative \(f'(x)\).

  2. Find the critical points by solving \(f'(x)=0\).

  3. Determine whether the critical point gives a maximum or minimum.

  4. Check the behaviour of the function over the complete domain.

🧩 Solution
  1. Differentiate the Function
    \[f(x)=x^2\]
  2. Therefore,
    \[f'(x)=2x\]
  3. Find the Critical Point A critical point occurs where \(f'(x)=0\) or where \(f'(x)\) does not exist. Since \(f'(x)=2x\) exists for every real number, we solve:
    \[2x=0\]
    \[x=0\]
    📝 Thus, the only critical point is \(x=0\).
  4. Determine Whether \(x=0\) Gives a Maximum or Minimum Consider the sign of \(f'(x)=2x\):
    • For \(x<0\), \(f'(x)<0\), so \(f(x)\) is decreasing.
    • For \(x>0\), \(f'(x)>0\), so \(f(x)\) is increasing.
    📝 Thus, the function changes from decreasing to increasing at \(x=0\). Hence, \(x=0\) gives an absolute minimum.
  5. Its value is
    \[f(0)=0^2=0\]
  6. Check Whether a Maximum Exists As \(x\) becomes arbitrarily large in magnitude, \(x^2\) also becomes arbitrarily large:
    \[\lim_{x\to\infty}x^2=\infty\]
    \[\lim_{x\to-\infty}x^2=\infty\]
    📝 Therefore, \(f(x)=x^2\) has no upper bound on \(\mathbb{R}\). Hence, the function has no maximum value on \(\mathbb{R}\).
🔍 Graphical Interpretation
The graph of \(y=x^2\) is an upward-opening parabola with vertex at \((0,0)\). The vertex is the lowest point of the graph, while the graph continues indefinitely upward in both directions.
🌟 Important Exam Note

The statement “maximum value could not be found” should be written more precisely as “the maximum value does not exist”. This is because \(x^2\) is unbounded above on \(\mathbb{R}\), not because the maximum is merely difficult to calculate.

Significance for Board and Competitive Exams

This example illustrates an important distinction between a critical point and an absolute extremum. Finding \(f'(x)=0\) identifies a candidate point, but the complete domain and the behaviour of the function must also be examined before concluding that a maximum or minimum exists.

🔑 Key Takeaway
∿

Example 15

📋
Table of Contents
7 sections
❓ Question
Find the maximum and minimum values, if any, of the function
\[f(x)=|x|,\qquad x\in\mathbb{R}\]
💡 Concept
🗺️ Roadmap

  1. Write the absolute value function in piecewise form.

  2. Find the minimum value using the non-negativity of \(|x|\).

  3. Check the behaviour of the function as \(x\to\pm\infty\).

  4. Conclude whether an absolute maximum exists.

🧩 Solution
  1. Write the Function in Piecewise Form
    \[f(x)=|x|=\begin{cases}-x, & x<0,\\x, & x\geq0\end{cases}\]
  2. For \(x < 0\),
    \[f(x)=-x\]
  3. and for \(x>0\),
    \[f(x)=x\]
  4. At \(x=0\)
    \[f(0)=|0|=0\]
  5. Find the Minimum Value For every real number \(x\),
    \[|x|\geq0\]
  6. Equality occurs when
    \[|x|=0\]
    \[x=0\]
  7. Therefore,
    \[f(0)=0\]
  8. Hence, the absolute minimum value is
    \[\boxed{0}\]
  9. and it occurs at
    \[\boxed{x=0}\]
  10. Check for a Maximum Value Consider positive values of \(x\). For \(x>0\),
    \[f(x)=x\]
  11. As \(x\) increases without bound,
    \[\lim_{x\to\infty}|x|=\infty\]
  12. Thus, \(f(x)=|x|\) is unbounded above on \(\mathbb{R}\). For example,
    \[f(10)=10,\qquad f(100)=100,\qquad f(1000)=1000\]
    📝 No real number can therefore serve as the largest value of \(f(x)\). Hence, an absolute maximum does not exist on \(\mathbb{R}\).
🔍 Graphical Interpretation
The graph of \(y=|x|\) is a V-shaped graph with its vertex at \((0,0)\). The vertex is the lowest point of the graph, while both arms extend indefinitely upward. Therefore, the graph has a minimum but no maximum.
🌟 Important Exam Note

The statement “maximum value could not be determined” is better expressed mathematically as “the maximum value does not exist.” The reason is that \(f(x)=|x|\) is unbounded above on \(\mathbb{R}\).

Also note that \(x=0\) is a critical point even though \(f'(0)\) does not exist. A critical point need not satisfy \(f'(x)=0\); a point where the derivative does not exist can also be a critical point, provided the point belongs to the domain.

Significance for Board and Competitive Exams

This example is important because it demonstrates why checking only \(f'(x)=0\) is insufficient. For functions such as \(f(x)=|x|\), an extremum can occur at a point where the derivative does not exist. Therefore, when solving maximum-minimum problems, critical points must include both points where \(f'(x)=0\) and points in the domain where \(f'(x)\) does not exist.

🔑 Key Takeaway
∿

Example 16

📋
Table of Contents
10 sections
❓ Question
Find the maximum and minimum values, if any, of the function
\[f(x)=x,\quad x\in(0,1)\]
💡 Concept
🗺️ Roadmap

  1. Determine whether the function is increasing or decreasing.

  2. Examine the endpoints of the domain.

  3. Check whether the endpoint values actually belong to the domain.

  4. Determine whether the function attains a maximum or minimum.

🧩 Solution
Given:
\[f(x)=x\]
  1. Determine the Monotonicity Differentiating given function
    \[f'(x)=1\]
  2. Since
    \[f'(x)=1>0\]
  3. for every \(x\in(0,1)\), the function is strictly increasing on \((0,1)\).
  4. Examine the Lower End of the Domain As \(x\) approaches \(0\) from the right,
    \[\lim_{x\to0^+}f(x)=\lim_{x\to0^+}x=0.\]
  5. However,
    \[0\notin(0,1)\]
    📝 Therefore, \(f(x)\) approaches \(0\), but never attains \(0\). Hence, the function has no minimum value.
  6. Nevertheless, \(0\) is the <strong>infimum</strong> of the range.
    \[\inf f((0,1))=0\]
  7. Examine the Upper End of the Domain As \(x\) approaches \(1\) from the left,
    \[\lim_{x\to1^-}f(x)=\lim_{x\to1^-}x=1\]
  8. But
    \[1\notin(0,1)\]
    📝 Therefore, the function approaches \(1\), but never attains \(1\). Hence, the function has no maximum value.
  9. However, \(1\) is the <strong>supremum</strong> of the range.
    \[\sup f((0,1))=1\]
  10. Determine the Range Since \(f(x)=x\) and \(x\in(0,1)\), the range is
    \[f((0,1))=(0,1)\]
    📝 Thus, the range itself contains neither \(0\) nor \(1\).
📌 Important Note on Monotonic Functions
📎 General Result: Extreme Value Theorem
A fundamental result in calculus is the Extreme Value Theorem:

Every continuous function defined on a closed and bounded interval \([a,b]\) attains both an absolute maximum value and an absolute minimum value on that interval.

In symbolic form, if \(f\) is continuous on \([a,b]\), then there exist \(c,d\in[a,b]\) such that
\[ f(c)\leq f(x)\leq f(d) \qquad\text{for every }x\in[a,b]. \]
Hence,
\[ f(c)=\min_{x\in[a,b]}f(x) \]
\[ f(d)=\max_{x\in[a,b]}f(x). \]
Comparison with a Closed Interval
Consider the same function on the closed interval \([0,1]\):
\[ f(x)=x,\qquad x\in[0,1].\]
Now both endpoints belong to the domain:
\[ f(0)=0 \]
\[ f(1)=1. \]
Therefore,
\[ \boxed{\min f(x)=0} \]
\[ \boxed{\max f(x)=1}. \]
⚡ Exam Tip
❌ Common Mistakes
  • Treating the limiting values \(0\) and \(1\) as the minimum and maximum even though they are not in the domain.
  • Saying that the maximum and minimum are “difficult to determine” instead of recognizing that they do not exist.
  • Forgetting that an open interval does not contain its endpoints.
  • Confusing infimum with minimum and supremum with maximum.
  • Applying the Extreme Value Theorem without checking continuity and whether the interval is closed and bounded.
🌟 Significance
This example tests a fundamental distinction between an extreme value and a limiting value. In maximum-minimum problems, students must examine the domain, endpoint inclusion, critical points, and whether candidate values are actually attained. This idea is also directly connected with the Extreme Value Theorem and the concepts of supremum and infimum.
🔑 Key Takeaway
∿

Local Maxima and Local Minima

Application of Derivatives NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Derivatives Rate of Change Increasing Functions Decreasing Functions Maxima and Minima Absolute Maximum Absolute Minimum Local Maxima Local Minima First Derivative Test Second Derivative Test Optimization Problems Related Rates Marginal Cost Marginal Revenue JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
12 sections
💡 Concept
📘 Point of Local Maximum
📘 Point of Local Minimum
🤔 Why Is \(c\) Required to Be an Interior Point?
The definition considers a two-sided neighbourhood
\[(c-h,c+h)\]
Therefore, \(c\) should be an interior point of the domain so that points of the domain exist on both sides of \(c\). This is the standard definition used for local extrema in the interior of the domain.
⚖️ Local Maximum Versus Absolute Maximum
A local maximum only compares \(f(c)\) with nearby values. An absolute maximum compares \(f(c)\) with every value of the function on its entire domain.
Feature Local Maximum Absolute Maximum
Comparison Only nearby points All points in the domain
Condition \(f(c)\geq f(x)\) near \(c\) \(f(c)\geq f(x)\) for every \(x\) in the domain
Nature Neighbourhood property Global property
⚖️ Local Minimum Versus Absolute Minimum
Feature Local Minimum Absolute Minimum
Comparison Only nearby points All points in the domain
Condition \(f(c)\leq f(x)\) near \(c\) \(f(c)\leq f(x)\) for every \(x\) in the domain
Nature Neighbourhood property Global property
✏️ Example
Local Maximum
Consider
\[f(x)=-(x-2)^2+5\]
At \(x=2\),
\[f(2)=5\]
For every \(x\neq2\),
\[(x-2)^2>0\]
and therefore
\[-(x-2)^2+5 < 5\]
Hence,
\[f(2)>f(x),\qquad x\neq2\]
Therefore, \(x=2\) is a local maximum point and \(5\) is the local maximum value. In fact, it is also the absolute maximum value because the inequality holds for every real \(x\).
✏️ Example
Local Minimum
Consider
\[f(x)=(x-3)^2+2\]
At \(x=3\),
\[f(3)=2\]
Since
\[(x-3)^2\geq0\]
we have
\[f(x)=(x-3)^2+2\geq2=f(3)\]
Hence, \(x=3\) is a local minimum point and \(2\) is the local minimum value. It is also the absolute minimum value on \(\mathbb{R}\).
👁️ Important Observation
🔗 Connection with Derivatives
If \(f\) is differentiable at an interior point \(c\) and \(c\) is a local maximum or local minimum point, then Fermat's theorem gives
\[\boxed{f'(c)=0}\]
However, the converse is not necessarily true. The condition
\[f'(c)=0\]
only identifies a possible critical point. It does not by itself prove that \(c\) is a local maximum or local minimum.

For example, consider
\[f(x)=x^3\]
Then
\[f'(x)=3x^2\]
\[f'(0)=0\]
Nevertheless, \(x=0\) is neither a local maximum nor a local minimum because the function is increasing on both sides of \(0\).
⚡ Exam Tip
🔑 Key Takeaway
∿

Theorem 2: Necessary Condition for a Local Maximum or Minimum

📋
Table of Contents
8 sections
🧮 Theorem
🧮 Theorem
Statement
Let \(f\) be a function defined on an open interval \(I\), and let \(c\in I\). If \(f\) has a local maximum or a local minimum at \(x=c\), then either
\[\boxed{f'(c)=0}\]
or \(f\) is not differentiable at \(x=c\).

In other words, if \(c\) is an interior point at which a local extremum occurs, then \(c\) must be a critical point.
Equivalent Form
A point \(c\) at which a local maximum or local minimum occurs can satisfy one of the following two conditions:
  1. The derivative exists and
    \[ f'(c)=0. \]
  2. The derivative \(f'(c)\) does not exist.
Therefore, while searching for local extrema, we must examine both types of critical points.
Proof
  1. Suppose \(f\) has a local maximum at \(x=c\).

    By the definition of local maximum, there exists \(h>0\) such that
    \[f(c)\geq f(x)\]
  2. for every \(x\in(c-h,c+h)\), \(x\neq c\).

    Consider a point \(x\) slightly greater than \(c\), so that
    \[c < x < c+h\]
  3. Since \(f(c)\geq f(x)\),
    \[f(x)-f(c)\leq0\]
  4. Because \(x-c>0\), we obtain
    \[\frac{f(x)-f(c)}{x-c}\leq0\]
  5. Taking the right-hand limit as \(x\to c^+\), if the right-hand derivative exists, then
    \[f'_+(c)\leq0.\]
  6. Now consider \(x < c\). Then
    \[f(c)\geq f(x)\]
  7. so
    \[f(x)-f(c)\leq0\]
  8. But now
    \[x-c < 0\]
  9. Therefore, dividing by \(x-c\) reverses the inequality:
    \[\frac{f(x)-f(c)}{x-c}\geq0\]
  10. Taking the left-hand limit as \(x\to c^-\), if the left-hand derivative exists, then
    \[f'_-(c)\geq0\]
  11. If \(f\) is differentiable at \(c\), then the left-hand and right-hand derivatives are equal:
    \[f'_-(c)=f'_+(c)=f'(c)\]
  12. Combining
    \[f'_-(c)\geq0\]
    and
    \[f'_+(c)\leq0\]
  13. gives
    \[f'(c)\geq0\]
    and
    \[f'(c)\leq0\]
  14. Hence,
    \[\boxed{f'(c)=0}\]
Thus, if \(f\) is differentiable at a local maximum, its derivative at that point must be zero.
🗒️ Local Minimum Case
The argument for a local minimum is analogous. If \(c\) is a local minimum, then
\[f(c)\leq f(x)\]
for all \(x\) sufficiently close to \(c\), \(x\neq c\). For \(x>c\),
\[f(x)-f(c)\geq0\]
and \(x-c>0\), so
\[\frac{f(x)-f(c)}{x-c}\geq0\]
Hence,
\[f'_+(c)\geq0\]
For \(x < c\),
\[f(x)-f(c)\geq0\]
while \(x-c<0\), giving
\[\frac{f(x)-f(c)}{x-c}\leq0\]
Hence,
\[f'_-(c)\leq0\]
If \(f\) is differentiable at \(c\), then
\[f'_-(c)=f'_+(c)=f'(c).\]
Therefore,
\[f'(c)\leq0\]
and
\[f'(c)\geq0\]
Consequently,
\[\boxed{f'(c)=0}\]
🤔 Why the Theorem Says “Either \(f'(c)=0\) or \(f\) Is Not Differentiable”
A local extremum does not necessarily occur at a point where the derivative exists. A sharp corner or cusp can also produce a local maximum or minimum.

For example, consider
\[f(x)=|x|\]
At \(x=0\),
\[f(0)=0\]
and
\[|x|\geq0=f(0)\]
Hence, \(x=0\) is a local minimum point. However,
\[f'_-(0)=-1\]
and
\[f'_+(0)=1\]
Since
\[f'_-(0)\neq f'_+(0)\]
\(f'(0)\) does not exist. Thus this example satisfies the second possibility of the theorem.
🗒️ Important Consequence

The theorem provides a necessary condition, not a sufficient condition.

Therefore,

\[ f'(c)=0 \]

does not necessarily mean that \(c\) is a local maximum or local minimum.

For example, let

\[ f(x)=x^3. \]

Then

\[ f'(x)=3x^2 \]
\[ f'(0)=0. \]

However, \(f(x)=x^3\) is increasing on the entire real line. Therefore, \(x=0\) is neither a local maximum nor a local minimum.

🔍 Graphical Interpretation
At a differentiable local maximum or minimum, the graph generally has a horizontal tangent:
\[f'(c)=0\]
However, a local extremum may also occur at a sharp point where no tangent with a finite derivative exists, as in \(f(x)=|x|\) at \(x=0\).
📜 Exam-Oriented Rule
📜 Rules
To locate possible local maxima and minima of a function, find all points \(c\) in the interior of its domain satisfying either
\[ \boxed{f'(c)=0} \]
or
\[ \boxed{f'(c)\text{ does not exist}}. \]
These points are candidates for local extrema. They must then be tested using the first derivative test, second derivative test, or another appropriate method.
❌ Common Mistakes
  • Assuming that every point satisfying \(f'(c)=0\) is a local maximum or minimum.
  • Ignoring points where \(f'(c)\) does not exist.
  • Confusing a necessary condition with a sufficient condition.
  • Forgetting that the theorem concerns an interior point of the domain.
  • Checking only \(f'(c)=0\) when the function contains absolute values, piecewise definitions, or other possible non-differentiable points.
🔑 Key Takeaway
∿

Theorem 3: First Derivative Test

📋
Table of Contents
5 sections
🧮 Theorem
🧮 Theorem
Statement
Let \(f\) be a function defined on an open interval \(I\), and let \(c\in I\) be a critical point at which \(f\) is continuous. The sign of \(f'(x)\) immediately to the left and right of \(c\) determines whether \(c\) is a local maximum, a local minimum, or neither.
Proof
  1. Condition for a Local Maximum
  2. If \(f'(x)\) changes sign from positive to negative as \(x\) increases through \(c\), that is,
    \[f'(x)>0 \quad \text{for }x < c\text{ sufficiently close to }c\]
    and
    \[ f'(x) < 0 \quad \text{for }x>c\text{ sufficiently close to }c, \]
  3. then \(f\) is increasing immediately to the left of \(c\) and decreasing immediately to the right of \(c\).

    Therefore,
    \(f\) attains a local maximum at \(x=c\).
    \[ \boxed{+\ \longrightarrow\ -\quad\Rightarrow\quad \text{Local Maximum}} \]
    Why Does \(+\to-\) Give a Maximum?
    Reason:

    If \(f'(x)>0\) just before \(c\), the function is increasing as \(x\) approaches \(c\). Thus, the function values rise toward \(f(c)\).

    If \(f'(x)<0\) immediately after \(c\), the function is decreasing as \(x\) moves away from \(c\). Thus, the function values fall after \(c\).

    Hence, \(f(c)\) is greater than nearby function values, so \(c\) is a local maximum point.

  4. Condition for a Local Minimum
  5. If \(f'(x)\) changes sign from negative to positive as \(x\) increases through \(c\), that is,
    \[ f'(x) < 0 \quad\text{for }x and
    \[ f'(x)>0 \quad\text{for }x>c\text{ sufficiently close to }c, \]
  6. then \(f\) is decreasing immediately to the left of \(c\) and increasing immediately to the right of \(c\).

    Therefore,
    \(f\) attains a local minimum at \(x=c\).
    \[ \boxed{-\ \longrightarrow\ +\quad\Rightarrow\quad \text{Local Minimum}} \]
    Why Does \(-\to+\) Give a Minimum?

    If \(f'(x)<0\) just before \(c\), the function is decreasing toward \(c\). Thus, the function values fall as \(x\) approaches \(c\).

    If \(f'(x)>0\) immediately after \(c\), the function is increasing away from \(c\). Thus, the function values rise after \(c\).

    Hence, \(f(c)\) is smaller than nearby function values, so \(c\) is a local minimum point.

  7. When \(f'(x)\) Does Not Change Sign
  8. If \(f'(x)\) does not change sign as \(x\) passes through \(c\), then the function continues to move in the same direction on both sides of \(c\).

    There are two important possibilities:
    • \(f'(x)>0\) on both sides of \(c\): the function is increasing on both sides.
    • \(f'(x)<0\) on both sides of \(c\): the function is decreasing on both sides.
    In either case, \(c\) is neither a local maximum nor a local minimum.
    \[ \boxed{+\ \longrightarrow\ +\quad\Rightarrow\quad \text{Neither}} \]
    \[ \boxed{-\ \longrightarrow\ -\quad\Rightarrow\quad \text{Neither}} \]
  9. A critical point where the derivative does not change sign may be a stationary point of inflection when \(f'(c)=0\). However, the statement that every such point is automatically a “point of inflection” needs qualification: a point of inflection is properly characterized by a change in concavity, not merely by the absence of a sign change in \(f'\).
🗒️ First Derivative Test Summary
Sign of \(f'(x)\) to the Left Sign of \(f'(x)\) to the Right Conclusion at \(c\)
\(+\) \(-\) Local maximum
\(-\) \(+\) Local minimum
\(+\) \(+\) Neither maximum nor minimum
\(-\) \(-\) Neither maximum nor minimum
✏️ Example
1
Question
Find Local Maximum of
\[f(x)=x^3-3x\]
  1. Differentiate: given function
    \[f'(x)=3x^2-3\]
    \[f'(x)=3(x-1)(x+1)\]
  2. Therefore, the critical points are
    \[x=-1,\quad x=1\]
  3. Consider the sign of \(f'(x)\):
    Interval Sign of \(f'(x)\) Behaviour of \(f\)
    \((-\infty,-1)\) \(+\) Increasing
    \((-1,1)\) \(-\) Decreasing
    \((1,\infty)\) \(+\) Increasing
  4. At \(x=-1\),
    \[+\longrightarrow-\]
  5. Therefore, \(x=-1\) is a local maximum point.
    \[f(-1)=(-1)^3-3(-1)=2\]
  6. Hence, the local maximum value is
    \[\boxed{2}\]
  7. At \(x=1\),
    \[-\longrightarrow+\]
  8. Therefore, \(x=1\) is a local minimum point.
    \[f(1)=1-3=-2\]
  9. Hence, the local minimum value is
    \[\boxed{-2}\]
2
Example
Critical Point That Is Not an Extremum
  1. Consider
    \[f(x)=x^3\]
  2. Then
    \[f'(x)=3x^2\]
  3. The only critical point is
    \[x=0\]
  4. But
    \[f'(x)>0\]
  5. for both \(x<0\) and \(x>0\), except at \(x=0\). Therefore, the sign pattern is
    \[+\longrightarrow+\]
    📝 The derivative does not change sign. Hence, \(x=0\) is neither a local maximum nor a local minimum.

    In fact, \(x=0\) is a stationary point of inflection because the concavity changes at \(x=0\).
3
Example
Non-Differentiable Local Minimum
  1. Consider
    \[f(x)=|x|\]
  2. For \(x < 0\),
    \[f'(x)=-1<0\]
  3. while for \(x>0\),
    \[f'(x)=1>0\]
  4. Thus the sign changes as
    \[-\longrightarrow+\]
    📝 Therefore, \(x=0\) is a local minimum point. Notice that \(f'(0)\) does not exist, so the first derivative test can still be applied by examining the signs on either side.
  5. Step-by-Step Procedure for the First Derivative Test
    1. Find \(f'(x)\).
    2. Find all critical points by solving
      \[ f'(x)=0 \]
      and identifying points where \(f'(x)\) does not exist but \(f(x)\) is defined.
    3. Arrange the critical points in increasing order.
    4. Divide the domain into intervals using these critical points.
    5. Determine the sign of \(f'(x)\) in each interval.
    6. Compare the signs immediately before and after each critical point.
    7. Apply the sign-change rule:
      \[ +\to-\Rightarrow\text{local maximum} \]
      \[ -\to+\Rightarrow\text{local minimum} \]
      \[ +\to+\text{ or }-\to-\Rightarrow\text{neither} \]
  6. Important Distinction: First Derivative Test and Point of Inflection

    The statement “\(f'(x)\) does not change sign” should not, by itself, be taken as the definition of a point of inflection.

    A point of inflection is a point where the concavity of the graph changes. For example, \(f(x)=x^3\) has a point of inflection at \(x=0\), and its derivative remains positive on both sides.

    However, a critical point with no change in the sign of \(f'\) is not automatically a point of inflection in every possible function. The concavity must be checked when an inflection point is claimed.

⚡ Exam Tip
🔑 Key Takeaway
∿

Example 17

📋
Table of Contents
9 sections
❓ Question
Find all points of local maxima and local minima of the function
\[f(x)=x^3-3x+3,\qquad x\in\mathbb{R}.\]
💡 Concept
🗺️ Roadmap

  1. Find \(f'(x)\).

  2. Set \(f'(x)=0\) to obtain the critical points.

  3. Divide the real line using the critical points.

  4. Determine the sign of \(f'(x)\) in each interval.

  5. Apply the First Derivative Test.

  6. Calculate the corresponding function values.

🧩 Solution
Part (a)
  1. Find the First Derivative Differentiating \(f(x)=x^3-3x+3\;\) with respect to \(x\),
    \[f'(x)=3x^2-3\]
  2. Factorising,
    \[f'(x)=3(x^2-1)\]
    \[f'(x)=3(x-1)(x+1)\]
  3. Find the Critical Points Since \(f'(x)\) is defined for every real \(x\), the critical points are obtained from
    \[f'(x)=0\]
    \[3(x-1)(x+1)=0\]
  4. Therefore,
    \[x-1=0\]
    \[x=1\]
    or
    \[x+1=0\]
    \[x=-1\]
  5. Thus, the critical points are
    \[\boxed{x=-1,\;1}\]
    📝 These are the points where local maxima or minima may occur. We must now apply the First Derivative Test.
  6. Determine the Sign of \(f'(x)\) We have
    \[f'(x)=3(x-1)(x+1)\]
  7. The critical points \(-1\) and \(1\) divide the real line into three intervals:
    \[(-\infty,-1),\quad(-1,1),\quad(1,\infty)\]
  8. Consider a test point from each interval.

    Interval 1: \(x<-1\)     Take \(x=-2\):
    \[f'(-2)=3(-2-1)(-2+1)\]
    \[=3(-3)(-1)\]
    \[=9>0\]
    📝 Hence, \(f'(x)>0\) on \((-\infty,-1)\), so \(f\) is increasing there.
  9. Interval 2: \(-1 < x < 1\) Take \(x=0\)
    \[f'(0)=3(0-1)(0+1)\]
    \[=3(-1)(1)\]
    \[=-3<0\]
    📝 Hence, \(f'(x)<0\) on \((-1,1)\), so \(f\) is decreasing there.
  10. Interval 3: \(x>1\) Take \(x=2\):
    \[f'(2)=3(2-1)(2+1)\]
    \[=3(1)(3)\]
    \[=9>0\]
    📝 Hence, \(f'(x)>0\) on \((1,\infty)\), so \(f\) is increasing there.
  11. Apply the First Derivative Test
    Interval Sign of \(f'(x)\) Behaviour of \(f(x)\)
    \((-\infty,-1)\) \(+\) Increasing
    \((-1,1)\) \(-\) Decreasing
    \((1,\infty)\) \(+\) Increasing
  12. At \(x=-1\), the derivative changes sign as
    \[+\longrightarrow-\]
  13. Therefore, by the First Derivative Test,
    \[\boxed{x=-1\text{ is a point of local maximum}.}\]
  14. At \(x=1\), the derivative changes sign as
    \[-\longrightarrow+\]
  15. Therefore,
    \[\boxed{x=1\text{ is a point of local minimum}.}\]
  16. Find the Local Maximum Value The local maximum occurs at \(x=-1\). Therefore,
    \[f(-1)=(-1)^3-3(-1)+3\]
    \[=-1+3+3\]
    \[=5\]
  17. Hence, the local maximum value is
    \[\boxed{5}\]
  18. The corresponding local maximum point on the graph is
    \[\boxed{(-1,5)}\]
  19. Find the Local Minimum Value The local minimum occurs at \(x=1\). Therefore,
    \[f(1)=1^3-3(1)+3\]
    \[=1-3+3\]
    \[=1\]
  20. Hence, the local minimum value is
    \[\boxed{1}\]
  21. The corresponding local minimum point on the graph is
    \[\boxed{(1,1)}\]
🔍 Graphical Interpretation
The function first increases up to \(x=-1\), then decreases between \(x=-1\) and \(x=1\), and finally increases after \(x=1\). Therefore, the graph rises to the local maximum point \((-1,5)\), falls to the local minimum point \((1,1)\), and then rises again.
⚡ Exam Tip
❌ Common Mistakes
  • Stopping after finding \(x=\pm1\).
  • Forgetting to determine the sign of \(f'(x)\) on each interval.
  • Confusing a critical point \(x=c\) with the corresponding graph point \((c,f(c))\).
  • Reporting \(x=-1\) as the “maximum value” instead of reporting \(f(-1)=5\) as the maximum value.
  • Reporting \(x=1\) as the “minimum value” instead of reporting \(f(1)=1\) as the minimum value.
🌟 Significance
Significance for Board and Competitive Exams
This example is a standard application of the First Derivative Test. It demonstrates the complete workflow used in maximum-minimum problems: finding critical points, constructing intervals, determining derivative signs, classifying critical points, and calculating the corresponding function values.
🔑 Key Takeaway
∿

Example 18

📋
Table of Contents
9 sections
❓ Question
Find all the points of local maxima and local minima of the function
\[f(x)=2x^3-6x^2+6x+5,\qquad x\in\mathbb{R}.\]
💡 Concept
🗺️ Roadmap

  1. Find \(f'(x)\).

  2. Find the critical points from \(f'(x)=0\).

  3. Check the sign of \(f'(x)\) on both sides of each critical point.

  4. Apply the First Derivative Test.

  5. Calculate the function value at the critical point.

🧩 Solution
Given:
\[f(x)=2x^3-6x^2+6x+5\]
  1. Find the First Derivative of given function Differentiating term by term,
    \[f'(x)=6x^2-12x+6\]
  2. Factorising,
    \[f'(x)=6(x^2-2x+1)\]
    \[f'(x)=6(x-1)^2\]
  3. Find the Critical Point Since \(f'(x)\) exists for every real \(x\), critical points are obtained from
    \[f'(x)=0\]
  4. Therefore,
    \[6(x-1)^2=0\]
    \[(x-1)^2=0\]
    \[x-1=0\]
    \[\boxed{x=1}\]
    📝 Thus, \(x=1\) is the only critical point.
  5. Examine the Sign of \(f'(x)\) We have
    \[f'(x)=6(x-1)^2\]
  6. Since a square is always non-negative,
    \[(x-1)^2\geq0\]
  7. Hence,
    \[f'(x)\geq0\]
  8. for every real \(x\), with equality only at \(x=1\).<br><br>More specifically:
    • For \(x<1\),
      \[f'(x)>0.\]
      Therefore, \(f\) is increasing on \((-\infty,1)\).
    • At \(x=1\),
      \[f'(1)=0.\]
    • For \(x>1\),
      \[f'(x)>0.\]
      Therefore, \(f\) is increasing on \((1,\infty)\).
  9. Thus, the sign pattern is
    \[+\longrightarrow0\longrightarrow+\]
    📝 There is no change of sign in \(f'(x)\) as \(x\) passes through \(1\).
  10. Apply the First Derivative Test Since
    \[ +\longrightarrow+\]
    📝 around \(x=1\), the derivative does not change sign from positive to negative or from negative to positive.
  11. Therefore, \(x=1\) is neither a local maximum nor a local minimum.
    \[\boxed{x=1\text{ is neither a local maximum nor a local minimum.}}\]
  12. Find the Function Value at the Critical Point Although \(x=1\) is not a local extremum, its function value is useful for identifying the corresponding stationary point:
    \[f(1)=2(1)^3-6(1)^2+6(1)+5\]
    \[=2-6+6+5\]
    \[=7\]
  13. Hence, the critical point on the graph is
    \[\boxed{(1,7)}\]
🤔 Further Analysis: Why Is There No Local Maximum or Minimum?
The derivative is positive on both sides of \(x=1\):
\[f'(x)>0\quad\text{for }x < 1\]
and
\[f'(x)>0\quad\text{for }x>1\]
Therefore, the function continues to increase through \(x=1\). It does not turn from increasing to decreasing, nor from decreasing to increasing.
br> Hence, \(x=1\) is a stationary point, but it is not a local extremum.
⚡ Exam Tip
❌ Common Mistakes
  • Concluding that \(x=1\) is a maximum or minimum merely because \(f'(1)=0\).
  • Failing to check the sign of \(f'(x)\) on both sides of \(x=1\).
  • Assuming that every critical point is a local extremum.
  • Using the second derivative test alone when \(f''(1)=0\).
  • Confusing the critical point \(x=1\) with the graph point \((1,7)\).
🌟 Significance for Board and Competitive Exams
This example is particularly important because it demonstrates that solving \(f'(x)=0\) is not the end of a maximum-minimum problem. The critical point must be classified. It also illustrates a common competitive-exam pattern in which a repeated factor of \(f'(x)\) produces a stationary point but no local extremum.
🔑 Key Takeaway
∿

Theorem 4: Second Derivative Test

📋
Table of Contents
11 sections
🧮 Theorem
(i) Condition for a Local Maximum
If
\[f'(c)=0\]
and
\[f''(c)<0\]
then \(x=c\) is a point of local maximum. The corresponding local maximum value is \(f(c)\).
\[\boxed{f'(c)=0,\quad f''(c)<0\quad\Rightarrow\quad\text{Local Maximum at }x=c}\]
Why Does \(f''(c)<0\) Give a Maximum?
The condition
\[f''(c)<0\]
indicates that the graph is locally concave downward near \(x=c\). If the tangent is horizontal because
\[f'(c)=0\]
the graph bends downward from that horizontal tangent, producing a local maximum.
(ii) Condition for a Local Minimum
If
\[f'(c)=0\]
and
\[f''(c)>0\]
then \(x=c\) is a point of local minimum. The corresponding local minimum value is \(f(c)\).
\[\boxed{f'(c)=0,\quad f''(c)>0\quad\Rightarrow\quad\text{Local Minimum at }x=c}\]
Why Does \(f''(c)>0\) Give a Minimum?
The condition
\[f''(c)>0\]
indicates that the graph is locally concave upward. If the tangent is horizontal because
\[f'(c)=0\]
the graph bends upward from that horizontal tangent, producing a local minimum.
(iii) When the Second Derivative Test Fails
If
\[f'(c)=0\]
and
\[f''(c)=0\]
the Second Derivative Test is inconclusive.
\[\boxed{f'(c)=0,\quad f''(c)=0\quad\Rightarrow\quad\text{Second Derivative Test is inconclusive}}\]
This does not mean that \(c\) is neither a maximum nor a minimum. It means only that the second derivative test cannot decide the nature of the critical point.

In this situation, use the First Derivative Test, higher derivatives, or another suitable method
Second Derivative Test at a Glance
Condition Conclusion
\(f'(c)=0,\ f''(c)<0\) Local maximum at \(x=c\)
\(f'(c)=0,\ f''(c)>0\) Local minimum at \(x=c\)
\(f'(c)=0,\ f''(c)=0\) Test is inconclusive
🧮 Theorem
Statement
Let \(f\) be a function defined on an interval \(I\), and let \(c\in I\). Suppose that \(f\) is twice differentiable at \(x=c\). If
\[f'(c)=0\]
then the second derivative can be used to determine the nature of the critical point \(x=c\).
(i) Condition for a Local Maximum
If
\[f'(c)=0\]
and
\[f''(c)<0\]
then \(x=c\) is a point of local maximum. The corresponding local maximum value is \(f(c)\).
\[\boxed{f'(c)=0,\quad f''(c)<0\quad\Rightarrow\quad\text{Local Maximum at }x=c}\]
Why Does \(f''(c)<0\) Give a Maximum?
The condition
\[f''(c)<0\]
indicates that the graph is locally concave downward near \(x=c\). If the tangent is horizontal because
\[f'(c)=0\]
the graph bends downward from that horizontal tangent, producing a local maximum.
(ii) Condition for a Local Minimum
If
\[f'(c)=0\]
and
\[f''(c)>0\]
then \(x=c\) is a point of local minimum. The corresponding local minimum value is \(f(c)\).
\[\boxed{f'(c)=0,\quad f''(c)>0\quad\Rightarrow\quad\text{Local Minimum at }x=c}\]
Why Does \(f''(c)>0\) Give a Minimum?
The condition
\[f''(c)>0\]
indicates that the graph is locally concave upward. If the tangent is horizontal because
\[f'(c)=0\]
the graph bends upward from that horizontal tangent, producing a local minimum.
(iii) When the Second Derivative Test Fails
If
\[f'(c)=0\]
and
\[f''(c)=0\]
the Second Derivative Test is inconclusive.
\[\boxed{f'(c)=0,\quad f''(c)=0\quad\Rightarrow\quad\text{Second Derivative Test is inconclusive}}\]
This does not mean that \(c\) is neither a maximum nor a minimum. It means only that the second derivative test cannot decide the nature of the critical point.

In this situation, use the First Derivative Test, higher derivatives, or another suitable method
Second Derivative Test at a Glance
Condition Conclusion
\(f'(c)=0,\ f''(c)<0\) Local maximum at \(x=c\)
\(f'(c)=0,\ f''(c)>0\) Local minimum at \(x=c\)
\(f'(c)=0,\ f''(c)=0\) Test is inconclusive
🧮 Proof of the Second Derivative Test
🧮 Theorem
Proof
  1. The result can be understood using the behaviour of the first derivative near \(c\).

    Case 1: \(f''(c)<0\)
  2. Since
    \[f''(c)<0\]
    the derivative \(f'(x)\) is decreasing near \(x=c\), under the usual continuity assumptions used with the test.
  3. Since
    \[f'(c)=0\]
  4. we have, immediately to the left of \(c\),
    \[f'(x)>0\]
  5. and immediately to the right of \(c\),
    \[f'(x) < 0\]
  6. Thus the sign changes as
    \[+\longrightarrow-\]
  7. By the First Derivative Test, \(x=c\) is a local maximum.
  8. Case 2: \(f''(c)>0\)
  9. Since
    \[f''(c)>0\]
    the derivative \(f'(x)\) is increasing near \(x=c\), under the usual continuity assumptions.
  10. Since \(f'(c)=0\), immediately to the left of \(c\),
    \[f'(x) < 0\]
  11. and immediately to the right of \(c\),
    \[f'(x)>0\]
  12. Thus the sign changes as
    \[-\longrightarrow+\]
  13. By the First Derivative Test, \(x=c\) is a local minimum.
✏️ Example
Second Derivative Test Gives a Local Maximum
Consider
\[f(x)=-x^2+4x+1\]
First derivative:
\[f'(x)=-2x+4\]
For a critical point,
\[-2x+4=0\]
\[x=2\]
The second derivative is
\[f''(x)=-2\]
Therefore,
\[f''(2)=-2 < 0\]
Hence, by the Second Derivative Test, \(x=2\) is a local maximum point.

The local maximum value is
\[f(2)=-(2)^2+4(2)+1\]
\[=-4+8+1\]
\[=5\]
\[\boxed{\text{Local maximum value}=5}\]
The corresponding point on the graph is
\[\boxed{(2,5)}\]
✏️ Example
Second Derivative Test Gives a Local Minimum
Consider
\[f(x)=x^2-6x+10\]
First derivative:
\[f'(x)=2x-6\]
Setting \(f'(x)=0\),
\[2x-6=0\]
\[x=3\]
The second derivative is
\[f''(x)=2\]
Therefore,
\[f''(3)=2>0\]
Hence, \(x=3\) is a local minimum point.

The local minimum value is
\[f(3)=3^2-6(3)+10\]
\[=9-18+10\]
\[=1\]
\[\boxed{\text{Local minimum value}=1}\]
The corresponding point on the graph is
\[\boxed{(3,1)}\]
✏️ Example
Second Derivative Test Is Inconclusive
Consider
\[f(x)=x^3\]
Then
\[f'(x)=3x^2\]
Therefore,
\[f'(0)=0\]
The second derivative is
\[f''(x)=6x\]
Hence,
\[f''(0)=0\]
Therefore, the Second Derivative Test is inconclusive.

We must use another method. From
\[ f'(x)=3x^2>0 \qquad\text{for }x\neq0, \]
the function is increasing on both sides of \(0\). Hence, \(x=0\) is neither a local maximum nor a local minimum.
📌 Note
Important Distinction: Inconclusive Does Not Mean “No Extremum”
🔗 Relationship with the First Derivative Test
The two tests can be summarised as follows:
Method Condition Conclusion
First Derivative Test \(+\to-\) Local maximum
First Derivative Test \(-\to+\) Local minimum
First Derivative Test No sign change Neither local maximum nor local minimum
Second Derivative Test \(f'(c)=0,\ f''(c)<0\) Local maximum
Second Derivative Test \(f'(c)=0,\ f''(c)>0\) Local minimum
Second Derivative Test \(f'(c)=0,\ f''(c)=0\) Inconclusive
📎 Exam-Oriented Procedure
  1. Find \(f'(x)\).
  2. Solve
    \[ f'(x)=0 \]
    to obtain the critical points.
  3. Calculate \(f''(x)\).
  4. Substitute each critical point \(c\) into \(f''(x)\).
  5. If
    \[ f''(c)<0, \]
    conclude local maximum.
  6. If
    \[ f''(c)>0, \]
    conclude local minimum.
  7. If
    \[ f''(c)=0, \]
    use the First Derivative Test or another appropriate method.
❌ Common Mistakes
  • Applying the Second Derivative Test without first checking \(f'(c)=0\).
  • Thinking that \(f''(c)=0\) means there is no maximum or minimum.
  • Forgetting to calculate \(f(c)\) after identifying the extremum point.
  • Confusing a local maximum/minimum point \(x=c\) with its value \(f(c)\).
  • Using the Second Derivative Test when \(f\) is not twice differentiable at the point under consideration.
🌟 Significance for Board and Competitive Exams
The Second Derivative Test provides a fast method for classifying differentiable critical points. It is especially useful when the first derivative has several critical points and constructing a complete sign chart would be lengthy. However, when \(f''(c)=0\), the First Derivative Test remains an important alternative.
🔑 Key Takeaway
∿

Example 19

📋
Table of Contents
11 sections
❓ Question
Find the local minimum value of the function
\[f(x)=3+|x|,\qquad x\in\mathbb{R}\]
💡 Concept
🗺️ Roadmap

  1. Write the absolute-value function in piecewise form.

  2. Check differentiability at \(x=0\).

  3. Find \(f'(x)\) to the left and right of \(0\).

  4. Apply the First Derivative Test.

  5. Calculate the corresponding minimum value.

🧩 Solution
  1. Write the Function in Piecewise Form We know that
    \[ |x|= \begin{cases} -x, & x < 0,\\ x, & x\geq0. \end{cases} \]
  2. Therefore,
    \[ f(x)=3+|x| = \begin{cases} 3-x, & x < 0,\\ 3+x, & x\geq0. \end{cases} \]
  3. Check Differentiability at \(x=0\) In particular,
    \[f(0)=3+|0|=3\]
  4. For \(x < 0\),
    \[f(x)=3-x\]
  5. Hence,
    \[f'(x)=-1,\qquad x < 0\]
  6. For \(x>0\),
    \[f(x)=3+x\]
  7. Hence,
    \[f'(x)=1,\quad x>0\]
  8. Therefore, the right-hand derivative at \(x=0\) is
    \[f'_+(0)=1\]
  9. Since
    \[f'_-(0)\neq f'_+(0),\]
  10. we conclude that
    \[\boxed{f'(0)\text{ does not exist}.}\]
  11. Apply the First Derivative Test On the left side of \(0\)
    \[f'(x)=-1<0\]
    📝 Therefore, \(f(x)\) is decreasing as \(x\) approaches \(0\) from the left.
  12. On the right side of \(0\),
    \[f'(x)=1>0\]
    📝 Therefore, \(f(x)\) is increasing as \(x\) moves to the right of \(0\).
  13. Thus, the sign of the derivative changes as
    \[-\longrightarrow+\]
  14. By the First Derivative Test,
    \[\boxed{x=0\text{ is a point of local minimum}.}\]
  15. Find the Local Minimum Value The local minimum value is obtained by substituting \(x=0\):
    \[f(0)=3+|0|\]
    \[=3\]
  16. Therefore,
    \[\boxed{\text{Local minimum value}=3}\]
  17. The corresponding point on the graph is
    \[\boxed{(0,3)}\]
🔍 Graphical Interpretation

The graph of \(f(x)=3+|x|\) is the graph of \(y=|x|\) shifted upward by \(3\) units. It has a sharp corner at \((0,3)\).

The graph decreases toward \((0,3)\) from the left and increases away from \((0,3)\) on the right. Hence, the corner is a minimum point even though the ordinary derivative does not exist there.

🗒️ First Derivative Sign Chart
Interval \(f'(x)\) Behaviour of \(f\)
\((-\infty,0)\) \(-1<0\) Decreasing
\(x=0\) Does not exist Critical point
\((0,\infty)\) \(1>0\) Increasing
🌟 Why This Example Is Important
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(3+x\) incorrectly as \(x-3\) for \(x>0\).
  • Ignoring the non-differentiability at \(x=0\).
  • Searching only for \(f'(x)=0\).
  • Writing the sign change incorrectly.
  • Confusing the local minimum point \(x=0\) with the local minimum value \(3\).
🌟 Significance for Board and Competitive Exams
This is a useful example of a non-differentiable critical point. It reinforces the First Derivative Test and shows why points where the derivative does not exist must be included when searching for extrema. It is especially relevant to questions involving absolute-value and piecewise-defined functions.
🔑 Key Takeaway
∿

Example 20

📋
Table of Contents
7 sections
❓ Question
Find the local maximum and local minimum values of the function
\[f(x)=3x^4+4x^3-12x^2+12\]
💡 Concept
🗺️ Roadmap

  1. Differentiate \(f(x)\).

  2. Solve \(f'(x)=0\).

  3. Find the second derivative \(f''(x)\).

  4. Apply the second derivative test.

  5. Calculate the corresponding function values to obtain the local maximum and minimum values.

🧩 Solution
Part (a)
  1. Find the First Derivative Using the power rule,
    \[\begin{aligned}f'(x)&=\frac{d}{dx}\left(3x^4+4x^3-12x^2+12\right)\\ &=12x^3+12x^2-24x\end{aligned}\]
  2. For critical points, put
    \[f'(x)=0\]
  3. Therefore,
    \[12x^3+12x^2-24x=0\]
  4. Taking \(12x\) common,
    \[12x(x^2+x-2)=0\]
  5. Factorising the quadratic,
    \[ x^2+x-2=(x+2)(x-1)\]
  6. Hence,
    \[12x(x+2)(x-1)=0\]
  7. Therefore,
    \[x=0,\quad x=-2,\quad x=\]
  8. Thus, the critical points are
    \[\boxed{x=-2,\;0,\;1}\]
  9. Find the Second Derivative of \(f'(x)=12x^3+12x^2-24x\) Differentiating again,
    \[\begin{aligned}f''(x)&=36x^2+24x-24\end{aligned}\]
  10. Apply the Second Derivative Test At \(x=0\)
    \[ \begin{aligned} f''(0) &=36(0)^2+24(0)-24\\ &=-24 \end{aligned} \]
  11. Since
    \[f''(0) < 0\]
    📝 the function has a local maximum at \(x=0\)
  12. Now calculate the corresponding function value:
    \[ \begin{aligned} f(0) &=3(0)^4+4(0)^3-12(0)^2+12\\ &=12 \end{aligned} \]
  13. Therefore,
    \[\boxed{\text{Local maximum value}=12}\]
  14. at
    \[\boxed{x=0}\]
  15. At \(x=-2\)
    \[ \begin{aligned} f''(-2) &=36(-2)^2+24(-2)-24\\ &=36(4)-48-24\ &=144-48-24\\ &=72. \end{aligned} \]
  16. Since
    \[f''(-2)>0\]
    📝 the function has a local minimum at \(x=-2\).
  17. Now calculate the corresponding function value:
    \[ \begin{aligned} f(-2) &=3(-2)^4+4(-2)^3-12(-2)^2+12\\ &=3(16)+4(-8)-12(4)+12\\ &=48-32-48+12\\ &=-20. \end{aligned} \]
  18. Therefore,
    \[\boxed{\text{Local minimum value}=-20}\]
  19. at
    \[\boxed{x=-2}\]
  20. At \(x=1\)
    \[ \begin{aligned} f''(1) &=36(1)^2+24(1)-24\ \ &=36+24-24\\ &=36. \end{aligned} \]
  21. Since
    \[f''(1)>0\]
    📝 the function has a local minimum at \(x=1\).
  22. Now calculate the corresponding function value:
    \[ \begin{aligned} f(1) &=3(1)^4+4(1)^3-12(1)^2+12\\ &=3+4-12+12\\ &=7 \end{aligned} \]
  23. Therefore,
    \[\boxed{\text{Local minimum value}=7}\]
  24. at
    \[\boxed{x=1}\]
  25. Summary Table
    Critical point \(x\) \(f''(x)\) Classification Function value
    \(-2\) \(72>0\) Local minimum \(-20\)
    \(0\) \(-24<0\) Local maximum \(12\)
    \(1\) \(36>0\) Local minimum \(7\)
🗒️ Board Exam Insight
A common mistake is to stop after solving \(f'(x)=0\). Finding \(f'(x)=0\) only identifies possible locations of local extrema. It does not tell us whether each critical point is a maximum, a minimum, or neither.

For the second derivative test, remember:
\[ \boxed{ \begin{aligned} f''(c)<0 &\Rightarrow \text{local maximum at }x=c,\\ f''(c)>0 &\Rightarrow \text{local minimum at }x=c,\\ f''(c)=0 &\Rightarrow \text{test is inconclusive.} \end{aligned}} \]
In particular, \(f''(c)=0\) does not mean that the point is neither a maximum nor a minimum. Another test, such as the first derivative test, must then be used.
🌟 Significance for Board and Competitive Examinations
This example illustrates the standard workflow for extrema questions: differentiate, locate critical points, classify them, and finally calculate the corresponding function values. The same approach is frequently used in questions involving polynomial functions, optimization, graphs, and application-based calculus.
🔑 Key Takeaway
∿

Example 21

Application of Derivatives NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Derivatives Rate of Change Increasing Functions Decreasing Functions Maxima and Minima Absolute Maximum Absolute Minimum Local Maxima Local Minima First Derivative Test Second Derivative Test Optimization Problems Related Rates Marginal Cost Marginal Revenue JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
6 sections
❓ Question
Find all the points of local maxima and local minima of the function
\[f(x)=2x^3-6x^2+6x+5\]
💡 Concept
🗺️ Roadmap

  1. Find \(f'(x)\).

  2. Set \(f'(x)=0\) to obtain the critical point.

  3. Calculate \(f''(x)\).

  4. Since \(f''(1)=0\), the second derivative test is inconclusive.

  5. Use the first derivative test to classify the critical point.

  6. Check the concavity if identifying a possible point of inflection.

🧩 Solution
  1. Find the First Derivative of
    \[f(x)=2x^3-6x^2+6x+5\]
  2. Differentiating,
    \[ \begin{aligned} f'(x) &=6x^2-12x+6\\ &=6(x^2-2x+1)\\ &=6(x-1)^2. \end{aligned} \]
  3. For critical points,
    \[f'(x)=0\]
  4. Therefore,
    \[6(x-1)^2=0\]
  5. Hence,
    \[\boxed{x=1}\]
    📝 Thus, \(x=1\) is the only critical point.
  6. Apply the Second Derivative Test Differentiating \(f'(x)\)
    \[f''(x)=12x-12\]
  7. At \(x=1\),
    \[ \begin{aligned} f''(1) &=12(1)-12\\ &=0 \end{aligned} \]
  8. Therefore,
    \[\boxed{f''(1)=0}\]
    📝 The second derivative test is therefore inconclusive. We cannot conclude from \(f''(1)=0\) alone that \(x=1\) is not a local maximum or minimum.
  9. Use the First Derivative Test We have
    \[f'(x)=6(x-1)^2\]
  10. Since
    \[(x-1)^2>0\]
  11. for every \(x\neq1\),
    \[f'(x)>0\]
  12. on both sides of \(x=1\). Therefore,
    \[ \begin{array}{c|ccc} x & (-\infty,1) & 1 & (1,\infty)\\ \hline f'(x) & + & 0 & + \end{array} \]
  13. Hence, \(f(x)\) is increasing on both
    \[(-\infty,1)\]
    and
    \[(1,\infty)\]
  14. The sign of \(f'(x)\) changes as
    \[+\longrightarrow+\]
    through \(x=1\).
    📝 Since there is no change in sign of \(f'(x)\), the first derivative test shows that \(x=1\) is neither a local maximum nor a local minimum.
  15. Find the Corresponding Function Value Although \(x=1\) is not a local extremum, its corresponding function value is useful for locating the stationary point on the graph.
    \[\begin{aligned}f(1)&=2(1)^3-6(1)^2+6(1)+5\\&=2-6+6+5\\&=7\end{aligned}\]
  16. Therefore, the critical point on the graph is
    \[\boxed{(1,7)}\]
  17. Is \((1,7)\) a Point of Inflection? We have
    \[f''(x)=12x-12=12(x-1)\]
  18. Consider the sign of \(f''(x)\):
    • For \(x<1\), \(f''(x)<0\), so the graph is concave downward.
    • For \(x>1\), \(f''(x)>0\), so the graph is concave upward.
    Thus, the concavity changes from downward to upward at \(x=1\).
  19. Therefore,
    \[\boxed{(1,7)\text{ is a point of inflection}.}\]
  20. Moreover, since
    \[f'(1)=0\]
    📝 the tangent at this point is horizontal. Hence, \((1,7)\) is specifically a stationary point of inflection.
⚡ Exam Tip
🔑 Key Takeaway
∿

Example 22

Application of Derivatives NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Derivatives Rate of Change Increasing Functions Decreasing Functions Maxima and Minima Absolute Maximum Absolute Minimum Local Maxima Local Minima First Derivative Test Second Derivative Test Optimization Problems Related Rates Marginal Cost Marginal Revenue JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
7 sections
❓ Question
Find two positive numbers whose sum is \(15\) and the sum of whose squares is minimum.
💡 Concept
🗺️ Roadmap

  1. Let one positive number be \(x\).

  2. Express the second number using the given sum.

  3. Form the sum of the squares as a function \(S(x)\).

  4. Find \(S'(x)\) and solve \(S'(x)=0\).

  5. Use \(S''(x)\) to confirm that the critical point gives a minimum.

  6. Find the two required numbers and the minimum value.

🧩 Solution
  1. Represent the Two Numbers Let the first positive number be \(x\).
    \[x>0\]
  2. Since their sum is \(15\), the second number is
    \[15-x\]
  3. Because the second number is also positive,
    \[15-x>0\]
  4. and hence
    \[0 < x < 15\]
    📝 Therefore, the problem is to minimize the sum of squares over the interval \(0 < x < 15\)
  5. Form the Function to Be Minimized Let \(S\) denote the sum of the squares of the two numbers.
    \[S=x^2+(15-x)^2\]
  6. Expanding,
    \[ \begin{aligned} S &=x^2+225-30x+x^2\\ &=2x^2-30x+225 \end{aligned} \]
  7. Thus,
    \[\boxed{S(x)=2x^2-30x+225}\]
  8. Find the Critical Point Differentiate \(S(x)\):
    \[\begin{aligned}S'(x)&=4x-30\end{aligned}\]
  9. For a possible minimum,
    \[S'(x)=0\]
  10. Therefore,
    \[\begin{aligned}4x-30&=0\\4x&=30\\x&=\frac{15}{2} \end{aligned}\]
  11. Verify the Minimum Using the Second Derivative Since
    \[S'(x)=4x-30\]
  12. Therefore,
    \[S''(x)=4\]
  13. Thus,
    \[S''\left(\frac{15}{2}\right)=4>0\]
  14. By the second derivative test,
    \[\boxed{x=\frac{15}{2}}\]
    📝 gives a minimum value of \(S\).
  15. Find the Two Numbers The first number is
    \[x=\frac{15}{2}\]
  16. The second number is
    \[15-x=15-\frac{15}{2}=\frac{15}{2}\]
  17. Therefore, the two numbers are
    \[\boxed{\frac{15}{2}\text{ and }\frac{15}{2}}.\]
  18. Find the Minimum Sum of Squares Substituting \(x=\frac{15}{2}\),
    \[ \begin{aligned} S_{\min} &=\left(\frac{15}{2}\right)^2+ \left(\frac{15}{2}\right)^2\\ &=2\left(\frac{225}{4}\right)\\ &=\frac{225}{2} \end{aligned} \]
  19. Hence,
    \[\boxed{S_{\min}=\frac{225}{2}}\]
🌟 Exam Significance
This is a standard application of derivatives involving optimization under a constraint. The same framework is used in questions involving minimum cost, maximum area, minimum distance, maximum volume, and other optimization problems.
❌ Common Mistakes
  • Forgetting the constraint \(0 < x < 15\) arising from the requirement that both numbers are positive.
  • Writing the second number as \(15+x\) instead of \(15-x\).
  • Differentiating \((15-x)^2\) incorrectly. The correct derivative is
    \[ -2(15-x). \]
  • Stopping after obtaining \(S'(x)=0\) without verifying whether the critical point gives a minimum.
  • Reporting only \(x=\frac{15}{2}\) instead of giving both numbers.
🔑 Key Takeaway
∿

Example 23

Application of Derivatives NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Derivatives Rate of Change Increasing Functions Decreasing Functions Maxima and Minima Absolute Maximum Absolute Minimum Local Maxima Local Minima First Derivative Test Second Derivative Test Optimization Problems Related Rates Marginal Cost Marginal Revenue JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
7 sections
❓ Question
Find the shortest distance of the point
\[ (0,c) \]
from the parabola
\[ y=x^2, \]
where
\[ 1\leq c\leq5. \]
💡 Concept
🗺️ Roadmap

  1. Take a general point \(P=(x,x^2)\) on the parabola.

  2. Form the square of the distance from \(P\) to \((0,c)\).

  3. Differentiate the squared distance.

  4. Find the critical points.

  5. Use the second derivative test to identify the minimum.

  6. Calculate the corresponding shortest distance.

🧩 Solution
  1. Take a General Point on the Parabola The equation of the parabola is
    \[y=x^2\]
  2. Therefore, a general point on the parabola is
    \[ P=(x,x^2)\]
  3. The fixed point is
    \[A=(0,c)\]
  4. Form the Distance Function Using the distance formula, the distance \(d\) between \(A=(0,c)\) and \(P=(x,x^2)\) is
    \[d=\sqrt{(x-0)^2+(x^2-c)^2}\]
  5. Therefore,
    \[d=\sqrt{x^2+(x^2-c)^2}\]
  6. Instead of minimizing \(d\), minimize its square:
    \[D=d^2=x^2+(x^2-c)^2\]
  7. Expanding,
    \[\begin{aligned}D&=x^2+x^4-2cx^2+c^2\\&=x^4+(1-2c)x^2+c^2.\end{aligned}\]
  8. Thus,
    \[\boxed{D(x)=x^4+(1-2c)x^2+c^2}\]
  9. Find the Critical Points Differentiate \(D(x)\):
    \[ \begin{aligned} D'(x) &=4x^3+2(1-2c)x\\ &=4x^3+2x-4cx\\ &=2x(2x^2+1-2c). \end{aligned} \]
  10. For a critical point,
    \[D'(x)=0\]
  11. Hence,
    \[2x(2x^2+1-2c)=0\]
  12. Therefore,
    \[x=0\]
    or
    \[2x^2+1-2c=0\]
  13. Thus,
    \[2x^2=2c-1\]
    and
    \[x^2=c-\frac12\]
  14. Hence,
    \[\boxed{x=\pm\sqrt{c-\frac12}}\]
  15. Since,
    \[1\leq c\leq5,\]
  16. we have
    \[c-\frac12>0,\]
    📝 so these two critical points are real.
  17. Classify the Critical Points Find the second derivative:
    \[ \begin{aligned} D'(x) &=4x^3+2(1-2c)x,\\ D''(x) &=12x^2+2(1-2c) \end{aligned} \]
  18. At \(x=0\)
    \[ \begin{aligned} D''(0) &=2(1-2c). \end{aligned} \]
  19. Since
    \[1\leq c\leq5\]
  20. we have
    \[1-2c<0\]
  21. Therefore,
    \[D''(0)<0\]
    📝 Hence, \(x=0\) gives a local maximum of the squared distance.

    Thus, \(x=0\) cannot give the shortest distance.
  22. At \(x=\pm\sqrt{c-\frac12}\) Since
    \[x^2=c-\frac12\]
  23. substitute into \(D''(x)\):
    \[ \begin{aligned} D''(x) &=12\left(c-\frac12\right)+2(1-2c)\\ &=12c-6+2-4c\\ &=8c-4\\ &=4(2c-1). \end{aligned} \]
  24. Since \(c\geq1\),
    \[2c-1>0\]
  25. Therefore,
    \[D''(x)>0\]
  26. Hence,
    \[\boxed{x=\pm\sqrt{c-\frac12}}\]
    📝 give minimum values of the squared distance.
  27. Find the Corresponding Points on the Parabola We have
    \[x^2=c-\frac12\]
  28. Since the parabola is \(y=x^2\),
    \[y=c-\frac12\]
  29. Therefore, the two points on the parabola nearest to \((0,c)\) are
    \[\boxed{\left(\sqrt{c-\frac12},\,c-\frac12\right)}\]
    and
    \[\boxed{\left(-\sqrt{c-\frac12},\,c-\frac12\right)}\]
  30. Find the Shortest Distance Recall that
    \[D=x^2+(x^2-c)^2\]
  31. At the minimum,
    \[x^2=c-\frac12\]
  32. Therefore,
    \[x^2-c=c-\frac12-c=-\frac12\]
  33. Hence,
    \[\begin{aligned}D_{\min}&=\left(c-\frac12\right)+\left(-\frac12\right)^2\\ &=c-\frac12+\frac14\\ &=c-\frac14\end{aligned}\]
  34. Since \(D=d^2\),
    \[d_{\min}=\sqrt{D_{\min}}\]
  35. Therefore,
    \[\boxed{d_{\min}=\sqrt{c-\frac14}}\]
🔍 Geometrical Interpretation
Because the parabola \(y=x^2\) is symmetric about the \(y\)-axis and the point \((0,c)\) lies on the \(y\)-axis, the two nearest points occur symmetrically on either side of the \(y\)-axis.

The two nearest points are therefore
\[\left(\sqrt{c-\frac12},c-\frac12\right)\]
and
\[\left(-\sqrt{c-\frac12},c-\frac12\right)\]
❌ Common Mistakes
  • Using an arbitrary point \((x,y)\) without imposing \(y=x^2\).
  • Minimizing the distance without first considering the simpler squared distance.
  • Forgetting that \(x^2=c-\frac12\) gives two values of \(x\).
  • Reporting only one nearest point even though the parabola is symmetric.
  • Confusing the minimum value of \(d^2\), which is \(c-\frac14\), with the minimum distance, which is \(\sqrt{c-\frac14}\).
  • Assuming that every critical point gives a minimum. Here \(x=0\) gives a local maximum of the squared-distance function.
🌟 Significance for Board and Competitive Examinations
  • For a point on \(y=x^2\), use \(P=(x,x^2)\).
  • Minimize \(d^2\) instead of \(d\) whenever possible.
  • The critical points are \(x=0\) and \(x=\pm\sqrt{c-\frac12}\).
  • The points \(x=\pm\sqrt{c-\frac12}\) give the minimum distance.
  • The shortest distance is \(\boxed{\sqrt{c-\frac14}}\).
  • The two nearest points are symmetric about the \(y\)-axis.
∿

Example 24

📋
Table of Contents
9 sections
❓ Question
Let \(AP\) and \(BQ\) be two vertical poles at points \(A\) and \(B\), respectively. If
\[AP=16\text{ m},\qquad BQ=22\text{ m},\qquad AB=20\text{ m},\]
then find the distance of a point \(R\) on \(AB\) from the point \(A\) such that
\[RP^2+RQ^2\]
is minimum.
💡 Concept
🗺️ Roadmap

  1. Let \(AR=x\).

  2. Express \(BR\) in terms of \(x\).

  3. Use the Pythagorean theorem to obtain \(RP^2\) and \(RQ^2\).

  4. Form \(S(x)=RP^2+RQ^2\).

  5. Find the critical point using \(S'(x)=0\).

  6. Use \(S''(x)>0\) to confirm the minimum.

🧩 Solution
  1. Introduce the Variable Let the distance of \(R\) from \(A\) be
    \[AR=x\]
  2. Since
    \[AB=20\text{ m}\]
  3. we have
    \[BR=AB-AR=20-x\]
  4. Since \(R\) lies on the segment \(AB\),
    \[\boxed{0\leq x\leq20}\]
  5. Find \(RP^2\) Since \(AP\) is vertical and \(AR\) is horizontal, triangle \(APR\) is a right triangle.<br><br>We know
    \[AP=16,\quad AR=x\]
  6. Therefore, by Pythagoras,
    \[ RP^2=AP^2+AR^2\]
  7. Hence,
    \[RP^2=16^2+x^2\]
  8. and therefore
    \[\boxed{RP^2=x^2+256}\]
  9. Find \(RQ^2\) Similarly, triangle \(BRQ\) is a right triangle because \(BQ\) is vertical and \(BR\) is horizontal. We know
    \[BQ=22,\quad BR=20-x\]
  10. Therefore,
    \[RQ^2=BQ^2+BR^2\]
  11. Thus,
    \[RQ^2=22^2+(20-x)^2\]
  12. Hence,
    \[\boxed{RQ^2=484+(20-x)^2}\]
  13. Form the Function to Be Minimized Let
    \[S=RP^2+RQ^2\]
  14. Substituting the expressions obtained above,
    \[ \begin{aligned} S &=x^2+256+484+(20-x)^2\\ &=x^2+(20-x)^2+740 \end{aligned} \]
  15. Expanding,
    \[ \begin{aligned} S &=x^2+400-40x+x^2+740\\ &=2x^2-40x+1140 \end{aligned} \]
  16. Therefore,
    \[\boxed{S(x)=2x^2-40x+1140}\]
  17. Find the Critical Point Differentiate \(S(x)\):
    \[S'(x)=4x-40\]
  18. For a possible minimum,
    \[S'(x)=0\]
  19. Therefore,
    \[4x-40=0\]
  20. Hence,
    \[4x=40\]
    and
    \[\boxed{x=10}\]
  21. Verify the Minimum Differentiate once more:
    \[S''(x)=4\]
  22. Therefore,
    \[S''(10)=4>0\]
  23. By the second derivative test, \(S\) has a minimum at
    \[x=10\]
  24. Since
    \[x=AR\]
  25. we obtain
    \[\boxed{AR=10\text{ m}}\]
🔍 Geometrical Interpretation
The two poles have different heights, \(16\) m and \(22\) m, but the vertical components do not affect the location of the minimum directly. The constant contribution from the pole heights is
\[16^2+22^2=256+484=740\]
Thus, minimizing \(RP^2+RQ^2\) is equivalent to minimizing
\[x^2+(20-x)^2\]
This expression is minimized when the horizontal distances \(AR\) and \(BR\) are equal:
\[AR=BR=10\text{ m}\]
Hence, \(R\) is the midpoint of \(AB\).
\[\boxed{R\text{ is the midpoint of }AB}\]
👁️ Important Observation
❌ Common Mistakes
  • Using \(BR=x\) instead of \(BR=20-x\).
  • Forgetting to square the pole heights when applying Pythagoras.
  • Using \(RP=16+x\), which is geometrically incorrect.
  • Minimizing \(RP+RQ\) instead of the quantity explicitly given, \(RP^2+RQ^2\).
  • Stopping at \(S'(x)=0\) without confirming that the critical point gives a minimum.
🌟 Significance for Board and Competitive Examinations
This is a standard geometric optimization problem. It combines the Pythagorean theorem with differentiation and illustrates how a geometric constraint can be converted into a one-variable optimization problem. The same technique is useful in shortest-distance, minimum-energy, minimum-cost, and related optimization questions.
🔑 Key Takeaway
∿

Example 25

📋
Table of Contents
8 sections
❓ Question
If the lengths of the three sides of a trapezium other than the base are equal to \(10\) cm, find the maximum possible area of the trapezium.
🗒️ Concpet
This is a problem of maximization using derivatives. Since three sides of the trapezium are fixed at \(10\) cm and only one base is variable, we express the area in terms of the variable base and then maximize it.

Let the three equal sides be the two non-parallel sides and one base. Thus, the trapezium has:
\[\text{one base}=10\text{ cm},\quad\text{other base}=x\text{ cm},\]
\[\text{each non-parallel side}=10\text{ cm}.\]
🗺️ Roadmap

  1. Let the variable base be \(x\) cm.

  2. Express the height of the trapezium in terms of \(x\).

  3. Form the area as a function of \(x\).

  4. Differentiate and find the critical point.

  5. Use the second derivative test to confirm the maximum.

  6. Calculate the maximum area.

🧩 Solution
  1. Geometrical Setup Let the parallel sides of the trapezium be \(10\) cm and \(x\) cm, and let each of the two non-parallel sides be \(10\) cm.
  2. ⭐ Drop perpendiculars from the endpoints of the shorter base to the longer base. For the symmetric configuration giving the maximum area, the two right triangles formed at the sides are congruent.
  3. The difference between the two bases is
    \[10-x\]
  4. Hence, each of the two right triangles has horizontal side
    \[\frac{10-x}{2}\]
  5. Since the slant side is \(10\) cm, the height \(h\) satisfies
    \[h^2+\left(\frac{10-x}{2}\right)^2=10^2\]
  6. Therefore,
    \[h^2=100-\frac{(10-x)^2}{4}\]
  7. Thus,
    \[\boxed{h=\frac12\sqrt{400-(10-x)^2}}\]
  8. Form the Area Function The area of a trapezium is
    \[A=\frac12(\text{sum of parallel sides})\times h\]
  9. Therefore,
    \[A=\frac12(10+x)h\]
  10. Substituting the expression for \(h\),
    \[ A =\frac12(10+x) \left[\frac12\sqrt{400-(10-x)^2}\right]. \]
  11. Hence,
    \[ \boxed{ A(x)=\frac14(10+x)\sqrt{400-(10-x)^2} }. \]
  12. Simplify the Area Function To avoid differentiating the square-root expression directly, maximize \(A^2\), since \(A>0\).
    \[A^2=\frac1{16}(10+x)^2\left[400-(10-x)^2\right].\]
  13. Now,
    \[400-(10-x)^2=400-(100-20x+x^2).\]
  14. Therefore,
    \[400-(10-x)^2=300+20x-x^2\]
  15. Hence,
    \[A^2=\frac1{16}(10+x)^2(300+20x-x^2)\]
  16. Find the Critical Point Differentiating \(A^2\) directly is possible, but logarithmic differentiation makes the calculation shorter.<br><br> Since
    \[A^2\propto(10+x)^2(300+20x-x^2),\]
  17. take logarithms:
    \[\ln(A^2)=2\ln(10+x)+\ln(300+20x-x^2)+\text{constant}\]
  18. Differentiating,
    \[\frac{2}{10+x}+\frac{20-2x}{300+20x-x^2}=0\]
  19. Therefore,
    \[\frac{2}{10+x}+\frac{2(10-x)}{300+20x-x^2}=0\]
  20. Multiplying by the positive denominators,
    \[2(300+20x-x^2)+2(10-x)(10+x)=0\]
  21. Dividing by \(2\),
    \[300+20x-x^2+100-x^2=0\]
  22. Hence,
    \[400+20x-2x^2=0\]
  23. Dividing by \(2\),
    \[200+10x-x^2=0\]
  24. Therefore,
    \[x^2-10x-200=0\]
  25. Factorising,
    \[(x-20)(x+10)=0\]
  26. Hence,
    \[x=20\quad\text{or}\quad x=-10.\]
  27. Since \(x\) represents a length,
    \[x>0\]
  28. Therefore,
    \[\boxed{x=20\text{ cm}}\]
  29. Find the Height When \(x=20\),
    \[\frac{10-x}{2}=\frac{10-20}{2}=-5\]
  30. The negative sign merely indicates which base is longer; the horizontal length of each right triangle is \(5\) cm. Thus,
    \[h^2=10^2-5^2\]
  31. Hence,
    \[h^2=100-25=75\]
    and
    \[\boxed{h=5\sqrt3\text{ cm}}\]
  32. Find the Maximum Area
    The parallel sides are \(10\) cm and \(20\) cm, and the height is \(5\sqrt3\) cm.
  33. Therefore,
    \[ \begin{aligned} A_{\max} &=\frac12(10+20)(5\sqrt3)\\ &=\frac12(30)(5\sqrt3)\\ &=75\sqrt3. \end{aligned} \]
  34. Hence,
    \[\boxed{A_{\max}=75\sqrt3\text{ cm}^2}\]
👁️ Important Geometrical Observation
❌ Common Mistakes
  • Using the area formula without first finding the height.
  • Assuming that the variable base is \(10+x\) instead of introducing it independently.
  • Forgetting that the two horizontal portions formed by the perpendiculars are equal in the maximum symmetric configuration.
  • Rejecting \(x=20\) merely because it is larger than the fixed \(10\)-cm base. Either base may be the longer parallel side.
  • Reporting the maximum value of \(A^2\) instead of the maximum area \(A\).
🌟 Significance for Board and Competitive Examinations
This problem combines coordinate-style geometric reasoning with differential optimization. It is a useful model for maximum-area questions in which a geometric constraint determines the height as a function of a variable side.
🔑 Key Takeaway
∿

Example 26

📋
Table of Contents
8 sections
❓ Question
Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of the radius of the cone.
💡 Concept
🗺️ Roadmap

  1. Let \(R\) and \(H\) be the radius and height of the given cone.

  2. Let \(r\) and \(h\) be the radius and height of the inscribed cylinder.

  3. Use similar triangles to establish a relation between \(r\) and \(h\).

  4. Express the curved surface area of the cylinder as a function of \(r\).

  5. Differentiate and find the critical point.

  6. Use the second derivative test to prove that the critical point gives the greatest curved surface area.

🧩 Solution
  1. Define the Cone and Cylinder Let the given right circular cone have:
    \[\text{radius}=R,\quad \text{height}=H.\]
  2. Let the inscribed right circular cylinder have:
    \[\text{radius}=r,\quad \text{height}=h\]
    The cylinder is positioned with its circular base on the base of the cone and its upper circular rim touching the lateral surface of the cone.
  3. Establish the Relation Between \(r\) and \(h\) Consider the axial cross-section of the cone and cylinder. The cone becomes an isosceles triangle, while the cylinder becomes a rectangle.
    The small triangle above the cylinder has height
    \[H-h\]
    and radius
    \[r\]
    The complete cone has height \(H\) and radius \(R\).
  4. By similarity of these triangles,
    \[\frac{r}{R}=\frac{H-h}{H}\]
  5. Therefore,
    \[r=R\left(\frac{H-h}{H}\right)\]
  6. Hence,
    \[\frac{r}{R}=1-\frac{h}{H}\]
  7. Solving for \(h\),
    \[\frac{h}{H}=1-\frac{r}{R}\]
  8. Therefore,
    \[\boxed{h=H\left(1-\frac{r}{R}\right)}\]
  9. Equivalently,
    \[\boxed{h=H-\frac{H}{R}r}\]
  10. Form the Curved Surface Area The curved surface area of the cylinder is
    \[A=2\pi rh\]
  11. Substituting
    \[h=H\left(1-\frac{r}{R}\right)\]
  12. we get
    \[A=2\pi rH\left(1-\frac{r}{R}\right)\]
  13. Therefore,
    \[A=2\pi H\left(r-\frac{r^2}{R}\right)\]
  14. Thus,
    \[\boxed{A(r)=2\pi H\left(r-\frac{r^2}{R}\right)}\]
  15. Differentiate
    \[A(r)=2\pi H\left(r-\frac{r^2}{R}\right).\]
  16. Differentiating with respect to \(r\),
    \[\begin{aligned}\frac{dA}{dr}&=2\pi H\left(1-\frac{2r}{R}\right)\end{aligned}\]
  17. For a maximum or minimum,
    \[\frac{dA}{dr}=0\]
  18. Therefore,
    \[2\pi H\left(1-\frac{2r}{R}\right)=0\]
  19. Since \(2\pi H\neq0\),
    \[1-\frac{2r}{R}=0\]
  20. Hence,
    \[\frac{2r}{R}=1\]
  21. Therefore,
    \[\boxed{r=\frac{R}{2}}\]
  22. Prove That This Gives the Greatest Curved Surface Area Find the second derivative:
    \[ \begin{aligned} \frac{d^2A}{dr^2} &= 2\pi H\left(-\frac{2}{R}\right)\\ &=-\frac{4\pi H}{R} \end{aligned} \]
  23. Since \(H>0\) and \(R>0\),
    \[-\frac{4\pi H}{R}<0\]
  24. Therefore,
    \[\boxed{\frac{d^2A}{dr^2}<0}\]
  25. By the second derivative test, the curved surface area is maximum at
    \[\boxed{r=\frac{R}{2}}\]
  26. Hence, the radius of the right circular cylinder having the greatest curved surface area is half the radius of the given cone.
    \[\boxed{\text{Cylinder radius}=\frac12(\text{cone radius})}\]
  27. Height of the Maximum-CSA Cylinder From
    \[h=H\left(1-\frac{r}{R}\right),\]
  28. substituting
    \[r=\frac{R}{2}\]
  29. gives
    \[ \begin{aligned} h &=H\left(1-\frac12\right)\\ &=\frac{H}{2}. \end{aligned} \]
  30. Therefore,
    \[\boxed{h=\frac{H}{2}}\]
  31. Thus, the cylinder of maximum curved surface area has both its radius and height equal to half those of the cone:
    \[\boxed{r=\frac{R}{2},\quad h=\frac{H}{2}}\]
  32. Maximum Curved Surface Area At
    \[r=\frac{R}{2},\quad h=\frac{H}{2},\]
  33. the maximum curved surface area is
    \[ \begin{aligned} A_{\max} &=2\pi rh\\ &=2\pi\left(\frac{R}{2}\right) \left(\frac{H}{2}\right)\\ &=\frac{\pi RH}{2}. \end{aligned} \]
  34. Hence,
    \[\boxed{A_{\max}=\frac{\pi RH}{2}}\]
🌟 Important Exam Insight
The key step is the similar-triangle relation
\[\frac{r}{R}=\frac{H-h}{H}\]
This converts the two-variable geometric problem into a one-variable optimization problem. Once
\[h=H\left(1-\frac rR\right)\]
is obtained, the curved surface area becomes a quadratic function of \(r\), making differentiation straightforward.
❌ Common Mistakes
  • Using \(h/H=r/R\) instead of the correct relation
    \[ \frac rR=\frac{H-h}{H}. \]
  • Using the total surface area \(2\pi r(r+h)\) when the question asks for curved surface area \(2\pi rh\).
  • Forgetting to verify that the stationary point gives a maximum.
  • Confusing the radius of the cone \(R\) with the radius of the cylinder \(r\).
  • Assuming the result \(r=R/2\) without deriving the constraint imposed by the cone.
🌟 Significance for Board and Competitive Examinations
This is a fundamental three-dimensional optimization problem. It demonstrates how similar triangles, a geometrical constraint, and the second derivative test work together to solve an extremum problem.
🔑 Key Takeaway
∿

Maximum and Minimum Values of a Function in a Closed Interval

📋
Table of Contents
9 sections
💡 Concept
✏️ Example
1
Example
Consider the function
\[f(x)=x+2\]
Part (a)
  1. Case 1: Domain \(x\in(0,1)\) On the open interval
    \[(0,1)\]
  2. the function is continuous and strictly increasing because
    \[f'(x)=1>0\]
    📝 Therefore, as \(x\) increases, \(f(x)\) increases. However, neither endpoint \(0\) nor \(1\) belongs to the domain.
  3. Consequently, the endpoint values
    \[f(0)=2\]
    and
    \[f(1)=3\]
    are not actually attained.
  4. More precisely,
    \[\lim_{x\to0^+}f(x)=2\]
    and
    \[\lim_{x\to1^-}f(x)=3\]
  5. But \(x=0\) and \(x=1\) are excluded. Hence,
    \[\boxed{\text{No minimum value exists on }(0,1)}\]
    and
    \[\boxed{\text{No maximum value exists on }(0,1)}.\]
  6. Range on \((0,1)\) Since \(f(x)=x+2\), the range is
    \[\boxed{(2,3)}\]
    📝 Thus, \(2\) is the infimum and \(3\) is the supremum, but neither is a value of the function on this domain.
  7. Local Maximum and Local Minimum on \((0,1)\) Since
    \[f'(x)=1>0\]
    📝 throughout \((0,1)\), the function is strictly increasing everywhere in its domain. Therefore, there is no interior point at which the function changes from increasing to decreasing or from decreasing to increasing.

    Hence, \(f\) has neither a local maximum nor a local minimum on \((0,1)\).
  8. Case 2: Domain \(x\in[0,1]\) Now extend the domain to the closed interval
    \[[0,1]\]
  9. The function remains continuous:
    \[f(x)=x+2\]
    📝 But this time the endpoints \(0\) and \(1\) are included in the domain.
    Therefore, the endpoint values are actually attained.
  10. At \(x=0\)
    \[f(0)=0+2=2\]
  11. Since \(f\) is increasing on the entire interval,
    \[f(x)\geq2\]
    for every \(x\in[0,1]\).
  12. Therefore,
    \[\boxed{\text{Absolute minimum value}=2}\]
  13. attained at
    \[\boxed{x=0}\]
  14. At \(x=1\)
    \[f(1)=1+2=3\]
  15. Since \(f\) is increasing on \([0,1]\),
    \[f(x)\leq3\]
    for every \(x\in[0,1]\).
  16. Therefore,
    \[\boxed{\text{Absolute maximum value}=3}\]
  17. attained at
    \[\boxed{x=1}.\]
⚖️ Important Distinction: Absolute vs Local Extrema
The terms absolute maximum and absolute minimum compare the function value with every value in the entire domain.

In contrast, a local maximum or local minimum is determined by comparing the function with values in a sufficiently small neighbourhood of the point.

At an endpoint of a closed interval, the terminology requires care. Under the standard relative-to-the-domain definition, an endpoint can be a local extremum using a one-sided neighbourhood within the domain. Regardless of that convention, the important conclusion here is that \(x=0\) and \(x=1\) are the absolute extrema of \(f\) on \([0,1]\).
⚖️ Comparison of the Two Domains
Domain Minimum Maximum Range
\((0,1)\) Does not exist Does not exist \((2,3)\)
\([0,1]\) \(2\) at \(x=0\) \(3\) at \(x=1\) \([2,3]\)
🗒️ Extreme Value Theorem
The above example illustrates the important Extreme Value Theorem.

Theorem: If a function \(f\) is continuous on a closed and bounded interval \([a,b]\), then \(f\) attains both an absolute maximum value and an absolute minimum value on \([a,b]\).

In other words, there exist points \(c,d\in[a,b]\) such that
\[f(c)\leq f(x)\leq f(d)\]
for every \(x\in[a,b]\).

For the present example, \(f(x)=x+2\) is continuous on \([0,1]\), so the Extreme Value Theorem guarantees that both an absolute maximum and an absolute minimum exist.

We find them to be
\[\boxed{f(0)=2}\]
and
\[\boxed{f(1)=3}\]
🗒️ How to Find Absolute Maximum and Minimum on \([a,b]\)
For a continuous function on a closed interval \([a,b]\), use the following systematic procedure:
  1. Find \(f'(x)\).
  2. Find all critical points in the open interval \((a,b)\).
  3. Evaluate \(f(x)\) at every critical point.
  4. Evaluate \(f(a)\) and \(f(b)\).
  5. Compare all these function values.
  6. The greatest value is the absolute maximum; the least value is the absolute minimum.
🛠️ Application to the Present Example
Here,
\[f'(x)=1\]
Therefore, there are no critical points in \((0,1)\). We only need to evaluate the endpoints:
\[f(0)=2,\quad f(1)=3.\]
Comparing the values,
\[2 < 3\]
Hence,
\[\boxed{\text{Absolute minimum}=2\text{ at }x=0}\]
\[\boxed{\text{Absolute maximum}=3\text{ at }x=1}.\]
🌟 Important Exam Insight

Do not ignore the endpoints when finding absolute extrema on a closed interval. Even when

\[ f'(x)\neq0 \]
throughout the interior, the absolute maximum or minimum may occur at an endpoint.

In fact, for a monotonic function on a closed interval, the absolute extrema occur at the endpoints.

🔑 Key Takeaway
∿

Theorem 5: Existence of Absolute Maximum and Minimum

📋
Table of Contents
12 sections
🧮 Theorem
Meaning of the Theorem
The theorem gives a guarantee about the existence of absolute extrema. If a function satisfies both of the following conditions:
  1. \(f\) is continuous, and
  2. the domain is a closed and bounded interval \([a,b]\),
then the function cannot merely approach its greatest or least value without attaining it. It must actually attain both values somewhere in the interval.
🧮 Theorem
Statement
Let \(f\) be a continuous function on the closed interval
\[ I=[a,b]. \]
Then \(f\) attains both an absolute maximum value and an absolute minimum value in \(I\).

In other words, there exist points \(c,d\in[a,b]\) such that
\[f(c)\leq f(x)\leq f(d)\]
for every \(x\in[a,b]\).

Thus,
\[\boxed{f(d)=\max_{x\in[a,b]}f(x)}\]
and
\[\boxed{f(c)=\min_{x\in[a,b]}f(x)}\]
Name of the Theorem

This result is commonly known as the Extreme Value Theorem (EVT).

Meaning of the Theorem
The theorem gives a guarantee about the existence of absolute extrema. If a function satisfies both of the following conditions:
  1. \(f\) is continuous, and
  2. the domain is a closed and bounded interval \([a,b]\),
then the function cannot merely approach its greatest or least value without attaining it. It must actually attain both values somewhere in the interval.
🔍 Mathematical Interpretation
An absolute maximum value \(f(d)\) means
\[f(x)\leq f(d)\quad\text{for every }x\in[a,b]\]
An absolute minimum value \(f(c)\) means
\[f(x)\geq f(c)\quad\text{for every }x\in[a,b]\]
Therefore,
\[\boxed{f(c)\leq f(x)\leq f(d)\quad\forall x\in[a,b]}\]
🤔 Why the Conditions Matter
Continuity is Required
A discontinuous function on a closed interval need not attain its supremum or infimum.

For example, define
\[f(x)=\begin{cases}x, & 0\leq x < 1,\\ 0, & x=1\end{cases}\]
The function is defined on the closed interval \([0,1]\), but it is not continuous at \(x=1\). Its values approach \(1\) as \(x\to1^{-}\), but \(f(x)\) never equals \(1\). Hence, the absolute maximum value does not exist.

This demonstrates why continuity is an essential hypothesis.
The Interval Must Be Closed
Even a continuous function may fail to attain its maximum or minimum on an open interval.

Consider
\[f(x)=x+2,\quad x\in(0,1)\]
The function is continuous, but
\[2 < f(x) < 3\]
The values \(2\) and \(3\) are approached but never attained because \(0\) and \(1\) are not included in the domain.

Hence, \(f\) has neither an absolute maximum nor an absolute minimum on \((0,1)\).

This shows why the closed-interval condition is essential.
✏️ Example
1
Example
Consider
\[f(x)=x^2,\qquad x\in[-2,3]\]
Continuous Function on a Closed Interval
  1. The function is continuous on the closed interval \([-2,3]\). Therefore, by the Extreme Value Theorem, it must attain both an absolute maximum and an absolute minimum. The critical point is obtained from
    \[f'(x)=2x=0\]
  2. giving
    \[x=0\]
  3. Evaluate the function at the critical point and endpoints:
    \[f(-2)=4\]
    \[f(0)=0\]
    \[f(3)=9\]
  4. Therefore,
    \[\boxed{\text{Absolute minimum value}=0\text{ at }x=0}\]
    and
    \[\boxed{\text{Absolute maximum value}=9\text{ at }x=3}\]
2
Example
Consider
\[f(x)=x+2,\qquad x\in[0,1]\]
A Monotonic Function
  1. Differentiating given function
    \[f'=1\]
  2. Since,
    \[f'(x)=1>0\]
  3. the function is strictly increasing. Therefore, its smallest value occurs at the left endpoint and its greatest value occurs at the right endpoint:
    \[f(0)=2,\quad f(1)=3\]
  4. Hence,
    \[\boxed{\text{Absolute minimum}=2}\]
  5. and
    \[\boxed{\text{Absolute maximum}=3}\]
🤔 How to Find Absolute Extrema in Practice
The Extreme Value Theorem guarantees existence, but it does not directly tell us where the extrema occur. For a differentiable function on \([a,b]\), use the following procedure:
  1. Find \(f'(x)\).
  2. Find all critical points in the open interval \((a,b)\).
  3. Evaluate \(f\) at every critical point.
  4. Evaluate \(f(a)\) and \(f(b)\).
  5. Compare all the resulting values.
  6. The largest value is the absolute maximum and the smallest value is the absolute minimum.
📌 Note
Critical Points Are Not the Whole Story
🔍 Geometrical Interpretation
Imagine the graph of a continuous function drawn without lifting the pencil over the closed interval \([a,b]\). The graph has a highest point and a lowest point somewhere on that interval.

These points may occur at an interior turning point or at an endpoint. The theorem guarantees that both are actually reached.
⚖️ Important Distinction
Concept Meaning
Absolute maximum Largest value of \(f(x)\) over the entire domain under consideration.
Absolute minimum Smallest value of \(f(x)\) over the entire domain under consideration.
Local maximum Maximum relative to a sufficiently small neighbourhood.
Local minimum Minimum relative to a sufficiently small neighbourhood.
🌟 Important Exam Insight

The Extreme Value Theorem is an existence theorem. It tells us that absolute maximum and minimum values exist, but it does not tell us their locations.

Once existence is guaranteed, derivatives can be used to locate the extrema.

\[ \boxed{ \text{Continuity on }[a,b] \Longrightarrow \text{absolute maximum and minimum exist}. } \]

For calculation:

\[ \boxed{ \text{Critical points in }(a,b) +\text{ endpoints} \longrightarrow \text{compare function values}. } \]
❌ Common Mistakes
  • Applying the theorem to an open interval such as \((a,b)\).
  • Applying the theorem to an unbounded interval such as \([a,\infty)\).
  • Forgetting that continuity is required.
  • Assuming the theorem identifies the point where the maximum or minimum occurs.
  • Checking only \(f'(x)=0\) and ignoring the endpoints.
  • Confusing an absolute extremum with a local extremum.
🔑 Key Takeaway
📎 Exam-Ready Statement

If \(f\) is continuous on a closed and bounded interval \([a,b]\), then there exist \(c,d\in[a,b]\) such that

\[ \boxed{ f(c)\leq f(x)\leq f(d) \quad\text{for every }x\in[a,b]. } \]

Thus, \(f(c)\) is the absolute minimum value and \(f(d)\) is the absolute maximum value of \(f\) on \([a,b]\).

∿

Theorem 6: Necessary Condition for Absolute Extrema

📋
Table of Contents
11 sections
🧮 Theorem
Meaning of the Theorem
Suppose an absolute maximum or minimum occurs at an interior point of a closed interval and the function is differentiable there. Then the tangent to the graph at that point must be horizontal.
\[ \boxed{\text{Interior differentiable absolute extremum} \;\Longrightarrow\; f'(c)=0} \]
Such a point is a critical point.
🧮 Theorem
Statement
Let \(f\) be a differentiable function on a closed interval
\[ I=[a,b], \]
and let \(c\) be an interior point of \(I\), that is,
\[ c\in(a,b). \]
Then:
  1. If \(f\) attains its absolute maximum value at \(x=c\), then
    \[ \boxed{f'(c)=0}. \]
  2. If \(f\) attains its absolute minimum value at \(x=c\), then
    \[ \boxed{f'(c)=0}. \]
Meaning of the Theorem
Suppose an absolute maximum or minimum occurs at an interior point of a closed interval and the function is differentiable there. Then the tangent to the graph at that point must be horizontal.
\[ \boxed{\text{Interior differentiable absolute extremum} \;\Longrightarrow\; f'(c)=0} \]
Such a point is a critical point.
🧭 Important Qualification
🧭
Condition The condition
\[ f'(c)=0 \]
is a necessary condition, not a sufficient condition, for an interior differentiable extremum.

In other words,
\[f'(c)=0\]
does not guarantee that \(c\) is a maximum or minimum.

For example, consider
\[f(x)=x^3\]
Then
\[f'(x)=3x^2\]
so
\[f'(0)=0\]
However, \(x=0\) is neither a local maximum nor a local minimum because \(x^3\) is increasing on both sides of \(0\).
🧮 Theorem
🧮 Theorem
Proof of Part (i): Absolute Maximum
  1. Suppose \(f\) attains its absolute maximum at the interior point \(x=c\). Then
    \[ f(c)\geq f(x) \quad\text{for every }x\in[a,b]. \]
  2. In particular, for sufficiently small \(h>0\), both \(c+h\) and \(c-h\) belong to the interval.
  3. Since \(c\) is an absolute maximum,
    \[f(c+h)\leq f(c)\]
  4. Therefore,
    \[f(c+h)-f(c)\leq0\]
  5. Since \(h>0\),
    \[\frac{f(c+h)-f(c)}{h}\leq0.\]
  6. Taking the right-hand limit as \(h\to0^+\),
    \[f'_+(c)\leq0\]
  7. Now consider \(h<0\). Since \(c+h\) is also in the interval,
    \[f(c+h)\leq f(c)\]
  8. Hence,
    \[f(c+h)-f(c)\leq0\]
  9. But \(h<0\), so dividing by \(h\) reverses the inequality:
    \[\frac{f(c+h)-f(c)}{h}\geq0\]
  10. Taking the left-hand limit as \(h\to0^-\),
    \[f'_-(c)\geq0\]
  11. Since \(f\) is differentiable at \(c\),
    \[f'_-(c)=f'_+(c)=f'(c)\]
  12. Therefore,
    \[f'(c)\leq0\]
    and
    \[f'(c)\geq0\]
  13. Consequently,
    \[\boxed{f'(c)=0}\]
Proof of Part (ii): Absolute Minimum
  1. Suppose \(f\) attains its absolute minimum at the interior point \(x=c\). Then
    \[f(c)\leq f(x)\quad\text{for every }x\in[a,b]\]
  2. For \(h>0\),
    \[f(c+h)\geq f(c)\]
  3. Therefore,
    \[f(c+h)-f(c)\geq0\]
  4. Dividing by \(h>0\),
    \[\frac{f(c+h)-f(c)}{h}\geq0\]
  5. Taking \(h\to0^+\)
    \[f'_+(c)\geq0\]
  6. For \(h<0\),
    \[f(c+h)\geq f(c),\]
  7. so
    \[f(c+h)-f(c)\geq0\]
  8. Dividing by the negative number \(h\) reverses the inequality:
    \[\frac{f(c+h)-f(c)}{h}\leq0\]
  9. Taking \(h\to0^-\),
    f'_-(c)\leq0.
  10. Since \(f\) is differentiable at \(c\),
    \[f'_-(c)=f'_+(c)=f'(c)\]
  11. Thus,
    \[f'(c)\geq0\]
    and
    \[f'(c)\leq0\]
  12. Hence,
    \[\boxed{f'(c)=0}\]
🔍 Geometrical Interpretation
At an interior differentiable maximum or minimum, the graph must have a horizontal tangent.
\[ \boxed{\text{Maximum or minimum at }x=c \quad\Longrightarrow\quad \text{tangent slope}=f'(c)=0} \]
This explains why solving
\[ f'(x)=0 \]
is an important first step in finding extrema.
🤔 Why the Endpoints Must Also Be Checked
The theorem applies only to an interior point \(c\in(a,b)\). It does not say that an absolute maximum or minimum at an endpoint must satisfy \(f'(a)=0\) or \(f'(b)=0\).

For example, consider
\[f(x)=x,\quad x\in[0,2]\]
3 We have
\[f'(x)=1\]
There is no point in \((0,2)\) where \(f'(x)=0\), yet
\[f(0)=0\]
is the absolute minimum and
\[f(2)=2\]
is the absolute maximum.
Both extrema occur at endpoints, where the theorem above does not require the derivative to be zero.
📜 Working Rule for Finding Absolute Maximum and Minimum Values
📜 Rules
Let \(f\) be continuous on the closed interval
\[[a,b]\]
By the Extreme Value Theorem, \(f\) is guaranteed to attain both an absolute maximum and an absolute minimum on \([a,b]\). To actually find them, use the following procedure.
Step 1: Find All Critical Points
Find all points \(x\in(a,b)\) where either
\[f'(x)=0\]
or \(f'(x)\) does not exist, provided \(f\) is defined at that point.
These are the critical points that can potentially give interior absolute extrema.
Step 2: Include the Endpoints
Always include
\[\boxed{x=a\quad\text{and}\quad x=b}\]
. Therefore, the complete list of candidates consists of:
\[\boxed{\text{critical points in }(a,b)\;+\;\text{endpoints }a,b}\]
Step 3: Evaluate the Function
Calculate \(f(x)\) at every candidate point.
\[f(a),\quad f(b),\quad f(c_1),\quad f(c_2),\quad\ldots \]
where \(c_1,c_2,\ldots\) are the critical points in \((a,b)\).
Step 4: Compare the Values
Compare all the calculated function values.
  • The largest value is the absolute maximum value.
  • The smallest value is the absolute minimum value.
Working Rule in Compact Form
\[ \boxed{ \begin{aligned} &\text{Step 1: Find critical points in }(a,b).\\ &\text{Step 2: Add the endpoints }a\text{ and }b.\\ &\text{Step 3: Evaluate }f\text{ at all these points}.\\ &\text{Step 4: Largest value}\Rightarrow\text{absolute maximum}.\\ &\text{Step 4: Smallest value}\Rightarrow\text{absolute minimum}. \end{aligned} } \]
✏️ Example
1
Question
Find the absolute maximum and minimum values of
\[f(x)=x^2-4x+5,\quad x\in[0,5]\]
  1. Find Critical Points Differenting given function
    \[f'(x)=2x-4\]
  2. Setting \(f'(x)=0\)
    \[2x-4=0\]
  3. gives
    \[x=2\]
  4. Include the Endpoints The candidate points are
    \[x=0,\quad2,\quad5\]
  5. Evaluate \(f\)
    \[f(0)=5,\]
    \[f(2)=4-8+5=1,\]
    \[f(5)=25-20+5=10\]
  6. Compare
    \[1 < 5 < 10\]
  7. Therefore,
    \[\boxed{\text{Absolute minimum value}=1\text{ at }x=2}\]
    and
    \[\boxed{\text{Absolute maximum value}=10\text{ at }x=5}\]
⚖️ Important Distinction: Critical Point vs Absolute Extremum
Term Meaning
Critical point A point in the domain where \(f'(x)=0\) or \(f'(x)\) does not exist.
Absolute maximum point A point where \(f\) has its greatest value over the entire interval.
Absolute minimum point A point where \(f\) has its least value over the entire interval.
A critical point is only a candidate for an interior extremum. It need not actually be an extremum.
🌟 Important Exam Insights
  • An interior differentiable absolute maximum or minimum must satisfy
    \[ f'(c)=0. \]
  • The converse is not necessarily true:
    \[ f'(c)=0 \]
    does not guarantee an extremum.
  • Absolute extrema can occur at endpoints even when \(f'(a)\neq0\) or \(f'(b)\neq0\).
  • For a closed interval, never omit the endpoints.
  • If \(f'(x)\) does not exist at an interior point where \(f\) is defined, that point must also be considered.
  • The Extreme Value Theorem guarantees existence for a continuous function on a closed and bounded interval; the working rule locates the extrema.
❌ Common Mistakes
  • Checking only the points where \(f'(x)=0\).
  • Ignoring points where \(f'(x)\) does not exist.
  • Ignoring the endpoints \(a\) and \(b\).
  • Assuming every critical point is an absolute maximum or minimum.
  • Confusing the extremum point \(x=c\) with the extremum value \(f(c)\).
  • Using the theorem at an endpoint even though its hypothesis requires an interior point.
🔑 Key Takeaway
∿

Example 27

Application of Derivatives NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Derivatives Rate of Change Increasing Functions Decreasing Functions Maxima and Minima Absolute Maximum Absolute Minimum Local Maxima Local Minima First Derivative Test Second Derivative Test Optimization Problems Related Rates Marginal Cost Marginal Revenue JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
8 sections
❓ Question
Absolute Maximum and Minimum on a Closed Interval
Find the absolute maximum and minimum values of the function
\[f(x)=2x^3-15x^2+36x+1\]
on the interval \([1,5]\)
💡 Concept
🗺️ Roadmap

  1. Find \(f'(x)\).

  2. Solve \(f'(x)=0\) to obtain critical points.

  3. Retain only the critical points lying in \((1,5)\).

  4. Evaluate \(f(x)\) at the critical points and at \(x=1,5\).

  5. Compare all values.

  6. Identify the absolute maximum and absolute minimum.

🧩 Solution
  1. Find the First Derivative
    \[f(x)=2x^3-15x^2+36x+1\]
  2. Differentiating,
    \[ \begin{aligned} f'(x) &=6x^2-30x+36\\ &=6(x^2-5x+6)\\ &=6(x-2)(x-3) \end{aligned} \]
  3. Find the Critical Points Set
    \[f'(x)=0\]
  4. Therefore,
    \[6(x-2)(x-3)=0\]
  5. Hence,
    \[x=2\]
    or
    \[x=3\]
  6. Both points lie inside the interval \((1,5)\). Therefore, the critical points that must be considered are
    \[\boxed{x=2,\;3}\]
  7. Include the Endpoints Since the domain is the closed interval \([1,5]\), we must also consider the endpoints
    \[\boxed{x=1,\;5}\]
  8. Thus, the complete set of candidates is
    \[\boxed{x=1,\;2,\;3,\;5}\]
  9. Evaluate the Function at All Candidate Points At \(x=1\)
    \[ \begin{aligned} f(1) &=2(1)^3-15(1)^2+36(1)+1\\ &=2-15+36+1\\ &=24 \end{aligned} \]
  10. Therefore,
    \[\boxed{f(1)=24}\]
  11. At \(x=2\)
    \[ \begin{aligned} f(2) &=2(2)^3-15(2)^2+36(2)+1\\ &=16-60+72+1\\ &=29 \end{aligned} \]
  12. Therefore,
    \[\boxed{f(2)=29}\]
  13. At \(x=3\)
    \[ \begin{aligned} f(3) &=2(3)^3-15(3)^2+36(3)+1\\ &=54-135+108+1\\ &=28 \end{aligned} \]
  14. Therefore,
    \[\boxed{f(3)=28}\]
  15. At \(x=5\)
    \[ \begin{aligned} f(5) &=2(5)^3-15(5)^2+36(5)+1\\ &=250-375+180+1\\ &=56 \end{aligned} \]
  16. Therefore,
    \[\boxed{f(5)=56}\]
  17. Compare the Values
    Candidate point \(x\) \(f(x)\)
    \(1\) \(24\)
    \(2\) \(29\)
    \(3\) \(28\)
    \(5\) \(56\)
  18. The smallest value is
    \[24\]
  19. and the largest value is
    \[56\]
  20. Therefore,
    \[\boxed{\text{Absolute minimum value}=24}\]
  21. attained at
    \[\boxed{x=1}\]
  22. and
    \[\boxed{\text{Absolute maximum value}=56}\]
  23. attained at
    \[\boxed{x=5}\]
🌟 Important Exam Insight

This example demonstrates why the working rule for absolute extrema on a closed interval requires both critical points and endpoints.

Here,

\[f(2)=29\]

is a local maximum, but it is not the absolute maximum because

\[f(5)=56>29\]

Similarly,

\[f(3)=28\]

is a local minimum, but it is not the absolute minimum because

\[f(1)=24<28\]

Thus, the endpoint values determine both absolute extrema in this example.

❌ Common Mistakes
  • Finding \(f'(x)=0\) and reporting \(x=2\) as the absolute maximum without checking the endpoints.
  • Reporting \(f(3)=28\) as the absolute minimum without checking \(f(1)=24\).
  • Forgetting that \(x=1\) and \(x=5\) belong to the closed interval.
  • Confusing local maximum/minimum values with absolute maximum/minimum values.
  • Giving only the value of \(x\) instead of the required maximum or minimum value of \(f\).
🌟 Significance for Board and Competitive Examinations
This is a standard application of the Extreme Value Theorem and the critical-point method. It tests whether a student can distinguish local extrema from absolute extrema and correctly include endpoints in a closed interval.
🔑 Key Takeaway
∿

Example 28

📋
Table of Contents
8 sections
❓ Question
Find the absolute maximum and minimum values of the function
\[f(x)=12x^{\frac{4}{3}}-6x^{\frac{1}{3}},\quad x\in[-1,1]\]
💡 Concept
🗺️ Roadmap

  1. Find \(f'(x)\).

  2. Determine where \(f'(x)=0\).

  3. Also determine where \(f'(x)\) does not exist.

  4. Evaluate \(f(x)\) at all critical points and endpoints.

  5. Compare all the resulting values.

🧩 Solution
Given:
\[f(x)=12x^{\frac{4}{3}}-6x^{\frac{1}{3}}\]
  1. Find the Derivative Differentiating term by term,
    \[ \begin{aligned} f'(x) &=12\left(\frac{4}{3}\right)x^{\frac{1}{3}} -6\left(\frac{1}{3}\right)x^{-\frac{2}{3}}\\ &=16x^{\frac{1}{3}}-2x^{-\frac{2}{3}}. \end{aligned} \]
  2. Taking \(2x^{-\frac{2}{3}}\) common,
    \[ \begin{aligned} f'(x) &=2x^{-\frac{2}{3}}(8x-1)\\ &=\frac{2(8x-1)}{x^{\frac{2}{3}}}. \end{aligned} \]
  3. ⭐ A critical point occurs where \(f'(x)=0\) or where \(f'(x)\) does not exist, provided the point belongs to the domain.
  4. Find the Critical Points

    Case 1: \(f'(x)=0\)
    \[\frac{2(8x-1)}{x^{\frac{2}{3}}}=0\]
  5. A fraction is zero when its numerator is zero. Hence,
    \[8x-1=0\]
  6. Therefore,
    \[x=\frac{1}{8}\]
  7. Case 2: \(f'(x)\) does not exist The denominator of \(f'(x)\) is \(x^{2/3}\), which becomes zero at \(x=0\). Thus \(f'(0)\) does not exist.However, \(f(0)\) itself exists:
    \[f(0)=12(0)^{\frac43}-6(0)^{\frac13}=0\]
    Therefore, \(x=0\) is also a critical point
  8. Hence, the critical points in \((-1,1)\) are
    \[\boxed{x=0,\quad x=\frac18}\]
  9. Include the Endpoints Since the interval is closed, we must also consider
    \[\boxed{x=-1,\quad x=1}\]
  10. Therefore, all candidate points are
    \[\boxed{-1,\quad 0,\quad \frac18,\quad 1}\]
  11. Evaluate \(f(x)\) at All Candidate Points At \(x=-1\)
    \[\begin{aligned}f(-1)&=12(-1)^{\frac43}-6(-1)^{\frac13}\end{aligned}\]
  12. Since
    \[(-1)^{\frac13}=-1\]
    and
    \[ (-1)^{\frac43} =\left((-1)^{\frac13}\right)^4 =(-1)^4=1, \]
  13. we get
    \[ \begin{aligned} f(-1) &=12(1)-6(-1)\\ &=12+6\\ &=18 \end{aligned} \]
    \[ \boxed{f(-1)=18} \]
  14. At \(x=0\)
    \[ \begin{aligned} f(0) &=12(0)^{\frac43}-6(0)^{\frac13}\\ &=0. \end{aligned} \]
    \[ \boxed{f(0)=0} \]
  15. At \(x=\frac18\). First calculate the required fractional powers:
    \[\left(\frac18\right)^{\frac13}=\frac12\]
    and
    \[ \left(\frac18\right)^{\frac43} =\left[\left(\frac18\right)^{\frac13}\right]^4 =\left(\frac12\right)^4 =\frac1{16}. \]
  16. Therefore,
    \[ \begin{aligned} f\left(\frac18\right) &=12\left(\frac1{16}\right) -6\left(\frac12\right)\\ &=\frac{12}{16}-3\\ &=\frac34-3\\ &=-\frac94 \end{aligned} \]
  17. Hence,
    \[\boxed{f\left(\frac18\right)=-\frac94}\]
  18. At \(x=1\)
    \[ \begin{aligned} f(1) &=12(1)^{\frac43}-6(1)^{\frac13}\\ &=12-6\\ &=6 \end{aligned} \]
    \[ \boxed{f(1)=6} \]
  19. Compare the Values
    Candidate point \(x\) \(f(x)\)
    \(-1\) \(18\)
    \(0\) \(0\)
    \(\frac18\) \(-\frac94\)
    \(1\) \(6\)
  20. The largest value among \(18,\quad 0,\quad -\frac94,\quad 6\) is
    \[18\]
  21. The smallest value is
    \[-\frac94\]
🌟 Important Mathematical Insight

This example is particularly important because the derivative does not exist at \(x=0\), yet \(x=0\) lies in the domain. Therefore, \(x=0\) must be included among the critical points.

This illustrates the general rule:

\[ \boxed{ \text{Critical points occur where }f'(x)=0 \text{ or where }f'(x)\text{ does not exist, provided }x\text{ is in the domain.} } \]

However, being a critical point does not automatically make a point an absolute maximum or minimum. Every candidate must be evaluated and compared.

❌ Common Mistakes
  • Ignoring \(x=0\) because \(f'(0)\) does not exist.
  • Assuming that only points satisfying \(f'(x)=0\) are critical points.
  • Forgetting the endpoints \(x=-1\) and \(x=1\).
  • Writing \((-1)^{4/3}=-1\); the correct value is \(1\).
  • Confusing a critical point with an absolute extremum.
  • Reporting the \(x\)-coordinate instead of the required maximum or minimum value of \(f\).
🌟 Significance for Board and Competitive Examinations
This example combines three important ideas from Applications of Derivatives: absolute extrema on a closed interval, critical points where the derivative is undefined, and systematic comparison of function values. Such questions are especially useful for testing whether the complete critical-point procedure has been applied rather than merely solving \(f'(x)=0\).
🔑 Key Takeaway
∿

Example 29

📋
Table of Contents
7 sections
❓ Question
An Apache helicopter of an enemy is flying along the curve \(y=x^2+7\). A soldier, placed at the point \((3,7)\). wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.
💡 Concept
🗺️ Roadmap

  1. Take a general point \(P=(x,x^2+7)\) on the curve.

  2. Write the squared distance from \(P\) to \((3,7)\).

  3. Differentiate the squared distance.

  4. Find the critical point.

  5. Verify that it gives the minimum distance.

  6. Calculate the corresponding minimum distance.

🧩 Solution
  1. Take a General Point on the Curve The helicopter is flying along
    \[y=x^2+7\]
  2. Therefore, a general position of the helicopter is
    \[P=(x,x^2+7)\]
  3. The soldier is at
    \[S=(3,7)\]
  4. Form the Squared Distance By the distance formula,
    \[d^2=(x-3)^2+\left[(x^2+7)-7\right]^2\]
  5. Simplifying,
    \[ \begin{aligned} d^2 &=(x-3)^2+(x^2)^2\\ &=(x-3)^2+x^4. \end{aligned} \]
  6. Expanding,
    \[ \begin{aligned} d^2 &=x^2-6x+9+x^4\\ &=x^4+x^2-6x+9. \end{aligned} \]
  7. Let
    \[D(x)=d^2=x^4+x^2-6x+9\]
  8. Differentiate
    \[ \begin{aligned} D'(x) &=4x^3+2x-6\\ &=2(2x^3+x-3). \end{aligned} \]
  9. For a critical point,
    \[D'(x)=0\]
  10. Hence,
    \[2x^3+x-3=0\]
  11. Find the Critical Point We observe that \(x=1\) is a root:
    \[2(1)^3+1-3=0\]
  12. Therefore,
    \[2x^3+x-3=(x-1)(2x^2+2x+3)\]
  13. Hence,
    \[(x-1)(2x^2+2x+3)=0\]
  14. The first factor gives
    \[x=1\]
  15. For the second factor,
    \[2x^2+2x+3=0\]
  16. Its discriminant is
    \[\Delta=(2)^2-4(2)(3)=4-24=-20 < 0\]
    📝 Therefore, it has no real roots.
  17. Hence, the only real critical point is
    \[\boxed{x=1}\]
  18. Verify That the Distance Is Minimum We have
    \[D'(x)=2(x-1)(2x^2+2x+3)\]
  19. Notice that
    \[2x^2+2x+3=2\left(x+\frac12\right)^2+\frac52>0\]
    for every real \(x\).
  20. Therefore, the sign of \(D'(x)\) is determined by \(x-1\):
    Interval Sign of \(D'(x)\) Behaviour of \(D(x)\)
    \((-\infty,1)\) \(-\) Decreasing
    \((1,\infty)\) \(+\) Increasing
  21. Thus, \(D(x)\) decreases up to \(x=1\) and increases after \(x=1\). Hence, \(x=1\) gives the absolute minimum of the squared distance.
    \[\boxed{x=1}\]
  22. Find the Nearest Point on the Curve At \(x=1\),
    \[y=x^2+7=1+7=8\]
  23. Therefore, the helicopter is nearest to the soldier when it is at
    \[\boxed{(1,8)}\]
  24. Calculate the Nearest Distance At \(x=1\),
    \[ \begin{aligned} d^2 &=(1-3)^2+(8-7)^2\\ &=(-2)^2+1^2\\ &=4+1\\ &=5. \end{aligned} \]
    📝 Therefore,
    \[d=\sqrt{5}\]
  25. Hence,
    \[\boxed{\text{Nearest distance}=\sqrt{5}\text{ units}}.\]
❌ Common Mistakes
  • Using \(y=x^2+7\) incorrectly when forming the distance.
  • Forgetting that the soldier's coordinates are \((3,7)\).
  • Minimizing \(x^2+y^2\) instead of the distance from \((3,7)\).
  • Taking the derivative of \(d\) directly when minimizing \(d^2\) is simpler.
  • Finding \(x=1\) but forgetting to calculate the actual nearest distance.
  • Reporting \((1,8)\) as the distance; \((1,8)\) is the nearest point, whereas the distance is \(\sqrt5\).
🌟 Significance for Board and Competitive Examinations
This problem is a standard application of derivatives to a geometrical optimization problem. It tests the ability to convert a minimum-distance condition into a function-minimization problem and reinforces the highly useful technique of minimizing the square of a distance.
🔑 Key Takeaway
∿

Example 30

📋
Table of Contents
8 sections
❓ Question
A car starts from a point \(P\) at time \(t=0\) seconds and stops at point \(Q\). The distance \(x\), in metres, covered by it in \(t\) seconds is given by \(x=t^2\left(2-\frac{t}{3}\right)\;\). Find:
  1. the time taken by the car to reach \(Q\), and
  2. the distance between \(P\) and \(Q\).
💡 Concept
🗺️ Roadmap

  1. Differentiate \(x\) with respect to \(t\) to obtain velocity.

  2. Set the velocity equal to zero.

  3. Discard the initial time \(t=0\), because the car starts from \(P\) then.

  4. Use the remaining time to calculate the distance \(PQ\).

  5. Verify that the car is moving forward until that time.

🧩 Solution
  1. Write the Distance Function
    \[x=t^2\left(2-\frac{t}{3}\right)\]
  2. Expanding,
    \[x=2t^2-\frac{t^3}{3}\]
  3. Find the Velocity Velocity is the rate of change of distance with respect to time:
    \[v=\frac{dx}{dt}\]
  4. Therefore,
    \[ \begin{aligned} v &=\frac{d}{dt}\left(2t^2-\frac{t^3}{3}\right)\\ &=4t-t^2\\ &=t(4-t) \end{aligned} \]
  5. Thus,
    \[\boxed{v=t(4-t)}\]
  6. Find the Time at Which the Car Stops At the stopping point \(Q\),
    \[v=0\]
  7. Hence,
    \[t(4-t)=0\]
  8. Therefore,
    \[t=0\]
    or
    \[t=4\]
    📝 The value \(t=0\) represents the instant at which the car starts from \(P\), not the time at which it reaches \(Q\)
  9. Therefore, the required time is
    \[\boxed{t=4\text{ seconds}}\]
  10. Find the Distance Between \(P\) and \(Q\) The car reaches \(Q\) at \(t=4\) seconds. Therefore, substitute \(t=4\) into the distance function:
    \[x=2t^2-\frac{t^3}{3}\]
  11. Thus,
    \[ \begin{aligned} x(4) &=2(4)^2-\frac{(4)^3}{3}\\ &=2(16)-\frac{64}{3}\\ &=32-\frac{64}{3}\\ &=\frac{96-64}{3}\\ &=\frac{32}{3} \end{aligned} \]
  12. Hence,
    \[\boxed{PQ=\frac{32}{3}\text{ m}}\]
  13. Verify That the Car Actually Moves from \(P\) to \(Q\) The velocity is
    \[v=t(4-t)\]
  14. For
    \[0 < t < 4\]
  15. we have
    \[t>0\]
    and
    \[4-t>0\]
  16. Therefore,
    \[v>0\]
    📝 Hence, the distance \(x\) continuously increases during \(0 < t < 4\). The car therefore moves forward from \(P\) until it reaches \(Q\) at \(t=4\) seconds.
🌟 Important Exam Insight
The phrase “the car stops at \(Q\)” is the key to the problem. It means that the velocity at \(Q\) is zero:
\[\boxed{v=\frac{dx}{dt}=0}\]
The equation \(v=0\) gives two values, \(t=0\) and \(t=4\). The first is the starting instant and must not be confused with the later stopping instant.
❌ Common Mistakes
  • Using \(x=0\) to determine when the car stops. Position being zero does not mean velocity is zero.
  • Forgetting that velocity is \(\frac{dx}{dt}\), not \(x\) itself.
  • Taking \(t=0\) as the time of reaching \(Q\).
  • Substituting the wrong time into the distance equation.
  • Reporting \(4\) metres as the distance; \(4\) is the time in seconds.
🌟 Significance for Board and Competitive Examinations
This problem is a direct application of derivatives to motion. It tests the relationship between displacement or distance, velocity and acceleration:
\[\boxed{v=\frac{dx}{dt},\quad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}}\]
The problem also illustrates an important optimization idea: a stopping point corresponds to a zero of the velocity function, which is the derivative of the distance function.
🔑 Key Takeaway
∿

Example 31

📋
Table of Contents
8 sections
❓ Question
A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is
\[\tan^{-1}(0.5)\]
Water is poured into it at a constant rate of \(5\) cubic metres per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is \(4\) m.
💡 Concept
🗺️ Roadmap

  1. Relate the radius \(r\) and depth \(h\) using the semi-vertical angle.

  2. Express the volume \(V\) entirely in terms of \(h\).

  3. Differentiate the volume equation with respect to time.

  4. Substitute \(dV/dt=5\) and \(h=4\).

  5. Calculate \(dh/dt\), the rate at which the water level rises.

🧩 Solution
🌟 Important Exam Insight
The most important step in this problem is to establish the geometric relation between \(r\) and \(h\) before differentiating. The semi-vertical angle gives
\[\tan\theta=\frac rh\]
Since the angle is fixed, the ratio \(r/h\) is constant:
\[\frac rh=\frac12\]
Consequently,
\[r=\frac h2\]
Substituting this relation into the volume formula reduces the problem to a single variable, \(h\).
❌ Common Mistakes
  • Using \(\tan\theta=h/r\) instead of \(\tan\theta=r/h\).
  • Forgetting to use the semi-vertical angle to establish \(r=h/2\).
  • Using the volume formula without expressing \(r\) in terms of \(h\).
  • Differentiating \(V\) with respect to \(h\) instead of with respect to time.
  • Forgetting the chain-rule factor \(\frac{dh}{dt}\).
  • Giving the answer in \(\text{m}^3/\text{h}\) instead of \(\text{m}/\text{h}\).
🌟 Significance for Board and Competitive Examinations
This is a standard related-rates application involving a cone. It combines trigonometric geometry, the volume formula for a cone, and implicit differentiation with respect to time. The key skill is recognizing that several quantities vary with time but are connected by a fixed geometric relation.
🔑 Key Takeaway
∿

Example 32

📋
Table of Contents
7 sections
❓ Question
A man of height \(2\) metres walks at a uniform speed of \(5\) km/h away from a lamp post which is \(6\) metres high. Find the rate at which the length of his shadow increases.
💡 Concept
🗺️ Roadmap

  1. Let \(x\) be the distance of the man from the lamp post.

  2. Let \(y\) be the length of his shadow.

  3. Use similar triangles to obtain a relation between \(x\) and \(y\).

  4. Differentiate the relation with respect to time.

  5. Substitute the man's walking speed.

  6. State the rate of increase of the shadow length with correct units.

🧩 Solution
  1. Define the Variables Let
    \[x=\text{distance of the man from the lamp post},\]
    \[y=\text{length of the man's shadow}\]
  2. The lamp post is \(6\) m high and the man is \(2\) m high. Therefore, the distance from the lamp post to the tip of the shadow is
    \[x+y\]
  3. Apply Similar Triangles Consider the large triangle formed by:
    • the lamp post of height \(6\) m,
    • the ground distance \(x+y\), and
    • the line from the top of the lamp to the tip of the shadow.
  4. The smaller triangle is formed by:
    • the man of height \(2\) m,
    • his shadow of length \(y\), and
    • the line from the top of the man's head to the tip of his shadow.
  5. These two triangles are similar. Hence,
    \[\frac{6}{x+y}=\frac{2}{y}\]
  6. Cross-multiplying,
    \[6y=2(x+y)\]
  7. Therefore,
    \[6y=2x+2y\]
  8. Hence,
    \[4y=2x\]
  9. Dividing by \(4\)
    \[\boxed{y=\frac{x}{2}}\]
  10. Differentiate with Respect to Time Since both \(x\) and \(y\) vary with time, differentiate \[y=\frac{x}{2}\] with respect to \(t\):
    \[\frac{dy}{dt}=\frac12\frac{dx}{dt}\]
  11. The man walks away from the lamp post at \(5\) km/h. Therefore,
    \[\frac{dx}{dt}=5\text{ km/h}\]
  12. Thus,
    \[\frac{dy}{dt}=\frac12(5)=\frac52\text{ km/h}\]
  13. Therefore,
    \[\boxed{\frac{dy}{dt}=2.5\text{ km/h}}\]
❌ Common Mistakes
  • Using \(x\) instead of \(x+y\) as the base of the larger triangle.
  • Writing \(\frac{6}{x}=\frac{2}{y}\), which gives an incorrect geometric relationship.
  • Forgetting that \(x\) represents the distance from the lamp to the man, while \(y\) represents the shadow length.
  • Differentiating the geometric equation without treating \(x\) and \(y\) as functions of time.
  • Forgetting to convert \(5\) km/h if the final answer is required in m/s.
  • Confusing the rate at which the man moves with the rate at which his shadow length increases.
🌟 Significance for Board and Competitive Examinations
This is a standard related-rates problem involving similar triangles. It tests the ability to translate a geometrical relationship into an equation involving changing quantities and then differentiate that equation with respect to time.
🔑 Key Takeaway
∿

Example 33

📋
Table of Contents
8 sections
❓ Question
Find the intervals in which the function \(f(x)=\frac{3}{10}x^4-\frac45x^3-3x^2+\frac{36}{5}x+11\;\) is
  1. increasing
  2. decreasing
💡 Concept
🗺️ Roadmap

  1. Differentiate \(f(x)\).

  2. Factorize \(f'(x)\).

  3. Find the critical points by solving \(f'(x)=0\).

  4. Divide the real line at the critical points.

  5. Determine the sign of \(f'(x)\) on each interval.

  6. State the intervals of increase and decrease.

🧩 Solution
  1. Differentiate the Function Differentiating term by term,
    \[\begin{aligned}f'(x)&=\frac{3}{10}(4x^3)-\frac45(3x^2)-3(2x)+\frac{36}{5}\\&=\frac65x^3-\frac{12}{5}x^2-6x+\frac{36}{5}\end{aligned}\]
  2. Taking \(\frac65\) common,
    \[f'(x)=\frac65\left(x^3-2x^2-5x+6\right)\]
  3. Factorise \(f'(x)\) We factorise
    \[x^3-2x^2-5x+6\]
  4. By grouping,
    \[ \begin{aligned} x^3-2x^2-5x+6 &=x^2(x-2)-3(x-2)\\ &=(x-2)(x^2-3). \end{aligned} \]
  5. However, this grouping does not correctly reproduce the given polynomial. The correct factorisation is obtained by taking \(x=1\) as a root:
    \[1-2-5+6=0\]
  6. Hence, \(x-1\) is a factor. Dividing by \(x-1\),
    \[x^3-2x^2-5x+6=(x-1)(x^2-x-6)\]
  7. Now,
    \[x^2-x-6=(x-3)(x+2)\]
  8. Therefore,
    \[\boxed{f'(x)=\frac65(x-1)(x-3)(x+2)}\]
  9. Find the Critical Points Since \(f'(x)\) is a polynomial, it exists for every real \(x\). Thus, critical points occur where
    \[f'(x)=0\]
  10. Therefore,
    \[\frac65(x-1)(x-3)(x+2)=0\]
  11. Hence,
    \[x=1,\quad x=3,\quad x=-2\]
  12. Thus, the critical points are
    \[\boxed{x=-2,\;1,\;3}\]
  13. Divide the Real Line into Intervals The critical points divide the real number line into four intervals:
    \[(-\infty,-2),\quad(-2,1),\quad(1,3),\quad(3,\infty)\]
  14. Determine the Sign of \(f'(x)\) Since
    \[f'(x)=\frac65(x+2)(x-1)(x-3),\]
  15. and
    \[\frac65>0,\]
    the sign of \(f'(x)\) is determined by the three linear factors. We can use a sign chart.
    Interval \(x+2\) \(x-1\) \(x-3\) \(f'(x)\) Behaviour of \(f\)
    \((-\infty,-2)\) \(-\) \(-\) \(-\) \(-\) Decreasing
    \((-2,1)\) \(+\) \(-\) \(-\) \(+\) Increasing
    \((1,3)\) \(+\) \(+\) \(-\) \(-\) Decreasing
    \((3,\infty)\) \(+\) \(+\) \(+\) \(+\) Increasing
  16. State the Required Intervals

    (a) Intervals in Which \(f(x)\) Is Increasing     The function is increasing wherever
    \[f'(x)>0\]
  17. From the sign chart,
    \[\boxed{f(x)\text{ is increasing on }(-2,1)\cup(3,\infty)}\]
  18. (b) Intervals in Which \(f(x)\) Is Decreasing The function is decreasing wherever
    \[f'(x)<0\]
  19. From the sign chart,
    \[\boxed{f(x)\text{ is decreasing on }(-\infty,-2)\cup(1,3)}\]
  20. ⭐
    \[\boxed{\begin{aligned}\text{Increasing:}\quad &(-2,1)\cup(3,\infty),\\[4pt] \text{Decreasing:}\quad &(-\infty,-2)\cup(1,3)\end{aligned}}\]
🌟 Important Exam Insight
For a polynomial function, \(f'(x)\) exists everywhere. Therefore, while finding intervals of increase and decrease, we only need to solve \(f'(x)=0\) and use those points to divide the real line.

The critical points themselves are not normally included in the open intervals of increase or decrease because the derivative is zero there. The standard answer is therefore written as
\[\boxed{(-2,1)\cup(3,\infty)}\]
for increasing intervals and
\[\boxed{(-\infty,-2)\cup(1,3)}\]
for decreasing intervals.
❌ Common Mistakes
  • Making an algebraic error while factorising \(f'(x)\).
  • Missing the critical point \(x=-2\).
  • Using the sign of \(f(x)\) instead of the sign of \(f'(x)\).
  • Checking only one test point for the entire real line.
  • Reversing the interpretation of \(f'(x)>0\) and \(f'(x)<0\).
  • Writing the critical points themselves as complete increasing or decreasing intervals.
🌟 Significance for Board and Competitive Examinations
This example is a standard application of the first-derivative test. Questions of this type frequently test factorisation, critical-point identification, sign analysis, and correct interval notation.
🔑 Key Takeaway
∿

Example 34

📋
Table of Contents
6 sections
❓ Question
Show that the function \(f(x)=\tan^{-1}(\sin x+\cos x),\quad x>0\;\) is always an increasing function in the interval \(\left(0,\frac{\pi}{4}\right)\)
💡 Concept
🗺️ Roadmap

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    solution Array
❌ Common Mistakes
  • Forgetting the chain-rule factor \(\cos x-\sin x\).
  • Writing the derivative of \(\tan^{-1}u\) as \(1/u\), which is incorrect.
  • Trying to determine the sign of the entire expression without first observing that the denominator is always positive.
  • Using \(\sin x>\cos x\) in the interval \(\left(0,\frac{\pi}{4}\right)\); the correct relation is \(\cos x>\sin x\).
  • Including \(x=\frac{\pi}{4}\) when claiming \(f'(x)>0\) throughout the specified open interval.
🌟 Significance \for Board and Competitive Examinations
This example is a direct application of the first-derivative criterion for increasing functions. It also tests differentiation of an inverse trigonometric function, the chain rule, and trigonometric sign analysis.
🔑 Key Takeaway
∿

Exercise 35

📋
Table of Contents
9 sections
❓ Question
A circular disc of radius \(3\) cm is being heated. Due to expansion, its radius increases at the rate of \(0.05\) cm/s. Find the rate at which its area is increasing when the radius is \(3.2\) cm.
💡 Concept
🗺️ Roadmap

  1. Write the area of the circular disc in terms of its radius.

  2. Differentiate the area equation with respect to time.

  3. Substitute \(r=3.2\) cm and \(\frac{dr}{dt}=0.05\) cm/s.

  4. Calculate the required rate of increase of area.

🧩 Solution
  1. Write the Area Formula The area of a circular disc of radius \(r\) is
    \[A=\pi r^2\]
  2. Differentiate with Respect to Time Since the radius changes with time, the area also changes with time. Differentiating both sides with respect to \(t\), using the chain rule, gives
    \[\frac{dA}{dt}=\pi\frac{d}{dt}(r^2)\]
  3. Therefore,
    \[\boxed{\frac{dA}{dt}=2\pi r\frac{dr}{dt}}\]
  4. Substitute the Given Values At the instant under consideration,
    \[r=3.2\text{ cm}\]
    and
    \[r=3.2\text{ cm}\]
  5. Therefore,
    \[ \begin{aligned} \frac{dA}{dt} &=2\pi(3.2)(0.05)\\ &=2\pi(0.16)\\ &=0.32\pi \end{aligned} \]
  6. Hence,
    \[\boxed{\frac{dA}{dt}=0.32\pi\text{ cm}^2/\text{s}}\]
  7. Approximate Value
    Using \(\pi\approx3.14159\)
    \[0.32\pi\approx1.0053\]
  8. Therefore,
    \[\boxed{\frac{dA}{dt}\approx1.005\text{ cm}^2/\text{s}}\]
👁️ Important Observation
🔍 Geometrical Interpretation
Since
\[\frac{dA}{dr}=2\pi r,\]
the rate of change of area with respect to radius is equal to the circumference of the disc.

Hence,
\[\frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}=(2\pi r)\frac{dr}{dt}\]
At \(r=3.2\) cm,
\[2\pi r=6.4\pi\text{ cm},\]
so
\[\frac{dA}{dt}=(6.4\pi)(0.05)=0.32\pi\text{ cm}^2/\text{s}\]
❌ Common Mistakes
  • Using the initial radius \(3\) cm instead of the required instantaneous radius \(3.2\) cm.
  • Forgetting the factor \(\frac{dr}{dt}\) while differentiating \(A=\pi r^2\) with respect to time.
  • Writing \(\frac{dA}{dt}=2\pi r\) without multiplying by \(\frac{dr}{dt}\).
  • Giving the answer in cm/s instead of cm²/s.
  • Confusing the rate of increase of radius with the rate of increase of area.
🌟 Significance for Board and Competitive Examinations
This example illustrates the fundamental related-rates principle
\[\boxed{\frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}}\]
It is a standard application of the chain rule and is useful for problems involving expanding circles, spheres, cylinders, cones and other geometrical figures.
🔑 Key Takeaway
∿

Exercise 36

📋
Table of Contents
8 sections
❓ Question
An open-topped box is to be constructed by removing equal squares from each corner of a \(3\) metre by \(8\) metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.
💡 Concept
🗺️ Roadmap

  1. Let \(x\) be the side length of each square removed.

  2. Determine the dimensions of the resulting box.

  3. Form the volume function \(V(x)\).

  4. Determine the permissible domain of \(x\).

  5. Differentiate \(V(x)\) and find the critical points.

  6. Identify the critical point giving the maximum volume.

  7. Calculate the maximum volume.

🧩 Solution
  1. Let
    \[x=\text{side length of each square removed}\]
  2. After removing the four squares and folding the remaining portions upward, the height of the box is
    \[\boxed{x\text{ m}}\]
  3. Determine the Dimensions of the Box The original rectangular sheet has dimensions
    \[3\text{ m}\times8\text{ m}\]
  4. Since a square of side \(x\) is removed from both ends of the \(8\) m side, the length of the base becomes
    \[8-2x\]
  5. Similarly, removing squares from both ends of the \(3\) m side gives the breadth
    \[3-2x\]
  6. Thus, the dimensions of the open box are
    \[\boxed{\text{Length}=8-2x,\quad\text{Breadth}=3-2x,\quad\text{Height}=x}\]
  7. Form the Volume Function The volume of a rectangular box is
    \[V=\text{length}\times\text{breadth}\times\text{height}\]
  8. Therefore,
    \[V(x)=x(8-2x)(3-2x)\]
  9. Expanding,
    \[ \begin{aligned} V(x) &=x(24-16x-6x+4x^2)\\ &=x(24-22x+4x^2)\\ &=24x-22x^2+4x^3 \end{aligned} \]
  10. Hence,
    \[\boxed{V(x)=4x^3-22x^2+24x}\]
  11. Determine the Domain For a physically meaningful box, all three dimensions must be positive. In particular,
    \[x>0\]
    and
    \[3-2x>0\]
  12. From
    \[3-2x>0\]
  13. we obtain
    \[x < \frac32\]
  14. Therefore,
    \[\boxed{0 < x < \frac32}\]
    📝 The condition \(8-2x>0\) gives \(x<4\), but the stronger restriction \(x<3/2\) comes from the \(3\) m side.
  15. Find the Critical Points Differentiate
    \[V(x)=4x^3-22x^2+24x\]
  16. Thus,
    \[ \begin{aligned} \frac{dV}{dx} &=12x^2-44x+24\\ &=4(3x^2-11x+6). \end{aligned} \]
  17. For a critical point,
    \[\frac{dV}{dx}=0\]
  18. Therefore,
    \[3x^2-11x+6=0\]
  19. Factorising,
    \[3x^2-11x+6=(3x-2)(x-3)\]
  20. Hence,
    \[(3x-2)(x-3)=0\]
  21. Therefore,
    \[x=\frac23\]
    or
    \[x=3\]
  22. However, the permissible domain is
    \[0 < x < \frac32\]
  23. Therefore, \(x=3\) is not physically possible. The only admissible critical point is
    \[\boxed{x=\frac23\text{ m}}\]
  24. Verify That the Volume Is Maximum The second derivative is
    \[\frac{d^2V}{dx^2}=24x-44\]
  25. At \(x=\frac23\) we get
    \[\begin{aligned}\frac{d^2V}{dx^2}\bigg|_{x=\frac23}&=24\left(\frac23\right)-44\\ &=16-44\\ &=-28<0\end{aligned}\]
  26. Since
    \[\frac{d^2V}{dx^2} < 0\]
  27. the volume is maximum at
    \[\boxed{x=\frac23\text{ m}}\]
  28. Calculate the Maximum Volume At $x=\frac23\;$ the dimensions of the box are
    \[ \begin{aligned} \text{Length} &=8-2\left(\frac23\right)\\ &=8-\frac43\\ &=\frac{20}{3}\text{ m}, \end{aligned} \]
    \[ \begin{aligned} \text{Breadth} &=3-2\left(\frac23\right)\\ &=3-\frac43\\ &=\frac53\text{ m}, \end{aligned} \]
    and
    \[ \text{Height}=\frac23\text{ m}. \]
  29. Therefore,
    \[ \begin{aligned} V_{\max} &= \left(\frac{20}{3}\right) \left(\frac53\right) \left(\frac23\right)\\ &=\frac{20\times5\times2}{27}\\ &=\frac{200}{27} \end{aligned} \]
  30. Hence,
    \[\boxed{V_{\max}=\frac{200}{27}\text{ m}^3}\]
  31. ⭐ Final Answer
    \[ \boxed{ \begin{aligned} \text{Side of each square removed} &=\frac23\text{ m},\\ \text{Maximum volume} &=\frac{200}{27}\text{ m}^3. \end{aligned} } \]
🌟 Important Exam Insight
In optimization problems involving physical dimensions, every algebraic critical point must be checked against the domain imposed by the geometry.

Here, the domain
\[0 < x < \frac32\]
is essential. Merely solving \(V'(x)=0\) is not sufficient.
❌ Common Mistakes
  • Using \(8-x\) and \(3-x\) instead of \(8-2x\) and \(3-2x\).
  • Forgetting that the height of the box is \(x\).
  • Accepting \(x=3\) as a valid critical point without checking the domain.
  • Forgetting that the original sheet has width \(3\) m, which restricts \(x\) to \(0
  • Finding the critical point but not verifying that it gives a maximum.
  • Reporting the dimensions of the box instead of the requested maximum volume.
🌟 Significance for Board and Competitive Examinations
This is a classic geometrical optimization problem from Applications of Derivatives. It demonstrates how a physical construction problem can be converted into a single-variable volume function and solved systematically using derivatives.
🔑 Key Takeaway
∿

Example 37

📋
Table of Contents
7 sections
❓ Question
A manufacturer can sell \(x\) items at a price of \(\left(5-\frac{x}{100}\right)\;\) rupees each. The cost price of \(x\) items is \(\left(\frac{x}{5}+500\right)\;\) rupees. Find the number of items he should sell to earn maximum profit.
💡 Concept
🗺️ Roadmap

  1. Find the revenue from selling \(x\) items.

  2. Write the cost function.

  3. Form the profit function.

  4. Differentiate the profit function.

  5. Set the first derivative equal to zero to obtain the critical point.

  6. Use the second derivative to verify that the profit is maximum.

  7. State the required number of items.

🧩 Solution
  1. Find the Revenue Function The selling price of each item is
    \[p=5-\frac{x}{100}\]
  2. If \(x\) items are sold, the total revenue is
    \[R(x)=x\left(5-\frac{x}{100}\right)\]
  3. Therefore,
    \[\begin{aligned}R(x)&=5x-\frac{x^2}{100}\end{aligned}\]
  4. Hence,
    \[\boxed{R(x)=5x-\frac{x^2}{100}}\]
  5. Write the Cost Function The cost price of \(x\) items is given by
    \[C(x)=\frac{x}{5}+500\]
  6. Thus,
    \[\boxed{C(x)=\frac{x}{5}+500}\]
  7. Form the Profit Function Profit is the difference between revenue and cost:
    \[P(x)=R(x)-C(x)\]
  8. Therefore,
    \[ \begin{aligned} P(x) &=\left(5x-\frac{x^2}{100}\right) -\left(\frac{x}{5}+500\right)\\ &=5x-\frac{x}{5}-\frac{x^2}{100}-500\\ &=\frac{24x}{5}-\frac{x^2}{100}-500 \end{aligned} \]
  9. Hence,
    \[\boxed{P(x)=\frac{24x}{5}-\frac{x^2}{100}-500}\]
  10. Find the Critical Point Differentiate the profit function with respect to \(x\):
    \[ \begin{aligned} P'(x) &=\frac{24}{5}-\frac{2x}{100}\\ &=\frac{24}{5}-\frac{x}{50}. \end{aligned} \]
  11. For maximum profit,
    \[P'(x)=0\]
  12. Hence,
    \[\frac{24}{5}-\frac{x}{50}=0\]
  13. Multiplying by \(50\)
    \[240-x=0\]
  14. Therefore,
    \[\boxed{x=240}\]
❌ Common Mistakes
  • Using the selling price per item directly as the revenue.
  • Forgetting to multiply the price per item by \(x\) to obtain total revenue.
  • Using cost minus revenue instead of revenue minus cost for profit.
  • Solving \(P'(x)=0\) but not verifying that the point gives maximum profit.
  • Reporting the maximum profit \(\text{₹}76\) instead of the requested number of items.
🌟 Significance for Board and Competitive Examinations
This example connects differential calculus with an economic optimization problem. It illustrates the construction of revenue, cost and profit functions and the use of the first and second derivative tests to determine the production or sales level that maximizes profit.
🔑 Key Takeaway
· Updated
NCERT Class XIIMathematicsChapter 6
Application of Derivatives

Three big ideas, one powerful tool. Learn how derivatives measure change, reveal where a function rises or falls, and find the best possible value — with a step-by-step solver, worked practice and live experiments.

Rate of ChangeIncreasing & DecreasingMaxima & Minima
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NCERT Class 12 Application of Derivatives Notes — Complete Notes & Solutions · academia-aeternum.com
NCERT Class 12 Mathematics Chapter 6, Application of Derivatives, is an essential chapter for understanding how derivatives are used to analyse and solve real-world and mathematical problems. This chapter covers important concepts such as rate of change, increasing and decreasing functions, tangents, maxima and minima, and optimization problems. Students will learn how to apply the first and second derivative tests to determine local and absolute extrema and solve practical problems involving…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    application-of-derivatives — Learning Resources

    Frequently Asked Questions

    Application of Derivatives is NCERT Class 12 Mathematics Chapter 6, covering rate of change, increasing and decreasing functions, maxima and minima, and optimization problems.

    The rate of change of y with respect to x is given by dy/dx. It measures how rapidly one quantity changes with respect to another.

    A function is increasing where f'(x) > 0 and decreasing where f'(x) < 0 on the given interval.

    Critical points are points in the domain where f'(x) = 0 or where f'(x) does not exist, provided the function is defined at those points.

    If f'(x) changes from positive to negative at a critical point, the function has a local maximum. If it changes from negative to positive, the function has a local minimum.

    If f'(c) = 0 and f''(c) < 0, the function has a local maximum at c. If f''(c) > 0, it has a local minimum at c. If f''(c) = 0, the test is inconclusive.

    Find all critical points in the interior of the interval, include both endpoints, evaluate the function at all candidate points, and compare the values.

    Related rates problems involve quantities that change with time and are connected by a mathematical relation. Differentiating that relation with respect to time gives the required rate.

    Derivatives help identify values that maximize or minimize quantities such as area, volume, distance, cost, revenue and profit, subject to given conditions.

    Application of Derivatives provides important calculus concepts used in optimization, monotonicity, maxima-minima, related rates and other mathematical problems in competitive entrance examinations.

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