Differentiation and its Applications

Overview

Welcome to the chapter on Differentiation and its Applications, a cornerstone of calculus essential for any aspiring data scientist. This module is meticulously designed to equip you with the fundamental analytical tools necessary to understand and interpret rates of change, sensitivity, and optimization problems that are ubiquitous in data science. For your CMI examinations, a solid grasp of differentiation is not merely about solving equations; it's about applying these concepts to model behavior, evaluate algorithm performance, and make informed decisions from data.

In the realm of data science, differentiation underpins a vast array of critical techniques. From understanding the gradients in machine learning loss functions to optimizing model parameters, and from analyzing the sensitivity of financial models to interpreting the elasticity of demand in economic data, the principles covered here are directly applicable. Mastering these concepts will empower you to tackle complex problems efficiently, making them a high-yield area for both your academic success and professional competence.

This chapter will guide you through the core mechanics of differentiation before moving into its powerful applications in finding maxima and minima. The ability to identify optimal points is crucial for tasks like minimizing error in predictive models, maximizing profit functions, or finding the most efficient resource allocations. Prepare to develop a robust understanding that will serve as a foundational pillar for advanced topics in machine learning, statistical inference, and mathematical optimization.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Fundamentals of Differentiation | Grasp rates of change and tangent lines. |
| 2 | Maxima and Minima | Identify optimal points for function behavior. |

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Learning Objectives

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Now let's begin with Fundamentals of Differentiation...

Part 1: Fundamentals of Differentiation

Introduction

Differentiation is a fundamental concept in calculus that quantifies the rate at which a quantity changes with respect to another. In simpler terms, it allows us to find the instantaneous rate of change of a function. This powerful tool is crucial for understanding how functions behave, whether they are increasing or decreasing, where they reach their maximum or minimum values, and how their curvature changes.

For students pursuing a Masters in Data Science, a strong grasp of differentiation is indispensable. It forms the backbone for various advanced topics such as optimization algorithms (e.g., gradient descent in machine learning), understanding model sensitivity, error propagation, and statistical inference. CMI frequently tests these foundational concepts, requiring a deep understanding of limits, continuity, differentiability, and their applications. This unit will provide a comprehensive overview, equipping you with the necessary theoretical knowledge and problem-solving skills.

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Key Concepts

1. Limits of Functions

The concept of a limit is foundational to calculus. It describes the behavior of a function as its input approaches a certain value, without necessarily evaluating the function at that exact point.

One-Sided Limits:

• Left-hand limit: $\lim_{x \to c^-} f(x)$ (as $x$ approaches $c$ from values less than $c$).


• Right-hand limit: $\lim_{x \to c^+} f(x)$ (as $x$ approaches $c$ from values greater than $c$).

For a limit to exist at $c$, both one-sided limits must exist and be equal: $\lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = L = \lim_{x \to c^+} f(x)$.

Properties of Limits:
If $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$, then:

Sum/Difference Rule: $\lim_{x \to c} [f(x) \pm g(x)] = L \pm M$

Product Rule: $\lim_{x \to c} [f(x) \cdot g(x)] = L \cdot M$

Quotient Rule: $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}$, provided $M \neq 0$.

Constant Multiple Rule: $\lim_{x \to c} k \cdot f(x) = k \cdot L$

Power Rule: $\lim_{x \to c} [f(x)]^n = L^n$

Methods for Evaluation of Limits:

* Direct Substitution: If $f(x)$ is a polynomial or a rational function (where the denominator is non-zero at $c$), the limit can be found by direct substitution: $\lim_{x \to c} f(x) = f(c)$.

* Factorization/Cancellation: If direct substitution results in an indeterminate form like $\frac{0}{0}$, factorize the numerator and denominator to cancel common factors.

* Rationalization: For limits involving square roots, multiply the numerator and denominator by the conjugate.

* Special Trigonometric Limits:
* $\lim_{x \to 0} \frac{\sin x}{x} = 1$
* $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
* $\lim_{x \to 0} \frac{\tan x}{x} = 1$

* Limits at Infinity: To evaluate $\lim_{x \to \pm \infty} f(x)$ for rational functions, divide the numerator and denominator by the highest power of $x$ in the denominator. For polynomials, the limit is determined by the term with the highest power.

Example: $\lim_{x \to \infty} \frac{ax^n + \dots}{bx^m + \dots}$
- If $n > m$, the limit is $\pm \infty$.
- If $n < m$, the limit is $0$.
- If $n = m$, the limit is $\frac{a}{b}$.

* L'Hôpital's Rule: This rule applies to indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$. If $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then

provided the latter limit exists. This rule can be applied multiple times. (PYQ 4, 6)

* Taylor Series Expansion for Limits: For complex indeterminate forms, especially involving exponential, trigonometric, or logarithmic functions, expanding functions into their Taylor series around the point $x=c$ (or $x=0$ if $c=0$) can simplify the limit evaluation.
* $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
* $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
* $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
* $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$

Worked Example: L'Hôpital's Rule

Problem: Evaluate

Solution:

Step 1: Check the form of the limit by direct substitution.

As $x \to 0$,

This is an indeterminate form $\frac{0}{0}$, so L'Hôpital's Rule can be applied.

Step 2: Apply L'Hôpital's Rule for the first time. Differentiate numerator and denominator separately.

Derivative of numerator:

Derivative of denominator:

Step 3: Check the form again.

As $x \to 0$,

Still an indeterminate form $\frac{0}{0}$, so apply L'Hôpital's Rule again.

Step 4: Apply L'Hôpital's Rule for the second time.

Derivative of numerator:

Derivative of denominator:

Step 5: Check the form again.

As $x \to 0$,

The limit is now a determinate form.

Step 6: Evaluate the limit.

Answer: $\boxed{\frac{5}{2}}$

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2. Continuity of Functions

Continuity is a property of a function where its graph can be drawn without lifting the pen. Informally, a function is continuous if small changes in the input result in small changes in the output.

Continuity on an Interval: A function is continuous on an open interval $(a, b)$ if it is continuous at every point in the interval. It is continuous on a closed interval $[a, b]$ if it is continuous on $(a, b)$, and also $\lim_{x \to a^+} f(x) = f(a)$ and $\lim_{x \to b^-} f(x) = f(b)$.

Types of Discontinuities: (PYQ 2)

Removable Discontinuity: Occurs when $\lim_{x \to c} f(x)$ exists but is not equal to $f(c)$, or $f(c)$ is undefined. Graphically, this is a "hole" in the graph. It's called "removable" because the discontinuity can be eliminated by redefining $f(c)$.

Example: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$. $\lim_{x \to 1} f(x) = 2$, but $f(1)$ is undefined.

Jump Discontinuity: Occurs when the left-hand limit and the right-hand limit at $c$ both exist but are not equal. Graphically, there's a "jump" in the function value.

Example: $f(x) = \lfloor x \rfloor$ (floor function) at integer points.

Infinite (Essential) Discontinuity: Occurs when one or both of the one-sided limits at $c$ are $\pm \infty$. Graphically, this is a vertical asymptote.

Example: $f(x) = \frac{1}{x}$ at $x=0$.

Properties of Continuous Functions: (PYQ 1)
If $f(x)$ and $g(x)$ are continuous at $x=c$, then:

$f(x) \pm g(x)$ is continuous at $x=c$.

$f(x) \cdot g(x)$ is continuous at $x=c$.

$\frac{f(x)}{g(x)}$ is continuous at $x=c$, provided $g(c) \neq 0$.

$k \cdot f(x)$ is continuous at $x=c$ for any constant $k$.

$f(g(x))$ (composition) is continuous at $x=c$ if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$.

Important Implications for Discontinuous Functions:
* If $f(x)$ is continuous and $g(x)$ is discontinuous at $c$, then $f(x) + g(x)$ is necessarily discontinuous at $c$. (PYQ 1)
* If $f(x)$ is continuous and $g(x)$ is discontinuous at $c$, then $f(x) \cdot g(x)$ can be continuous or discontinuous at $c$. For example, if $f(x)=0$ for all $x$, then $f(x)g(x)=0$ which is continuous, even if $g(x)$ is discontinuous. (PYQ 1)
* If $f(x)$ and $g(x)$ are both discontinuous at $c$, then $f(x) + g(x)$ can be continuous or discontinuous at $c$. For example, $f(x) = \frac{1}{x}$ and $g(x) = -\frac{1}{x}$ are both discontinuous at $x=0$, but $f(x)+g(x)=0$ (continuous). (PYQ 1)
* If $f(x)$ and $g(x)$ are both discontinuous at $c$, then $f(x) \cdot g(x)$ can be continuous or discontinuous at $c$. (PYQ 1)

Worked Example: Removable Discontinuity

Problem: Determine if $f(x) = \frac{x^3 - x}{x^3 + x}$ has a removable discontinuity at $x=0$.

Solution:

Step 1: Check if $f(0)$ is defined.

$f(0) = \frac{0^3 - 0}{0^3 + 0} = \frac{0}{0}$, which is undefined. So, there is a discontinuity at $x=0$.

Step 2: Evaluate the limit as $x \to 0$.

Factor out $x$ from numerator and denominator:

Cancel the common factor $x$ (since $x \neq 0$ for the limit):

Substitute $x=0$:

Step 3: Compare the limit with the function value.

Since $\lim_{x \to 0} f(x) = -1$ exists, but $f(0)$ is undefined, $f(x)$ has a removable discontinuity at $x=0$.

Answer: Yes, $f(x)$ has a removable discontinuity at $x=0$.

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3. Differentiability of Functions

Differentiability signifies that a function has a well-defined tangent line at every point in its domain. It implies a "smooth" curve without sharp corners, cusps, or vertical tangents.

Geometric Meaning: The derivative $f'(x)$ represents the slope of the tangent line to the graph of $f(x)$ at the point $(x, f(x))$.

Physical Meaning: If $f(t)$ represents the position of an object at time $t$, then $f'(t)$ represents its instantaneous velocity.

Relationship between Differentiability and Continuity:
If a function $f(x)$ is differentiable at a point $c$, then it must be continuous at $c$.
The converse is not true: a function can be continuous at a point but not differentiable (e.g., $f(x) = |x|$ at $x=0$).

Differentiable & Continuous

Continuous, Not Differentiable (Corner)

The derivative of a function $f(x)$ can itself be a function, $f'(x)$. The continuity of $f'(x)$ implies that the original function $f(x)$ is "smooth" without abrupt changes in slope. (PYQ 8)

Rules of Differentiation:

Derivatives of Standard Functions:
* $\frac{d}{dx}(\sin x) = \cos x$
* $\frac{d}{dx}(\cos x) = -\sin x$
* $\frac{d}{dx}(\tan x) = \sec^2 x$
* $\frac{d}{dx}(e^x) = e^x$
* $\frac{d}{dx}(a^x) = a^x \ln a$
* $\frac{d}{dx}(\ln x) = \frac{1}{x}$
* $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$
* $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$
* $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$

Implicit Differentiation: Used when $y$ cannot be easily expressed as an explicit function of $x$. Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the chain rule to terms involving $y$. (PYQ 10, 11)

Higher Order Derivatives: The derivative of $f'(x)$ is the second derivative $f''(x)$, and so on. $f^{(n)}(x)$ denotes the $n$-th derivative.

Constant Derivative Implies Constant Function: If $f'(x) = 0$ for all $x$ in an interval $(a, b)$, then $f(x)$ is a constant function on that interval. This is a direct consequence of the Mean Value Theorem. If $f'(x) = 0$ for all rational numbers $q$, and $f(x)$ is twice differentiable (implying $f'(x)$ is continuous), then $f'(x)$ must be 0 for all real $x$. Therefore, $f(x)$ is a constant function. (PYQ 9)

Worked Example: Implicit Differentiation and Chain Rule

Problem: Find $\frac{dy}{dx}$ if $x^2 + y^2 = 25$.

Solution:

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

Step 2: Apply differentiation rules. For $y^2$, use the chain rule (treating $y$ as a function of $x$).

Step 3: Isolate $\frac{dy}{dx}$.

Answer: $\frac{dy}{dx} = -\frac{x}{y}$

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4. Applications of Derivatives

a. Equation of Tangents and Normals

The derivative provides the slope of the tangent line to a curve at a given point.

Worked Example: Tangent Line and X-intercept

Problem: Let $f(x) = \sqrt{x}$. Draw a tangent to the curve $y = f(x)$ at the point whose $x$-coordinate is $4$. Where does this tangent intersect the $X$-axis?

Solution:

Step 1: Find the $y$-coordinate of the point.

Given $x_0 = 4$, $y_0 = f(4) = \sqrt{4} = 2$.
The point is $(4, 2)$.

Step 2: Find the derivative $f'(x)$.

Step 3: Calculate the slope of the tangent at $x=4$.

Step 4: Write the equation of the tangent line.

Using $y - y_0 = m_T(x - x_0)$:

Step 5: Find the $X$-intercept.

The $X$-axis is where $y=0$. Substitute $y=0$ into the tangent equation:

Answer: The tangent intersects the $X$-axis at $x = -4$.

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b. Related Rates

Related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. The key is to use implicit differentiation with respect to time ($t$). (PYQ 10, 11)

Strategy for Related Rates:

Identify variables: List all quantities that are changing and those that are constant.

Identify rates: List known rates of change ($\frac{dV}{dt}$, $\frac{dr}{dt}$, etc.) and the rate to be found.

Formulate an equation: Find an equation that relates the quantities from step 1. This often comes from geometry (area, volume, Pythagorean theorem).

Differentiate implicitly: Differentiate both sides of the equation with respect to time $t$, using the chain rule.

Substitute and solve: Substitute all known values into the differentiated equation and solve for the unknown rate.

Worked Example: Related Rates (Volume of a Sphere)

Problem: A spherical ball of ice of radius $20m$ is dropped in a vat of hot water. The ice melts in such a way that (i) the shape of the ball remains spherical, and (ii) the radius of the ball decreases at a constant rate of $0.5ms^{-1}$. At what rate does the volume of the ice ball decrease, when the radius of the ball is $15m$?

Solution:

Step 1: Identify variables and rates.

• Let $V$ be the volume of the sphere and $r$ be its radius.


• Given rate of change of radius: $\frac{dr}{dt} = -0.5 \, \text{ms}^{-1}$ (negative because the radius is decreasing).


• We need to find the rate of change of volume: $\frac{dV}{dt}$ when $r = 15 \, \text{m}$.

Step 2: Formulate an equation relating $V$ and $r$.
The volume of a sphere is given by:

Step 3: Differentiate implicitly with respect to time $t$.

Apply the chain rule:

Step 4: Substitute known values and solve.
We need $\frac{dV}{dt}$ when $r = 15 \, \text{m}$ and $\frac{dr}{dt} = -0.5 \, \text{ms}^{-1}$.

The negative sign indicates that the volume is decreasing. The rate of decrease is $450\pi \, \text{m}^3\text{s}^{-1}$.

Answer: The volume of the ice ball decreases at a rate of $450\pi \, \text{m}^3\text{s}^{-1}$.

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c. Monotonicity and Extrema

Derivatives are used to determine where a function is increasing or decreasing and to find its local maximum or minimum values.

* Increasing/Decreasing Functions:
* If $f'(x) > 0$ on an interval, $f(x)$ is increasing on that interval.
* If $f'(x) < 0$ on an interval, $f(x)$ is decreasing on that interval.
* If $f'(x) = 0$ on an interval, $f(x)$ is constant on that interval.

* Critical Points: A point $c$ in the domain of $f$ is a critical point if $f'(c) = 0$ or $f'(c)$ is undefined. Local extrema (maxima or minima) can only occur at critical points.

* First Derivative Test:
1. Find critical points.
2. Test the sign of $f'(x)$ in intervals around each critical point.
3. If $f'(x)$ changes from positive to negative at $c$, $f(c)$ is a local maximum.
4. If $f'(x)$ changes from negative to positive at $c$, $f(c)$ is a local minimum.
5. If $f'(x)$ does not change sign, $f(c)$ is neither a local maximum nor minimum.

* Second Derivative Test:
1. Find critical points where $f'(c) = 0$.
2. Calculate $f''(x)$.
3. If $f''(c) > 0$, $f(c)$ is a local minimum.
4. If $f''(c) < 0$, $f(c)$ is a local maximum.
5. If $f''(c) = 0$, the test is inconclusive (use the First Derivative Test).

Worked Example: Local Extrema

Problem: Find the local extrema of $f(x) = x^3 - 6x^2 + 9x + 1$.

Solution:

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

Step 2: Find critical points by setting $f'(x) = 0$.

Divide by 3:

Factor:

Critical points are $x=1$ and $x=3$.

Step 3: Use the Second Derivative Test. Find $f''(x)$.

Step 4: Evaluate $f''(x)$ at each critical point.

For $x=1$:

Since $f''(1) < 0$, there is a local maximum at $x=1$.

Local maximum value is $5$ at $x=1$.

For $x=3$:

Since $f''(3) > 0$, there is a local minimum at $x=3$.

Local minimum value is $1$ at $x=3$.

Answer: $\boxed{\text{Local maximum at } (1, 5) \text{ and local minimum at } (3, 1)}$

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d. Concavity and Inflection Points

The second derivative helps determine the concavity of a function's graph and identify inflection points.

* Concavity:
* If $f''(x) > 0$ on an interval, the graph of $f(x)$ is concave up (opens upwards) on that interval.
* If $f''(x) < 0$ on an interval, the graph of $f(x)$ is concave down (opens downwards) on that interval.

Inflection Point: A point $(c, f(c))$ on the graph of $f(x)$ is an inflection point if the concavity of the graph changes at $c$. This typically occurs where $f''(c) = 0$ or $f''(c)$ is undefined, and* $f''(x)$ changes sign around $c$. (PYQ 7)

Worked Example: Inflection Points

Problem: Given the function $f(x) = \frac{x^5}{20} - \frac{x^4}{2} + 3x + 1$, identify its inflection points.

Solution:

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

Step 2: Find the second derivative $f''(x)$.

Step 3: Set $f''(x) = 0$ to find potential inflection points.

Factor out $x^2$:

Potential inflection points are $x=0$ and $x=6$.

Step 4: Analyze the sign of $f''(x)$ around these points to check for a change in concavity.

* For $x=0$:
* Choose a test value slightly less than $0$, e.g., $x=-1$:

(concave down).
* Choose a test value slightly greater than $0$, e.g., $x=1$:

(concave down).
Since $f''(x)$ does not change sign around $x=0$, $x=0$ is not an inflection point.

* For $x=6$:
* Choose a test value slightly less than $6$, e.g., $x=5$:

(concave down).
* Choose a test value slightly greater than $6$, e.g., $x=7$:

(concave up).
Since $f''(x)$ changes sign from negative to positive at $x=6$, $x=6$ is an inflection point.

Answer: $\boxed{x=6 \text{ is the only inflection point.}}$

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $f(x) = \begin{cases} x^2+1, & x \le 1 \\ ax+b, & x > 1 \end{cases}$. For $f(x)$ to be differentiable at $x=1$, what must be the values of $a$ and $b$?" options=["$a=2, b=1$","$a=1, b=2$","$a=2, b=0$","$a=0, b=2$"] answer="$a=2, b=0$" hint="For differentiability, the function must first be continuous. Then, the left-hand derivative must equal the right-hand derivative." solution="Step 1: For continuity at $x=1$, the left-hand limit, right-hand limit, and function value must be equal.

So,

Step 2: For differentiability at $x=1$, the left-hand derivative must equal the right-hand derivative.

So,

Step 3: Substitute $a=2$ into the continuity equation.

Therefore, $a=2$ and $b=0$.
Answer: \boxed{a=2, b=0}"
:::

:::question type="NAT" question="A ladder 13 meters long is leaning against a wall. The base of the ladder is pulled away from the wall at a rate of $0.6 \text{ m/s}$. How fast is the top of the ladder sliding down the wall when the base is 5 meters from the wall? (Provide the speed in m/s, positive value)." answer="0.25" hint="Use the Pythagorean theorem to relate the distance of the ladder's base from the wall, its height on the wall, and its length. Then differentiate implicitly with respect to time." solution="Step 1: Define variables and known rates.
Let $L$ be the length of the ladder, $x$ be the distance of the base from the wall, and $y$ be the height of the top of the ladder on the wall.
Given: $L = 13 \text{ m}$ (constant).
Rate at which the base is pulled away: $\frac{dx}{dt} = 0.6 \text{ m/s}$.
We need to find $\frac{dy}{dt}$ (speed at which the top slides down) when $x = 5 \text{ m}$.

Step 2: Formulate an equation relating $x$, $y$, and $L$.
By the Pythagorean theorem:

Step 3: Find $y$ when $x=5 \text{ m}$.

Step 4: Differentiate the equation $x^2 + y^2 = 169$ implicitly with respect to time $t$.

Divide by 2:

Step 5: Substitute the known values and solve for $\frac{dy}{dt}$.
Substitute $x=5$, $y=12$, and $\frac{dx}{dt}=0.6$:

The negative sign indicates that the height $y$ is decreasing, meaning the top of the ladder is sliding down. The speed is the absolute value of this rate.

The speed is $0.25 \text{ m/s}$.
Answer: \boxed{0.25}"
:::

:::question type="MSQ" question="Which of the following functions have a local maximum at $x=0$?" options=["$f(x) = -x^2$","$g(x) = \cos x$","$h(x) = x^3$","$k(x) = e^{-x^2}$"] answer="A,B,D" hint="Use the first or second derivative test. For a local maximum at $x=0$, $f'(0)=0$ and $f''(0) < 0$ (or $f'(x)$ changes from positive to negative at $x=0$). " solution="Let's analyze each option:
A. $f(x) = -x^2$

Since $f'(0)=0$ and $f''(0)<0$, $f(x)$ has a local maximum at $x=0$. Correct.

B. $g(x) = \cos x$

Since $g'(0)=0$ and $g''(0)<0$, $g(x)$ has a local maximum at $x=0$. Correct.

C. $h(x) = x^3$

The second derivative test is inconclusive. Use the first derivative test:
For $x<0$, $h'(x) = 3x^2 > 0$.
For $x>0$, $h'(x) = 3x^2 > 0$.
Since $h'(x)$ does not change sign (it's positive on both sides of $x=0$), $x=0$ is an inflection point, not a local extremum. Incorrect.

D. $k(x) = e^{-x^2}$

Since $k'(0)=0$ and $k''(0)<0$, $k(x)$ has a local maximum at $x=0$. Correct.
Answer: \boxed{A,B,D}"
:::

:::question type="SUB" question="Find the interval(s) where the function $f(x) = x^3 - 3x^2 - 9x + 5$ is decreasing." answer="$(-1, 3)$" hint="A function is decreasing where its first derivative is negative." solution="Step 1: Find the first derivative of $f(x)$.

Step 2: Find the critical points by setting $f'(x) = 0$.

Divide by 3:

Factor the quadratic:

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

Step 3: Create a sign table for $f'(x)$ using the critical points to define intervals.
The intervals are $(-\infty, -1)$, $(-1, 3)$, and $(3, \infty)$.

* Interval $(-\infty, -1)$: Choose test value $x=-2$.

So, $f(x)$ is increasing on $(-\infty, -1)$.

* Interval $(-1, 3)$: Choose test value $x=0$.

So, $f(x)$ is decreasing on $(-1, 3)$.

* Interval $(3, \infty)$: Choose test value $x=4$.

So, $f(x)$ is increasing on $(3, \infty)$.

Step 4: Identify the interval(s) where $f(x)$ is decreasing.
The function $f(x)$ is decreasing on the interval where $f'(x) < 0$.

Therefore, $f(x)$ is decreasing on $(-1, 3)$.
Answer: \boxed{(-1, 3)}"
:::

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Summary

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What's Next?

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Part 2: Maxima and Minima

Introduction

The concepts of maxima and minima are fundamental in calculus, allowing us to determine the largest and smallest values a function can attain. These values are crucial for understanding the behavior of functions and have extensive applications in various fields, including economics, physics, engineering, and data science, where optimizing processes or models is often required. In the CMI exam, problems involving maxima and minima frequently appear, testing your ability to identify critical points, classify them as maxima or minima, and apply these principles to solve complex real-world optimization challenges. This section will equip you with the necessary tools and techniques to master these concepts.

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Key Concepts

1. Critical Points

Critical points are the candidates for local maxima or minima. They are points where the function's rate of change is zero or undefined.

Explanation:
At a local maximum or minimum, the tangent line to the curve is horizontal, meaning its slope (the first derivative) is zero. Alternatively, if the function has a sharp turn or a vertical tangent, the derivative might be undefined.

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2. Tests for Local Extrema

To classify a critical point as a local maximum, local minimum, or neither, we use derivative tests.

2.1 First Derivative Test

This test examines the sign of the first derivative around a critical point.

Worked Example:

Problem: Find the local extrema of $f(x) = x^3 - 3x$.

Solution:

Step 1: Find the first derivative and critical points.

Set $f'(x) = 0$:

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

Step 2: Apply the First Derivative Test.

Consider $x=-1$:

• For $x < -1$ (e.g., $x=-2$),

($f$ is increasing)

• For $-1 < x < 1$ (e.g., $x=0$),

($f$ is decreasing)
Since $f'(x)$ changes from positive to negative at $x=-1$, there is a local maximum at $x=-1$.

Consider $x=1$:

• For $-1 < x < 1$ (e.g., $x=0$),

($f$ is decreasing)

• For $x > 1$ (e.g., $x=2$),

($f$ is increasing)
Since $f'(x)$ changes from negative to positive at $x=1$, there is a local minimum at $x=1$.

Answer: Local maximum at $(-1, 2)$ and local minimum at $(1, -2)$.

---

2.2 Second Derivative Test

This test uses the sign of the second derivative at a critical point to classify it.

Explanation:
The second derivative tells us about the concavity of the function. If $f''(c) < 0$, the function is concave down at $c$, suggesting a peak (maximum). If $f''(c) > 0$, the function is concave up, suggesting a valley (minimum).

Worked Example:

Problem: Use the Second Derivative Test to find the local extrema of $f(x) = x^3 - 3x$.

Solution:

Step 1: Find the first derivative and critical points (from previous example).

Critical points are $x=1$ and $x=-1$.

Step 2: Find the second derivative.

Step 3: Evaluate $f''(x)$ at each critical point.

For $x=-1$:

Since $f''(-1) < 0$, there is a local maximum at $x=-1$.

For $x=1$:

Since $f''(1) > 0$, there is a local minimum at $x=1$.

Answer: Local maximum at $(-1, 2)$ and local minimum at $(1, -2)$.

---

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#
## 3. Extrema on Closed Intervals

For a continuous function $f(x)$ on a closed interval $[a,b]$, the Extreme Value Theorem guarantees that both a global maximum and a global minimum exist.

Procedure to find global extrema on $[a,b]$:

Find all critical points of $f(x)$ in the open interval $(a,b)$.

Evaluate $f(x)$ at each critical point found in step 1.

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

The largest of these values is the global maximum, and the smallest is the global minimum.

Worked Example:

Problem: Find the global maximum and minimum of $f(x) = x^3 - 3x$ on the interval $[0, 2]$.

Solution:

Step 1: Find critical points in $(0,2)$.
From previous examples, critical points are $x=-1$ and $x=1$. Only $x=1$ lies in $(0,2)$.

Step 2: Evaluate $f(x)$ at the critical point $x=1$.

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

Step 3: Evaluate $f(x)$ at the endpoints $x=0$ and $x=2$.

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

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

Step 4: Compare values.
Values are $f(1)=-2$, $f(0)=0$, $f(2)=2$.
The largest value is $2$, and the smallest value is $-2$.

Answer: Global maximum is $2$ at $x=2$. Global minimum is $-2$ at $x=1$.

---

#
## 4. Concavity and Inflection Points

Concavity describes the direction in which the curve bends. Inflection points are where the concavity changes.

Explanation:

• If $f''(x) > 0$, the slope $f'(x)$ is increasing, meaning the curve is bending upwards.


• If $f''(x) < 0$, the slope $f'(x)$ is decreasing, meaning the curve is bending downwards.


• An inflection point is where this behavior switches. Note that $f''(c)=0$ is a necessary condition, but not sufficient; $f''(x)$ must change sign. For example, for $f(x)=x^4$, $f''(0)=0$, but $f''(x)=12x^2 \ge 0$ for all $x$, so concavity doesn't change, and $x=0$ is not an inflection point.

Worked Example:

Problem: Find the intervals of concavity and inflection points for $f(x) = x^4 - 4x^3$.

Solution:

Step 1: Find the first and second derivatives.

$f'(x) = 4x^3 - 12x^2$

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

Step 2: Find points where $f''(x) = 0$ or is undefined.

$12x^2 - 24x = 0$

$12x(x - 2) = 0$

So, $x=0$ and $x=2$ are potential inflection points.

Step 3: Test intervals for the sign of $f''(x)$.

• For $x < 0$ (e.g., $x=-1$), $f''(-1) = 12(-1)(-1-2) = 12(-1)(-3) = 36 > 0$. (Concave Up)


• For $0 < x < 2$ (e.g., $x=1$), $f''(1) = 12(1)(1-2) = 12(-1) = -12 < 0$. (Concave Down)


• For $x > 2$ (e.g., $x=3$), $f''(3) = 12(3)(3-2) = 36(1) = 36 > 0$. (Concave Up)

Step 4: Identify inflection points.
Concavity changes at $x=0$ and $x=2$.
$f(0) = 0^4 - 4(0)^3 = 0$
$f(2) = 2^4 - 4(2)^3 = 16 - 32 = -16$

Answer:

• Concave up on $(-\infty, 0)$ and $(2, \infty)$.


• Concave down on $(0, 2)$.


• Inflection points at $(0, 0)$ and $(2, -16)$.

---

#
## 5. Optimization in Real-World Problems

Many practical problems involve finding the maximum or minimum value of a quantity (e.g., profit, volume, cost, distance).

General Strategy:

Understand the Problem: Read carefully, identify what needs to be maximized or minimized.

Draw a Diagram (if applicable): Visual representation helps.

Define Variables: Assign symbols to quantities.

Formulate the Objective Function: Write an equation for the quantity to be optimized in terms of the variables.

Identify Constraints: Write equations or inequalities relating the variables.

Reduce to a Single Variable: Use constraint equations to express the objective function in terms of a single independent variable.

Determine the Domain: Find the possible values for the independent variable.

Apply Calculus: Find critical points using the first derivative and classify them using the First or Second Derivative Test. Consider endpoints if the domain is a closed interval.

Interpret the Result: State the answer in the context of the problem.

Worked Example:

Problem: A farmer wants to fence a rectangular plot of land adjacent to a river. No fence is needed along the river. If the farmer has 200 meters of fencing, what is the maximum area of the land he can enclose?

Solution:

Step 1: Understand the Problem - Maximize area, with a fixed perimeter (fencing length).

Step 2: Draw a Diagram.




River
Area
$l$
$w$
$w$

Step 3: Define Variables.
Let $l$ be the length of the side parallel to the river.
Let $w$ be the width of the sides perpendicular to the river.

Step 4: Formulate the Objective Function.
$A = l \cdot w$

Step 5: Identify Constraints.
$2w + l = 200$

Step 6: Reduce to a Single Variable.
From the constraint,
$l = 200 - 2w$
Substitute this into the area formula:
$A(w) = (200 - 2w)w = 200w - 2w^2$

Step 7: Determine the Domain.
Since $w$ and $l$ must be positive:
$w > 0$
$l = 200 - 2w > 0 \implies 200 > 2w \implies w < 100$
So, the domain for $w$ is $(0, 100)$.

Step 8: Apply Calculus.
Find the first derivative of $A(w)$:
$A'(w) = 200 - 4w$
Set $A'(w) = 0$ to find critical points:
$200 - 4w = 0$
$4w = 200$
$w = 50$
This critical point $w=50$ is within the domain $(0, 100)$.

Use the Second Derivative Test:
$A''(w) = -4$
Since $A''(50) = -4 < 0$, this confirms that $w=50$ corresponds to a local maximum. Since it's the only critical point in an open interval and the function is a downward-opening parabola, it's also the global maximum.

Step 9: Interpret the Result.
When $w=50$ meters,
$l = 200 - 2(50) = 200 - 100 = 100 \text{ meters}$
The maximum area is
$A(50) = 100 \cdot 50 = 5000 \text{ square meters}$

Answer: The maximum area is $5000 \text{ m}^2$, achieved when the width is $50 \text{ m}$ and the length is $100 \text{ m}$.

---

#
## 6. Proving Inequalities using Calculus

Calculus can be a powerful tool to prove inequalities by analyzing the monotonicity or extrema of a related function.

General Strategy:
To prove $f(x) \ge g(x)$ for a given interval, consider the function $h(x) = f(x) - g(x)$. If you can show that the global minimum of $h(x)$ on that interval is non-negative (i.e., $\min h(x) \ge 0$), then the inequality holds.

Worked Example:

Problem: Prove that $e^x \ge 1+x$ for all real $x$.

Solution:

Step 1: Define a new function.
$h(x) = e^x - (1+x) = e^x - x - 1$
We need to show that $h(x) \ge 0$ for all real $x$.

Step 2: Find the derivative of $h(x)$.

$h'(x) = e^x - 1$

Step 3: Find critical points.
Set $h'(x) = 0$:

$e^x - 1 = 0$

$e^x = 1$

$x = 0$

Step 4: Use the Second Derivative Test to classify the critical point.

$h''(x) = e^x$

$h''(0) = e^0 = 1$
Since $h''(0) = 1 > 0$, there is a local minimum at $x=0$.

Step 5: Evaluate $h(x)$ at the minimum.

$h(0) = e^0 - 0 - 1 = 1 - 0 - 1 = 0$
Since $x=0$ is the only critical point and it corresponds to a local minimum, and $h''(x) = e^x > 0$ for all $x$, the function $h(x)$ is concave up everywhere, meaning this local minimum is also the global minimum.

Step 6: Conclude the inequality.
The global minimum value of $h(x)$ is $0$, which occurs at $x=0$. Therefore, $h(x) \ge 0$ for all real $x$.
This means $e^x - x - 1 \ge 0$, which implies $e^x \ge 1+x$.

Answer: The inequality is proven.

---
```

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#
## 7. Optimization of Functions with Specific Forms (e.g., $\frac{\ln x}{x}$)

Functions like $f(x) = \frac{\ln x}{x}$ frequently appear in CMI for proving inequalities. Analyzing its behavior is key.

Worked Example:

Problem: Analyze the function $f(x) = \frac{\ln x}{x}$ for $x>0$. Find its maximum value and use it to compare $a^b$ and $b^a$ for $a, b > 0$.

Solution:

Part 1: Analyze $f(x) = \frac{\ln x}{x}$.

Step 1: Find the first derivative.

Step 2: Find critical points.
Set $f'(x) = 0$:

The critical point is $x=e$.

Step 3: Use the First Derivative Test (or Second Derivative Test).
For $x < e$ (e.g., $x=1$), $f'(1) = \frac{1 - \ln 1}{1^2} = \frac{1 - 0}{1} = 1 > 0$. ($f$ is increasing)
For $x > e$ (e.g., $x=e^2$), $f'(e^2) = \frac{1 - \ln e^2}{(e^2)^2} = \frac{1 - 2}{e^4} = -\frac{1}{e^4} < 0$. ($f$ is decreasing)
Since $f'(x)$ changes from positive to negative at $x=e$, there is a local maximum at $x=e$.

Step 4: Evaluate the maximum value.

Since it's the only critical point for $x>0$ and $f(x) \to -\infty$ as $x \to 0^+$ and $f(x) \to 0$ as $x \to \infty$, this local maximum is also the global maximum.

Part 2: Use this to compare $a^b$ and $b^a$.

Consider the inequality $a^b < b^a$.
This is equivalent to $\ln(a^b) < \ln(b^a)$.

This means $f(a) < f(b)$.

We know that $f(x) = \frac{\ln x}{x}$ has a global maximum at $x=e$.

• For $x > e$, $f(x)$ is a decreasing function.


• For $0 < x < e$, $f(x)$ is an increasing function.

If $a > b \ge 3$:
Since $e \approx 2.718$, both $a$ and $b$ are greater than $e$.
In the interval $(e, \infty)$, $f(x)$ is a decreasing function.
If $a > b$, then $f(a) < f(b)$ because $f$ is decreasing.
So, $\frac{\ln a}{a} < \frac{\ln b}{b}$.
This implies $b \ln a < a \ln b$, which further implies $\ln(a^b) < \ln(b^a)$, and thus $a^b < b^a$.

Answer: The function $f(x) = \frac{\ln x}{x}$ has a global maximum of $\frac{1}{e}$ at $x=e$. For $a > b \ge 3$, since $a, b > e$ and $f(x)$ is decreasing for $x>e$, we have $f(a) < f(b)$, which proves $a^b < b^a$.

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#
## 8. Optimization of Composite Functions

When a function is a composition, $f(g(x))$, we can often find its extrema by finding the extrema of the inner function $g(x)$ or by substitution.

Worked Example:

Problem: Find the range of the function $h(x) = \sin^2 x - \sin x + 2$.

Solution:

Step 1: Use substitution.
Let $u = \sin x$.
Since $-1 \le \sin x \le 1$, the variable $u$ is restricted to the interval $[-1, 1]$.
The function becomes a quadratic in $u$: $g(u) = u^2 - u + 2$.

Step 2: Find the extrema of $g(u)$ on the interval $[-1, 1]$.
Find the derivative of $g(u)$:

Set $g'(u) = 0$:

This critical point $u=\frac{1}{2}$ lies within the interval $[-1, 1]$.

Step 3: Evaluate $g(u)$ at the critical point and the endpoints.

• At $u=\frac{1}{2}$:

• At endpoint $u=-1$:

• At endpoint $u=1$:

Step 4: Determine the maximum and minimum values.
The values are $\frac{7}{4}$, $4$, and $2$.
The global minimum is $\frac{7}{4}$.
The global maximum is $4$.

Answer: The range of the function $h(x)$ is $\left[\frac{7}{4}, 4\right]$.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $f(x) = x^4 - 8x^2$. Which of the following statements about its local extrema is true?" options=["$f(x)$ has a local maximum at $x=0$ and local minima at $x=\pm 2$.","$f(x)$ has a local minimum at $x=0$ and local maxima at $x=\pm 2$.","$f(x)$ has local minima at $x=0$ and $x=\pm 2$.","$f(x)$ has a local maximum at $x=\pm 2$ and a local minimum at $x=0$."] answer="$f(x)$ has a local maximum at $x=0$ and local minima at $x=\pm 2$." hint="Find $f'(x)$ and $f''(x)$ to classify critical points." solution="Step 1: Find the first derivative.

Step 2: Find critical points by setting $f'(x)=0$.



Critical points are $x=0, x=2, x=-2$.
Step 3: Find the second derivative.

Step 4: Apply the Second Derivative Test.

• For $x=0$: $f''(0) = 12(0)^2 - 16 = -16 < 0$. So, $f(x)$ has a local maximum at $x=0$.


• For $x=2$: $f''(2) = 12(2)^2 - 16 = 12(4) - 16 = 48 - 16 = 32 > 0$. So, $f(x)$ has a local minimum at $x=2$.


• For $x=-2$: $f''(-2) = 12(-2)^2 - 16 = 12(4) - 16 = 48 - 16 = 32 > 0$. So, $f(x)$ has a local minimum at $x=-2$.

The correct statement is that $f(x)$ has a local maximum at $x=0$ and local minima at $x=\pm 2$.
Answer: \boxed{f(x) \text{ has a local maximum at } x=0 \text{ and local minima at } x=\pm 2}"
:::

:::question type="NAT" question="A company's daily profit $P$ (in thousands of rupees) from selling $x$ units of a product is given by $P(x) = -0.002x^2 + 10x - 500$. What is the maximum daily profit (in thousands of rupees) the company can achieve?" answer="12000" hint="Find the critical point and verify it's a maximum." solution="Step 1: Find the first derivative of the profit function.

Step 2: Set $P'(x) = 0$ to find the critical point.



Step 3: Verify it's a maximum using the Second Derivative Test.

Since $P''(x)$ is always negative, the critical point $x=2500$ corresponds to a global maximum.
Step 4: Calculate the maximum profit.




The maximum daily profit is 12000 thousand rupees.
Answer: \boxed{12000}"
:::

:::question type="MSQ" question="Let $f(x) = x \ln x$ for $x > 0$. Which of the following statements is/are true?" options=["$f(x)$ has a local minimum at $x = e^{-1}$.","$f(x)$ has a local maximum at $x = e^{-1}$.","$f(x)$ is concave up for all $x > 0$.","$f(x)$ has an inflection point at $x = e^{-1}$."] answer="A,C" hint="Calculate $f'(x)$ and $f''(x)$." solution="Step 1: Find the first derivative.

Step 2: Find critical points by setting $f'(x)=0$.



Step 3: Find the second derivative.

Step 4: Apply the Second Derivative Test at $x=e^{-1}$.

Since $f''(e^{-1}) = e > 0$, $f(x)$ has a local minimum at $x = e^{-1}$. So, option A is true, and option B is false.
Step 5: Analyze concavity.
Since $f''(x) = \frac{1}{x}$ and $x > 0$, we have $f''(x) > 0$ for all $x > 0$. This means $f(x)$ is concave up for all $x > 0$. So, option C is true.
Since $f''(x)$ never changes sign, there are no inflection points. So, option D is false.
Answer: \boxed{A,C}"
:::

:::question type="SUB" question="A cylindrical can is to be made to hold $1000 \text{ cm}^3$ of liquid. Find the dimensions (radius and height) of the can that will minimize the amount of metal used (i.e., minimize the surface area)." answer="Radius $r = \sqrt[3]{\frac{500}{\pi}} \text{ cm}$, Height $h = 2\sqrt[3]{\frac{500}{\pi}} \text{ cm}$" hint="Formulate surface area and volume equations, then minimize surface area in terms of one variable." solution="Step 1: Define variables and formulas.
Let $r$ be the radius of the base and $h$ be the height of the cylindrical can.
Volume $V = \pi r^2 h$.
Surface Area $A = 2\pi r^2 + 2\pi r h$ (includes top and bottom).

Step 2: Use the constraint to reduce the objective function to a single variable.
Given $V = 1000 \text{ cm}^3$.

Express $h$ in terms of $r$:

Substitute $h$ into the surface area formula:


The domain for $r$ is $(0, \infty)$.

Step 3: Find the first derivative of $A(r)$.

Step 4: Set $A'(r) = 0$ to find critical points.

Step 5: Use the Second Derivative Test to verify it's a minimum.

For $r = \sqrt[3]{\frac{500}{\pi}}$, $r^3 = \frac{500}{\pi}$.

Since $A''(r) > 0$, this critical point corresponds to a local minimum. As it's the only critical point and $A(r) \to \infty$ as $r \to 0^+$ and $r \to \infty$, it's the global minimum.

Step 6: Calculate the corresponding height $h$.

Alternatively, from $4\pi r^3 = 2000$, we have $2\pi r^3 = 1000$.
Since $h = \frac{1000}{\pi r^2}$, we can write $h = \frac{2\pi r^3}{\pi r^2} = 2r$.
So, $h = 2\sqrt[3]{\frac{500}{\pi}}$.

The dimensions that minimize the surface area are $r = \sqrt[3]{\frac{500}{\pi}} \text{ cm}$ and $h = 2\sqrt[3]{\frac{500}{\pi}} \text{ cm}$.
Answer: \boxed{\text{Radius } r = \sqrt[3]{\frac{500}{\pi}} \text{ cm}, \text{ Height } h = 2\sqrt[3]{\frac{500}{\pi}} \text{ cm}}"
:::

:::question type="SUB" question="Prove that for $x > 0$, $x - \frac{x^2}{2} < \ln(1+x)$. (Hint: Consider the function $f(x) = \ln(1+x) - x + \frac{x^2}{2}$)" answer="Proof that $f(x) > 0$" hint="Analyze the monotonicity of $f(x)$ by checking its derivatives." solution="Step 1: Define the function to analyze.
Let $f(x) = \ln(1+x) - x + \frac{x^2}{2}$. We want to show $f(x) > 0$ for $x > 0$.

Step 2: Evaluate $f(x)$ at the boundary $x=0$.

Step 3: Find the first derivative of $f(x)$.

Step 4: Analyze the sign of $f'(x)$ for $x > 0$.
To do this, let's find the derivative of $f'(x)$, i.e., $f''(x)$.

Set $f''(x) = 0$:




Since we are interested in $x > 0$, consider $x=0$.

Now, let's check the sign of $f''(x)$ for $x > 0$:
For $x > 0$, $(1+x)^2 > 1$.
So, $\frac{1}{(1+x)^2} < 1$.
Therefore, $f''(x) = 1 - \frac{1}{(1+x)^2} > 0$ for all $x > 0$.

Step 5: Conclude about $f'(x)$.
Since $f''(x) > 0$ for $x > 0$, $f'(x)$ is strictly increasing for $x > 0$.
Now, evaluate $f'(x)$ at $x=0$:

Since $f'(x)$ is strictly increasing for $x > 0$ and $f'(0)=0$, it means $f'(x) > 0$ for all $x > 0$.

Step 6: Conclude about $f(x)$.
Since $f'(x) > 0$ for $x > 0$, $f(x)$ is strictly increasing for $x > 0$.
We know $f(0) = 0$. Since $f(x)$ is strictly increasing for $x > 0$, it must be that $f(x) > f(0)$ for $x > 0$.
Thus, $f(x) > 0$ for $x > 0$.
Substituting back the definition of $f(x)$:

This proves the inequality.
Answer: \boxed{\text{Proof that } f(x) > 0}"
:::

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Summary

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What's Next?

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Let $f(x) = x e^{-x^2/2}$. Which of the following statements is true about the local extrema of $f(x)$?" options=["$f(x)$ has a local minimum at $x=-1$ and a local maximum at $x=1$.","$f(x)$ has a local maximum at $x=-1$ and a local minimum at $x=1$.","$f(x)$ has a local minimum at $x=0$ only.","$f(x)$ has no local extrema."] answer="A" hint="Use the product rule and chain rule to find $f'(x)$. Then, set $f'(x)=0$ to find critical points and apply the First Derivative Test." solution="To find the local extrema, we first need to find the first derivative $f'(x)$.
Given $f(x) = x e^{-x^2/2}$.
Using the product rule $(uv)' = u'v + uv'$ where $u=x$ and $v=e^{-x^2/2}$:
$u' = 1$

So,

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

Since $e^{-x^2/2}$ is always positive, we must have $1 - x^2 = 0$.

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

Now, we use the First Derivative Test to classify these critical points by examining the sign of $f'(x)$ around them:
* For $x < -1$ (e.g., $x=-2$): $f'(-2) = e^{-(-2)^2/2}(1 - (-2)^2) = e^{-2}(1-4) = -3e^{-2} < 0$. So $f(x)$ is decreasing.
* For $-1 < x < 1$ (e.g., $x=0$): $f'(0) = e^{-(0)^2/2}(1 - (0)^2) = e^0(1-0) = 1 > 0$. So $f(x)$ is increasing.
* For $x > 1$ (e.g., $x=2$): $f'(2) = e^{-(2)^2/2}(1 - (2)^2) = e^{-2}(1-4) = -3e^{-2} < 0$. So $f(x)$ is decreasing.

Based on the sign changes:
* At $x=-1$, $f'(x)$ changes from negative to positive. Thus, $f(x)$ has a local minimum at $x=-1$.
* At $x=1$, $f'(x)$ changes from positive to negative. Thus, $f(x)$ has a local maximum at $x=1$.

Therefore, option A is true.

The final answer is $\boxed{A}$"
:::

:::question type="NAT" question="A cylindrical can without a top is to be made to hold $1000 \pi$ cm³ of liquid. Find the radius of the can (in cm) that minimizes the surface area. (Enter your answer as a plain number)." answer="10" hint="Define the volume and surface area equations. Use the volume constraint to express the surface area as a function of a single variable (radius or height), then differentiate and find the minimum." solution="Let $r$ be the radius of the cylindrical can and $h$ be its height.
The volume of the can is given by $V = \pi r^2 h$.
We are given $V = 1000 \pi$ cm³.
So,

This implies $r^2 h = 1000$, from which we can express $h$ in terms of $r$:

The surface area of the can without a top consists of the area of the base and the lateral surface area.
Area of the base $= \pi r^2$.
Lateral surface area $= 2 \pi r h$.
Total surface area $A = \pi r^2 + 2 \pi r h$.

Substitute the expression for $h$ into the surface area equation:

To minimize $A(r)$, we need to find its derivative with respect to $r$ and set it to zero.

Set $A'(r) = 0$:

Divide both sides by $2 \pi$:

To confirm this is a minimum, we can use the Second Derivative Test:

Now, evaluate $A''(10)$:

Since $A''(10) = 6 \pi > 0$, the surface area is indeed minimized at $r=10$ cm.

The final answer is $\boxed{10}$"
:::

:::question type="MCQ" question="Let $f(x)$ be a differentiable function whose derivative is given by $f'(x) = -(x-1)(x-3)^2$. Which of the following statements about $f(x)$ is true?" options=["$f(x)$ has a local minimum at $x=1$.","$f(x)$ has a local maximum at $x=1$.","$f(x)$ has a local maximum at $x=3$.","$f(x)$ is increasing on the interval $(1, 3)$."] answer="B" hint="Identify the critical points where $f'(x)=0$. Then, analyze the sign changes of $f'(x)$ around these points to determine the nature of the extrema and intervals of increase/decrease." solution="The critical points of $f(x)$ are the values of $x$ for which $f'(x)=0$ or $f'(x)$ is undefined.
Given $f'(x) = -(x-1)(x-3)^2$.
Setting $f'(x)=0$:

This gives us critical points at $x=1$ and $x=3$.

Now, we analyze the sign of $f'(x)$ in intervals around these critical points to apply the First Derivative Test:

Interval $(-\infty, 1)$: Choose a test point, e.g., $x=0$.

Since $f'(x) > 0$, $f(x)$ is increasing on $(-\infty, 1)$.

Interval $(1, 3)$: Choose a test point, e.g., $x=2$.

Since $f'(x) < 0$, $f(x)$ is decreasing on $(1, 3)$.

Interval $(3, \infty)$: Choose a test point, e.g., $x=4$.

Since $f'(x) < 0$, $f(x)$ is decreasing on $(3, \infty)$.

Based on this analysis:
* At $x=1$: $f'(x)$ changes from positive to negative. This indicates that $f(x)$ has a local maximum at $x=1$. (Option B is true).
* At $x=3$: $f'(x)$ does not change sign (it is negative before $x=3$ and remains negative after $x=3$). Therefore, $f(x)$ has neither a local maximum nor a local minimum at $x=3$. It is a point where the function's decrease flattens out temporarily.
* Option A is false because $x=1$ is a local maximum, not a local minimum.
* Option C is false because $x=3$ is not a local maximum (or minimum).
* Option D is false because $f(x)$ is decreasing on the interval $(1, 3)$.

The final answer is $\boxed{B}$"
:::

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What's Next?