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Applied Mathematics for Class 11th & 12th (Concepts and Questions)
11th Concepts Questions
12th Concepts Questions

Applied Maths Class 12th Chapters (Concepts)
1. Numbers, Quantification and Numerical Applications 2. Matrices 3. Differentiation and Its Applications
4. Integration and Its Application 5. Differential Equations and Modeling 6. Probability Distribution
7. Inferential Statistics 8. Index Numbers and Time Based Data 9. Financial Mathematics
10. Linear Programming

Content On This Page
Higher Order Derivatives Application of Derivatives Marginal Cost and Marginal Revenue using Derivatives
Increasing /Decreasing Functions Maxima and Minima


Chapter 3 Differentiation and Its Applications (Concepts)

Welcome to this pivotal chapter exploring the fundamentals of Differential Calculus and its powerful applications, particularly within the context of Applied Mathematics. Differential calculus is the mathematical study of rates of change and slopes of curves. It provides indispensable tools for analyzing how quantities change in relation to one another, optimizing processes, and modeling dynamic systems across diverse fields like economics, business, finance, physics, engineering, and biology. While we may briefly revisit the foundational limit definition, the primary goal here is to achieve proficiency in calculating derivatives using established rules and, more importantly, to understand and apply the derivative concept to solve meaningful real-world problems, especially those involving economic analysis and optimization.

We begin by reinforcing the interpretation of the derivative, denoted as $f'(x)$ or $\frac{dy}{dx}$, as the instantaneous rate of change of a function $y=f(x)$ with respect to its variable $x$, or geometrically, as the slope of the tangent line to the graph of the function at a point $x$. While the definition using the first principle ($f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$) underpins the concept, practical differentiation relies heavily on mastering a set of powerful rules of differentiation. Our focus will be on the efficient and accurate application of:

We will extensively practice applying these rules to find derivatives of standard function types crucial in applications: polynomial, exponential ($e^x, a^x$), and logarithmic ($\ln x, \log_a x$) functions, often combined in complex expressions. The chapter may also introduce techniques for implicit differentiation (used when the relationship between $x$ and $y$ is given implicitly by an equation) and potentially differentiation of functions defined in parametric form.

Understanding higher-order derivatives, particularly the second derivative ($f''(x)$ or $\frac{d^2y}{dx^2}$), is introduced. Its significance often relates to analyzing the concavity of a function's graph (whether it curves upwards or downwards) and is a key component of the second derivative test for classifying extrema.

The true power of differential calculus lies in its applications, which form the core of this chapter. We explore:

This chapter equips you with the essential calculus tools for analyzing change and finding optimal solutions, critical skills for quantitative analysis in numerous applied disciplines.



Higher Order Derivatives


Differentiation is a fundamental concept in calculus that allows us to find the instantaneous rate of change of a function. Given a function $y = f(x)$, the process of differentiation with respect to the independent variable $x$ yields the first derivative, denoted by $f'(x)$, $\frac{dy}{dx}$, or $y'$. This first derivative represents the slope of the tangent line to the curve $y = f(x)$ at any point $(x, y)$, or the rate at which $y$ is changing with respect to $x$.

Often, we need to understand how the rate of change itself is changing. This leads to the concept of higher-order derivatives, which are obtained by repeatedly differentiating the function. We differentiate the first derivative to get the second derivative, differentiate the second derivative to get the third derivative, and so on.


Second Order Derivative

The derivative of the first derivative $f'(x)$ with respect to $x$ is called the second derivative of the function $f(x)$. It measures the rate of change of the first derivative. In physical terms, if the first derivative represents velocity, the second derivative represents acceleration.

Notation for the second derivative of $y = f(x)$:

Mathematically, the second derivative is defined as the derivative of the first derivative:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

... (i)


Third Order Derivative

The derivative of the second derivative $f''(x)$ with respect to $x$ is called the third derivative of the function $f(x)$. It measures the rate of change of the second derivative.

Notation for the third derivative of $y = f(x)$:

Mathematically, the third derivative is defined as the derivative of the second derivative:

$\frac{d^3y}{dx^3} = \frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)$

... (ii)


n-th Order Derivative

We can continue this process to find derivatives of any positive integer order $n$. The n-th order derivative of a function $f(x)$, denoted by $f^{(n)}(x)$, $\frac{d^ny}{dx^n}$, or $y^{(n)}$, is obtained by differentiating the function $n$ times successively.

For derivatives beyond the third order (i.e., $n \ge 4$), the prime notation ($f', f'', f'''$) becomes cumbersome and is usually replaced by placing the order of the derivative in parentheses as a superscript, e.g., $f^{(4)}(x)$, $f^{(5)}(x)$, $f^{(n)}(x)$.

The n-th derivative is formally defined recursively as the derivative of the $(n-1)$-th derivative:

$\frac{d^ny}{dx^n} = \frac{d}{dx}\left(\frac{d^{n-1}y}{dx^{n-1}}\right)$, for $n \ge 2$

... (iii)

where $\frac{d^{1}y}{dx^{1}} = \frac{dy}{dx}$ is the first derivative.


Physical Interpretation of Higher Order Derivatives

Higher-order derivatives have direct interpretations in various physical contexts. One common example is in kinematics, which deals with the motion of objects.

If $s(t)$ represents the position of an object at time $t$:

The first derivative, $\frac{ds}{dt} = v(t)$, represents the velocity of the object. Velocity is the rate of change of position with respect to time.

The second derivative, $\frac{d^2s}{dt^2} = \frac{d}{dt}\left(\frac{ds}{dt}\right) = \frac{dv}{dt} = a(t)$, represents the acceleration of the object. Acceleration is the rate of change of velocity with respect to time. It describes how quickly the velocity is changing.

The third derivative, $\frac{d^3s}{dt^3} = \frac{d}{dt}\left(\frac{d^2s}{dt^2}\right) = \frac{da}{dt} = j(t)$, is sometimes called the jerk. Jerk represents the rate of change of acceleration.

Higher derivatives describe increasingly complex aspects of motion. For instance, the fourth derivative is sometimes called 'snap' or 'jounce', the fifth 'crackle', and the sixth 'pop'. These terms are less commonly used in basic kinematics but illustrate the concept of higher-order rates of change.


Examples

Example 1. Find the second derivative of $y = x^4 + 3x^2 - 5x + 1$.

Answer:

Given the function $y = x^4 + 3x^2 - 5x + 1$.

Step 1: Find the first derivative, $\frac{dy}{dx}$. We differentiate $y$ with respect to $x$ using the power rule and linearity of differentiation.

$\frac{dy}{dx} = \frac{d}{dx}(x^4 + 3x^2 - 5x + 1)$

$= \frac{d}{dx}(x^4) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(5x) + \frac{d}{dx}(1)$

$= 4x^{4-1} + 3(2x^{2-1}) - 5(1x^{1-1}) + 0$

$\frac{dy}{dx} = 4x^3 + 6x - 5$

... (i)

Step 2: Find the second derivative, $\frac{d^2y}{dx^2}$. We differentiate the result from Step 1, $\frac{dy}{dx}$, with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(4x^3 + 6x - 5)$

$= \frac{d}{dx}(4x^3) + \frac{d}{dx}(6x) - \frac{d}{dx}(5)$

$= 4(3x^{3-1}) + 6(1x^{1-1}) - 0$

$\frac{d^2y}{dx^2} = 12x^2 + 6$

... (ii)

The second derivative of $y = x^4 + 3x^2 - 5x + 1$ is $12x^2 + 6$.


Example 2. If $y = \sin(ax+b)$, find the second derivative $\frac{d^2y}{dx^2}$.

Answer:

Given the function $y = \sin(ax+b)$.

Step 1: Find the first derivative, $\frac{dy}{dx}$. We differentiate $y$ with respect to $x$ using the chain rule, where the outer function is $\sin(u)$ and the inner function is $u = ax+b$.

$\frac{dy}{dx} = \frac{d}{dx}(\sin(ax+b))$

$= \cos(ax+b) \cdot \frac{d}{dx}(ax+b)$

$= \cos(ax+b) \cdot (a \frac{d}{dx}(x) + \frac{d}{dx}(b))$

$= \cos(ax+b) \cdot (a \cdot 1 + 0)$

$\frac{dy}{dx} = a \cos(ax+b)$

... (i)

Step 2: Find the second derivative, $\frac{d^2y}{dx^2}$. We differentiate the result from Step 1, $\frac{dy}{dx} = a \cos(ax+b)$, with respect to $x$. Again, we use the chain rule, where the outer function is $\cos(u)$ and the inner function is $u = ax+b$. The scalar $a$ is a constant multiplier.

$\frac{d^2y}{dx^2} = \frac{d}{dx}(a \cos(ax+b))$

$= a \frac{d}{dx}(\cos(ax+b))$

$= a (-\sin(ax+b) \cdot \frac{d}{dx}(ax+b))$

[Chain Rule]

$= a (-\sin(ax+b) \cdot a)$

[Differentiating $ax+b$]

$\frac{d^2y}{dx^2} = -a^2 \sin(ax+b)$

... (ii)

The second derivative of $y = \sin(ax+b)$ is $-a^2 \sin(ax+b)$.



Application of Derivatives


Derivatives are not just abstract mathematical concepts; they are powerful tools with extensive applications in diverse fields. The fundamental idea behind the derivative is the instantaneous rate of change, which allows us to model and analyse dynamic situations where quantities are changing. This section explores some of the key applications of derivatives, with a focus on concepts relevant to Applied Mathematics.


Rate of Change of Quantities

One of the most direct applications of the derivative is in determining the rate at which one quantity changes in response to another. If a quantity $y$ is functionally dependent on another quantity $x$, expressed as $y = f(x)$, then the derivative $\frac{dy}{dx}$ represents the instantaneous rate of change of $y$ with respect to $x$.

If the quantities are changing over time, and $y$ is a function of time $t$, i.e., $y = f(t)$, then $\frac{dy}{dt}$ represents the instantaneous rate of change of $y$ with respect to time $t$. This is often what is meant by "rate of change" in physical or economic contexts.

Example: Consider the area of a circle, $A$, which depends on its radius, $r$. The formula is $A = \pi r^2$. To find the rate at which the area changes with respect to the radius, we differentiate $A$ with respect to $r$:

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

Using the constant multiple rule and the power rule for differentiation:

$\frac{dA}{dr} = \pi \frac{d}{dr}(r^2) = \pi (2r^{2-1}) = 2\pi r$

So, $\frac{dA}{dr} = 2\pi r$. This result means that at any given radius $r$, the area of the circle is changing at a rate of $2\pi r$ square units per unit change in radius. For example, when $r = 5$ cm, the area is changing at a rate of $10\pi$ sq cm/cm. This rate increases as the radius increases, which makes intuitive sense – increasing the radius of a larger circle adds more area than increasing the radius of a smaller circle.


Tangent and Normal to a Curve

Geometrically, the derivative of a function at a specific point gives the slope of the tangent line to the curve at that point.

Let $y = f(x)$ be a differentiable curve. For a point $(x_0, y_0)$ on the curve (where $y_0 = f(x_0)$), the slope of the tangent line to the curve at $(x_0, y_0)$ is given by the value of the first derivative evaluated at $x_0$.

$m_{tangent} = \left(\frac{dy}{dx}\right)_{(x_0, y_0)} = f'(x_0)$

Assuming the tangent is not a vertical line, its equation can be found using the point-slope form of a linear equation: $y - y_0 = m(x - x_0)$. Substituting the slope of the tangent:

$y - y_0 = f'(x_0)(x - x_0)$

... (iv)

The normal line to the curve at the point $(x_0, y_0)$ is the line perpendicular to the tangent line at that same point. If two lines are perpendicular and neither is vertical, the product of their slopes is $-1$.

Slope of the normal at $(x_0, y_0)$ is $m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{f'(x_0)}$, provided $f'(x_0) \ne 0$.

Assuming the normal is not a vertical line, its equation is:

$y - y_0 = m_{normal}(x - x_0)$

$y - y_0 = -\frac{1}{f'(x_0)}(x - x_0)$

... (v)

Special Cases:


Approximations (Differentials)

Derivatives can be used to approximate the change in the value of a function ($\Delta y$) for a small change in the independent variable ($\Delta x$). This is based on the idea that for small changes, the curve is very close to its tangent line.

Recall the definition of the derivative: $\frac{dy}{dx} = \lim\limits_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$. For a sufficiently small $\Delta x$, the ratio $\frac{\Delta y}{\Delta x}$ is approximately equal to the derivative at $x$.

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

Multiplying by $\Delta x$, we get an approximation for the change in $y$:

$\Delta y \approx f'(x) \Delta x$

Since $\Delta y = f(x + \Delta x) - f(x)$, we can write:

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

Rearranging this gives the linear approximation formula:

$\mathbf{f(x + \Delta x) \approx f(x) + f'(x) \Delta x}$

... (vi)

This formula allows us to estimate the value of $f$ at a point $x + \Delta x$ if we know the value of $f$ and its derivative at a nearby point $x$.

Example: Use differentials to approximate $\sqrt{25.3}$. We want to approximate the value of the function $f(x) = \sqrt{x}$ at $x = 25.3$. We know the exact value of $\sqrt{x}$ at a point near 25.3, which is $x=25$. Let $f(x) = \sqrt{x}$. We need to approximate $f(25.3)$. Choose the known point $x_0 = 25$ and the change $\Delta x = 25.3 - 25 = 0.3$. We need $f(x_0)$ and $f'(x_0)$.

$f(x_0) = f(25) = \sqrt{25} = 5$.

Now, find the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

Evaluate the derivative at $x_0 = 25$:

$f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{2 \times 5} = \frac{1}{10} = 0.1$

Now, use the approximation formula $f(x_0 + \Delta x) \approx f(x_0) + f'(x_0) \Delta x$:

$f(25 + 0.3) \approx f(25) + f'(25) \times 0.3$

$f(25.3) \approx 5 + (0.1) \times (0.3)$

$f(25.3) \approx 5 + 0.03$

$f(25.3) \approx 5.03$

Therefore, $\sqrt{25.3}$ is approximately $5.03$.


Beyond these fundamental applications, derivatives are crucial in various other areas:

In the subsequent sections of this chapter, we will delve deeper into some of these applications, specifically focusing on Marginal Cost and Marginal Revenue, Increasing/Decreasing Functions, and Maxima/Minima, as they are important topics in Applied Mathematics.



Marginal Cost and Marginal Revenue using Derivatives


In the study of economics and business, understanding how costs, revenues, and profits change in response to variations in production or sales levels is crucial for making optimal decisions. The concept of "marginal" is central to this analysis, referring to the change in a total quantity resulting from a one-unit change in output or sales. When dealing with cost, revenue, or profit functions that are continuous and differentiable, calculus, specifically differentiation, provides a precise way to calculate these marginal quantities as instantaneous rates of change.


Cost Function, Revenue Function, and Profit Function

To analyse marginal concepts, we first need to define the basic economic functions:

Let $x$ represent the number of units of a product produced and/or sold. Typically, $x$ is assumed to be a non-negative quantity.


Marginal Concepts using Derivatives

In economics, the marginal value of a quantity is the change in that quantity resulting from a one-unit increase in output or input. When these quantities are described by continuous functions, the marginal concept at a specific level of output $x$ is precisely the instantaneous rate of change of the total function with respect to $x$. This is where differentiation plays a key role. The derivative of a total function gives the corresponding marginal function.

Marginal Cost (MC)

Marginal cost is the rate of change of the total cost with respect to the number of units produced. It tells us the approximate additional cost incurred by producing one more unit beyond the current level of production.

Mathematically, if $C(x)$ is the total cost function, the marginal cost function, $MC(x)$, is given by its first derivative:

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

... (iii)

The marginal cost at a specific production level, say $x_0$ units, is $MC(x_0) = C'(x_0)$.

Marginal Revenue (MR)

Marginal revenue is the rate of change of the total revenue with respect to the number of units sold. It tells us the approximate additional revenue generated by selling one more unit beyond the current sales level.

Mathematically, if $R(x)$ is the total revenue function, the marginal revenue function, $MR(x)$, is given by its first derivative:

$\mathbf{MR(x) = \frac{dR}{dx} = R'(x)}$

... (iv)

The marginal revenue at a specific sales level, say $x_0$ units, is $MR(x_0) = R'(x_0)$.

Marginal Profit (MP)

Marginal profit is the rate of change of the total profit with respect to the number of units sold or produced. It tells us the approximate additional profit obtained from producing and selling one more unit beyond the current level.

Mathematically, if $P(x)$ is the total profit function, the marginal profit function, $MP(x)$, is given by its first derivative:

$\mathbf{MP(x) = \frac{dP}{dx} = P'(x)}$

... (v)

We can also derive a relationship between marginal profit, marginal revenue, and marginal cost directly from the profit definition $P(x) = R(x) - C(x)$. Differentiating both sides with respect to $x$:

$\frac{dP}{dx} = \frac{d}{dx}(R(x) - C(x))$

Using the difference rule for differentiation, $\frac{d}{dx}(f(x) - g(x)) = \frac{df}{dx} - \frac{dg}{dx}$:

$\frac{dP}{dx} = \frac{dR}{dx} - \frac{dC}{dx}$

Substituting the definitions of marginal revenue and marginal cost:

$\mathbf{MP(x) = MR(x) - MC(x)}$

... (vi)

This equation means that the additional profit from one more unit is the additional revenue from that unit minus the additional cost of that unit.


Profit Maximization

A key application of marginal analysis is finding the production level $x$ that maximizes total profit. From calculus, a function reaches its maximum (or minimum) at critical points where its first derivative is zero. For the profit function $P(x)$, the critical points occur where $P'(x) = 0$.

Since $P'(x) = MP(x)$, profit is maximized (or minimized) when the marginal profit is zero:

$MP(x) = 0$

Using the relationship $MP(x) = MR(x) - MC(x)$, this condition becomes:

$MR(x) - MC(x) = 0$

$\mathbf{MR(x) = MC(x)}$

... (vii)

Thus, profit is maximized (or minimized) at the production level where marginal revenue equals marginal cost.

To distinguish between a maximum and a minimum profit, we use the second derivative test. For maximum profit, the second derivative of the profit function at the critical point must be negative, i.e., $P''(x) < 0$. Since $P'(x) = R'(x) - C'(x)$, the second derivative is $P''(x) = R''(x) - C''(x)$. So, for maximum profit, at the output level where $MR = MC$, we must have $R''(x) - C''(x) < 0$, which means $R''(x) < C''(x)$. This condition ensures that the marginal revenue curve is steeper than the marginal cost curve at the point of intersection (when viewed from left to right, the slope of MR is less than the slope of MC).


Examples

Example 1. The total cost function of a firm is given by $C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000$, where $x$ is the number of units produced. Find the marginal cost when 100 units are produced.

Answer:

Given the total cost function $C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000$.

The marginal cost (MC) is the first derivative of the total cost function with respect to $x$.

$MC(x) = \frac{dC}{dx} = \frac{d}{dx}(0.005x^3 - 0.02x^2 + 30x + 5000)$

Using the power rule, constant multiple rule, sum/difference rule, and the derivative of a constant:

$MC(x) = 0.005 \cdot (3x^{3-1}) - 0.02 \cdot (2x^{2-1}) + 30 \cdot (1x^{1-1}) + 0$

$MC(x) = 0.015x^2 - 0.04x + 30$

... (i)

We need to find the marginal cost when $x = 100$ units are produced. Substitute $x=100$ into the marginal cost function $MC(x)$.

$MC(100) = 0.015(100)^2 - 0.04(100) + 30$

$= 0.015(10000) - 4 + 30$

$= 150 - 4 + 30$

$= 146 + 30 = 176$

The marginal cost when 100 units are produced is $\textsf{₹}\$176$. This means that producing the 101st unit will approximately add $\textsf{₹}\$176$ to the total cost.


Example 2. The total revenue function is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$.

Answer:

Given the total revenue function $R(x) = 13x^2 + 26x + 15$.

The marginal revenue (MR) is the first derivative of the total revenue function with respect to $x$.

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

Using the power rule, constant multiple rule, sum rule, and the derivative of a constant:

$MR(x) = 13 \cdot (2x^{2-1}) + 26 \cdot (1x^{1-1}) + 0$

$MR(x) = 26x + 26$

... (i)

We need to find the marginal revenue when $x = 7$ units are sold. Substitute $x=7$ into the marginal revenue function $MR(x)$.

$MR(7) = 26(7) + 26$

Calculate the product and sum:

$26 \times 7 = 182$

$MR(7) = 182 + 26$

$MR(7) = 208$

The marginal revenue when $x = 7$ is $\textsf{₹}\$208$. This means that selling the 8th unit will approximately add $\textsf{₹}\$208$ to the total revenue.



Increasing /Decreasing Functions


One of the significant applications of the first derivative of a function is to determine the nature of the function's behaviour – specifically, where it is increasing or decreasing. This property of a function is called monotonicity. Geometrically, a function is increasing over an interval if its graph rises as you move from left to right along that interval. Conversely, it is decreasing if its graph falls as you move from left to right.

The sign of the first derivative of a function is directly related to its monotonicity. If the derivative is positive on an interval, the function is increasing on that interval. If the derivative is negative, the function is decreasing. If the derivative is zero, the function is constant.


Definitions of Increasing and Decreasing Functions

Let $f$ be a real-valued function defined on an interval $I$ (which can be open, closed, or half-open).

Functions that are either increasing or decreasing on an interval are called monotonic functions. If they are strictly increasing or strictly decreasing, they are called strictly monotonic functions.


First Derivative Test for Monotonicity

This test provides a direct link between the sign of the derivative and the monotonic behaviour of a differentiable function.

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

For non-strict monotonicity, the conditions are:

The points where the first derivative $f'(x)$ is equal to zero or where $f'(x)$ is undefined are called critical points of the function $f$. These critical points are important because they are the only points where the function's monotonicity can change from increasing to decreasing or vice versa. They divide the domain of the function into intervals on which the derivative has a constant sign (either positive or negative, assuming $f'$ is continuous between critical points).


Procedure to Find Intervals of Monotonicity

To find the intervals on which a function $f(x)$ is increasing or decreasing:

  1. Find the first derivative of the function, $f'(x)$.
  2. Find the critical points of $f(x)$ by determining the values of $x$ where $f'(x) = 0$ or where $f'(x)$ is undefined.
  3. Use the critical points to divide the domain of the function into open intervals. Arrange the critical points in ascending order and mark them on a number line. These points partition the number line into intervals.
  4. Choose a test value within each open interval determined in Step 3.
  5. Evaluate the sign of $f'(x)$ at each test value. The sign of $f'(x)$ in each interval tells us whether the function is strictly increasing or strictly decreasing on that interval.
    • If $f'(x) > 0$ in an interval, $f(x)$ is strictly increasing on that open interval.
    • If $f'(x) < 0$ in an interval, $f(x)$ is strictly decreasing on that open interval.
  6. To state the intervals where the function is simply increasing or decreasing (non-strictly), include the critical points as endpoints of the intervals, provided the function is defined and continuous at those points.

Examples

Example 1. Find the intervals on which the function $f(x) = x^2 - 4x + 6$ is increasing or decreasing.

Answer:

Given the function $f(x) = x^2 - 4x + 6$. The domain of this function is all real numbers, $(-\infty, \infty)$.

Step 1: Find the first derivative $f'(x)$.

$f'(x) = \frac{d}{dx}(x^2 - 4x + 6)$

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

... (i)

Step 2: Find the critical points by setting $f'(x) = 0$ or finding points where $f'(x)$ is undefined. $f'(x) = 2x - 4$ is a polynomial, so it is defined for all real numbers. We set $f'(x) = 0$:

$2x - 4 = 0$

... (ii)

$2x = 4$

$x = 2$

... (iii)

The only critical point is $x = 2$.

Step 3: Use the critical point to divide the domain $(-\infty, \infty)$ into intervals. The critical point $x=2$ divides the number line into two open intervals: $(-\infty, 2)$ and $(2, \infty)$.

Step 4: Choose test values in each interval and evaluate the sign of $f'(x)$.

  • For the interval $(-\infty, 2)$, choose a test value, say $x=0$. Evaluate $f'(0) = 2(0) - 4 = -4$.
  • For the interval $(2, \infty)$, choose a test value, say $x=3$. Evaluate $f'(3) = 2(3) - 4 = 6 - 4 = 2$.

Step 5: Determine the monotonicity based on the sign of $f'(x)$.

  • In the interval $(-\infty, 2)$, $f'(x) = -4 < 0$. Therefore, $f(x)$ is strictly decreasing on $(-\infty, 2)$.
  • In the interval $(2, \infty)$, $f'(x) = 2 > 0$. Therefore, $f(x)$ is strictly increasing on $(2, \infty)$.

Since the function is continuous at $x=2$, we can include this endpoint for non-strict monotonicity:

  • The function is increasing on $[2, \infty)$.
  • The function is decreasing on $(-\infty, 2]$.

Summary: $f(x)$ is strictly decreasing on the interval $(-\infty, 2)$. $f(x)$ is strictly increasing on the interval $(2, \infty)$.


Example 2. Find the intervals on which $f(x) = 2x^3 - 9x^2 + 12x + 1$ is increasing or decreasing.

Answer:

Given the function $f(x) = 2x^3 - 9x^2 + 12x + 1$. The domain of this function is all real numbers, $(-\infty, \infty)$.

Step 1: Find the first derivative $f'(x)$.

$f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x + 1)$

$f'(x) = 2(3x^2) - 9(2x) + 12(1) + 0$

$f'(x) = 6x^2 - 18x + 12$

... (i)

Step 2: Find the critical points by setting $f'(x) = 0$. $f'(x) = 6x^2 - 18x + 12$ is defined for all real numbers. We set $f'(x) = 0$:

$6x^2 - 18x + 12 = 0$

... (ii)

Divide the equation by the common factor 6:

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

Factor the quadratic equation:

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

Setting each factor to zero gives the critical points:

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

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

The critical points are $x = 1$ and $x = 2$.

Step 3: Use the critical points to divide the domain $(-\infty, \infty)$ into intervals. The critical points $x=1$ and $x=2$ divide the number line into three open intervals: $(-\infty, 1)$, $(1, 2)$, and $(2, \infty)$.

Step 4: Choose test values in each interval and evaluate the sign of $f'(x)$. We use the factored form $f'(x) = 6(x-1)(x-2)$ or the expanded form $f'(x) = 6x^2 - 18x + 12$.

  • For the interval $(-\infty, 1)$, choose a test value, say $x=0$. $f'(0) = 6(0)^2 - 18(0) + 12 = 12$. The sign is positive ($+$).
  • For the interval $(1, 2)$, choose a test value, say $x=1.5$. $f'(1.5) = 6(1.5-1)(1.5-2) = 6(0.5)(-0.5) = -1.5$. The sign is negative ($-$ ).
  • For the interval $(2, \infty)$, choose a test value, say $x=3$. $f'(3) = 6(3-1)(3-2) = 6(2)(1) = 12$. The sign is positive ($+$).

We can summarise the signs in a table:

Interval Test Value $x$ Sign of $f'(x)$ Monotonicity of $f(x)$
$(-\infty, 1)$ $0$ $+$ ($f'(0)=12$) Strictly Increasing
$(1, 2)$ $1.5$ $-$ ($f'(1.5)=-1.5$) Strictly Decreasing
$(2, \infty)$ $3$ $+$ ($f'(3)=12$) Strictly Increasing

Step 5: Determine the monotonicity based on the sign of $f'(x)$.

  • $f(x)$ is strictly increasing on the intervals $(-\infty, 1)$ and $(2, \infty)$.
  • $f(x)$ is strictly decreasing on the interval $(1, 2)$.

Since the function is continuous at the critical points $x=1$ and $x=2$, we can include these endpoints for non-strict monotonicity:

  • The function is increasing on $(-\infty, 1] \cup [2, \infty)$.
  • The function is decreasing on $[1, 2]$.

Summary: $f(x)$ is strictly increasing on $(-\infty, 1)$ and $(2, \infty)$. $f(x)$ is strictly decreasing on $(1, 2)$.



Maxima and Minima


Derivatives are powerful tools used to find the extreme values (maximum and minimum values) of a function. These extreme values are fundamental in solving optimization problems, which involve finding the best possible outcome, such as maximizing profit, minimizing cost, finding the shortest distance, or finding the largest area.

The extreme values of a function can be classified into two types: local (or relative) extrema and global (or absolute) extrema.


Local (Relative) Extrema

A function $f$ is said to have a local maximum at a point $c$ if there exists an open interval $(a, b)$ containing $c$ such that $f(x) \le f(c)$ for all $x$ in $(a, b)$. This means that $f(c)$ is the largest value of the function in some small neighborhood around $c$.

A function $f$ is said to have a local minimum at a point $c$ if there exists an open interval $(a, b)$ containing $c$ such that $f(x) \ge f(c)$ for all $x$ in $(a, b)$. This means that $f(c)$ is the smallest value of the function in some small neighborhood around $c$.

Local maxima and local minima are collectively called local extrema (plural of extremum). A function can have multiple local maxima and minima.


Critical Points

A critical point of a function $f$ is a point $c$ in the domain of $f$ where either its first derivative $f'(c)$ is equal to zero, or $f'(c)$ is undefined.

The significance of critical points is established by Fermat's Theorem:
If a function $f$ has a local extremum (local maximum or local minimum) at a point $c$, and if $f'(c)$ exists at that point, then $f'(c)$ must be equal to $0$.

This theorem tells us that local extrema can only occur at critical points (either where $f'(x) = 0$ or where $f'(x)$ is undefined). However, the converse is not always true: a critical point does not guarantee a local extremum. A function can have a critical point that is neither a local maximum nor a local minimum, such as at an inflection point with a horizontal tangent (e.g., $f(x) = x^3$ at $x=0$, where $f'(0)=0$).


Tests for Local Extrema

Since local extrema occur only at critical points, we use tests to determine whether a critical point corresponds to a local maximum, a local minimum, or neither.

First Derivative Test

The First Derivative Test uses the sign of the first derivative around a critical point to determine the nature of the extremum. Suppose $c$ is a critical point of a continuous function $f$.

Second Derivative Test

The Second Derivative Test uses the concavity of the function at a critical point where the tangent is horizontal ($f'(c) = 0$) to determine the nature of the extremum. Suppose $f''(x)$ exists in an open interval containing $c$, and $f'(c) = 0$.


Global (Absolute) Extrema

While local extrema describe the highest or lowest points in a limited region of the function's graph, global extrema (also called absolute extrema) describe the absolute highest or lowest points over the entire domain or a specified subset of the domain.

A function $f$ is said to have an absolute maximum value $f(c)$ on a domain $D$ at a point $c \in D$ if $f(x) \le f(c)$ for all $x \in D$.

A function $f$ is said to have an absolute minimum value $f(c)$ on a domain $D$ at a point $c \in D$ if $f(x) \ge f(c)$ for all $x \in D$.

A key result concerning global extrema is the Extreme Value Theorem, which states that if a function $f$ is continuous on a closed interval $[a, b]$, then $f$ must attain both an absolute maximum value and an absolute minimum value on that interval.

Finding Absolute Extrema on a Closed Interval

For a continuous function $f$ on a closed interval $[a, b]$, the absolute maximum and minimum values must occur at either the critical points of $f$ that lie within the open interval $(a, b)$ or at the endpoints $a$ and $b$ of the interval.

Procedure to find the absolute extrema of a continuous function $f$ on $[a, b]$:

  1. Find all critical points of $f$ in the open interval $(a, b)$.
  2. Evaluate the function $f(x)$ at all the critical points found in Step 1.
  3. Evaluate the function $f(x)$ at the endpoints $a$ and $b$ of the interval.
  4. Compare all the function values obtained in Steps 2 and 3. The largest of these values is the absolute maximum value of $f$ on $[a, b]$, and the smallest value is the absolute minimum value of $f$ on $[a, b]$.

Examples

Example 1. Find the local maximum and local minimum values of the function $f(x) = x^3 - 6x^2 + 9x + 15$.

Answer:

Given the function $f(x) = x^3 - 6x^2 + 9x + 15$.

Step 1: Find the first derivative $f'(x)$ and the critical points.

$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 15)$

$f'(x) = 3x^2 - 12x + 9$

... (i)

To find critical points, set $f'(x) = 0$. $f'(x)$ is defined for all real $x$.

$3x^2 - 12x + 9 = 0$

... (ii)

Divide the equation by 3:

$x^2 - 4x + 3 = 0$

Factor the quadratic equation:

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

Setting each factor to zero gives the critical points:

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

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

The critical points are $x = 1$ and $x = 3$.

Step 2: Use the Second Derivative Test to classify the critical points. Find the second derivative $f''(x)$.

$f''(x) = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12$

... (iii)

Evaluate $f''(x)$ at each critical point:

At $x = 1$:

$f''(1) = 6(1) - 12 = 6 - 12 = -6$

... (iv)

Since $f'(1) = 0$ and $f''(1) = -6 < 0$, by the Second Derivative Test, the function has a local maximum at $x=1$.

The local maximum value is $f(1)$.

$f(1) = (1)^3 - 6(1)^2 + 9(1) + 15 = 1 - 6 + 9 + 15 = 19$

At $x = 3$:

$f''(3) = 6(3) - 12 = 18 - 12 = 6$

... (v)

Since $f'(3) = 0$ and $f''(3) = 6 > 0$, by the Second Derivative Test, the function has a local minimum at $x=3$.

The local minimum value is $f(3)$.

$f(3) = (3)^3 - 6(3)^2 + 9(3) + 15 = 27 - 6(9) + 27 + 15 = 27 - 54 + 27 + 15 = 15$

The local maximum value is 19, which occurs at $x=1$. The local minimum value is 15, which occurs at $x=3$.

Alternate Method (Using the First Derivative Test)

The critical points are $x=1$ and $x=3$. These points divide the real line into three intervals: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$. We examine the sign of $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$ in each interval.

Interval Test Value $x$ Sign of $x-1$ Sign of $x-3$ Sign of $f'(x)=3(x-1)(x-3)$ Conclusion (Monotonicity)
$(-\infty, 1)$ $0$ $-$ $-$ $3(-) (-) = +$ Increasing
$(1, 3)$ $2$ $+$ $-$ $3(+) (-) = -$ Decreasing
$(3, \infty)$ $4$ $+$ $+$ $3(+) (+) = +$ Increasing

Applying the First Derivative Test:

  • At $x=1$, the sign of $f'(x)$ changes from positive to negative as $x$ increases through 1. Thus, $x=1$ corresponds to a local maximum. The local maximum value is $f(1) = 19$.
  • At $x=3$, the sign of $f'(x)$ changes from negative to positive as $x$ increases through 3. Thus, $x=3$ corresponds to a local minimum. The local minimum value is $f(3) = 15$.

Both methods yield the same results for the local extrema.


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

Answer:

Given the function $f(x) = x^3 - 3x + 2$ on the closed interval $[0, 2]$. Since $f(x)$ is a polynomial, it is continuous on this interval. Therefore, the absolute maximum and minimum values exist and occur at critical points within $(0, 2)$ or at the endpoints 0 and 2.

Step 1: Find the critical points in the open interval $(0, 2)$. Find the first derivative $f'(x)$.

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

Set $f'(x) = 0$ to find critical points where the tangent is horizontal:

$3x^2 - 3 = 0$

... (i)

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

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

The critical points are $x = 1$ and $x = -1$. We are only interested in critical points that lie within the open interval $(0, 2)$. The point $x=1$ is in the interval $(0, 2)$. The point $x=-1$ is not in the interval $(0, 2)$. So, the only critical point in the interval $(0, 2)$ is $x = 1$.

Step 2: Evaluate $f(x)$ at the critical point(s) found in the interval.

At $x = 1$:

$f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0$

Step 3: Evaluate $f(x)$ at the endpoints of the interval $[0, 2]$. The endpoints are $x=0$ and $x=2$.

At $x = 0$:

$f(0) = (0)^3 - 3(0) + 2 = 0 - 0 + 2 = 2$

At $x = 2$:

$f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4$

Step 4: Compare all the function values obtained in Steps 2 and 3. The values are $f(1) = 0$, $f(0) = 2$, and $f(2) = 4$.

Comparing these values: Maximum value = $\max\{0, 2, 4\} = 4$. Minimum value = $\min\{0, 2, 4\} = 0$.

The absolute maximum value of $f(x)$ on the interval $[0, 2]$ is 4, which occurs at $x=2$. The absolute minimum value of $f(x)$ on the interval $[0, 2]$ is 0, which occurs at $x=1$.