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Chapter 6 Application of Derivatives (Class 12 - Maths NCERT Concept Notes)

Welcome to Chapter 6: Application of Derivatives! This chapter moves beyond the mechanics of differentiation to explore the practical utility of the derivative. We use the derivative $\frac{dy}{dx}$ to measure the rate of change of quantities, such as how volume varies with time ($\frac{dV}{dt}$) or area with respect to radius ($\frac{dA}{dr}$). By analyzing the sign of $f'(x)$, we can determine if a function is strictly increasing ($f'(x) > 0$) or strictly decreasing ($f'(x) < 0$).

Geometrically, the derivative $f'(x_0)$ provides the slope of the tangent to a curve at a specific point. We also learn to find the equations of normals, which are lines perpendicular to the tangent at the point of contact. Furthermore, derivatives allow us to make approximations and find differentials to estimate small changes in function values.

The most significant application involves finding Maxima and Minima. By identifying critical points and using the First and Second Derivative Tests, we solve complex optimization problems, such as maximizing profit or minimizing material costs.

To enhance the understanding of these concepts, this page includes visualizations, flowcharts, mindmaps, and practical examples. This page is prepared by learningspot.co to provide a structured and comprehensive learning experience for every student, ensuring a deep mastery of differential calculus applications.

Content On This Page
Derivatives as a Rate Measure Tangents and Normals Approximations, Errors and Differentials
Increasing and Decreasing Functions Maxima and Minima Local Maximum and Local Minimum Values
Practical Problems on Maxima and Minima


Derivatives as a Rate Measure

The concept of a derivative is the cornerstone of calculus, acting as a mathematical bridge that connects static functions to dynamic change. In various disciplines such as Engineering, Sciences, Social Sciences, and Economics, we rarely encounter systems that remain constant. Instead, we observe quantities that evolve, fluctuate, or respond to shifts in other variables. The derivative provides a precise, quantitative measure of this responsiveness, known as the rate-measure.


The Mathematical Foundation of Rate of Change

To understand the derivative as a rate measure, we must distinguish between the average rate of change and the instantaneous rate of change. Consider a function $y = f(x)$, where $y$ is the dependent variable (the effect) and $x$ is the independent variable (the cause).

Average Rate of Change

If the independent variable $x$ changes by a small amount $\delta x$, the dependent variable $y$ will subsequently change by an amount $\delta y$. The average rate of change is simply the ratio of these two changes over a finite interval.

$\text{Average Rate} = \frac{\delta y}{\delta x} = \frac{f(x + \delta x) - f(x)}{\delta x}$

Instantaneous Rate of Change

In most scientific applications, we are interested in the rate of change at a specific point rather than over an interval. As we make the interval $\delta x$ smaller and smaller, approaching zero ($\delta x \to 0$), the average rate transforms into the instantaneous rate of change.

Mathematically, this is the definition of the derivative:

$\frac{dy}{dx} = \lim\limits_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$

[Definition of Derivative]


Elaboration Across Various Disciplines

The utility of the derivative as a rate measure extends far beyond pure mathematics. Below is an elaboration on how this concept is applied in different fields:

A. Physical Sciences and Engineering

In physics, the most common application is Kinematics. If $s$ represents the displacement of a particle and $t$ represents time, then:

1. Velocity ($v$): The rate of change of displacement with respect to time.

$v = \frac{ds}{dt}$

2. Acceleration ($a$): The rate of change of velocity with respect to time.

$a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

B. Economics and Commerce

In the context of Indian trade and industry, businesses use derivatives to optimize production and pricing. This is often referred to as Marginal Analysis. For instance, if $C(x)$ is the total cost in $\textsf{₹}$ for producing $x$ units of a commodity:

Marginal Cost (MC)

It represents the additional cost incurred by producing one more unit. It is the instantaneous rate of change of total cost with respect to the output quantity.

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

C. Social Sciences and Biology

Derivatives are used to model the growth rate of populations. If $P$ is the population of a region at time $t$, then $\frac{dP}{dt}$ represents the rate at which the population is growing or declining at any given instant.


Geometric Interpretation

Geometrically, the derivative $\frac{dy}{dx}$ at a point $P(x_0, y_0)$ represents the slope of the tangent to the curve $y = f(x)$ at that point. This slope tells us how steeply the "output" variable $y$ is rising or falling for every unit increase in the "input" variable $x$ at that exact location on the graph.

If $\frac{dy}{dx} > 0$, the quantity is increasing. Conversely, if $\frac{dy}{dx} < 0$, the quantity is decreasing. If $\frac{dy}{dx} = 0$, the quantity has reached a stationary point, such as a local maximum or minimum.


Definition and Derivation of Rate of Change

To establish the derivative as a mathematical tool for measuring change, let us consider a continuous function $y = f(x)$. Here, $x$ is the independent variable (such as time, length, or quantity) and $y$ is the dependent variable that changes in response to $x$.

The Concept of Increments

Suppose the independent variable $x$ undergoes a small change, denoted by the increment $\delta x$. Because $y$ is a function of $x$, this change produces a corresponding increment in $y$, denoted by $\delta y$.

$y + \delta y = f(x + \delta x)$

To find the change in $y$ alone, we subtract the original function $y = f(x)$ from equation (i):

$\delta y = f(x + \delta x) - f(x)$

[Incremental change in $y$]

Average Rate of Change

The average rate of change of $y$ with respect to $x$ over the interval from $x$ to $x + \delta x$ is defined as the ratio of the change in the dependent variable to the change in the independent variable.

$\text{Average Rate} = \frac{\delta y}{\delta x} = \frac{f(x + \delta x) - f(x)}{\delta x}$

Derivation of Instantaneous Rate of Change

The average rate provides an overview over an interval, but in physics and engineering, we require the instantaneous rate of change—the rate at a specific, exact moment. To find this, we allow the interval $\delta x$ to become infinitely small, approaching zero.

Taking the limit as $\delta x \to 0$ on both sides of the above equation:

$\lim\limits_{\delta x \to 0} \frac{\delta y}{\delta x} = \lim\limits_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$

By the definition of the derivative in calculus, this limit is denoted as $\frac{dy}{dx}$ or $f'(x)$. Thus, the instantaneous rate of change of $y$ with respect to $x$ is:

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

If we wish to calculate this rate at a specific point, say $x = x_0$, we denote it as:

$\left. \frac{dy}{dx} \right]_{x = x_0} \text{ or } f'(x_0)$


Related Rates and the Chain Rule

In many real-world scenarios, we encounter situations where two or more variables are not directly linked but are both dependent on a third variable, most commonly time ($t$). For instance, as the radius of a spherical balloon increases over time, its volume also increases over time. In such cases, the rates of change of these variables are "related" to each other.

Mathematical Derivation

Consider two variables $x$ and $y$ that are expressed as functions of a parameter $t$. Let:

$y = f(t) \text{ and } x = g(t)$

To find the rate of change of $y$ with respect to $x$, we apply the Chain Rule of differentiation. The Chain Rule states that if $y$ is a function of $x$ and $x$ is a function of $t$, then the derivative of $y$ with respect to $t$ is:

$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

[By Chain Rule]

By rearranging the terms in the above equation to isolate $\frac{dy}{dx}$, we obtain the formula for related rates:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

[Provided $\frac{dx}{dt} \neq 0$]

This derivation shows that the instantaneous rate of change of $y$ with respect to $x$ is the ratio of their respective rates of change with respect to the common parameter $t$.


Geometric and Physical Significance of the Sign

The sign of the derivative $\frac{dy}{dx}$ is crucial for understanding the behavior of the relationship between $x$ and $y$.

1. Positive Rate of Change ($\frac{dy}{dx} > 0$)

If $\frac{dy}{dx}$ is positive, it indicates that the variables $x$ and $y$ move in the same direction. That is, as $x$ increases, $y$ also increases. Conversely, if $x$ decreases, $y$ also decreases.

Example: In a circle, as the radius $r$ increases, the area $A$ also increases, so $\frac{dA}{dr}$ is positive.

2. Negative Rate of Change ($\frac{dy}{dx} < 0$)

If $\frac{dy}{dx}$ is negative, it indicates that the variables move in opposite directions. As $x$ increases, $y$ decreases. This is often seen in problems involving distance (e.g., a ladder sliding down a wall).

Example: If a vessel is being emptied, the height $h$ of the liquid decreases as time $t$ increases, making $\frac{dh}{dt}$ negative.


Commonly Used Formulas in Related Rates

Competitive exams often feature problems based on geometric shapes. The following table summarizes the formulas where derivatives are frequently applied as rate measures:

Shape Variable (y) Formula Derivative (dy/dx)
Circle Area ($A$) $A = \pi r^2$ $\frac{dA}{dr} = 2\pi r$
Sphere Volume ($V$) $V = \frac{4}{3}\pi r^3$ $\frac{dV}{dr} = 4\pi r^2$
Sphere Surface Area ($S$) $S = 4\pi r^2$ $\frac{dS}{dr} = 8\pi r$
Cube Volume ($V$) $V = x^3$ $\frac{dV}{dx} = 3x^2$
Cone Volume ($V$) $V = \frac{1}{3}\pi r^2 h$ Variable (depends on $r$ and $h$)

Important Remarks

1. Unit Consistency: Always ensure that all quantities are in the same system of units (e.g., if the radius is in $cm$ and time in $seconds$, the rate must be in $cm^2/s$ or $cm^3/s$).

2. Instantaneous vs. Average: If the question asks for the rate "at $x = 5$", it refers to the instantaneous rate. If it asks "between $x=2$ and $x=5$", it refers to the average rate.


Applications in Economics and Commerce

In the modern economic landscape, decision-making relies heavily on understanding how small changes in production affect financial outcomes. Calculus, specifically the derivative, allows economists to calculate marginal values. These values represent the instantaneous rate of change of a total quantity (like cost or revenue) with respect to the number of units produced or sold.

The Concept of Marginal Analysis

Marginal analysis is the study of the additional benefits or costs of an activity. If $x$ is the number of units of a commodity, then any function $f(x)$ representing a total quantity has a "marginal" counterpart $f'(x)$.

1. Marginal Cost (MC)

Marginal Cost is defined as the instantaneous rate of change of the Total Cost (C) with respect to the number of items produced ($x$). In simple terms, it is the approximate cost of producing one additional unit of the product.

Let $C(x)$ be the total cost function. The marginal cost $MC$ is given by the derivative:

$MC = \frac{dC}{dx} = f'(x)$

Note: In many problems, $C(x)$ is expressed as a polynomial: $C(x) = ax^2 + bx + k$, where $k$ represents the Fixed Cost (cost when $x=0$) and $ax^2 + bx$ represents the Variable Cost.

2. Marginal Revenue (MR)

Marginal Revenue is the instantaneous rate of change of the Total Revenue (R) with respect to the number of items sold ($x$). It represents the additional income generated by selling one more unit of a product.

If $p$ is the price per unit (demand function) and $x$ is the number of units sold, then the Total Revenue $R(x)$ is calculated as:

$R(x) = p \times x$

The marginal revenue $MR$ is the derivative of this revenue function:

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

Related Economic Measures

To provide a comprehensive understanding for competitive exams, we must also consider Average Cost (AC) and Average Revenue (AR):

1. Average Cost (AC): It is the cost per unit of production.

$AC = \frac{C(x)}{x}$

2. Average Revenue (AR): It is the revenue per unit sold, which is identical to the price $p$.

$AR = \frac{R(x)}{x} = \frac{p \cdot x}{x} = p$


Example 1. The total revenue received from the sale of $x$ units of a product is given by $R(x) = 3x^2 + 36x + 5$. Find the marginal revenue when $x = 5$ in $\textsf{₹}$.

Answer:

Given: Total Revenue function $R(x) = 3x^2 + 36x + 5$

To Find: Marginal Revenue ($MR$) at $x = 5$.

Solution:

We know that Marginal Revenue is the derivative of the Total Revenue function with respect to $x$:

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

Differentiating $R(x)$ with respect to $x$:

$MR = \frac{d}{dx}(3x^2 + 36x + 5)$

$MR = 6x + 36$

Now, substituting the value $x = 5$:

$MR = 6(5) + 36$

$MR = 30 + 36 = 66$

Hence, the marginal revenue when 5 units are sold is $\textsf{₹} \ 66$.


Example 2. The total cost $C(x)$ in Rupees ($\textsf{₹}$) associated with the production of $x$ units of an item is given by $C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000$.

Find the marginal cost when 17 units are produced.

Answer:

Given the total cost function:

$C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000$

We know that Marginal Cost ($MC$) is the derivative of total cost with respect to $x$:

$MC = \frac{dC}{dx} = \frac{d}{dx}(0.007x^3 - 0.003x^2 + 15x + 4000)$

Differentiating with respect to $x$:

$MC = 0.021x^2 - 0.006x + 15$

When $x = 17$:

$MC = 0.021(17)^2 - 0.006(17) + 15$

$MC = 0.021(289) - 0.102 + 15$

$MC = 6.069 - 0.102 + 15$

$MC = 20.967$

Hence, the marginal cost is $\textsf{₹} \ 20.97$ (approximately).


Example 3. Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 5 \ cm$.

Answer:

The area $A$ of a circle with radius $r$ is given by:

$A = \pi r^2$

The rate of change of the area $A$ with respect to its radius $r$ is:

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

When $r = 5 \ cm$:

$\frac{dA}{dr} = 2\pi (5) = 10\pi$

Thus, the area of the circle is changing at the rate of $10\pi \ cm^2/cm$.


Example 4. The volume of a sphere is increasing at the rate of $8\ cm^3/s$. How fast is its surface area increasing when the radius of the sphere is $12\ cm$?

Answer:

Given: Rate of change of volume with respect to time $t$, $\frac{dV}{dt} = 8\ cm^3/s$.

To Find: Rate of change of surface area $\frac{dS}{dt}$ when $r = 12\ cm$.

Solution:

Let $V$ be the volume, $S$ be the surface area, and $r$ be the radius of the sphere at any time $t$.

We know the volume of a sphere is:

$V = \frac{4}{3}\pi r^3$

Differentiating both sides with respect to $t$:

$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt}$

$8 = 4\pi r^2 \frac{dr}{dt}$

Solving for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{8}{4\pi r^2} = \frac{2}{\pi r^2}$

... (i)

Now, the surface area of the sphere is:

$S = 4\pi r^2$

Differentiating surface area with respect to $t$:

$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$

Substituting the value of $\frac{dr}{dt}$ from equation (i):

$\frac{dS}{dt} = 8\pi r \left(\frac{2}{\pi r^2}\right)$

$\frac{dS}{dt} = \frac{16}{r}$

When radius $r = 12\ cm$:

$\frac{dS}{dt} = \frac{16}{12} = \frac{4}{3}$

Hence, the surface area is increasing at the rate of $\frac{4}{3}\ cm^2/s$.


Example 5. An edge of a variable cube is increasing at the rate of $3\ cm/s$. How fast is the volume of the cube increasing when the edge is $10\ cm$ long?

Answer:

Let $x$ be the length of the edge and $V$ be the volume of the cube.

Given: $\frac{dx}{dt} = 3\ cm/s$ and $x = 10\ cm$.

The volume of a cube is given by:

$V = x^3$

Differentiating with respect to time $t$:

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

Substituting the given values:

$\frac{dV}{dt} = 3(10)^2 \cdot (3)$

$\frac{dV}{dt} = 3(100) \cdot 3 = 900$

Thus, the volume of the cube is increasing at the rate of $900\ cm^3/s$.


Example 6. The total revenue in Rupees ($\textsf{₹}$) received from the sale of $x$ units of a product is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$.

Answer:

Given: $R(x) = 13x^2 + 26x + 15$

We know that Marginal Revenue ($MR$) is the rate of change of total revenue with respect to the number of items sold ($x$):

$MR = \frac{dR}{dx} = \frac{d}{dx}(13x^2 + 26x + 15)$

$MR = 26x + 26$

When $x = 7$:

$MR = 26(7) + 26$

$MR = 182 + 26 = 208$

Hence, the marginal revenue is $\textsf{₹}\ 208$.


Example 7. A stone is dropped into a quiet lake and waves move in circles at a speed of $4\ cm/s$. At the instant when the radius of the circular wave is $10\ cm$, how fast is the enclosed area increasing?

Answer:

Given: The radius $r$ is increasing at the rate of $4\ cm/s$. Therefore, $\frac{dr}{dt} = 4\ cm/s$.

The area $A$ of a circle is:

$A = \pi r^2$

The rate of change of area with respect to time $t$ is:

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

When $r = 10\ cm$:

$\frac{dA}{dt} = 2\pi (10)(4)$

$\frac{dA}{dt} = 80\pi$

Thus, the enclosed area is increasing at the rate of $80\pi\ cm^2/s$.



Tangents and Normals

The study of Tangents and Normals is one of the most direct geometric applications of differential calculus. By calculating the derivative of a function at a specific point, we gain precise information about the direction and orientation of the curve at that exact location. This is often referred to as finding the gradient of the curve.


Geometric Interpretation of the Derivative

Consider a continuous curve defined by the function $y = f(x)$. Let $P(x_1, y_1)$ be a fixed point on this curve. If we draw a straight line that just touches the curve at point $P$, this line is called the Tangent to the curve at $P$.

The Normal to a curve at a point $P(x_1, y_1)$ is a straight line that passes through $P$ and is perpendicular to the tangent at that same point.

Tangent and Normal to a curve at point P

Suppose the tangent line makes an angle $\psi$ with the positive direction of the $x$-axis. The slope ($m$) is defined as follows:

$m = \tan \psi = \left. \frac{dy}{dx} \right|_{(x_1, y_1)}$

[Slope of the Tangent]

For the Normal line, using the condition of perpendicularity ($m_1 \cdot m_2 = -1$):

$m_N = -\frac{1}{\left( \frac{dy}{dx} \right)_{(x_1, y_1)}}$

[Slope of the Normal]


Special Geometric Cases

Case I: Tangent Parallel to the X-axis (Horizontal Tangent)

When the tangent is parallel to the $x$-axis, its inclination $\psi = 0^\circ$. This usually occurs at the turning points (maxima or minima) of a curve.

Tangent parallel to X-axis

$\tan 0^\circ = 0 \implies \frac{dy}{dx} = 0$

Case II: Tangent Parallel to the Y-axis (Vertical Tangent)

When the tangent is parallel to the $y$-axis, it is perpendicular to the $x$-axis. Here, $\psi = 90^\circ$. The slope becomes infinite, which mathematically implies the denominator of the derivative is zero.

Tangent parallel to Y-axis

$\tan 90^\circ = \infty \implies \frac{dx}{dy} = 0$

Case III: Tangent Equally Inclined to the Axes

If a tangent is equally inclined to the coordinate axes, it makes an angle of $45^\circ$ or $135^\circ$ with the positive $x$-axis.

Tangent equally inclined to axes

$\tan 45^\circ = 1 \text{ or } \tan 135^\circ = -1$

$\frac{dy}{dx} = \pm 1$


The Concept of the Normal

The Normal to a curve at a point $P(x_1, y_1)$ is defined as a straight line that passes through $P$ and is perpendicular to the tangent at that same point.

According to the laws of perpendicularity in coordinate geometry, if the slope of the tangent is $m_T$ and the slope of the normal is $m_N$, then their product must be $-1$.

$m_T \cdot m_N = -1$

(Perpendicular Condition)

By substituting the derivative for $m_T$, we derive the slope of the normal:

$m_N = -\frac{1}{\left( \frac{dy}{dx} \right)_{(x_1, y_1)}} = -\left( \frac{dx}{dy} \right)_{(x_1, y_1)}$


Summary Table for Quick Reference

Orientation of Tangent Angle ($\psi$) Mathematical Condition
Horizontal (Parallel to X-axis) $0^\circ$ $\frac{dy}{dx} = 0$
Vertical (Parallel to Y-axis) $90^\circ$ $\frac{dx}{dy} = 0$
Equally Inclined $45^\circ / 135^\circ$ $\frac{dy}{dx} = \pm 1$
Perpendicular to a line with slope $m$ --- $\frac{dy}{dx} = -\frac{1}{m}$

Equations of Tangent and Normal

1. Equation of the Tangent

To find the equation of a tangent to the curve $y = f(x)$ at a given point $P(x_1, y_1)$, we utilize the Point-Slope Form of a straight line from coordinate geometry. This form states that a line passing through $(x_1, y_1)$ with a known slope $m$ is represented by the equation $y - y_1 = m(x - x_1)$.

In the context of calculus, the slope $m$ of the tangent at point $P$ is exactly the derivative of the function evaluated at that point.

$m = \left. \frac{dy}{dx} \right|_{(x_1, y_1)}$

[Gradient at point $P$]

Substituting this value into the point-slope formula, we obtain the Equation of the Tangent:

$y - y_1 = \left( \frac{dy}{dx} \right)_{(x_1, y_1)} (x - x_1)$

Special Remark

If the derivative $\frac{dy}{dx}$ does not exist (becomes infinite) at the point $(x_1, y_1)$, the tangent is vertical (parallel to the $y$-axis). In such a case, the equation of the tangent simplifies to:

$x = x_1$

2. Equation of the Normal

The Normal to a curve at a point $P(x_1, y_1)$ is defined as the line perpendicular to the tangent at that point. From the properties of perpendicular lines in coordinate geometry, we know that the product of the slopes of two perpendicular lines is $-1$.

$m_{tangent} \cdot m_{normal} = -1$

Thus, the slope of the normal ($m_n$) can be derived as:

$m_n = -\frac{1}{m_{tangent}} = -\frac{1}{\left( \frac{dy}{dx} \right)_{(x_1, y_1)}}$

[Condition for Perpendicularity]

By applying the point-slope form again, we find the Equation of the Normal at $(x_1, y_1)$:

$y - y_1 = -\frac{1}{\left( \frac{dy}{dx} \right)_{(x_1, y_1)}} (x - x_1)$

This equation can also be written in an alternative linear form to avoid fractions:

$(x - x_1) + \left( \frac{dy}{dx} \right)_{(x_1, y_1)} (y - y_1) = 0$


Angle of Intersection of Two Curves

The angle of intersection between two curves is a fundamental concept in coordinate geometry and calculus. It is defined as the angle between the tangents drawn to the two curves at their common point of intersection. If the two tangents are perpendicular, the curves are said to intersect orthogonally.

Geometric Representation

Let two curves be represented by the functions $y = f(x)$ and $y = g(x)$. Suppose these curves intersect at a point $P(x_1, y_1)$. At this point, we can draw a tangent $T_1$ to the first curve and a tangent $T_2$ to the second curve. The angle $\theta$ between these two tangents is the angle of intersection of the curves.

Angle of intersection between two curves showing tangents at point P

Formula and Derivation

Let $m_1$ be the slope of the tangent to the curve $y = f(x)$ at point $P$, and $m_2$ be the slope of the tangent to the curve $y = g(x)$ at point $P$.

$m_1 = \left( \frac{dy}{dx} \right)_{\text{for curve 1 at } P}$

$m_2 = \left( \frac{dy}{dx} \right)_{\text{for curve 2 at } P}$

From trigonometry, the acute angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Special Conditions for Intersection

1. Orthogonal Intersection

Two curves are said to cut each other orthogonally (at right angles) if the tangents at their point of intersection are perpendicular to each other. In this case, $\theta = 90^\circ$.

$m_1 \cdot m_2 = -1$

[Condition for Orthogonality]

2. Tangential Curves (Touching Curves)

Two curves are said to touch each other if the tangents at their point of intersection coincide. In this case, the angle $\theta = 0^\circ$.

$m_1 = m_2$

[Condition for Touching Curves]

Step-by-Step Procedure to find Angle of Intersection

To find the angle between two curves, follow these steps:

Step I: Solve the equations of the two given curves simultaneously to find the Point of Intersection $(x_1, y_1)$. If there are multiple points, calculate the angle at each point separately.

Step II: Find the derivative $\frac{dy}{dx}$ for both curves separately.

Step III: Calculate the numerical values of the slopes $m_1$ and $m_2$ by substituting the coordinates of the intersection point into the respective derivatives.

Step IV: Substitute $m_1$ and $m_2$ into the $\tan \theta$ formula to find the acute angle.


Example 1. Show that the curves $y^2 = x$ and $x^2 = y$ intersect at $(1, 1)$ and find the angle between them at this point.

Answer:

Given Curves: $y^2 = x$ (Curve 1) and $x^2 = y$ (Curve 2).

Verification of Point: At $(1, 1)$, for Curve 1: $1^2 = 1$ (True). For Curve 2: $1^2 = 1$ (True). Thus, $(1, 1)$ is a point of intersection.

Slope of Curve 1 ($m_1$):

$2y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y}$

At $(1, 1)$, $m_1 = \frac{1}{2(1)} = \frac{1}{2}$.

Slope of Curve 2 ($m_2$):

$2x = \frac{dy}{dx}$

At $(1, 1)$, $m_2 = 2(1) = 2$.

Angle of Intersection ($\theta$):

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{1/2 - 2}{1 + (1/2)(2)} \right|$

$\tan \theta = \left| \frac{-3/2}{2} \right| = \left| \frac{-3}{4} \right| = \frac{3}{4}$

Therefore, the angle of intersection is $\theta = \tan^{-1}\left(\frac{3}{4}\right)$.


Example 2. Find the equations of the tangent and normal to the curve $x^{2/3} + y^{2/3} = 2$ at the point $(1, 1)$.

Answer:

Given: Equation of the curve $x^{2/3} + y^{2/3} = 2$.

Step 1: Differentiate the equation with respect to $x$.

$\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$

$\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left( \frac{y}{x} \right)^{1/3}$

Step 2: Find the slope of the tangent ($m$) at $(1, 1)$.

$m = -\left( \frac{1}{1} \right)^{1/3} = -1$

Step 3: Equation of the Tangent at $(1, 1)$:

$y - 1 = -1(x - 1)$

$y - 1 = -x + 1 \implies x + y - 2 = 0$

Step 4: Equation of the Normal at $(1, 1)$:

Slope of normal $m_n = -\frac{1}{m} = -\frac{1}{-1} = 1$.

$y - 1 = 1(x - 1)$

$y - 1 = x - 1 \implies x - y = 0$

The equation of tangent is $x + y = 2$ and the normal is $x = y$.



Approximations, Errors and Differentials

The concept of Approximation using derivatives is based on the idea of linearization. When we look at a curve very closely at a specific point, it behaves almost like a straight line (the tangent). Differentials allow us to use this straight-line behavior to estimate changes in a function without performing complex calculations.


Conceptual Definition of Differentials

Let $y = f(x)$ be a differentiable function. Suppose $x$ is an independent variable that undergoes a small change, denoted by $\delta x$. This change in $x$ causes a corresponding change in the dependent variable $y$, which we denote as $\delta y$.

The Mathematical Derivation

By the first principle of derivatives, the derivative of $y$ with respect to $x$ is defined as the limit of the ratio of these changes:

$\frac{dy}{dx} = \lim\limits_{\delta x \to 0} \frac{\delta y}{\delta x}$

For a non-zero but very small increment $\delta x$, the ratio $\frac{\delta y}{\delta x}$ is approximately equal to the derivative $\frac{dy}{dx}$. We can express this relationship by introducing a small quantity $\epsilon$ that vanishes as $\delta x$ approaches zero:

$\frac{\delta y}{\delta x} = \frac{dy}{dx} + \epsilon$

[where $\epsilon \to 0$ as $\delta x \to 0$]

Multiplying both sides by $\delta x$:

$\delta y = \frac{dy}{dx} \cdot \delta x + \epsilon \cdot \delta x$

Since $\epsilon \cdot \delta x$ is the product of two extremely small quantities, it becomes negligible. Therefore, for practical calculation and approximation, we have:

$\delta y \approx \frac{dy}{dx} \cdot \delta x$

[Fundamental Approximation]


Defining the Differentials $dx$ and $dy$

To make the notation consistent with the symbol $\frac{dy}{dx}$, we formally define the differentials as separate mathematical entities:

A. Differential of the Independent Variable ($dx$)

The differential of the independent variable $x$, denoted by $dx$, is defined to be exactly equal to its increment $\delta x$. It represents an arbitrary change in $x$.

$dx = \delta x$

B. Differential of the Dependent Variable ($dy$)

The differential of the dependent variable $y$, denoted by $dy$, is defined as the product of the derivative $f'(x)$ and the differential $dx$.

$dy = f'(x) dx = \left( \frac{dy}{dx} \right) dx$


Critical Distinction: $\delta y$ vs $dy$

It is crucial for students to understand that $\delta y$ and $dy$ are not the same, although they are very close in value when $\delta x$ is small.

1. Actual Change ($\delta y$): This is the true difference in the function's value, $f(x + \delta x) - f(x)$. Geometrically, it represents the change in height along the curve.

2. Differential Change ($dy$): This is the principal part of the change. Geometrically, it represents the change in height along the tangent line to the curve at point $x$.

In most approximation problems, we treat $dy$ as a reliable substitute for $\delta y$. This leads to the general formula for approximating a value near a known point:

$f(x + \delta x) \approx f(x) + dy$

$f(x + \delta x) \approx f(x) + f'(x) \delta x$


Detailed Application to Approximations

To find the approximate value of a function $f$ at a point $x + \delta x$, where $\delta x$ is very small, we use the concept of linear approximation. The underlying principle is that for a very small interval, the curve of a function is nearly identical to its tangent line.

Derivation of the Approximation Formula

Let $y = f(x)$. If $x$ increases to $x + \delta x$, then the actual change in $y$ is denoted by $\delta y$. We can write this relationship as:

$y + \delta y = f(x + \delta x)$

By subtracting $y = f(x)$ from both sides, we isolate the change in $y$:

$\delta y = f(x + \delta x) - f(x)$

[Actual change in $y$]

We know from the definition of differentials that the actual change $\delta y$ is approximately equal to the differential $dy$. Therefore:

$\delta y \approx dy$

Substituting the expression for $dy = f'(x) \delta x$ into the above equation:

$f(x + \delta x) - f(x) \approx f'(x) \delta x$

Rearranging the formula gives us the final tool for calculation:

$f(x + \delta x) \approx f(x) + f'(x) \delta x$

[Approximation Formula]

Working Strategy: To use this formula, we split the given number into two parts: $x$ (a value whose function result is perfectly known) and $\delta x$ (the small remaining part or error).


Classification of Errors

In scientific measurements and engineering calculations, no measurement is perfectly precise. The difference between the measured value and the actual value is termed as an "error". Calculus helps us determine how an error in an independent variable propagates to the dependent variable.

Definitions of Error Types

Suppose $x$ is the measured value of a quantity and $\delta x$ is the error in its measurement:

Type of Error Mathematical Formula Description
Absolute Error $\delta x$ The actual magnitude of the error in measurement of $x$.
Relative Error $\frac{\delta x}{x}$ The ratio of the absolute error to the true value of the quantity.
Percentage Error $\frac{\delta x}{x} \times 100$ The relative error expressed as a part of 100.

Propagated Error in Dependent Variables

If $y = f(x)$ is a calculated quantity based on the measured value $x$, the absolute error in $y$ ($\delta y$) can be estimated using the derivative:

$\delta y = \left( \frac{dy}{dx} \right) \cdot \delta x$

Similarly, the Relative Error in $y$ is given by $\frac{\delta y}{y}$ and the Percentage Error in $y$ is $\frac{\delta y}{y} \times 100$.


Geometrical Interpretation of Differentials

Consider the graph of a differentiable function $y = f(x)$. Let $P(x, y)$ be a point on the curve. Suppose the independent variable $x$ increases by a small amount $\delta x$ to reach a new value $x + \delta x$. This corresponds to a new point $Q(x + \delta x, y + \delta y)$ on the curve.

The Construction

1. Draw a tangent to the curve at point $P$. Let this tangent make an angle $\psi$ with the positive $x$-axis.

2. From point $Q$, drop a perpendicular to the $x$-axis. Let it intersect the horizontal line through $P$ at point $M$ and the tangent line at point $R$.

3. Here, the distance $PM$ represents the change in $x$, so $PM = \delta x = dx$.

Geometric representation of dx, dy, delta x, and delta y on a curve

Formal Derivation of the Relationship

From the right-angled triangle $\triangle PRM$ in the figure above, we can define the slope of the tangent line:

$\tan \psi = \frac{RM}{PM}$

We know from the geometric definition of the derivative that the slope of the tangent at $P$ is $\frac{dy}{dx}$. Also, since $PM = dx$:

$\frac{dy}{dx} = \frac{RM}{dx}$

By cross-multiplying, we find the value of the vertical segment $RM$:

$RM = \left( \frac{dy}{dx} \right) dx$

By our mathematical definition, $dy = f'(x) dx$. Therefore, we can conclude that:

$RM = dy$

[Differential of $y$]

Distinguishing the Changes

Now, let us observe the segment $QM$ in the diagram:

1. Actual Increment ($\delta y$): The segment $QM$ represents the true change in the function's value as we move along the curve from $P$ to $Q$. Thus, $QM = \delta y$.

2. Differential ($dy$): The segment $RM$ represents the change in the vertical coordinate along the tangent line. Thus, $RM = dy$.

3. The Approximation Error: The small segment $QR$ represents the difference between the actual change and the differential. Mathematically, $\text{Error} = \delta y - dy$.

The Principle of Linearization

As the change in $x$ becomes smaller and smaller ($\delta x \to 0$), the point $Q$ moves closer to $P$ along the curve. The segment $QR$ shrinks much faster than the segments $QM$ or $RM$. Consequently, the tangent line becomes an almost perfect substitute for the curve in the immediate neighborhood of $P$.

This is why we can confidently state:

$\delta y \approx dy$

[As $\delta x \to 0$]


Example 1. Use differentials to approximate $\sqrt{36.6}$.

Answer:

Let the function be $f(x) = \sqrt{x}$.

We choose a value $x$ near $36.6$ whose square root is perfectly known. Let $x = 36$.

Then the increment $\delta x = 36.6 - 36 = 0.6$.

Differentiating $f(x)$:

$f'(x) = \frac{1}{2\sqrt{x}}$

Using the approximation formula $f(x + \delta x) \approx f(x) + f'(x) \delta x$:

$\sqrt{36.6} \approx \sqrt{36} + \frac{1}{2\sqrt{36}} (0.6)$

$\sqrt{36.6} \approx 6 + \frac{1}{2 \times 6} (0.6)$

$\sqrt{36.6} \approx 6 + \frac{0.6}{12}$

$\sqrt{36.6} \approx 6 + 0.05 = 6.05$

Hence, the approximate value of $\sqrt{36.6}$ is $6.05$.


Example 2. If the radius of a sphere is measured as $7 \ cm$ with an error of $0.01 \ cm$, find the approximate error in calculating its volume.

Answer:

Given: Radius $r = 7 \ cm$ and error in radius $\delta r = 0.01 \ cm$.

The volume $V$ of a sphere is given by:

$V = \frac{4}{3} \pi r^3$

The approximate error in volume $\delta V$ is given by $dV$:

$dV = \left( \frac{dV}{dr} \right) \delta r$

Differentiating $V$ with respect to $r$:

$\frac{dV}{dr} = \frac{4}{3} \pi (3r^2) = 4\pi r^2$

Substituting the values:

$dV = 4 \pi (7)^2 (0.01)$

$dV = 4 \pi (49) (0.01) = 1.96 \pi \ cm^3$

Thus, the approximate error in volume is $1.96 \pi \ cm^3$.



Increasing and Decreasing Functions

The concept of Monotonicity refers to the behavior of a function—whether it moves in an upward (increasing) or downward (decreasing) direction as the input value $x$ increases. In calculus, this behavior is determined by the rate of change of the function, which is represented by its derivative $f'(x)$.

If we consider a function $y = f(x)$, the derivative $f'(x)$ at any point represents the slope of the tangent to the curve at that point. The sign of this slope dictates the nature of the function:

$f'(x) > 0$

[Tangent makes an acute angle with positive x-axis]

$f'(x) < 0$

[Tangent makes an obtuse angle with positive x-axis]

The Mathematical Relationship

To determine where a function is increasing or decreasing, we must analyze the intervals where the derivative maintains a constant sign. This is achieved by solving polynomial or rational inequalities. The process generally involves the following steps:

1. Finding Critical Points: We find the values of $x$ where $f'(x) = 0$ or where $f'(x)$ is not defined. These points are where the function might change its behavior from increasing to decreasing or vice versa.

2. Dividing the Domain: These critical points divide the domain of the function into several disjoint open intervals.

3. Testing Signs: We check the sign of $f'(x)$ in each interval. If the derivative is positive throughout an interval, the function is strictly increasing there. If it is negative, the function is strictly decreasing.

Importance of Inequality Solving

Since the core of this section involves solving $f'(x) > 0$ and $f'(x) < 0$, mastering the Method of Intervals (also known as the Wavy Curve Method) is a prerequisite. Without a systematic way to solve these inequalities, it is impossible to accurately identify the intervals of monotonicity for complex functions like:

$f'(x) = \frac{(x-1)^3(x+2)}{(x-4)^2}$

Summary Table of Monotonicity

Condition of Derivative Nature of Function $f(x)$
$f'(x) > 0$ for all $x \in (a, b)$ Strictly Increasing
$f'(x) \geq 0$ for all $x \in (a, b)$ Increasing (Non-decreasing)
$f'(x) < 0$ for all $x \in (a, b)$ Strictly Decreasing
$f'(x) \leq 0$ for all $x \in (a, b)$ Decreasing (Non-increasing)
$f'(x) = 0$ for all $x \in (a, b)$ Constant Function

Method of Intervals (Wavy Curve Method)

The Method of Intervals, popularly known as the Wavy Curve Method, is a powerful strategy used to solve inequalities of the form $f(x) > 0, f(x) < 0, f(x) \geq 0$ or $f(x) \leq 0$, where $f(x)$ is a polynomial or a rational function. This method helps in finding the intervals where the function remains positive or negative without testing every single point.

Consider a general polynomial expression $f(x)$ which has been factorised into linear factors:

$f(x) = (x - a_1)^{k_1} (x - a_2)^{k_2} \dots (x - a_n)^{k_n}$

Where $a_1, a_2, \dots, a_n$ are distinct real roots (critical points) and $k_1, k_2, \dots, k_n$ are natural numbers representing the multiplicity (powers) of these roots. We assume the roots are arranged such that:

$a_1 < a_2 < a_3 < \dots < a_n$

Sign Determination Rules and Logic

The sign of the expression changes or stays the same based on the power of the factor as we cross a critical point on the number line.

Rule 1: Even Powers (Multiplicity is Even)

If a factor is raised to an even power, such as $(x - a)^{2}, (x - a)^{4}$, etc., the term $(x - a)^k$ is always non-negative for all real $x$.

$(x - a)^k \geq 0$

[Whenever $k \in \{2, 4, 6, \dots\}$]

Since a positive number does not change the sign of the product, the sign of $f(x)$ does not change as $x$ moves from one side of $a$ to the other. For the purpose of the wavy curve, we can effectively "ignore" such factors, but we must account for them if the inequality is $\geq 0$ or $\leq 0$ (as the expression becomes zero at $x = a$).

Rule 2: Odd Powers (Multiplicity is Odd)

If a factor is raised to an odd power, such as $(x - a)^{1}, (x - a)^{3}$, etc., the sign of $(x - a)^k$ is exactly the same as the sign of $(x - a)$.

$\text{sign}[(x - a)^k] = \text{sign}(x - a)$

[Whenever $k \in \{1, 3, 5, \dots\}$]

As $x$ crosses the root $a$, the factor $(x - a)$ changes from negative to positive. Consequently, the sign of the entire expression $f(x)$ changes (flips).

Step-by-Step Procedure for Wavy Curve

To implement this method efficiently, follow these structural steps:

Step 1: Standardise the Factors

Ensure that the coefficient of $x$ in every linear factor is positive (preferably $+1$). If a factor is $(a - x)$, rewrite it as $-(x - a)$ and adjust the inequality sign if necessary (multiplying by $-1$ reverses the inequality).

Step 2: Plot Critical Points

Identify all roots by setting each factor to zero. Mark these points $a_1, a_2, \dots, a_n$ on a horizontal real number line.

Number line showing intervals separated by critical points

Step 3: Assign Signs to Intervals

1. Rightmost Interval: For the interval to the extreme right (where $x > a_n$), the expression $f(x)$ will always be positive (+), provided all factors were standardised to the $(x - a)$ form. This is because every single factor in the product will be positive.

2. Moving Leftwards: As you move to the next interval by crossing a root:

$\bullet$ If the root comes from a factor with an odd power, change the sign (from $+$ to $-$ or vice versa).

$\bullet$ If the root comes from a factor with an even power, maintain the same sign.

Step 4: Identify the Solution Set

Based on the original inequality:

$\bullet$ If $f(x) > 0$, choose all intervals marked with a (+) sign.

$\bullet$ If $f(x) < 0$, choose all intervals marked with a (–) sign.

$\bullet$ For $\geq$ or $\leq$, include the roots themselves (closed brackets), but exclude any roots that make the denominator zero in the case of rational functions.


Rational Inequalities

A Rational Inequality involves a ratio of two polynomial functions, generally expressed in the form $Q(x) = \frac{f(x)}{g(x)}$. The fundamental objective is to determine the range of values for $x$ such that the quotient is either positive, negative, or zero. Solving these is very similar to solving polynomial inequalities, but with specific constraints regarding the denominator.

Let the rational expression be defined as:

$Q(x) = \frac{(x - a_1) (x - a_2) \dots (x - a_n)}{(x - b_1) (x - b_2) \dots (x - b_m)}$

... (i)

Where $a_i$ are the roots of the numerator and $b_j$ are the roots of the denominator.

The Principle of Sign Equivalence

The core logic behind solving rational inequalities is that the sign (positive or negative) of a quotient $\frac{a}{b}$ is identical to the sign of the product $a \cdot b$. This is because in both multiplication and division, the result is positive if the signs are the same and negative if the signs are opposite.

Mathematical Derivation

To convert a rational inequality into a polynomial-like form without changing the sense of the inequality, we can multiply both sides by the square of the denominator. Since the square of any real non-zero number is always positive, the inequality remains valid.

Consider the inequality:

$\frac{f(x)}{g(x)} > 0$

Multiply both sides by $[g(x)]^2$ (given $g(x) \neq 0$):

$\frac{f(x)}{g(x)} \cdot [g(x)]^2 > 0 \cdot [g(x)]^2$

[Multiplying by a positive quantity]

Simplifying the left-hand side:

$f(x) \cdot g(x) > 0$

Thus, the sign of the rational function $\frac{f(x)}{g(x)}$ is exactly the same as the sign of the product $f(x) \cdot g(x)$. This allows us to use the Wavy Curve Method on the combined roots of both the numerator and the denominator.

Critical Constraints and Domain

While the sign determination is similar to polynomials, we must strictly adhere to the following rules regarding the denominator:

1. Undefined Points: The denominator $g(x)$ can never be zero. Therefore, even if the inequality is non-strict ($\geq$ or $\leq$), the roots of $g(x)$ must always be excluded from the solution set using open intervals or parenthesis.

2. Common Factors: If there is a common factor $(x - c)$ in both the numerator and denominator, it cancels out for sign determination, but the point $x = c$ must still be excluded from the domain as the original function remains undefined at that point.

Comparison of Conditions

Quotient Form Equivalent Product Form Constraint
$\frac{f(x)}{g(x)} > 0$ $f(x) \cdot g(x) > 0$ $g(x) \neq 0$
$\frac{f(x)}{g(x)} \geq 0$ $f(x) \cdot g(x) \geq 0$ $g(x) \neq 0$ (Exclude $b_j$ roots)
$\frac{f(x)}{g(x)} < 0$ $f(x) \cdot g(x) < 0$ $g(x) \neq 0$

Example 1. Find the set of all real values of $x$ for which the function $f(x) = (x - 1)(x - 2)(x - 3)$ is strictly positive.

Answer:

Given:

$f(x) = (x - 1)(x - 2)(x - 3) > 0$

To Find: The intervals of $x$ where the above inequality holds true.

Solution:

Step 1: Identify Critical Points

The critical points (roots) are obtained by setting each linear factor to zero:

$x - 1 = 0 \implies x = 1$

$x - 2 = 0 \implies x = 2$

$x - 3 = 0 \implies x = 3$

Step 2: Plotting on Number Line and Interval Creation

Mark the points $1$, $2$, and $3$ on the real number line. These points divide the number line into four distinct intervals:

1. $(-\infty, 1)$

2. $(1, 2)$

3. $(2, 3)$

4. $(3, \infty)$

Number line with roots 1, 2, 3 showing signs + and -

Step 3: Sign Analysis using the Wavy Curve Logic

Since the leading coefficient of $x$ in all factors is positive ($+1$) and all factors have an odd power (power of $1$), we start from the rightmost interval with a positive sign and alternate:

$\bullet$ Interval $(3, \infty)$: Take a test point $x = 4$. Then $f(4) = (4-1)(4-2)(4-3) = 3 \times 2 \times 1 = 6$, which is positive (+).

$\bullet$ Interval $(2, 3)$: Take a test point $x = 2.5$. Then $f(2.5) = (1.5)(0.5)(-0.5) = -0.375$, which is negative (–).

$\bullet$ Interval $(1, 2)$: Take a test point $x = 1.5$. Then $f(1.5) = (0.5)(-0.5)(-1.5) = +0.375$, which is positive (+).

$\bullet$ Interval $(-\infty, 1)$: Take a test point $x = 0$. Then $f(0) = (-1)(-2)(-3) = -6$, which is negative (–).


Step 4: Selection of Intervals

We are looking for the region where $f(x) > 0$ (the positive regions). From the analysis above, these are the intervals $(1, 2)$ and $(3, \infty)$.

The final solution is the union of these intervals:

$x \in (1, 2) \cup (3, \infty)$

[Final Solution Set]


Example 2. Find the set of all real values of $x$ satisfying the inequality:

$f(x) = \frac{(x - 1)^{2} (x - 2)^{3} (x - 4)^{4}}{(x - 5)} \leq 0$

Answer:

Given:

$\frac{(x - 1)^{2} (x - 2)^{3} (x - 4)^{4}}{(x - 5)} \leq 0$

To Find: The interval(s) of $x$ for which the expression is non-positive.

Solution:

Step 1: Identify Critical Points and Multiplicity

The critical points are obtained from both the numerator and the denominator:

$\bullet$ From $(x - 1)^2$: $x = 1$ (Even power, multiplicity 2)

$\bullet$ From $(x - 2)^3$: $x = 2$ (Odd power, multiplicity 3)

$\bullet$ From $(x - 4)^4$: $x = 4$ (Even power, multiplicity 4)

$\bullet$ From $(x - 5)$: $x = 5$ (Odd power, multiplicity 1)

Step 2: Sign Analysis on the Number Line

We plot $1, 2, 4,$ and $5$ on the number line. We start from the extreme right ($x > 5$) with a positive (+) sign because all leading coefficients of $x$ are positive.

Sign scheme showing behavior at even and odd powers

1. At $x = 5$: The power is $1$ (Odd). The sign changes from $(+)$ to $(-)$.

2. At $x = 4$: The power is $4$ (Even). The sign remains $(-)$.

3. At $x = 2$: The power is $3$ (Odd). The sign changes from $(-)$ to $(+)$.

4. At $x = 1$: The power is $2$ (Even). The sign remains $(+)$.

Interval Sign Summary:

$\bullet$ $(5, \infty) \implies (+)$

$\bullet$ $(4, 5) \implies (-)$

$\bullet$ $(2, 4) \implies (-)$

$\bullet$ $(1, 2) \implies (+)$

$\bullet$ $(-\infty, 1) \implies (+)$

Step 3: Including Boundary Points

Since the inequality is $\leq 0$, we must include points where $f(x) = 0$ and exclude points where $f(x)$ is undefined.

$x = 1, 2, 4$

[Numerator roots: Included]

$x \neq 5$

[Denominator root: Excluded]

Final Conclusion

The negative regions are $(2, 4)$ and $(4, 5)$. Including the roots $x=2$ and $x=4$, these merge into the interval $[2, 5)$. Additionally, we must include the isolated point $x = 1$ because it makes the expression equal to zero.

$x \in [2, 5) \cup \{1\}$


Example 3. Solve the rational inequality: $\frac{x - 3}{x + 5} \leq 0$.

Answer:

Given:

A rational inequality $\frac{x - 3}{x + 5} \leq 0$.

To Find:

The set of real values (intervals) for $x$ that satisfy this condition.

Solution:

Step 1: Locate Critical Points

The points where the expression either becomes zero or undefined are called critical points.

$\bullet$ Setting the numerator to zero: $x - 3 = 0 \implies x = 3$

$\bullet$ Setting the denominator to zero: $x + 5 = 0 \implies x = -5$

Step 2: Interval Testing (Wavy Curve Method)

The critical points $-5$ and $3$ divide the real number line into three distinct intervals: $(-\infty, -5)$, $(-5, 3)$, and $(3, \infty)$.

Number line showing intervals for x-3 and x+5

Let $Q(x) = \frac{x - 3}{x + 5}$:

1. For $x > 3$: Let $x = 4$. $Q(4) = \frac{4-3}{4+5} = \frac{1}{9} > 0$. (Positive region)

2. For $-5 < x < 3$: Let $x = 0$. $Q(0) = \frac{0-3}{0+5} = \frac{-3}{5} < 0$. (Negative region)

3. For $x < -5$: Let $x = -6$. $Q(-6) = \frac{-6-3}{-6+5} = \frac{-9}{-1} = 9 > 0$. (Positive region)

Step 3: Applying Boundary Conditions

We require the expression to be less than or equal to zero ($\leq 0$).

$\bullet$ The inequality is satisfied in the interval where the sign is negative, i.e., $(-5, 3)$.

$\bullet$ Since the inequality includes "equal to" ($\leq$), we check the endpoints. At $x = 3$, $Q(x) = 0$, which is allowed. However, at $x = -5$, the expression is undefined.

$x \in (-5, 3]$

... (Final Result)


Increasing and Decreasing Functions

Let $f$ be a real-valued function defined in a domain $D$, where $D$ is a subset of the set of real numbers $\mathbb{R}$. We analyze the behavior of the function over an interval $D_1$ (which is a subset of $D$) based on how the output values $f(x)$ change relative to the input values $x$.

Increasing and Strictly Increasing Functions

In mathematical analysis, the behavior of a real-valued function $f(x)$ as the input $x$ moves from left to right on the real number line is referred to as its Monotonicity. When a function consistently maintains an upward trend or does not decrease, we classify it as an Increasing Function or a Strictly Increasing Function.

Let $f$ be a real-valued function defined on an interval $D \subseteq \mathbb{R}$, and let $D_1$ be a subset of $D$.

1. Increasing Function (Non-decreasing)

A function $f$ is said to be an increasing function on an interval $D_1$ if for any two points $x_1$ and $x_2$ in that interval, the order of the function values follows the order of the inputs, allowing for the possibility that the values may remain equal over some sub-interval.

Mathematically, the condition is defined as:

$x_1 < x_2 \implies f(x_1) \leq f(x_2)$

$\forall \ x_1, x_2 \in D_1$

Graphical Interpretation: The graph of an increasing function never moves downwards. It may move upwards or it may stay horizontal (flat) for some time. In terms of calculus, the slope of the tangent to such a curve is never negative, meaning the derivative satisfies:

$f'(x) \geq 0$

[Slope is non-negative]

2. Strictly Increasing Function

A function $f$ is strictly increasing on $D_1$ if a greater input always results in a strictly greater output. There are no "flat" or constant portions in the graph of a strictly increasing function.

The rigorous definition is:

$x_1 < x_2 \implies f(x_1) < f(x_2)$

$\forall \ x_1, x_2 \in D_1$

Graphical Interpretation: The graph always moves upwards as we move from left to right. There are no horizontal segments. For a differentiable function, this usually implies that the derivative is positive:

$f'(x) > 0$

[Slope is strictly positive]

Graphical representation of increasing vs strictly increasing functions

Key Differences and Summary

The fundamental difference lies in the strictness of the inequality. While every strictly increasing function is technically also an increasing function, the reverse is not always true.

Feature Increasing Function Strictly Increasing Function
Condition $f(x_1) \leq f(x_2)$ $f(x_1) < f(x_2)$
Horizontal Segments Permitted (Graph can stay flat) Not Permitted (Graph must rise)
Derivative ($f'$) $f'(x) \geq 0$ $f'(x) > 0$
Nature Non-decreasing Always increasing

Example. Prove that the function $f(x) = 3x + 17$ is strictly increasing on $\mathbb{R}$.

Answer:

To Prove: $f(x)$ is strictly increasing on $\mathbb{R}$.

Proof:

Let $x_1$ and $x_2$ be any two real numbers such that:

$x_1 < x_2$

Multiplying both sides by $3$ (a positive number, which preserves the inequality):

$3x_1 < 3x_2$

Adding $17$ to both sides:

$3x_1 + 17 < 3x_2 + 17$

This can be written as:

$f(x_1) < f(x_2)$

(By definition of $f$)

Since $x_1 < x_2 \implies f(x_1) < f(x_2)$ for all $x_1, x_2 \in \mathbb{R}$, the function is strictly increasing.


Decreasing and Strictly Decreasing Functions

In the study of monotonocity, when a function consistently maintains a downward trend or does not increase as the input $x$ grows, it is classified under Decreasing or Strictly Decreasing functions. This behavior indicates that the output $f(x)$ is inversely related to the input $x$.

Let $f$ be a real-valued function defined on an interval $D \subseteq \mathbb{R}$, and let $D_2$ be a sub-interval of $D$.

3. Decreasing Function (Non-increasing)

A function $f$ is termed a decreasing function on an interval $D_2$ if, for any two values in that interval, the larger input results in an output that is either smaller than or equal to the output of the smaller input. This allows for portions of the graph to be horizontal.

The formal mathematical definition is:

$x_1 < x_2 \implies f(x_1) \geq f(x_2)$

$\forall \ x_1, x_2 \in D_2$

Graphical Interpretation: The graph of a decreasing function never rises as it moves from left to right. It may descend or remain flat (constant). In the context of calculus, the slope of the tangent (derivative) at any point is either negative or zero:

$f'(x) \leq 0$

[Slope is non-positive]

4. Strictly Decreasing Function

A function $f$ is strictly decreasing on $D_2$ if every increase in the input $x$ results in a definite decrease in the output $f(x)$. Unlike a standard decreasing function, there are no constant or horizontal segments in its graph.

The formal mathematical definition is:

$x_1 < x_2 \implies f(x_1) > f(x_2)$

$\forall \ x_1, x_2 \in D_2$

Graphical Interpretation: The graph strictly falls as we move along the positive x-axis. For differentiable functions, the derivative is strictly negative:

$f'(x) < 0$

[Slope is strictly negative]

Comparison of Decreasing and Strictly Decreasing graphs

Comparison Table of Decreasing Nature

Function Type Inequality Condition Derivative Property Graph Behavior
Decreasing $f(x_1) \geq f(x_2)$ $f'(x) \leq 0$ Falls or stays flat
Strictly Decreasing $f(x_1) > f(x_2)$ $f'(x) < 0$ Always falls

Example. Show that the function $f(x) = \cos x$ is strictly decreasing in the interval $(0, \pi)$.

Answer:

To Prove: $f(x) = \cos x$ is strictly decreasing in $x \in (0, \pi)$.

Proof:

We use the First Derivative Test. Let us differentiate the function with respect to $x$:

$f'(x) = \frac{d}{dx}(\cos x)$

$f'(x) = -\sin x$

We know that for $x \in (0, \pi)$, i.e., in the first and second quadrants, the sine function is always positive:

$\sin x > 0$

$\forall \ x \in (0, \pi)$

Multiplying the inequality by $-1$ (which reverses the inequality sign):

$-\sin x < 0$

$\implies f'(x) < 0$

Since the derivative $f'(x)$ is strictly negative throughout the interval $(0, \pi)$, the function $f(x) = \cos x$ is strictly decreasing in this interval.


Monotonic Functions

In mathematical analysis, the term Monotonic is derived from the Greek words 'monos' (single) and 'tonos' (tone or direction). A function is considered monotonic over an interval if it preserves a consistent direction of movement—it either never decreases or never increases as the input $x$ moves across the interval.

Monotonicity is a fundamental property used to analyze the global and local behavior of functions, particularly in solving complex inequalities and optimization problems.

Classification of Monotonicity

A function $f$ defined on an interval $D$ is classified based on the strictness of its directional consistency:

1. Monotonic Function

A function is called monotonic in an interval if it is either increasing or decreasing throughout that interval. In this case, the function is allowed to remain constant (horizontal) over some parts of the interval.

$f(x_1) \leq f(x_2)$ or $f(x_1) \geq f(x_2)$

[For all $x_1 < x_2$]

2. Strictly Monotonic Function

A function is called strictly monotonic in an interval if it is either strictly increasing or strictly decreasing throughout that interval. Such functions never stay constant; they are always either rising or falling.

$f(x_1) < f(x_2)$ or $f(x_1) > f(x_2)$

[For all $x_1 < x_2$]

Summary of Monotonicity Conditions

The following table summarizes the relationship between the input values, the function outputs, and the visual behavior of the graph:

Nature of Function Condition for $x_1 < x_2$ Graphical Interpretation
Increasing (Non-decreasing) $f(x_1) \leq f(x_2)$ The graph moves upward or remains horizontal.
Strictly Increasing $f(x_1) < f(x_2)$ The graph moves upward only (no flat regions).
Decreasing (Non-increasing) $f(x_1) \geq f(x_2)$ The graph moves downward or remains horizontal.
Strictly Decreasing $f(x_1) > f(x_2)$ The graph moves downward only (no flat regions).

Mathematical Significance of Monotonicity

1. Inverse Functions: A function has an inverse (is invertible) if and only if it is strictly monotonic over its entire domain. This ensures that the function is "One-to-One" (Injective).

2. Derivative Relationship: For a differentiable function $f(x)$, monotonicity is determined by the sign of the first derivative $f'(x)$.

$f'(x) \geq 0 \implies \text{Increasing}$

$f'(x) \leq 0 \implies \text{Decreasing}$


Conditions for an Increasing or a Decreasing Function

The derivative of a function $f(x)$ serves as a mathematical tool to determine the intervals of monotonicity. Since the derivative $f'(x)$ represents the slope of the tangent at any point $P(x, y)$ on the curve, its sign indicates whether the function is ascending or descending.

Geometrical Interpretation

The derivative of a function at any given point provides the slope of the tangent to the curve at that point. By analyzing the angle that this tangent makes with the positive direction of the x-axis, we can geometrically determine whether a function is increasing or decreasing.

1. Strictly Increasing Behavior and Acute Angles

For a function $f(x)$ that is strictly increasing in an interval, the curve consistently rises as we move from left to right. Consequently, the tangent drawn at any point on such a curve will lean towards the right.

Let $\psi$ be the angle formed by the tangent with the positive direction of the x-axis. For a strictly increasing function, this angle $\psi$ is typically an acute angle (i.e., $0 < \psi < 90^\circ$).

From the principles of trigonometry, we know that the tangent of an acute angle is always positive:

$\tan \psi > 0$

[For $0 < \psi < \frac{\pi}{2}$]

Since the derivative $f'(x)$ is equal to $\tan \psi$, we conclude:

$f'(x) > 0$

[Condition for Strictly Increasing]

Graph showing tangent making an acute angle with the x-axis for an increasing function

2. Strictly Decreasing Behavior and Obtuse Angles

Conversely, if a function $f(x)$ is strictly decreasing, the curve falls as $x$ increases. The tangent drawn at any point on this curve leans towards the left, forming an obtuse angle ($\psi$) with the positive direction of the x-axis (i.e., $90^\circ < \psi < 180^\circ$).

The tangent of an obtuse angle (in the second quadrant) is always negative:

$\tan \psi < 0$

[For $\frac{\pi}{2} < \psi < \pi$]

This implies that the derivative must be negative for the function to be strictly decreasing:

$f'(x) < 0$

[Condition for Strictly Decreasing]

Graph showing tangent making an obtuse angle with the x-axis for a decreasing function

The Nuanced Case: Analyzing $f(x) = x^3$

While the sign of $f'(x)$ is a strong indicator, geometric intuition can be refined by observing functions where the derivative becomes zero at specific points without the function losing its strictly increasing nature.

Consider the cubic function $f(x) = x^3$. Let us analyze its behavior across the entire set of real numbers $\mathbb{R}$. For any two values $x_1, x_2 \in \mathbb{R}$:

$x_1 < x_2 \implies x_1^3 < x_2^3$

[By definition of cubes]

This confirms that $f(x) = x^3$ is strictly increasing on $\mathbb{R}$. However, let us look at its derivative:

$f'(x) = 3x^2$

At the origin ($x = 0$):

$f'(0) = 3(0)^2 = 0$

Geometrically, this means that at $x = 0$, the tangent is horizontal (parallel to the x-axis). Although the slope is not strictly positive at this single isolated point, the function does not "flatten out" over an interval. It merely has a point of inflection where the rate of change momentarily becomes zero before continuing its upward journey.

Graph of y=x^3 showing a horizontal tangent at the origin

Conclusion on Geometrical Conditions

Based on these observations, we refine our criteria. A function is strictly increasing if its tangent slope is generally positive. Even if the slope becomes zero at finite isolated points, the function remains strictly monotonic. However, if the slope remains zero over a continuous interval, the function becomes a constant function over that interval and is no longer "strictly" increasing.


Fundamental Theorems on Monotonicity

The relationship between the derivative of a function and its monotonic behavior is formalized through two key theorems. These theorems allow us to use the tools of calculus to determine whether a function is increasing or decreasing over a specific interval without testing every individual point.

For these theorems, we assume that the function $f$ satisfies two primary conditions:

1. $f$ is continuous in the closed interval $[a, b]$.

2. $f$ is derivable (differentiable) in the open interval $(a, b)$.

Theorem 1: Conditions for Increasing Nature

This theorem provides the criteria for identifying non-decreasing and strictly rising behavior of a function.

(i) Condition for Increasing (Non-decreasing) Function

If $f'(x) \geq 0$ for all $x \in (a, b)$, then $f$ is increasing in $[a, b]$. In this scenario, the function's value either rises or stays constant, but never falls.

(ii) Condition for Strictly Increasing Function

If $f'(x) > 0$ for all $x \in (a, b)$, then $f$ is strictly increasing in $[a, b]$. This implies that for every increase in $x$, there is a guaranteed increase in $f(x)$.

Theorem 2: Conditions for Decreasing Nature

Similarly, these conditions define when a function moves downward or stays non-rising.

(i) Condition for Decreasing (Non-increasing) Function

If $f'(x) \leq 0$ for all $x \in (a, b)$, then $f$ is decreasing in $[a, b]$. The function value either falls or remains horizontal.

(ii) Condition for Strictly Decreasing Function

If $f'(x) < 0$ for all $x \in (a, b)$, then $f$ is strictly decreasing in $[a, b]$. The function value consistently reduces as $x$ increases.

Derivation of the Theorems

The formal proof of these theorems is rooted in Lagrange’s Mean Value Theorem (LMVT). Let us derive the condition for a strictly increasing function.

To Prove: If $f'(x) > 0$ for all $x \in (a, b)$, then $f(x)$ is strictly increasing.

Proof:

Let $x_1$ and $x_2$ be any two points in the interval $[a, b]$ such that $x_1 < x_2$. Since $f(x)$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$, by applying Lagrange’s Mean Value Theorem, there exists at least one point $c \in (x_1, x_2)$ such that:

$\frac{f(x_2) - f(x_1)}{x_2 - x_1} = f'(c)$

…(i)

This can be rearranged as:

$f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$

…(ii)

Now, let us analyze the signs of the terms on the right-hand side:

$x_2 - x_1 > 0$

[As we assumed $x_1 < x_2$]

If $f'(x) > 0$ for all $x$ in the interval, then $f'(c) > 0$. Multiplying two positive quantities in equation (ii) gives:

$f(x_2) - f(x_1) > 0$

$\implies f(x_2) > f(x_1)$           ... (iv)

Since $x_1 < x_2 \implies f(x_1) < f(x_2)$, the function is strictly increasing. The same logic applies to decreasing functions by substituting $f'(c) < 0$.


The Discrete Zero Property

The Discrete Zero Property is a vital extension of the first derivative test. In many real-world and mathematical functions, the derivative might become zero at certain specific points without the function losing its "strictly" monotonic nature.

A set of points is called discrete or isolated if each point has a neighborhood that contains no other points from the set. In simpler terms, the points where the derivative is zero are "scattered" and do not form a continuous line segment.

Formal Statement of the Corollary

Let $f(x)$ be a function that is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$.

1. For Strictly Increasing Nature

If $f'(x) \geq 0$ for all $x \in (a, b)$ and $f'(x) = 0$ only at a finite number of isolated points, then $f(x)$ is strictly increasing on $[a, b]$.

2. For Strictly Decreasing Nature

If $f'(x) \leq 0$ for all $x \in (a, b)$ and $f'(x) = 0$ only at a finite number of isolated points, then $f(x)$ is strictly decreasing on $[a, b]$.

Why does this property hold?

The core logic lies in the definition of strictly increasing behavior: $x_1 < x_2 \implies f(x_1) < f(x_2)$. If the derivative is zero only at an isolated point (like a "momentary pause"), the function immediately continues its upward or downward journey after that point. Since the function does not stay constant over any interval, the output value $f(x)$ will always be greater for a larger $x$.

Mathematical Comparison

Derivative Behavior Monotonic Nature Example
$f'(x) > 0$ always Strictly Increasing $e^x$
$f'(x) \geq 0$ and $f'(x) = 0$ at isolated points Strictly Increasing $x^3$
$f'(x) = 0$ over a sub-interval $(c, d)$ Increasing (but not strictly) Step functions

Example. Prove that the function $f(x) = x - \sin x$ is strictly increasing for all $x \in \mathbb{R}$.

Answer:

Given:

$f(x) = x - \sin x$

To Prove: $f(x)$ is strictly increasing on $\mathbb{R}$.

Proof:

First, we find the derivative of the function with respect to $x$:

$f'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\sin x)$

$f'(x) = 1 - \cos x$

We know the range of the cosine function:

$-1 \leq \cos x \leq 1$

[Range of $\cos x$]

Subtracting all parts from $1$:

$1 - 1 \leq 1 - \cos x \leq 1 - (-1)$

$0 \leq f'(x) \leq 2$

Here, $f'(x) \geq 0$ for all $x$. Now, let's check where $f'(x) = 0$:

$1 - \cos x = 0 \implies \cos x = 1$

$x = 2n\pi$

[where $n \in \mathbb{Z}$]

The points $x = 0, \pm 2\pi, \pm 4\pi, \dots$ are isolated discrete points. The derivative is zero only at these points and positive everywhere else. It never remains zero over any interval.

Conclusion: By the Discrete Zero Property, $f(x) = x - \sin x$ is strictly increasing on $\mathbb{R}$.


Geometrical Concept: Point of Inflection

In cases where the derivative is zero at a discrete point but the function remains strictly monotonic, that point is often a point of inflection. At such a point, the tangent is horizontal, but the curve does not turn back; it continues its original trend. This is visually represented in the graph of $y = x^3$ at the origin.

Graph showing isolated zero derivative for strictly increasing function

Summary of Derivative Tests for Monotonicity

The First Derivative Test is the most efficient method to determine the monotonic behavior of a differentiable function. By analyzing the sign of $f'(x)$ over a specific interval, we can conclude whether the function is rising, falling, or remaining constant.

Comprehensive Derivative Test Table

The following table summarizes the relationship between the sign of the first derivative and the nature of the function $f(x)$ in an interval $(a, b)$ where $f(x)$ is continuous and differentiable.

Sign of $f'(x)$ Monotonic Nature of $f(x)$
$f'(x) > 0$ Strictly Increasing
$f'(x) \geq 0$ Increasing (Non-decreasing)
$f'(x) < 0$ Strictly Decreasing
$f'(x) \leq 0$ Decreasing (Non-increasing)
$f'(x) = 0$ Constant Function

Detailed Analysis of Conditions

1. Strictly Increasing Functions ($f'(x) > 0$)

When the derivative is strictly positive for all $x$ in an interval, the slope of the tangent is always positive. This means the function value $f(x)$ always increases as $x$ increases. There are no points where the graph "levels off" or stays horizontal.

2. Increasing/Non-decreasing Functions ($f'(x) \geq 0$)

If the derivative is greater than or equal to zero, the function never decreases. It may rise ($f'(x) > 0$) or it may remain constant ($f'(x) = 0$) over a sub-interval. This is a broader category that includes strictly increasing functions.

3. Strictly Decreasing Functions ($f'(x) < 0$)

A strictly negative derivative indicates that the tangent at every point makes an obtuse angle with the positive x-axis. As $x$ increases, $f(x)$ consistently decreases without any horizontal pauses.

4. Decreasing/Non-increasing Functions ($f'(x) \leq 0$)

If the derivative is less than or equal to zero, the function never rises. It can either fall or remain at a constant value for some duration. This category includes strictly decreasing functions.

5. Constant Functions ($f'(x) = 0$)

If the derivative is zero throughout an entire interval, the rate of change is zero. The function does not increase or decrease, resulting in a horizontal straight line.

$f(x) = k$

[Where $k$ is a constant]

Special Condition: The Discrete Zero Case

A critical refinement of the "Strictly Increasing" condition is used. If the derivative $f'(x)$ is zero only at isolated points (discrete points) and positive elsewhere, the function remains strictly increasing.

$f'(x) \geq 0$

[Provided $f'(x) = 0$ at finite points only]

This allows us to classify functions like $f(x) = x^3$ as strictly increasing even though $f'(0) = 0$. Because the zero is momentary and doesn't occur over an interval, the strict inequality $f(x_1) < f(x_2)$ is never violated.


Example 1. Find the intervals in which the function $f$ given by $f(x) = x^{2} - 4x + 6$ is:

(a) Strictly increasing

(b) Strictly decreasing

Answer:

Given:

$f(x) = x^{2} - 4x + 6$

Solution:

Differentiating the function with respect to $x$:

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

To find the critical points, we set $f'(x) = 0$:

$2x - 4 = 0 \implies x = 2$

The point $x = 2$ divides the real line into two disjoint intervals, namely $(-\infty, 2)$ and $(2, \infty)$.

Interval Analysis:

1. In the interval $(-\infty, 2)$, let us pick a test point $x = 0$:

$f'(0) = 2(0) - 4 = -4 < 0$

[Function is strictly decreasing]

2. In the interval $(2, \infty)$, let us pick a test point $x = 3$:

$f'(3) = 2(3) - 4 = 2 > 0$

[Function is strictly increasing]

Final Result:

(a) $f$ is strictly increasing in the interval $(2, \infty)$.

(b) $f$ is strictly decreasing in the interval $(-\infty, 2)$.


Example 2. Find the intervals in which the function $f(x) = 2x^{3} - 3x^{2} - 36x + 7$ is strictly increasing or strictly decreasing.

Answer:

Given:

$f(x) = 2x^{3} - 3x^{2} - 36x + 7$

Solution:

Differentiating $f(x)$ with respect to $x$:

$f'(x) = 6x^{2} - 6x - 36$

Factoring the derivative:

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

Setting $f'(x) = 0$ gives critical points $x = -2$ and $x = 3$. These points divide the domain into three intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$.

Sign of $f'(x)$ Table:

Interval Sign of $f'(x) = 6(x-3)(x+2)$ Nature of Function
$(-\infty, -2)$ $(-)(-) = \text{Positive}$ Strictly Increasing
$(-2, 3)$ $(-)(+) = \text{Negative}$ Strictly Decreasing
$(3, \infty)$ $(+)(+) = \text{Positive}$ Strictly Increasing

Conclusion:

The function $f(x)$ is strictly increasing in $(-\infty, -2) \cup (3, \infty)$ and strictly decreasing in $(-2, 3)$.



Maxima and Minima

In calculus, the concept of Maxima and Minima refers to finding the largest and smallest values of a function. These are collectively called the extrema of a function. Understanding these points is essential for optimizing real-world problems, such as finding the maximum profit in a business or the minimum material required to manufacture a container.

Maximum and Minimum Values (Absolute Extrema)

Let $f$ be a real-valued function defined on a domain $D$. We analyze the behavior of the function's output to determine if it reaches a peak or a trough within that domain.

(i) Maximum Value and Point of Maxima

A function $f$ is considered to have an Absolute Maximum value in its domain $D$ if there is a specific point $x = c$ where the function's value is greater than or equal to every other value of $f(x)$ for all $x$ in $D$.

$f(c) \geq f(x), \ \forall \ x \in D$

In this context:

1. The real number $f(c)$ is known as the absolute maximum value (or simply the maximum value) of the function.

2. The coordinate $c$ is the point of maxima. It tells us where the highest value occurs on the x-axis.

Graph showing a peak at point c

(ii) Minimum Value and Point of Minima

Conversely, a function $f$ has an Absolute Minimum value if there exists a point $x = d$ in $D$ such that the function's value at $d$ is the lowest possible output in the entire domain.

$f(d) \leq f(x), \ \forall \ x \in D$

In this context:

1. The real number $f(d)$ is called the absolute minimum value of $f$.

2. The coordinate $d$ is termed the point of minima.

Graph showing a trough at point d

Elaboration on Key Observations

1. Uniqueness of Values vs. Multiplicity of Points

It is important to distinguish between the value and the location. If a maximum exists, the value $f(c)$ is unique—a function cannot have two different "highest" values. However, the function can reach this same unique maximum value at multiple points.

Example: For the function $f(x) = \sin x$ on the interval $[0, 4\pi]$, the maximum value is $1$. This value is unique, but it is attained at two distinct points:

$x = \frac{\pi}{2} \text{ and } x = \frac{5\pi}{2}$

[Multiple points of maxima]

2. Existence of Extrema

Not every function is guaranteed to have a maximum or a minimum. This depends largely on the nature of the function and the type of domain ($D$).

Case A: Infinite growth/decay: Consider the linear function $f(x) = x$ on the domain $\mathbb{R}$. As $x$ increases, $f(x)$ increases without bound. As $x$ decreases, $f(x)$ decreases without bound. Thus, it has no maxima and no minima.

Case B: Open Intervals: Consider $f(x) = x$ on the open interval $(0, 1)$. Although the values are bounded between $0$ and $1$, the function never actually "reaches" $0$ or $1$. Therefore, it has no absolute maximum or minimum on this specific domain.

Summary Table of Extrema Definitions

Term Mathematical Condition Description
Point of Maxima $x = c$ The input value where the peak occurs.
Maximum Value $f(c)$ The highest output (y-value) attained.
Point of Minima $x = d$ The input value where the trough occurs.
Minimum Value $f(d)$ The lowest output (y-value) attained.

Examples and Graphical Interpretations

To understand the practical application of the definitions of maxima and minima, we examine various types of functions: trigonometric, polynomial, and quadratic. These examples illustrate how the domain and the nature of the function dictate the existence of extreme values.

Example 1: Trigonometric Function

In the study of Calculus and Trigonometry, understanding the behaviour of functions over a restricted interval is crucial for identifying local and absolute extrema. Here, we analyze the basic circular functions over the domain $D = [-\pi, 2\pi]$.

The domain specifies the set of all possible input values ($x$) for which the function is defined.

Analysis of the Sine Function ($f(x) = \sin x$)

Consider the function $f(x) = \sin x$ defined in the interval $[-\pi, 2\pi]$. The sine function is a periodic function with a period of $2\pi$.

Extrema of Sine Function

The Maximum Value of the sine function is $1$. In the interval $[-\pi, 2\pi]$, the function attains this value at:

$x = \frac{\pi}{2}$

[Point of Maxima]

$f\left(\frac{\pi}{2}\right) = 1$

(Maximum Value)

The Minimum Value of the sine function is $-1$. Within our specified domain, the function reaches its minimum at two distinct points:

$x = -\frac{\pi}{2}, \frac{3\pi}{2}$

[Points of Minima]

$f\left(-\frac{\pi}{2}\right) = f\left(\frac{3\pi}{2}\right) = -1$

(Minimum Value)

Graph Visualization

Graph of sin x from -pi to 2pi

Analysis of the Cosine Function ($g(x) = \cos x$)

Now, let us evaluate $g(x) = \cos x$ within the same domain $[-\pi, 2\pi]$. The cosine function is an even function, satisfying $g(-x) = g(x)$.

Extrema of Cosine Function

The maximum value of $1$ is achieved at the start, middle, and end of its typical cycles within this range:

$x = 0, 2\pi$

[Points of Maxima]

$g(0) = g(2\pi) = 1$

(Maximum Value)

The minimum value of $-1$ is achieved at:

$x = -\pi, \pi$

[Points of Minima]

$g(-\pi) = g(\pi) = -1$

(Minimum Value)

Graph Visualization of $\cos x$

Graph of cos x from -pi to 2pi showing peaks and zeros

Zeros of the function

The function crosses the x-axis where $\cos x = 0$:

$x = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}$

Tabular Comparison of Values

For quick reference in competitive exams, the following table summarizes the values of $\sin x$ and $\cos x$ at key angles within the domain $[-\pi, 2\pi]$.

Angle ($x$) $\sin x$ $\cos x$
$-\pi$ $0$ $-1$ (Min)
$-\frac{\pi}{2}$ $-1$ (Min) $0$
$0$ $0$ $1$ (Max)
$\frac{\pi}{2}$ $1$ (Max) $0$
$\pi$ $0$ $-1$ (Min)
$\frac{3\pi}{2}$ $-1$ (Min) $0$
$2\pi$ $0$ $1$ (Max)

Observation

Note that while the range of both functions is identical (i.e., $[-1, 1]$), the points at which they attain these values are phase-shifted by $\frac{\pi}{2}$.

Example 2: Polynomial Function ($f(x) = x^2$)

Consider the simplest power function $f(x) = x^2$ defined over the entire set of real numbers $\mathbb{R}$. Geometrically, this function represents a parabola opening upwards with its vertex at the origin $(0, 0)$. It is a classic example of an even function because the square of a negative number is equal to the square of its positive counterpart, i.e., $f(-x) = f(x)$.

Domain and Range Analysis

The Domain of a function represents the set of all possible input values ($x$) for which the function is defined. For the square function, every real number has a valid square, therefore:

$x \in (-\infty, \infty)$ or $\mathbb{R}$

[Domain]

The Range represents the set of all possible output values ($f(x)$). Since the square of any real number (whether positive, negative, or zero) is always non-negative, we have the following property:

$x^2 \geq 0$

[For all $x \in \mathbb{R}$]

Consequently, the output values start from $0$ and extend to infinity:

$f(x) \in [0, \infty)$

[Range]

Analysis of Minima

By observing the property established in the above equation, we can see that the smallest possible value of $f(x)$ is $0$. This value is attained only when the input $x$ is $0$.

$f(0) = 0^2 = 0$

[Minimum Value reached at $x=0$]

Thus, $x = 0$ is the point of minima and $0$ is the absolute minimum value of the function.

Analysis of Maxima

As the input $x$ increases towards $+\infty$ or decreases towards $-\infty$, the squared value $x^2$ grows without any upper bound. There is no real number $M$ such that $x^2 \leq M$ for every $x \in \mathbb{R}$.

$\lim\limits_{x \to \pm\infty} x^2 = \infty$

[Infinite growth]

Therefore, the function has no maxima in the set of real numbers.

Graph of f(x) = x^2 showing an upward opening parabola

Example 3: Quadratic Function with Maxima ($f(x) = 2x - x^2$)

In this section, we examine the quadratic function $f(x) = 2x - x^2$. Unlike the standard square function ($x^2$), this expression includes a linear term and a negative coefficient for the squared term. Geometrically, this function represents a parabola opening downwards.

Mathematical Transformation

To better understand the behaviour of this function, we can rewrite it by completing the square. This allows us to identify the vertex of the parabola easily.

$f(x) = -(x^2 - 2x)$

[Taking negative sign common]

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

[Adding and subtracting 1]

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

Domain and Range Analysis

Domain

Since $f(x)$ is a polynomial function, it is defined for all real values of $x$. In competitive exams, this is often stated as the set of all real numbers.

$x \in \mathbb{R}$

[Domain]

Range

From above equation, we observe that $(x - 1)^2$ is always non-negative. Therefore, $-(x - 1)^2$ must always be less than or equal to zero.

$-(x - 1)^2 \leq 0$

[For all $x \in \mathbb{R}$]

Adding $1$ to both sides of the inequality:

$1 - (x - 1)^2 \leq 1$

This shows that the maximum value the function can attain is $1$, and it can decrease indefinitely. Thus, the range is:

$f(x) \in (-\infty, 1]$

[Range]

Analysis of Maxima and Minima

Points of Maxima

The function reaches its absolute maximum value when the subtracted term $(x-1)^2$ becomes zero. This occurs at $x = 1$.

$f(1) = 2(1) - (1)^2 = 1$

[Maximum Value]

So, the point of maxima is $x = 1$.

Analysis of Minima

As $x$ moves towards $+\infty$ or $-\infty$, the term $x^2$ grows much faster than $2x$, and due to the negative sign, the function values decrease without bound.

$\lim\limits_{x \to \pm\infty} (2x - x^2) = -\infty$

Therefore, this function has no minimum value in the set of real numbers.

Intercepts and Graph

To sketch the graph, we find where the function crosses the x-axis (roots):

$2x - x^2 = 0 \implies x(2 - x) = 0$

[Factoring]

The roots are $x = 0$ and $x = 2$. The vertex is situated exactly midway between the roots at $(1, 1)$.

Graph of 2x - x^2 showing a downward opening parabola with vertex at (1,1)

Summary of Results

The following table provides a comprehensive summary of the domain, range, and extreme values for the trigonometric and polynomial functions analyzed in the previous sections.

Function $f(x)$ Domain ($D$) Point(s) of Extrema Extreme Value Type of Extrema
$\sin x$ $[-\pi, 2\pi]$ $x = \frac{\pi}{2}$ $1$ Absolute Maxima
$x = -\frac{\pi}{2}, \frac{3\pi}{2}$ $-1$ Absolute Minima
$\cos x$ $[-\pi, 2\pi]$ $x = 0, 2\pi$ $1$ Absolute Maxima
$x = -\pi, \pi$ $-1$ Absolute Minima
$x^2$ $\mathbb{R}$ $x = 0$ $0$ Absolute Minima
$x^2$ $\mathbb{R}$ None $\infty$ No Maxima
$2x - x^2$ $\mathbb{R}$ $x = 1$ $1$ Absolute Maxima
$2x - x^2$ $\mathbb{R}$ None $-\infty$ No Minima

Local Maxima and Local Minima

In mathematical analysis, particularly in the study of calculus, the behavior of a function $f$ in the immediate vicinity of a point is of paramount importance. Let $f$ be a real-valued function defined on a domain $D$ (where $D \subseteq \mathbb{R}$). We distinguish between the global behavior of a function and its local or relative behavior.


Mathematical Definitions

(i) Local Maxima

A function $f$ is said to have a local (or relative) maxima at $x = c$ (where $c \in D$) if and only if there exists a positive real number $\delta$ such that the value of the function at $c$ is the greatest among all values in its immediate neighbourhood.

$f(c) \geq f(x)$

$\forall \ x \in (c - \delta, \ c + \delta)$

In this case, $c$ is known as the point of local maxima and $f(c)$ is termed the local maximum value.

(ii) Local Minima

Conversely, a function $f$ is said to have a local (or relative) minima at $x = d$ (where $d \in D$) if and only if there exists a positive real number $\delta$ such that the function value at $d$ is the smallest compared to its neighbours.

$f(d) \leq f(x)$

$\forall \ x \in (d - \delta, \ d + \delta)$

Here, $d$ is called the point of local minima and $f(d)$ is called the local minimum value.

Geometric Interpretation

A function may exhibit several "peaks" and "troughs" within its domain. Identifying which of these are merely local and which is the absolute highest or lowest is a key analytical skill.

Geometrically, a point in the domain of the function $f$ is classified as follows:

Local Maxima (Relative Peaks)

A point $x = c$ is a point of local maxima if the graph forms a "peak" at that point. This means that in the immediate vicinity of $c$, no other point has a higher functional value. A function can have numerous local maxima.

$f(c) \geq f(x)$

[For $x \in (c - \delta, c + \delta)$]

Local Minima (Relative Troughs)

A point $x = d$ is a point of local minima if the graph forms a "trough" or a "cavity." It represents the lowest point in its specific neighbourhood. A function can have numerous local minima.

$f(d) \leq f(x)$

[For $x \in (d - \delta, d + \delta)$]

Absolute Extrema in the Presence of Multiple Local Extrema

While there may be many local peaks and troughs, we often look for the "extremes of the extremes":

Absolute Maxima

Among all the local maxima (peaks) and the values at the boundaries of the domain, the greatest value is called the Absolute Maxima. In a continuous function on a closed interval, there is exactly one absolute maximum value, though it may be attained at more than one point.

Absolute Minima

Similarly, among all the local minima (troughs) and the boundary values, the least value is called the Absolute Minima. There is exactly one absolute minimum value for a function in its entire domain.

Graph with numerous peaks and troughs showing one highest peak and one lowest trough

Tabular Summary of Multiple Extrema

Consider a function $f(x)$ analyzed over a domain where several critical points are identified:

Point on x-axis Geometric Feature Classification Status
$x = x_1$ Peak Local Maxima Ordinary peak
$x = x_2$ Trough Local Minima Ordinary trough
$x = x_3$ Highest Peak Local Maxima Absolute Maxima
$x = x_4$ Trough Local Minima Ordinary trough
$x = x_5$ Peak Local Maxima Ordinary peak
$x = x_6$ Lowest Trough Local Minima Absolute Minima

Important Insight

1. Between any two local maxima, there must lie at least one local minima.

2. The absolute maximum can be a local maximum or can occur at the endpoints of the interval.

$f(x)_{abs.max} = \max\{f(\text{endpoints}), f(\text{local max})\}$

$f(x)_{abs.min} = \min\{f(\text{endpoints}), f(\text{local min})\}$


Analysis of Critical, Turning, and Stationary Points

In the study of Applications of Derivatives, specific points on a curve where the rate of change is zero play a vital role in determining the local behavior of the function. These points are synonymous with the concepts of Stationary or Turning points.


Formal Definition

Let $f$ be a real-valued function defined on a domain $D \subseteq \mathbb{R}$. A point $c \in D$ is called a critical point (or turning or stationary point) of $f$ if and only if the following two conditions are satisfied:

1. The function $f$ is differentiable at $x = c$.

2. The first derivative of the function at that point is zero.

$f'(c) = 0$

Relationship Between Critical and Extreme Points

An extreme point (local maxima or minima) is not always a critical point, and a critical point is not necessarily an extreme point. We examine this through three distinct cases:

Case I: A Critical Point that is an Extreme Point

Consider the trigonometric function $f(x) = \sin x$ for $x \in \mathbb{R}$. The derivative is given by $f'(x) = \cos x$.

$f'\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$

[$\because x = \frac{\pi}{2}$ is a critical point]

Since $\sin\left(\frac{\pi}{2}\right) = 1$ is a local maximum value, here the critical point coincides with an extreme point. Similarly, at $x = -\frac{\pi}{2}$:

$f'\left(-\frac{\pi}{2}\right) = \cos\left(-\frac{\pi}{2}\right) = 0$

[Point of local minima]

Case II: A Critical Point that is NOT an Extreme Point

Consider the cubic function $f(x) = x^3$. This is a strictly increasing function for all $x \in \mathbb{R}$, meaning it has no peaks or troughs.

$f'(x) = 3x^2$

Evaluating the derivative at the origin:

$f'(0) = 3(0)^2 = 0$

[Critical point at $x=0$]

Although $x = 0$ is a critical point, the function continues to increase before and after this point. Hence, it is not an extreme point (this is often called a point of inflection).

Graph of x cubed showing stationary point at origin

Case III: An Extreme Point that is NOT a Critical Point

Consider the modulus function $f(x) = |x|$. This function has a clear local minima at $x = 0$ since $f(0) = 0$ and $f(x) > 0$ for all other $x$.

$f(0) = |0| = 0$

[Local Minimum Value]

However, the function $|x|$ is not differentiable at $x = 0$ because the left-hand and right-hand limits of the derivative do not match (creating a sharp corner). Since differentiability is a requirement for a critical point, $x = 0$ is not a critical point despite being an extreme point.

V-shaped graph of modulus x showing non-differentiable minima

Summary Table of Comparison

The following table summarizes the key differences observed in the examples above:

Function Point ($x$) Critical Point? Extreme Point? Reason
$\sin x$ $\frac{\pi}{2}$ Yes Yes $f'(x)=0$ and peak exists.
$x^3$ $0$ Yes No $f'(x)=0$ but function is strictly increasing.
$|x|$ $0$ No Yes Local minima exists, but function is not differentiable.

Crucial Remarks

1. Condition for Stationary Point: A "Stationary point" is strictly defined by the condition $f'(x) = 0$. Always check if the function is differentiable at that point before labeling it stationary.

2. Necessity vs Sufficiency: $f'(c) = 0$ is a necessary condition for an extreme point if the function is differentiable, but it is not sufficient to guarantee an extreme point (as seen in the $x^3$ example).

3. Search for Extrema: To find all local extrema, you must check both the critical points ($f'(x)=0$) and the points where the function is not differentiable.


Working Rules for Finding (Absolute) Maximum and Minimum

To determine the Absolute (or Global) maximum and minimum values of a function $f$ defined on a closed interval $[a, b]$, we must consider not only the stationary points where the derivative is zero but also the points where the function might not be smooth and the boundaries of the interval itself.


The Procedure

If a function $f$ is differentiable in $[a, b]$ except possibly at finitely many points, the absolute extrema can be found by following these steps:

Step (i): Stationary Points

Evaluate the function $f(x)$ at all points $x \in (a, b)$ where the first derivative is zero.

$f'(x) = 0$

[Critical Points]

Step (ii): Points of Non-Differentiability

Evaluate $f(x)$ at all points in the interval $(a, b)$ where the derivative $f'(x)$ fails to exist (such as corners, cusps, or points of discontinuity).

Step (iii): Boundary Values

Evaluate the function at the endpoints of the interval, namely $f(a)$ and $f(b)$.

Final Conclusion:

The Maximum of all the values collected in the steps above is the Absolute Maximum value, and the Minimum of these values is the Absolute Minimum value of the function $f$.


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

Answer:

Step 1: Finding points where $f'(x) = 0$

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

Setting $f'(x) = 0$:

$6(x^2 - 5x + 6) = 0$

$6(x - 2)(x - 3) = 0$

$\implies x = 2, 3$

Both $x = 2$ and $x = 3$ lie within the interval $[1, 5]$. Now, calculate values at these points:

$f(2) = 2(8) - 15(4) + 36(2) + 1 = 29$

$f(3) = 2(27) - 15(9) + 36(3) + 1 = 28$


Step 2: Check for points where derivative fails to exist

Since $f(x)$ is a polynomial, it is differentiable everywhere. No points to evaluate here.


Step 3: Evaluate at Endpoints

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

$f(5) = 2(125) - 15(25) + 36(5) + 1 = 56$


Conclusion:

Comparing $\{29, 28, 24, 56\}$, we find:

Absolute Maximum Value = $56$ (at $x = 5$)

Absolute Minimum Value = $24$ (at $x = 1$)


Example 2. Find the absolute maximum and minimum values of $f(x) = |x - 2| + 3$ in the interval $[0, 5]$.

Answer:

Step 1: Check $f'(x) = 0$

For $x \neq 2$, $f'(x)$ is either $1$ or $-1$. Thus, there are no points where $f'(x) = 0$.


Step 2: Points where derivative fails to exist

The function $f(x) = |x - 2| + 3$ has a sharp corner at $x = 2$, so $f'(2)$ does not exist.

$f(2) = |2 - 2| + 3 = 3$


Step 3: Evaluate at Endpoints

$f(0) = |0 - 2| + 3 = 5$

$f(5) = |5 - 2| + 3 = 6$


Conclusion:

Comparing values $\{3, 5, 6\}$:

Absolute Maximum Value = $6$ (at $x = 5$)

Absolute Minimum Value = $3$ (at $x = 2$)


Tabular Summary of the Procedure

Step Number Condition / Point Type Action
1 $f'(x) = 0$ Calculate $f(x)$ at these interior points.
2 $f'(x)$ is undefined Calculate $f(x)$ at these interior points.
3 Interval Endpoints Calculate $f(a)$ and $f(b)$.
Final Comparison Identify the largest and smallest results.


Local Maximum and Local Minimum Values

In this section, we address the mathematical problem of identifying the points at which a function achieves its local extreme values. Based on the behavior of the derivative, we classify these into two distinct cases: points where the derivative is non-existent and turning points where the derivative vanishes.


Case I: At points where the derivative fails to exist

In many real-world and mathematical scenarios, a function may have a "sharp corner" or a "cusp." At such points, the tangent cannot be uniquely defined, meaning $f'(c)$ does not exist. However, these points are often candidates for local maxima or local minima.

General Definition of Local Extremum at Non-Differentiable Points

Let $c$ be a point in the domain of a continuous function $f$. We say $f(c)$ is a local extremum if there exists an interval $(c - h, c + h)$ such that $f(c)$ is the greatest or least value in that interval.

1. Local Maxima at Non-Differentiable Points

If $c$ is a point of local maxima where $f'(c)$ is undefined, the function exhibits the following behavior:

1. The function is strictly increasing in the immediate left-neighbourhood of $c$. That is, for a very small $h > 0$:

$f'(x) > 0$

$\forall \ x \in (c - h, c)$

2. The function is strictly decreasing in the immediate right-neighbourhood of $c$. That is, for a very small $h > 0$:

$f'(x) < 0$

$\forall \ x \in (c, c + h)$

Consequently, as we move from left to right through $c$, the sign of the derivative $f'(x)$ changes from Positive (+ve) to Negative (-ve).

Graph showing local maxima at a sharp point

2. Local Minima at Non-Differentiable Points

If $c$ is a point of local minima where $f'(c)$ is undefined, the function exhibits the following behavior:

1. The function is strictly decreasing in the immediate left-neighbourhood of $c$. That is, for $h > 0$:

$f'(x) < 0$

$\forall \ x \in (c - h, c)$

2. The function is strictly increasing in the immediate right-neighbourhood of $c$. That is, for $h > 0$:

$f'(x) > 0$

$\forall \ x \in (c, c + h)$

Consequently, as we move from left to right through $c$, the sign of the derivative $f'(x)$ changes from Negative (-ve) to Positive (+ve).

Summary Table

Nature of Point ($c$) Sign of $f'(x)$ for $x < c$ Sign of $f'(x)$ for $x > c$ Change in Sign
Local Maxima Positive (+) Negative (-) + to -
Local Minima Negative (-) Positive (+) - to +

Example. Examine the function $f(x) = |x|$ for local maxima or local minima at $x = 0$.

Answer:

The given function is $f(x) = |x|$, which can be written as:

$f(x) = \begin{cases} -x & , & x < 0 \\ x & , & x \geq 0 \end{cases}$

Step 1: Check Differentiability at $x = 0$

The derivative $f'(x)$ for $x \neq 0$ is:

$f'(x) = \begin{cases} -1 & , & x < 0 \\ 1 & , & x > 0 \end{cases}$

Since the left-hand derivative ($-1$) is not equal to the right-hand derivative ($1$), $f'(0)$ does not exist.

Step 2: Apply the First Derivative Test (Case I)

To the left of $x = 0$ (in the neighbourhood):

$f'(x) = -1$

(Negative)

To the right of $x = 0$ (in the neighbourhood):

$f'(x) = 1$

(Positive)

Since $f'(x)$ changes sign from negative to positive as $x$ increases through $0$, by the definition of Case I:

$x = 0$ is a point of Local Minima.

The local minimum value is $f(0) = |0| = 0$.


Case II: At Turning Points (First Derivative Test)

In this case, we consider points where the function $f(x)$ is smooth and differentiable. A turning point (also known as a stationary point) is a point $c$ in the domain of $f$ where the tangent to the curve is horizontal. This occurs when the instantaneous rate of change is zero.

$f'(c) = 0$

[Condition for Stationary Point]

At these points, we apply the First Derivative Test to determine the nature of the extremum by observing the sign of $f'(x)$ in the immediate neighbourhood of $c$.

1. Local Maxima at Turning Points

If $c$ is a point of local maxima, the function reaches a peak at $x = c$. For this to happen:

(a) The function must be increasing before reaching $c$, meaning $f'(x) > 0$ for $x < c$.

(b) The function must be decreasing after passing $c$, meaning $f'(x) < 0$ for $x > c$.

Conclusion: $f'(x)$ changes its sign from positive to negative as $x$ increases through $c$.

Graph of a smooth curve reaching a maximum value where tangent is horizontal

2. Local Minima at Turning Points

If $c$ is a point of local minima, the function reaches a valley at $x = c$. For this to happen:

(a) The function must be decreasing before reaching $c$, meaning $f'(x) < 0$ for $x < c$.

(b) The function must be increasing after passing $c$, meaning $f'(x) > 0$ for $x > c$.

Conclusion: $f'(x)$ changes its sign from negative to positive as $x$ increases through $c$.

Point of Inflection

It is important to note that $f'(c) = 0$ is a necessary but not sufficient condition for a local extremum. If $f'(x)$ does not change sign as $x$ increases through $c$ (i.e., it remains positive on both sides or negative on both sides), then the point $c$ is called a Point of Inflection.

Summary of First Derivative Test (Turning Points)

Change in Sign of $f'(x)$ Nature of Point $c$ Geometrical Shape
Positive ($+$) $\rightarrow$ Zero ($0$) $\rightarrow$ Negative ($-$) Local Maxima Concave Downwards ($\cap$)
Negative ($-$) $\rightarrow$ Zero ($0$) $\rightarrow$ Positive ($+$) Local Minima Concave Upwards ($\cup$)
No Change (e.g., $+$ to $+$) Point of Inflection Stair-step like curve

Example. Find all points of local maxima and minima for the function $f(x) = x^3 - 3x + 3$.

Answer:

Step 1: Find the first derivative

$f(x) = x^3 - 3x + 3$

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

Step 2: Find the Turning Points

Set $f'(x) = 0$ to find critical values:

$3x^2 - 3 = 0$

$3(x^2 - 1) = 0$

$(x - 1)(x + 1) = 0$

$x = 1, -1$

[Turning Points]

Step 3: Test the point $x = -1$

Take a value slightly less than $-1$ (e.g., $-1.1$): $f'(-1.1) = 3(-1.1)^2 - 3 = 3(1.21) - 3 = 0.63$ (+ve).

Take a value slightly more than $-1$ (e.g., $-0.9$): $f'(-0.9) = 3(-0.9)^2 - 3 = 3(0.81) - 3 = -0.57$ (-ve).

Since $f'(x)$ changes from +ve to -ve, $x = -1$ is a Point of Local Maxima.

Step 4: Test the point $x = 1$

Take a value slightly less than $1$ (e.g., $0.9$): $f'(0.9) = 3(0.81) - 3 = -0.57$ (-ve).

Take a value slightly more than $1$ (e.g., $1.1$): $f'(1.1) = 3(1.21) - 3 = 0.63$ (+ve).

Since $f'(x)$ changes from -ve to +ve, $x = 1$ is a Point of Local Minima.


Summary of the First Derivative Test

The following table summarizes the behavior of a function near a critical point $c$ using the first derivative:

Sign of $f'(x)$ for $x < c$ Sign of $f'(x)$ for $x > c$ Nature of Point $c$
Positive ($+$) Negative ($-$) Local Maximum
Negative ($-$) Positive ($+$) Local Minimum
Positive ($+$) Positive ($+$) Point of Inflection
Negative ($-$) Negative ($-$) Point of Inflection

Second Derivative Test

The Second Derivative Test is a more efficient and systematic method used to determine the nature of critical points. Unlike the First Derivative Test, which requires checking the sign of the derivative on both sides of a point, this test utilizes the concavity of the function at the stationary point itself.

Theorem: Let $f$ be a real-valued function defined on an interval $I$ and let $c \in I$. Let $f$ be twice differentiable at $c$. Then:

(i) Condition for Local Maxima

If $f'(c) = 0$ and $f''(c) < 0$, then $x = c$ is a point of local maxima. The value $f(c)$ is the local maximum value of $f$.

(ii) Condition for Local Minima

If $f'(c) = 0$ and $f''(c) > 0$, then $x = c$ is a point of local minima. The value $f(c)$ is the local minimum value of $f$.

$f''(c) \neq 0$

[Condition for Conclusive Test]

Note: If $f'(c) = 0$ and $f''(c) = 0$, the test fails. In such a situation, the point $c$ could be a local maxima, a local minima, or a point of inflection. We must then revert to the First Derivative Test to investigate the sign change of $f'(x)$.

Derivation and Mathematical Intuition

The logic behind the second derivative test is rooted in the concept of the Rate of Change of the Slope.

1. Intuition for Local Maxima ($f''(c) < 0$)

At a local maximum, the graph is "concave downwards" (shaped like an inverted bowl $\cap$). This means the slope $f'(x)$ is decreasing as it passes through the point $c$. Since $f''(x)$ represents the derivative of $f'(x)$, a decreasing slope implies that its derivative must be negative.

$\frac{d}{dx}[f'(x)] < 0$

(At Maxima)

2. Intuition for Local Minima ($f''(c) > 0$)

At a local minimum, the graph is "concave upwards" (shaped like a bowl $\cup$). This means the slope $f'(x)$ is increasing as it passes through the point $c$ (changing from negative to zero to positive). An increasing slope implies that the second derivative is positive.

$\frac{d}{dx}[f'(x)] > 0$

(At Minima)

Example. Find the local maximum and local minimum values of the function $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.

Answer:

Step 1: Differentiation

Given function: $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$

First derivative:

$f'(x) = 12x^3 + 12x^2 - 24x$

Second derivative:

$f''(x) = 36x^2 + 24x - 24$

Step 2: Finding Critical Points

Set $f'(x) = 0$:

$12x^3 + 12x^2 - 24x = 0$

$12x(x^2 + x - 2) = 0$

$12x(x + 2)(x - 1) = 0$

$x = 0, -2, 1$

[Critical Points]

Step 3: Applying Second Derivative Test

Case (a): At $x = 0$

$f''(0) = 36(0)^2 + 24(0) - 24 = -24$

$f''(0) < 0$

$\therefore$ Local Maxima at $x=0$

Local Maximum Value $= f(0) = 3(0)^4 + 4(0)^3 - 12(0)^2 + 12 = \mathbf{12}$.

Case (b): At $x = -2$

$f''(-2) = 36(-2)^2 + 24(-2) - 24$

$f''(-2) = 36(4) - 48 - 24 = 144 - 72 = 72$

$f''(-2) > 0$

$\therefore$ Local Minima at $x=-2$

Local Minimum Value $= f(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 + 12$

$f(-2) = 3(16) + 4(-8) - 12(4) + 12 = 48 - 32 - 48 + 12 = \mathbf{-20}$.

Case (c): At $x = 1$

$f''(1) = 36(1)^2 + 24(1) - 24 = 36$

$f''(1) > 0$

$\therefore$ Local Minima at $x=1$

Local Minimum Value $= f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 + 12 $$ = 3 + 4 - 12 + 12 = \mathbf{7}$.


Points of Inflexion

In the study of curves, we have observed points where a function reaches its highest or lowest values (Local Maxima and Minima). However, there is another critical characteristic of a curve known as Concavity. A Point of Inflexion is a precise location on a curve where the nature of concavity changes—meaning the curve transitions from being "cup-shaped" (concave upwards) to "cap-shaped" (concave downwards), or vice versa.

Geometrical Interpretation

Geometrically, a point $P(c, f(c))$ is identified as a point of inflexion based on its relationship with the tangent line at that point. At a local maximum or minimum, the curve stays entirely on one side of the tangent. In contrast, at a point of inflexion:

1. The curve crosses its tangent at that specific point.

2. On one side of the point $P$, the curve lies below the tangent (Concave Downwards).

3. On the other side of the point $P$, the curve lies above the tangent (Concave Upwards).

A curve crossing its tangent line at a point of inflexion

The Mathematical Theorem for Inflexion

To identify these points analytically without relying solely on graphs, we use the Higher-Order Derivative Test. The theorem provides the sufficient conditions for a point to be an inflexion point.

Theorem Statement: A function $f$ (or the curve $y = f(x)$) possesses a point of inflexion at $x = c$ if and only if the following conditions are satisfied:

(i) The Horizontal Tangent Condition (For Stationary Inflexion)

The first derivative must be zero if we are looking for a stationary point of inflexion:

$f'(c) = 0$

(ii) The Concavity Condition

The second derivative, which represents the rate of change of the slope, must be zero at that point. This indicates a potential transition in concavity:

$f''(c) = 0$

[Necessary condition]

(iii) The Non-Zero Third Derivative Condition

For the concavity to actually change, the third derivative must not be zero. This ensures that the second derivative is not just zero but is actually changing sign at $x = c$:

$f'''(c) \neq 0$

[Sufficient condition]

Detailed Derivation and Logic

Why do we require $f''(c) = 0$ and $f'''(c) \neq 0$? Let's break down the logic:

1. The Role of the Second Derivative

The second derivative $f''(x)$ tells us about the concavity. If $f''(x) > 0$, the curve is concave up ($\cup$). If $f''(x) < 0$, it is concave down ($\cap$). For a point to be an inflexion point, $f''(x)$ must change sign. According to the Intermediate Value Theorem, for a continuous $f''(x)$ to change from positive to negative, it must pass through zero.

2. The Significance of the Third Derivative

If $f'''(c) \neq 0$, it means that $f''(x)$ is either increasing or decreasing at $x = c$. Since $f''(c) = 0$ and it is either increasing or decreasing, it must be negative on one side of $c$ and positive on the other. This guarantees the change in concavity required for an inflexion point.

Relationship with the Gradient (Slope)

An interesting property of the point of inflexion is its relationship with the "steepness" of the curve. Since $f''(c) = 0$ and $f'''(c) \neq 0$, the point $x = c$ is actually a local maximum or minimum for the first derivative $f'(x)$.

Conclusion: At the point of inflexion, the gradient (slope) of the curve $y = f(x)$ is either at its maximum possible value or its minimum possible value in that locality. In Indian competitive exams, this is often referred to as the point where the rate of change is extreme.

Comparison of Critical Points

Feature Local Maxima Local Minima Point of Inflexion
$f'(c)$ $0$ $0$ $0$ (for stationary)
$f''(c)$ $< 0$ (Negative) $> 0$ (Positive) $= 0$ (Zero)
$f'''(c)$ N/A N/A $\neq 0$ (Non-zero)
Concavity Concave Down Concave Up Changes Sign

Students should be aware that $f'(c) = 0$ is not mandatory for a point of inflexion. If $f''(c) = 0$ and $f'''(c) \neq 0$, the point is an inflexion point regardless of the value of the first derivative. These are called Non-Stationary Points of Inflexion (e.g., in the function $f(x) = \sin x$ at $x = \pi$).

The Odd-Order Derivative Rule

In a more general sense, if the first non-zero derivative at a point $c$ (where $f''(c)=0$) is of an odd order (like $3^{rd}, 5^{th}, 7^{th}$ etc.), then $x = c$ is a point of inflexion. If the first non-zero derivative is of an even order, it remains a local maximum or minimum.

Step-by-Step Identification Process:

1. Find $f''(x)$ and equate it to zero to find potential points.

2. Verify if $f''(x)$ changes sign when passing through $x = c$.

3. Alternatively, check if $f'''(c) \neq 0$.

Example. Examine the function $f(x) = x^3$ for points of inflexion.

Answer:

Given function: $f(x) = x^3$

Step 1: Successive Differentiation

Differentiating with respect to $x$:

$f'(x) = 3x^2$

Differentiating again:

$f''(x) = 6x$

Differentiating for the third time:

$f'''(x) = 6$

Step 2: Testing at $x = 0$

At $x = 0$, we observe the values of the derivatives:

$f'(0) = 3(0)^2 = 0$

$f''(0) = 6(0) = 0$

$f'''(0) = 6 \neq 0$

Step 3: Conclusion

Since $f'(0) = 0$, $f''(0) = 0$ and $f'''(0) \neq 0$, the conditions of the theorem are satisfied.

Therefore, $x = 0$ is a point of inflexion of the function $f(x) = x^3$.


Rules to Find Points of Local Maxima and Minima

To determine the extreme values of a function, we follow a systematic procedure that covers all possible candidates for maxima and minima. This process involves identifying critical points and testing them using various derivative methods.

Step 1: Identifying Candidate Points

The first step is to locate all points where the function $f$ is likely to have extreme values. These points are categorized into three groups:

(i) Points of Non-Differentiability: The points where the derivative $f'(x)$ fails to exist.

(ii) Turning Points: The critical points where the derivative is zero i.e., $f'(x) = 0$.

(iii) Endpoints: If the function $f$ is defined in a closed interval $[a, b]$, the endpoints $a$ and $b$ must be considered.

$x \in \{ \text{Non-diff points, Turning points, $$ Endpoints} \}$

[Set of potential extrema]

Step 2: Testing Points where $f'(c)$ Does Not Exist

If $f'(c)$ is undefined but $f'(x)$ exists in the immediate neighbourhood of $c$, we use the First Derivative Test to determine the nature of the point:

$x$ slightly $< c$ $x$ slightly $> c$ Nature of Point $c$
$f'(x)$ is +ve $f'(x)$ is -ve Maxima
$f'(x)$ is -ve $f'(x)$ is +ve Minima

Step 3: General Higher-Order Derivative Test

If $c$ is a turning point ($f'(c) = 0$), and we need a more robust test than the second derivative test (specifically when $f''(c) = 0$), we use the $n^{th}$ Derivative Test.

Let $n \geq 2$ be the smallest positive integer such that the $n^{th}$ derivative at $c$ is non-zero, i.e., $f^{(n)}(c) \neq 0$.

Value of $n$ Sign of $f^{(n)}(c)$ Nature of turning point $c$
Odd ($3, 5, 7...$) +ve or -ve Neither Maxima nor Minima (Inflexion)
Even ($2, 4, 6...$) +ve ($>0$) Minima
Even ($2, 4, 6...$) -ve ($<0$) Maxima

Step 4: Alternative First Derivative Test for Turning Points

If calculating higher-order derivatives is difficult, we analyze the sign of $f'(x)$ around the turning point $c$ ($f'(c) = 0$):

$f'(x)$ slightly $< c$ $f'(x)$ slightly $> c$ Nature of turning point $c$
+ve -ve Maxima
-ve +ve Minima
+ve +ve Neither Maxima nor Minima
-ve -ve Neither Maxima nor Minima

Step 5: Behaviour at Endpoints

When a function is restricted to a closed interval $[a, b]$, the boundaries can be extreme points even if the derivative isn't zero there.

I. Left Endpoint ($a$)

$x$ slightly $> a$ Nature of point $a$
$f'(x)$ is +ve Minima
$f'(x)$ is -ve Maxima

II. Right Endpoint ($b$)

$x$ slightly $< b$ Nature of point $b$
$f'(x)$ is +ve Maxima
$f'(x)$ is -ve Minima

Step 6: Absolute Maximum and Minimum

To find the Absolute (Global) extreme values in a closed interval $[a, b]$:

1. Calculate the values of $f(x)$ at all non-differentiable points, turning points, and endpoints.

2. Compare all these values.

3. The greatest value among them is the Absolute Maximum.

4. The least value among them is the Absolute Minimum.


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

Answer:

Step 1: Find $f'(x)$ and Critical Points

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

Setting $f'(x) = 0$:

$6(x^2 - 5x + 6) = 0 \implies 6(x-2)(x-3) = 0$

Turning points are $x = 2$ and $x = 3$.

Step 2: Identify all points to check

Points are: $1$ (Left endpoint), $2$ (Turning point), $3$ (Turning point), $5$ (Right endpoint).

Step 3: Evaluate $f(x)$ at these points

$f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 24$

$f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$

$f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$

$f(5) = 2(125) - 15(25) + 36(5) + 1 = 250 - 375 + 180 + 1 = 56$

Conclusion:

The Absolute Maximum value is $\mathbf{56}$ at $x = 5$.

The Absolute Minimum value is $\mathbf{24}$ at $x = 1$.


Example 2. Find the absolute maximum and absolute minimum values of the function $f(x) = \sin x + \cos x$ defined on the interval $[0, \pi]$.

Answer:

Step 1: Find the derivative

$f(x) = \sin x + \cos x$

$f'(x) = \cos x - \sin x$

Step 2: Locate Turning Points

For turning points, set $f'(x) = 0$:

$\cos x - \sin x = 0$

$\sin x = \cos x$

$\tan x = 1$

$x = \frac{\pi}{4}$

[Since $x \in [0, \pi]$]

Step 3: Evaluate function at critical points and endpoints

The points to check are $x = 0$ (left endpoint), $x = \frac{\pi}{4}$ (turning point), and $x = \pi$ (right endpoint).

1. At $x = 0$:

$f(0) = \sin 0 + \cos 0 = 0 + 1 = 1$

2. At $x = \frac{\pi}{4}$:

$f\left(\frac{\pi}{4}\right) = \sin\frac{\pi}{4} + \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$

3. At $x = \pi$:

$f(\pi) = \sin \pi + \cos \pi = 0 + (-1) = -1$

Conclusion:

Comparing the values $\{1, \sqrt{2}, -1\}$:

The Absolute Maximum value is $\mathbf{\sqrt{2}}$ at $x = \frac{\pi}{4}$.

The Absolute Minimum value is $\mathbf{-1}$ at $x = \pi$.


Example 3. Using the $n^{th}$ derivative test, find the nature of the turning point for $f(x) = (x - 2)^4$.

Answer:

Step 1: Successive Differentiation

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

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

$f''(x) = 12(x - 2)^2$

$f'''(x) = 24(x - 2)$

$f^{iv}(x) = 24$

Step 2: Find the turning point

Set $f'(x) = 0 \implies 4(x - 2)^3 = 0 \implies x = 2$.

Step 3: Apply $n^{th}$ Derivative Test

At $x = 2$:

$f''(2) = 12(2 - 2)^2 = 0$

$f'''(2) = 24(2 - 2) = 0$

$f^{iv}(2) = 24$

Here, the smallest $n$ for which $f^{(n)}(2) \neq 0$ is $n = 4$.

$n = 4$

(Even number)

$f^{iv}(2) = 24 > 0$

(Positive value)

Conclusion:

Since $n$ is even and $f^{(n)}(c) > 0$, the point $x = 2$ is a point of Local Minima.


Example 4. Examine the function $f(x) = |x - 3| + 2$ for absolute extrema on the interval $[1, 5]$.

Answer:

This function involves a point where the derivative fails to exist (Case I of working rules).

Step 1: Identify critical points

The function can be written as:

$f(x) = \begin{cases} -(x - 3) + 2 = 5 - x & , & 1 \leq x < 3 \\ (x - 3) + 2 = x - 1 & , & 3 \leq x \leq 5 \end{cases}$

The derivative $f'(x)$ is $-1$ for $x < 3$ and $+1$ for $x > 3$. Thus, $f'(3)$ does not exist.

Step 2: Evaluate at all candidate points

Candidates: Endpoints $\{1, 5\}$ and non-differentiable point $\{3\}$.

1. At $x = 1$ (Left endpoint):

$f(1) = |1 - 3| + 2 = 2 + 2 = 4$

2. At $x = 3$ (Critical point):

$f(3) = |3 - 3| + 2 = 0 + 2 = 2$

3. At $x = 5$ (Right endpoint):

$f(5) = |5 - 3| + 2 = 2 + 2 = 4$

Step 3: Analyze behaviour at $x = 3$ (First Derivative Test)

$x$ range $f'(x)$ sign
Slightly $< 3$ -ve ($-1$)
Slightly $> 3$ +ve ($1$)

Since $f'(x)$ changes from negative to positive, $x = 3$ is a local minimum.

Final Conclusion:

The Absolute Maximum value is $\mathbf{4}$ (attained at both $x = 1$ and $x = 5$).

The Absolute Minimum value is $\mathbf{2}$ (attained at $x = 3$).



Problems on Maxima and Minima

Practical problems involve real-world scenarios where we need to find the optimal solution—such as minimizing the cost of production, maximizing the volume of a container, or minimizing the material required for construction. These problems are solved by translating the given verbal conditions into mathematical functions.


The Single Extreme Point Theorem

A crucial rule often applied in practical optimization problems relates to functions defined on open intervals. In real-life applications, we often deal with physical dimensions (like length or radius) that must be positive, leading to open intervals such as $(0, \infty)$.

Theorem Statement: If a function $f$ is continuous in an open interval $(a, b)$ and it possesses only one extreme point (critical point) within that interval, then:

1. If the point is a local maxima, it is also the absolute maxima in $(a, b)$.

2. If the point is a local minima, it is also the absolute minima in $(a, b)$.

This theorem is exceptionally useful because it eliminates the need to check boundary values when only one critical point exists in the domain of interest.


Steps to Solve Practical Problems

To solve optimization word problems, follow these systematic steps:

Step 1: Identify the variable to be maximized or minimized (the objective function).

Step 2: Identify the given constant and the relationship between variables (the constraint).

Step 3: Express the objective function in terms of a single variable using the constraint.

Step 4: Differentiate the function and set it to zero ($f'(x) = 0$) to find the critical point.

Step 5: Use the Second Derivative Test ($f''(x)$) to confirm whether the point is a maximum or minimum.


Example 1. Show that of all the rectangles with a given perimeter, the square has the maximum area.

Answer:

Given: A rectangle with a fixed perimeter $P$. Let the sides be $x$ and $y$.

To Prove: The area $A$ is maximum when $x = y$ (i.e., the rectangle is a square).

Step 1: Establishing relationships

The perimeter $P$ is constant:

$P = 2(x + y)$

[Formula for Perimeter]           ... (i)

From equation (i), we can express $y$ in terms of $x$:

$y = \frac{P}{2} - x$

... (ii)

Step 2: Formulate the Objective Function

Area $A = x \cdot y$. Substituting the value of $y$ from (ii):

$A(x) = x \left( \frac{P}{2} - x \right) = \frac{Px}{2} - x^2$

... (iii)

Step 3: Differentiation for Maxima/Minima

Differentiating $A$ with respect to $x$:

$\frac{dA}{dx} = \frac{P}{2} - 2x$

... (iv)

Setting $\frac{dA}{dx} = 0$ for critical points:

$\frac{P}{2} - 2x = 0 \implies 2x = \frac{P}{2} \implies x = \frac{P}{4}$

Step 4: Verification using Second Derivative

Differentiating equation (iv) again:

$\frac{d^2A}{dx^2} = -2$

... (v)

Since $\frac{d^2A}{dx^2} = -2 < 0$, the area is maximum at $x = \frac{P}{4}$.

Step 5: Finding the other dimension

Substitute $x = \frac{P}{4}$ in equation (ii):

$y = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}$

Since $x = y = \frac{P}{4}$, the rectangle is a square.


Example 2. Find two positive numbers $x$ and $y$ such that their sum is $35$ and the product $x^2 y^5$ is maximum.

Answer:

Given: Sum of two numbers is $35$.

$x + y = 35$

(Given)

From the above, we can write:

$y = 35 - x$

... (i)

To Find: Values of $x$ and $y$ that maximize $P = x^2 y^5$.

Substituting $y$ from (i) into the product function:

$P(x) = x^2 (35 - x)^5$

Solution: Differentiating $P$ with respect to $x$ using product rule:

$P'(x) = 2x(35 - x)^5 + x^2 \cdot 5(35 - x)^4 \cdot (-1)$

$P'(x) = x(35 - x)^4 [2(35 - x) - 5x]$

$P'(x) = x(35 - x)^4 [70 - 7x]$

For maxima/minima, set $P'(x) = 0$:

$x(35 - x)^4 (70 - 7x) = 0$

This gives $x = 0$ (not possible for positive numbers), $x = 35$ (not possible), or $70 - 7x = 0 \implies x = 10$.

Applying the First Derivative Test at $x=10$:

When $x < 10$, $P'(x) > 0$ (Increasing)

When $x > 10$, $P'(x) < 0$ (Decreasing)

Since $P'(x)$ changes from positive to negative, $x = 10$ is a point of local maxima.

Now, find $y$:

$y = 35 - 10 = 25$

The two required numbers are 10 and 25.


Example 3. A square piece of tin of side $18 \text{ cm}$ is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is maximum?

Answer:

Let the side of the square to be cut off from each corner be $x \text{ cm}$.

The dimensions of the resulting open box will be:

Length ($L$) $= 18 - 2x$

Breadth ($B$) $= 18 - 2x$

Height ($H$) $= x$

Volume Function:

$V(x) = L \cdot B \cdot H$

$V(x) = (18 - 2x)^2 \cdot x$

$V(x) = x(324 - 72x + 4x^2) = 4x^3 - 72x^2 + 324x$

Differentiation:

$V'(x) = 12x^2 - 144x + 324$

$V''(x) = 24x - 144$

For critical points, $V'(x) = 0$:

$12(x^2 - 12x + 27) = 0$

$12(x - 9)(x - 3) = 0$

Therefore, $x = 9$ or $x = 3$.

Verification:

If $x = 9$, the side length $(18 - 2x)$ becomes $0$, which is impossible.

For $x = 3$:

$V''(3) = 24(3) - 144 = 72 - 144 = -72 < 0$

Since $V''(3) < 0$, the volume is maximum when $x = 3$.

The side of the square to be cut off is 3 cm.


Example 4. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water of volume $V$. The cost of the material for the base is $\textsf{₹} \ 60$ per sq. metre and for the sides is $\textsf{₹} \ 30$ per sq. metre. Find the least cost of construction.

Answer:

Let the side of the square base be $x$ and the height of the tank be $h$.

$V = x^2 h$

(Given constant Volume)

From this, we get:

$h = \frac{V}{x^2}$

... (i)

Cost Function ($C$):

Cost $=$ (Area of Base $\times$ Rate) $+$ (Area of 4 Sides $\times$ Rate)

$C = (x^2 \times 60) + (4xh \times 30)$

$C = 60x^2 + 120xh$

Substitute $h$ from equation (i):

$C(x) = 60x^2 + 120x \left( \frac{V}{x^2} \right)$

$C(x) = 60x^2 + \frac{120V}{x}$

Minimizing Cost:

$C'(x) = 120x - \frac{120V}{x^2}$

Set $C'(x) = 0$:

$120x = \frac{120V}{x^2} \implies x^3 = V \implies x = V^{1/3}$

Second Derivative Test:

$C''(x) = 120 + \frac{240V}{x^3}$

At $x = V^{1/3}$, $C''(x) = 120 + \frac{240V}{V} = 360 > 0$.

The cost is minimum when $x = V^{1/3}$ and $h = \frac{V}{(V^{1/3})^2} = V^{1/3}$.

Thus, for least cost, the dimensions are $x = h = \sqrt[3]{V}$.


Example 5. Show that a closed right circular cylinder of a given volume and least surface area is such that its height is equal to its diameter.

Answer:

Given: Volume $V$ is constant.

$V = \pi r^2 h$

[Fixed Volume]           ... (i)

To Minimize: Total Surface Area ($S$):

$S = 2\pi r^2 + 2\pi rh$

Substitute $h = \frac{V}{\pi r^2}$ from (i) into $S$:

$S(r) = 2\pi r^2 + 2\pi r \left( \frac{V}{\pi r^2} \right) = 2\pi r^2 + \frac{2V}{r}$

Differentiation:

$\frac{dS}{dr} = 4\pi r - \frac{2V}{r^2}$

For minimum $S$, set $\frac{dS}{dr} = 0$:

$4\pi r = \frac{2V}{r^2} \implies 2\pi r^3 = V$

Substitute $V = \pi r^2 h$ back:

$2\pi r^3 = \pi r^2 h$

$2r = h$

(Dividing by $\pi r^2$)

Verification:

$\frac{d^2S}{dr^2} = 4\pi + \frac{4V}{r^3} > 0$ for all $r > 0$.

Since the second derivative is positive, the surface area is minimum when height $h = 2r$ (Diameter).