Applications of derivatives

This chapter explores the practical utility of derivatives in analyzing function characteristics. It details methods for determining function monotonicity, locating extrema, and characterizing tangent and normal lines. A thorough understanding of these applications is fundamental for solving advanced calculus problems frequently encountered in the CMI examinations.

---

Chapter Contents

| Topic |

|---|-------| | 1 | Monotonicity | | 2 | Tangent and normal | | 3 | Maxima and minima | | 4 | Tangency conditions | | 5 | Root-location arguments |

---

We begin with Monotonicity.

Part 1: Monotonicity

Monotonicity

Overview

Monotonicity tells us whether a function keeps increasing or keeps decreasing on an interval. In calculus, this is one of the main applications of derivatives because the sign of

f'(x)

usually reveals the behaviour of

f(x)

. In CMI-style questions, monotonicity is not just a definition-based topic; it is used to study turning points, interval behaviour, uniqueness of roots, and qualitative graph shape. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Define increasing, decreasing, strictly increasing, and strictly decreasing functions.

Use the derivative sign to determine monotonicity on intervals.

Make and use sign charts for $f'(x)$ .

Distinguish local behaviour from global monotonicity.

Use monotonicity to infer number of roots and graph behaviour.

---

Core Definitions

📖 Increasing and Decreasing Functions

Let $f$ be defined on an interval $I$ .

$f$ is increasing on $I$ if for any $x_1 < x_2$ in $I$ , we have

\qquad f(x_1) \le f(x_2)

$f$ is strictly increasing on $I$ if for any $x_1 < x_2$ in $I$ , we have

\qquad f(x_1) < f(x_2)

$f$ is decreasing on $I$ if for any $x_1 < x_2$ in $I$ , we have

\qquad f(x_1) \ge f(x_2)

$f$ is strictly decreasing on $I$ if for any $x_1 < x_2$ in $I$ , we have

\qquad f(x_1) > f(x_2)

---

Derivative Test for Monotonicity

📐 First Derivative Criterion

Suppose $f$ is differentiable on an interval $I$ .

If $f'(x) > 0$ for all $x \in I$ , then $f$ is strictly increasing on $I$ .

If $f'(x) < 0$ for all $x \in I$ , then $f$ is strictly decreasing on $I$ .

If $f'(x) = 0$ for all $x \in I$ , then $f$ is constant on $I$ .

❗ Non-Strict Version

If $f'(x) \ge 0$ on an interval, then $f$ is increasing there.

If $f'(x) \le 0$ on an interval, then $f$ is decreasing there.

---

Critical Points and Sign Changes

📖 Critical Points

A point $x=a$ is called a critical point for monotonicity analysis if:

$f'(a)=0$ , or

$f'(a)$ does not exist, but $f(a)$ exists

These points are important because monotonicity can change only at such points.

💡 Sign Chart Method

To study monotonicity:

Compute $f'(x)$

Find where $f'(x)=0$ or is undefined

Split the real line into intervals using these points

Check the sign of $f'(x)$ on each interval

Positive means increasing, negative means decreasing

---

How Sign of Derivative Controls Behaviour

📐 Interpretation of

f'(x)

$f'(x) > 0$ means the graph is moving upward as $x$ increases

$f'(x) < 0$ means the graph is moving downward as $x$ increases

$f'(x) = 0$ means a horizontal tangent may occur, but not necessarily a maximum or minimum

---

Important Subtle Point

⚠️ Zero Derivative at a Point Does Not Decide Monotonicity

A single point where $f'(a)=0$ does not tell the full story.

Example:
For $f(x)=x^3$ ,
$\qquad f'(x)=3x^2$

At $x=0$ , we have
$\qquad f'(0)=0$

But $f(x)=x^3$ is increasing on all real numbers.

So derivative equal to zero at a point does not mean the function stops increasing globally.

---

Standard Procedure

💡 Monotonicity Procedure

To determine where a function is increasing or decreasing:

Differentiate the function

Solve $f'(x)=0$

Mark critical points on a number line

Test sign of $f'(x)$ on each interval

Write final increasing and decreasing intervals carefully

---

Minimal Worked Examples

Example 1 Determine the monotonicity of

f(x)=x^2

. We have

\qquad f'(x)=2x

Now:

if $x<0$ , then $f'(x)<0$

if $x>0$ , then $f'(x)>0$

So:

$f$ is decreasing on $\qquad (-\infty,0)$

$f$ is increasing on $\qquad (0,\infty)$

Hence

x^2

is not monotonic on all real numbers. --- Example 2 Determine the monotonicity of

f(x)=x^3-3x

. Differentiate:

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

Now:

for $x<-1$ , $f'(x)>0$

for $-1<x<1$ , $f'(x)<0$

for $x>1$ , $f'(x)>0$

Hence:

increasing on $\qquad (-\infty,-1)\cup(1,\infty)$

decreasing on $\qquad (-1,1)$

---

Monotonicity and Number of Roots

❗ Very Useful Consequence

If a function is strictly increasing on an interval, then it can take each value at most once on that interval.

So:

a strictly increasing function cannot have two different points with the same function value

this helps prove uniqueness of roots or intersections

For example, if

f

is strictly increasing and

f(a)<0<f(b)

, then the equation

f(x)=0

has exactly one root in

(a,b)

. ---

Relation with Local Maxima and Minima

📐 First Derivative Sign Change Test

If $f'(x)$ changes:

from positive to negative at $x=a$ , then $f$ has a local maximum at $a$

from negative to positive at $x=a$ , then $f$ has a local minimum at $a$

This is closely connected to monotonicity intervals.

---

Common Patterns in Questions

📐 Typical Exam Patterns

Find intervals of increase and decrease

Decide whether a function is monotonic on all real numbers

Find parameter values so that a function is increasing

Use monotonicity to prove uniqueness of a root

Use derivative sign to describe graph behaviour

---

Common Mistakes

⚠️ Avoid These Errors

❌ Looking only at points where $f'(x)=0$

✅ Also study the sign of

f'(x)

on intervals

❌ Saying $f'(a)=0$ implies maximum or minimum

✅ Check sign change before concluding

❌ Forgetting points where $f'(x)$ is undefined

✅ They can also split intervals

❌ Writing monotonicity on a closed interval without checking endpoints

✅ Usually write open intervals unless the question asks otherwise

❌ Mixing increasing with strictly increasing

✅ The inequalities in the definition matter

---

CMI Strategy

💡 How to Solve Monotonicity Questions

Differentiate first, simplify second.

Factor $f'(x)$ completely if possible.

Mark all sign-changing points carefully.

Build a sign chart instead of guessing from the graph.

Use monotonicity to make stronger conclusions like uniqueness of roots.

---

Practice Questions

:::question type="MCQ" question="On which interval is the function

f(x)=x^2-4x+1

decreasing?" options=["

(-\infty,2)

","

(2,\infty)

","

(-\infty,0)

","

(0,\infty)

"] answer="A" hint="Differentiate and check where the derivative is negative." solution="We have

\qquad f'(x)=2x-4=2(x-2)

The function is decreasing where

\qquad f'(x)<0

So,

\qquad 2(x-2)<0 \implies x<2

Hence the function is decreasing on

\qquad (-\infty,2)

Therefore the correct option is

\boxed{A}

." ::: :::question type="NAT" question="How many open intervals of increase does the function

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

have?" answer="2" hint="Find

f'(x)

and make a sign chart." solution="Differentiate:

\qquad f'(x)=3x^2-6x=3x(x-2)

Now check the sign:

if $x<0$ , then $f'(x)>0$

if $0<x<2$ , then $f'(x)<0$

if $x>2$ , then $f'(x)>0$

So the function is increasing on

\qquad (-\infty,0)

and

\qquad (2,\infty)

Thus the number of open intervals of increase is

\boxed{2}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If

f'(x)>0

on an interval, then

f

is strictly increasing there","If

f'(a)=0

, then

f

must have a local maximum or minimum at

a

","If

f'(x)<0

on an interval, then

f

is strictly decreasing there","A strictly increasing function can take the same value at two different points of the interval"] answer="A,C" hint="Separate derivative facts from common misconceptions." solution="1. True. Positive derivative on an interval implies strict increase.

False. For example,

f(x)=x^3

has

f'(0)=0

but no local maximum or minimum at

0

True. Negative derivative on an interval implies strict decrease.

False. A strictly increasing function cannot take the same value at two different points.

Hence the correct answer is

\boxed{A,C}

." ::: :::question type="SUB" question="Determine the intervals on which the function

f(x)=x^3-6x^2+9x

is increasing and decreasing." answer="Increasing on

(-\infty,1)

and

(3,\infty)

; decreasing on

(1,3)

" hint="Differentiate and factor completely." solution="Differentiate:

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

Factor:

\qquad f'(x)=3(x^2-4x+3)=3(x-1)(x-3)

Now study the sign:

for $x<1$ , both $(x-1)$ and $(x-3)$ are negative, so their product is positive

for $1<x<3$ , one factor is positive and the other negative, so the product is negative

for $x>3$ , both factors are positive, so the product is positive

Therefore:

$f$ is increasing on $\qquad (-\infty,1)$ and $(3,\infty)$

$f$ is decreasing on $\qquad (1,3)$

Hence the answer is

\boxed{\text{Increasing on }(-\infty,1)\text{ and }(3,\infty);\ \text{decreasing on }(1,3)}

." ::: ---

Summary

❗ Key Takeaways for CMI

Monotonicity is determined mainly by the sign of the derivative.

Positive derivative means increasing, negative derivative means decreasing.

Critical points help split the domain into monotonicity intervals.

A zero derivative at a point alone does not decide maximum, minimum, or monotonicity.

Monotonicity is a powerful tool for graph analysis and uniqueness arguments.

---

💡 Next Up

Proceeding to Tangent and normal.

---

Part 2: Tangent and normal

Tangent and Normal

Overview

The tangent and the normal are the two most basic geometric lines associated with a smooth curve at a point. The tangent captures the instantaneous direction of the curve, while the normal is the line perpendicular to the tangent at the same point. In CMI-style questions, this topic tests differentiation, line equations, slope logic, and geometric interpretation together. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Find the tangent line to a curve at a given point.

Find the normal line to a curve at a given point.

Use derivative information to determine slopes of tangent and normal.

Handle special cases such as horizontal tangents and vertical normals.

Solve parameter-based questions involving tangent and normal lines.

---

Core Idea

📖 Tangent and Normal at a Point

Let the curve be
$\qquad y=f(x)$

and let $P(a,f(a))$ be a point on the curve where the derivative exists.

The tangent at $P$ is the line with slope

\qquad f'(a)

The normal at $P$ is the line perpendicular to the tangent.

f'(a)\ne 0

, then the slope of the normal is

\qquad -\dfrac{1}{f'(a)}

---

Tangent Formula

📐 Equation of Tangent

At the point $P(a,f(a))$ , the tangent line is

$\qquad y-f(a)=f'(a)(x-a)$

This is the direct formula you should use when the point of contact is known. ---

Normal Formula

📐 Equation of Normal

If $f'(a)\ne 0$ , then the normal at $P(a,f(a))$ is

$\qquad y-f(a)=-\dfrac{1}{f'(a)}(x-a)$

This comes from the fact that perpendicular lines have slopes whose product is

-1

. ::: ---

Special Cases

❗ Very Important Special Cases

\qquad f'(a)=0

then the tangent is horizontal, so the tangent is

\qquad y=f(a)

and the normal is vertical, so the normal is

\qquad x=a

If the tangent is vertical, then the normal is horizontal.

In school-level questions on curves of the form

y=f(x)

, the most common special case is

f'(a)=0

---

Slope Logic

📐 Slope Relations

If tangent slope is $m_t$ and normal slope is $m_n$ , then

$\qquad m_n=-\dfrac{1}{m_t}$

provided $m_t\ne0$ .

Also,

if $m_t=0$ , normal is vertical

if normal is horizontal, tangent is vertical

---

Standard Procedure

💡 How to Solve Tangent and Normal Questions

Find the point on the curve.

Differentiate to get $f'(x)$ .

Evaluate $f'(a)$ at the required point.

Use point-slope form for the tangent.

Use perpendicular slope for the normal.

Check special cases before writing the normal slope.

---

Standard Curves

📐 Useful Ready Forms

For $y=x^2$ at $x=t$ :

point: $\qquad (t,t^2)$

tangent slope: $\qquad 2t$

tangent: $\qquad y-t^2=2t(x-t)$

normal: $\qquad y-t^2=-\dfrac{1}{2t}(x-t)$ for $t\ne0$

For

y=x^3

x=t

point: $\qquad (t,t^3)$

tangent slope: $\qquad 3t^2$

tangent: $\qquad y-t^3=3t^2(x-t)$

normal: $\qquad y-t^3=-\dfrac{1}{3t^2}(x-t)$ for $t\ne0$

---

Minimal Worked Examples

Example 1 Find the tangent and normal to

\qquad y=x^2

x=1

. We have

\qquad f(x)=x^2,\quad f'(x)=2x

x=1

\qquad f(1)=1,\quad f'(1)=2

So the tangent is

\qquad y-1=2(x-1)

\qquad y=2x-1

The slope of the normal is

\qquad -\dfrac{1}{2}

So the normal is

\qquad y-1=-\dfrac{1}{2}(x-1)

--- Example 2 Find the tangent and normal to

\qquad y=x^3

x=0

. We have

\qquad f'(x)=3x^2

\qquad f'(0)=0

Hence the tangent is horizontal:

\qquad y=0

and the normal is vertical:

\qquad x=0

---

Tangent and Normal Through a Given Point

❗ Parameter Method

If a tangent or normal is required to pass through a given point, let the point of contact on the curve be $x=t$ .

Then:

write the tangent or normal in terms of $t$

substitute the external point

solve for $t$

This is a very common exam method.

---

Tangent vs Normal

⚠️ Do Not Mix Them Up

Tangent slope is $\qquad f'(a)$

Normal slope is $\qquad -\dfrac{1}{f'(a)}$

A very common mistake is to write the normal using the tangent slope itself.

---

Common Mistakes

⚠️ Avoid These Errors

❌ Forgetting to check whether $f'(a)=0$

✅ In that case, the normal is vertical

❌ Using the same slope for tangent and normal

✅ The normal slope is the negative reciprocal

❌ Writing line equations without using the actual point

✅ Always use the point

(a,f(a))

❌ Differentiating incorrectly before substituting the point

✅ Differentiate first, then evaluate

---

CMI Strategy

💡 Fast Exam Strategy

First identify whether the question asks for tangent, normal, or both.

Compute the derivative cleanly.

At the point of contact, calculate the slope before writing any line.

Check whether the slope is zero.

Use point-slope form immediately to avoid algebra mistakes.

In parameter questions, let the contact point be $x=t$ and solve systematically.

---

Practice Questions

:::question type="MCQ" question="The normal to the curve

y=x^2

at the point

(1,1)

is" options=["

y=2x-1

","

y=-\dfrac{1}{2}x+\dfrac{3}{2}

","

y=-2x+3

","

x=1

"] answer="B" hint="First find the tangent slope, then take its negative reciprocal." solution="For

y=x^2

, we have

\qquad \dfrac{dy}{dx}=2x

. At

x=1

, the tangent slope is

\qquad 2

. So the normal slope is

\qquad -\dfrac{1}{2}

. Using point-slope form through

(1,1)

\qquad y-1=-\dfrac{1}{2}(x-1)

\qquad y=-\dfrac{1}{2}x+\dfrac{3}{2}

. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Find the slope of the normal to the curve

y=x^3

x=1

." answer="-1/3" hint="Differentiate first." solution="For

y=x^3

\qquad \dfrac{dy}{dx}=3x^2

. At

x=1

, the slope of the tangent is

\qquad 3

. Therefore the slope of the normal is

\qquad -\dfrac{1}{3}

. Hence the answer is

\boxed{-\dfrac{1}{3}}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If the tangent slope at a point is

m\ne0

, then the normal slope is

-\dfrac{1}{m}

","If

f'(a)=0

, then the tangent is horizontal","If

f'(a)=0

, then the normal is horizontal","The tangent to

y=f(x)

x=a

passes through

(a,f(a))

"] answer="A,B,D" hint="Use the definitions carefully." solution="1. True. For nonzero tangent slope, the normal slope is its negative reciprocal.

True. If derivative is zero, tangent is horizontal.

False. If tangent is horizontal, the normal is vertical.

True. The tangent touches the curve at the point

(a,f(a))

Therefore the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Find the tangent and the normal to the curve

y=x^2

at the point where

x=2

." answer="Tangent:

y=4x-4

, Normal:

y=-\dfrac{1}{4}x+\dfrac{9}{2}

" hint="Use the point

(2,4)

and slope

f'(2)

." solution="For the curve

\qquad y=x^2

, we have

\qquad \dfrac{dy}{dx}=2x

. At

x=2

\qquad y=4,\quad \dfrac{dy}{dx}=4

So the tangent slope is

4

, and the tangent through

(2,4)

\qquad y-4=4(x-2)

\qquad y=4x-4

The normal slope is

\qquad -\dfrac{1}{4}

So the normal through

(2,4)

\qquad y-4=-\dfrac{1}{4}(x-2)

\qquad y=-\dfrac{1}{4}x+\dfrac{9}{2}

Hence,

\qquad \text{Tangent } = \boxed{y=4x-4}

and

\qquad \text{Normal } = \boxed{y=-\dfrac{1}{4}x+\dfrac{9}{2}}

." ::: ---

Summary

❗ Key Takeaways for CMI

Tangent slope at $x=a$ is $f'(a)$ .

Normal slope is $-\dfrac{1}{f'(a)}$ when $f'(a)\ne0$ .

Tangent at $(a,f(a))$ is $y-f(a)=f'(a)(x-a)$ .

If $f'(a)=0$ , tangent is horizontal and normal is vertical.

Most errors in this topic come from mixing up tangent and normal slopes.

---

💡 Next Up

Proceeding to Maxima and minima.

---

Part 3: Maxima and minima

Maxima and Minima

Overview

Maxima and minima form the heart of applications of derivatives. In CMI-style questions, this topic is not limited to the textbook “find where

f'(x)=0

”. It appears in many forms:

locating local maxima and minima of differentiable functions

proving that a local extremum forces derivative conditions

finding maximum or minimum distance from a point to a curve

optimizing area or volume

counting or bounding stationary points of polynomials

deciding whether an extremum is attained on restricted domains

So this topic is really about turning a geometric or algebraic condition into a one-variable optimization problem and then solving it carefully. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Define local and global maxima/minima clearly.

Find stationary points and classify them using derivative tests.

Use first derivative sign changes to detect extrema.

Use the second derivative test when it is valid.

Solve optimization problems involving distance, area, and other quantities.

Handle extrema on closed intervals and on restricted domains.

Understand why polynomials with many real roots must have many stationary points.

---

Core Definitions

📖 Local Maximum and Local Minimum

Let $f$ be a function and let $a$ be a point in its domain.

$f(a)$ is a local maximum value if there exists an interval around $a$ such that

\qquad f(a) \ge f(x)

for all nearby

x

$f(a)$ is a local minimum value if there exists an interval around $a$ such that

\qquad f(a) \le f(x)

for all nearby

x

.

The point

a

is then called a point of local maximum or local minimum.

📖 Global Maximum and Global Minimum

On a set $S$ :

$f(a)$ is a global maximum if

\qquad f(a)\ge f(x)

for every

x\in S

$f(a)$ is a global minimum if

\qquad f(a)\le f(x)

for every

x\in S

---

Stationary and Critical Points

📐 Stationary Point

A stationary point is a point where

$\qquad f'(x)=0$

❗ Critical Point

A critical point is a point where either

$\qquad f'(x)=0$ , or

$\qquad f'(x)$ does not exist,

provided the function itself is defined there.

A local extremum may occur only at a critical point or at an endpoint of the interval under consideration. ---

Necessary Condition for Local Extremum

❗ Fermat-Type Result

If $f$ is differentiable at $a$ and has a local maximum or local minimum at $a$ , then

$\qquad f'(a)=0$

This is one of the most important facts in the topic. It is a necessary condition, not a sufficient one.

⚠️ Important Caution

The condition
$\qquad f'(a)=0$
does not guarantee a maximum or minimum.

Example:
$\qquad f(x)=x^3$

Then
$\qquad f'(0)=0$

but $x=0$ is neither a local maximum nor a local minimum.

---

First Derivative Test

📐 Sign Change Test

Suppose $f$ is differentiable near $a$ .

If $f'(x)$ changes from positive to negative at $a$ , then $f$ has a local maximum at $a$ .

If $f'(x)$ changes from negative to positive at $a$ , then $f$ has a local minimum at $a$ .

If there is no sign change, then there is no local extremum there.

This is often the safest classification method. ---

Second Derivative Test

📐 Second Derivative Test

Suppose
$\qquad f'(a)=0$

Then:

if $\qquad f''(a)>0$ , then $f$ has a local minimum at $a$

if $\qquad f''(a)<0$ , then $f$ has a local maximum at $a$

if $\qquad f''(a)=0$ , the test is inconclusive

⚠️ Do Not Overuse This

If $f''(a)=0$ , you cannot conclude anything immediately.

Examples:

$f(x)=x^3$ gives $f''(0)=0$ and no extremum

$f(x)=x^4$ gives $f''(0)=0$ but there is a local minimum

---

Absolute Extrema on Closed Intervals

❗ Closed Interval Method

To find the absolute maximum or minimum of $f$ on $[a,b]$ :

Compute $f'(x)$

Find all critical points inside $(a,b)$

Evaluate $f$ at:

- all those interior critical points
- the endpoints

a

and

b

Compare all obtained values

This method is essential for questions on bounded intervals. ---

Optimization Strategy

💡 General Optimization Method

For a word problem:

Choose a variable.

Express the target quantity in terms of that variable.

Determine the correct domain.

Differentiate.

Find critical points.

Compare values using derivative test or endpoint check.

Interpret the answer in the original context.

Typical target quantities are:

area

distance

squared distance

volume

perimeter

height or coordinate values

::: ---

Why Squared Distance Is Better

📐 Distance Optimization Trick

If the distance from a point to a curve is

$\qquad d=\sqrt{A(x)}$

then minimizing or maximizing $d$ is equivalent to minimizing or maximizing

$\qquad d^2=A(x)$

provided $A(x)\ge0$ .

This avoids square roots and makes differentiation easier.

This is a very common CMI trick in geometry-based maxima/minima problems. ---

Polynomial Roots and Stationary Points

❗ Rolle-Type Principle

If a polynomial has many distinct real roots, then between every two consecutive roots there must be a root of its derivative.

So if a polynomial has $n$ distinct real roots, it must have at least $n-1$ stationary points.

This comes from Rolle’s theorem and is a very important connection between:

real roots of $p(x)$

stationary points of $p(x)$

::: ---

Minimal Worked Examples

Example 1 Find local extrema of

\qquad f(x)=x^3-3x

We have

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

So the stationary points are

\qquad x=-1,\ 1

Now

\qquad f''(x)=6x

Thus

at $x=-1$ , $\qquad f''(-1)=-6<0$ , so local maximum

at $x=1$ , $\qquad f''(1)=6>0$ , so local minimum

So:

local maximum at $\qquad x=-1$

local minimum at $\qquad x=1$

--- Example 2 Find the minimum distance from

(0,1)

to the curve

\qquad y=x^2

. A point on the curve is

\qquad (x,x^2)

. So the squared distance is

\qquad D(x)=x^2+(x^2-1)^2

\qquad =x^4-x^2+1

Now

\qquad D'(x)=4x^3-2x=2x(2x^2-1)

So the critical points are

\qquad x=0,\ \pm \dfrac{1}{\sqrt{2}}

Checking values,

\qquad D(0)=1

and

\qquad D\left(\dfrac{1}{\sqrt2}\right)=\dfrac34

Hence the minimum squared distance is

\qquad \dfrac34

so the minimum distance is

\qquad \dfrac{\sqrt3}{2}

---

Maxima/Minima on Restricted Domains

❗ Very Important Domain Issue

A function may have:

a supremum but no maximum

an infimum but no minimum

This happens often on open intervals or on curves with endpoints excluded.

So always ask:

Is the extremal value actually attained?

Is the relevant point allowed by the domain?

Example: On

(0,1)

, the function

f(x)=x

has no maximum and no minimum. ---

Implicit or Constrained Extremum Problems

💡 When the Curve Is Given Indirectly

If the curve is given by an equation such as

$\qquad F(x,y)=0$

and you want the largest or least possible value of $y$ , then:

solve for $y$ if possible, or

differentiate implicitly and find where extremum in $y$ can occur, or

rewrite the relation into a one-variable expression and optimize that

These questions often test both algebra and calculus. ---

Common Mistakes

⚠️ Avoid These Errors

❌ Thinking $f'(a)=0$ is enough for extremum

✅ It is only a necessary condition in general.

❌ Forgetting endpoints in closed interval problems

✅ Always test endpoints.

❌ Minimizing distance directly when squared distance is easier

✅ Optimize

d^2

❌ Ignoring the actual domain

✅ Open interval and restricted curve questions are domain-sensitive.

❌ Using second derivative test when $f''(a)=0$ and still making a conclusion

✅ Then use another method.

❌ Forgetting that a stationary point need not be a local max/min

✅ Example:

x^3

---

CMI Strategy

💡 How to Think in Exam Conditions

First decide whether the question asks for local extremum or absolute extremum.

Find the correct variable and domain before differentiating.

For geometry questions, optimize squared distance instead of distance.

For area or volume, write the target quantity as a clean function first.

For polynomial-root questions, think of Rolle’s theorem immediately.

For open intervals or restricted curves, check whether the extremum is attained.

---

Practice Questions

:::question type="MCQ" question="For the function

f(x)=x^3-3x

, the local maximum occurs at" options=["

x=-1

","

x=0

","

x=1

","

x=3

"] answer="A" hint="Find stationary points and classify them." solution="We have

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

so stationary points are

x=-1

and

x=1

. Also

\qquad f''(x)=6x

. At

x=-1

f''(-1)<0

, so there is a local maximum. Hence the correct option is

\boxed{A}

." ::: :::question type="NAT" question="Find the minimum value of

x+\dfrac{4}{x}

for

x>0

." answer="4" hint="Differentiate or use AM-GM." solution="Let

\qquad f(x)=x+\dfrac4x

for

x>0

. Then

\qquad f'(x)=1-\dfrac4{x^2}

Set

f'(x)=0

\qquad 1=\dfrac4{x^2}

\qquad x=2

. Then

\qquad f(2)=2+\dfrac42=4

Therefore the minimum value is

\boxed{4}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If a differentiable function has a local maximum at

a

, then

f'(a)=0

","If

f'(a)=0

, then

a

must be a local maximum or local minimum","If a polynomial has

5

distinct real roots, then it must have at least

4

stationary points","A stationary point need not be a point of local maximum or local minimum"] answer="A,C,D" hint="Separate necessary conditions from sufficient conditions." solution="1. True by the standard differentiable-extremum result.

False. Example:

f(x)=x^3

x=0

True by Rolle's theorem.

True. Again

x=0

for

f(x)=x^3

is a counterexample.

Hence the correct answer is

\boxed{A,C,D}

." ::: :::question type="SUB" question="Find the maximum value of

x^2-x^3

on the interval

[0,2]

." answer="

\dfrac{4}{27}

" hint="Use derivative and endpoint check." solution="Let

\qquad f(x)=x^2-x^3

Then

\qquad f'(x)=2x-3x^2=x(2-3x)

Critical points in

[0,2]

are

\qquad x=0,\ \dfrac23

Now evaluate:

\qquad f(0)=0

\qquad f\left(\dfrac23\right)=\dfrac{4}{9}-\dfrac{8}{27}=\dfrac{4}{27}

\qquad f(2)=4-8=-4

Hence the maximum value on

[0,2]

\boxed{\dfrac{4}{27}}

." ::: ---

Summary

❗ Key Takeaways for CMI

A differentiable local extremum forces $f'(a)=0$ , but the converse is false.

First derivative sign change is often the safest way to classify extrema.

Second derivative test is useful only when it gives a nonzero value.

For absolute extrema on a closed interval, check both critical points and endpoints.

Geometry problems often become easier after optimizing squared distance.

Distinct real roots of a polynomial force stationary points in between.

Domain matters: a least upper bound need not be an attained maximum.

---

💡 Next Up

Proceeding to Tangency conditions.

---

Part 4: Tangency conditions

Tangency Conditions

Overview

Tangency conditions arise when a line or curve just touches another curve at a point without crossing it immediately there. In calculus, tangency questions usually combine common point conditions with equal slope conditions. In CMI-style problems, this topic is important because it tests algebra, differentiation, and geometric interpretation together. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Identify what it means for a line to be tangent to a curve.

Use point and slope conditions to form tangency equations.

Solve for unknown parameters using derivative-based conditions.

Distinguish tangency from mere intersection.

Handle tangency problems involving quadratics, cubics, and standard curves.

---

Core Idea

📖 Tangency at a Point

A line is tangent to a curve at $x=a$ if:

the line and the curve pass through the same point, and

the slope of the line equals the slope of the curve at that point.

If the curve is

y=f(x)

and the tangent point is

x=a

, then:

Point on curve: $\qquad (a,f(a))$

Slope of tangent: $\qquad f'(a)$

Tangent line: $\qquad y-f(a)=f'(a)(x-a)$

---

Tangent to a Curve from Calculus

📐 Equation of Tangent

If $y=f(x)$ and the tangent is drawn at $x=a$ , then

$\qquad y-f(a)=f'(a)(x-a)$

This is the standard tangent-line formula.

📐 Two Basic Tangency Conditions

Suppose the line is $y=mx+c$ and it is tangent to the curve $y=f(x)$ at $x=a$ . Then:

Same point condition:

\qquad f(a)=ma+c

Same slope condition:

\qquad f'(a)=m

These two equations are usually enough to determine the unknowns.

---

Tangency Versus Intersection

❗ Do Not Confuse These

If a line intersects a curve, they may have one or more common points.

If a line is tangent to a curve at $x=a$ , then at that point:

they share the same point

they share the same slope

So tangency is stronger than intersection.

⚠️ Common Mistake

Solving only $f(x)=mx+c$ gives common points, not tangency by itself.

You must also use
$\qquad f'(a)=m$
or an equivalent repeated-root condition.

---

Tangency as a Repeated Root

📐 Repeated Root View

If the line $y=mx+c$ is tangent to the curve $y=f(x)$ , then the equation

$\qquad f(x)=mx+c$

has a repeated root at the tangency point.

Equivalently, the equation

$\qquad f(x)-mx-c=0$

has a double root.

This viewpoint is especially useful when the curve is polynomial. For a quadratic, tangency often means the resulting quadratic has discriminant zero. ---

Tangency to a Parabola

📐 Tangency to

y=ax^2+bx+c

If the tangent is drawn at $x=t$ , then

$\qquad y-(at^2+bt+c)=(2at+b)(x-t)$

This is often simplified to a line in terms of $t$ .

Example form: For

y=x^2

x=t

\qquad y-t^2=2t(x-t)

So,

\qquad y=2tx-t^2

This is the tangent to

y=x^2

(t,t^2)

. ---

Tangency to Other Standard Curves

📐 Useful Standard Tangents

For $y=x^2$ at $x=t$ :

\qquad y=2tx-t^2

For $y=x^3$ at $x=t$ :

\qquad y-t^3=3t^2(x-t)

For $y=\dfrac{1}{x}$ at $x=t$ :

\qquad y-\dfrac{1}{t}=-\dfrac{1}{t^2}(x-t)

, where

t\ne0

For $y=\sqrt{x}$ at $x=t$ :
$\qquad y-\sqrt{t}=\dfrac{1}{2\sqrt{t}}(x-t)$ , where $t>0$

---

Standard Methods

💡 Method 1: Use Point and Slope

If the tangent point is known or assumed to be $x=a$ :

use $f(a)=ma+c$

use $f'(a)=m$

This is the cleanest method.

💡 Method 2: Parameter Method

For curves like $y=x^2$ , $y=x^3$ , etc., let the tangency point be $x=t$ . Then write the tangent directly in terms of $t$ , and use extra conditions to find $t$ .

💡 Method 3: Repeated Root / Discriminant

If a line is tangent to a polynomial curve, substitute the line into the curve equation. Tangency means the resulting polynomial has a repeated root.

For a quadratic equation:
$\qquad ax^2+bx+c=0$

repeated root condition is:
$\qquad b^2-4ac=0$

---

Minimal Worked Examples

Example 1 Find the tangent to

y=x^2

x=3

. We have

\qquad f(x)=x^2,\quad f'(x)=2x

x=3

\qquad f(3)=9,\quad f'(3)=6

So the tangent is

\qquad y-9=6(x-3)

\qquad y=6x-9

--- Example 2 Find the value of

k

such that the line

y=2x+k

is tangent to

y=x^2

. Tangency means

\qquad x^2=2x+k

So,

\qquad x^2-2x-k=0

For tangency, this quadratic must have equal roots. Hence,

\qquad (-2)^2-4(1)(-k)=0

\qquad 4+4k=0

\qquad k=-1

So the required value is

\boxed{-1}

. ---

Geometry Insight

❗ What Tangency Means Geometrically

Near the tangency point, the tangent line gives the best linear approximation to the curve.

So if a problem asks for a tangent:

the point tells you where to touch

the derivative tells you how steep the touch is

---

Common Patterns in Questions

📐 Patterns to Recognize

Find the tangent line at a given point.

Find a line tangent to a curve with given slope.

Find parameter values so that a given line is tangent.

Find common tangents between two curves.

Determine when two curves touch each other.

When two curves

y=f(x)

and

y=g(x)

touch at

x=a

, usually use:

$\qquad f(a)=g(a)$

$\qquad f'(a)=g'(a)$

---

Common Mistakes

⚠️ Avoid These Errors

❌ Using only the common-point condition

✅ Also use equality of slopes

❌ Forgetting to differentiate correctly

✅ Tangency depends directly on slope

❌ Treating every single intersection as a tangent point

✅ Tangency requires repeated contact, not just meeting

❌ Ignoring repeated-root logic in polynomial problems

✅ Use discriminant zero when appropriate

---

CMI Strategy

💡 How to Solve Tangency Problems

Decide whether the unknown is the tangent point, slope, or parameter.

Write the tangent formula or use same-point and same-slope equations.

If the curve is polynomial, also think in terms of repeated roots.

For standard curves, use parameter form to reduce algebra.

Always verify the final point really lies on the curve.

---

Practice Questions

:::question type="MCQ" question="The tangent to the curve

y=x^2

x=2

is" options=["

y=4x-4

","

y=2x+4

","

y=4x+4

","

y=2x-4

"] answer="A" hint="Use slope

f'(2)

and point

(2,4)

." solution="For

f(x)=x^2

, we have

f'(x)=2x

. At

x=2

, slope is

\qquad f'(2)=4

and the point is

\qquad (2,4)

. So the tangent is

\qquad y-4=4(x-2)

which gives

\qquad y=4x-4

. Hence the correct option is

\boxed{A}

." ::: :::question type="NAT" question="If the line

y=6x+c

is tangent to the curve

y=x^2

, then find

c

." answer="-9" hint="A tangent with slope

6

touches

y=x^2

where

2x=6

." solution="For

y=x^2

, the derivative is

\qquad y'=2x

. Tangency with slope

6

means

\qquad 2x=6 \implies x=3

. The point of tangency is

\qquad (3,9)

. Since the line is

y=6x+c

, substitute

(3,9)

\qquad 9=6\cdot 3+c

\qquad 9=18+c

\qquad c=-9

. Therefore the answer is

\boxed{-9}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If

y=mx+c

is tangent to

y=f(x)

x=a

, then

f(a)=ma+c

","If

y=mx+c

is tangent to

y=f(x)

x=a

, then

f'(a)=m

","If a line and a curve have one common point, then the line is tangent to the curve","If

f(x)-mx-c=0

has a repeated root, then

y=mx+c

is tangent to

y=f(x)

at that root"] answer="A,B,D" hint="Tangency means same point and same slope." solution="1. True. Tangency requires the point of contact to lie on both the line and the curve.

True. Tangency also requires equal slope.

False. A common point alone does not guarantee tangency.

True. A repeated root indicates repeated contact, which is tangency in these polynomial-style settings.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Find the value of

k

such that the line

y=4x+k

is tangent to the curve

y=x^2

." answer="

k=-4

" hint="Set

x^2=4x+k

and use repeated-root condition." solution="If the line is tangent to the parabola, then the equation

\qquad x^2=4x+k

must have equal roots. So,

\qquad x^2-4x-k=0

For equal roots,

\qquad (-4)^2-4(1)(-k)=0

\qquad 16+4k=0

\qquad 4k=-16

\qquad k=-4

Therefore the required value is

\boxed{-4}

." ::: ---

Summary

❗ Key Takeaways for CMI

Tangency means same point and same slope.

For $y=f(x)$ , tangent at $x=a$ is $y-f(a)=f'(a)(x-a)$ .

If a line is tangent to a polynomial curve, the intersection equation has a repeated root.

For quadratics, discriminant zero is a fast tangency test.

Tangency problems combine geometry, algebra, and differentiation.

---

💡 Next Up

Proceeding to Root-location arguments.

---

Part 5: Root-location arguments

Root-Location Arguments

Overview

Root-location arguments are used to prove that an equation has a solution, has a unique solution, or has a solution inside a specific interval. In differentiation-based problems, the key idea is to convert the equation into the form

\qquad f(x)=0

and then study the graph of

f

using continuity, derivatives, monotonicity, and sometimes the values of

f

at special points. In CMI-style questions, this topic often mixes existence, uniqueness, and relationships between roots of two different equations. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Prove existence of roots using sign change and continuity.

Prove uniqueness using monotonicity.

Count the possible number of roots using derivatives and turning points.

Locate roots inside intervals.

Relate roots of two different equations by transformation or substitution.

---

Core Idea

📖 Root-location viewpoint

To solve an equation, do not work directly with both sides all the time. Instead, bring everything to one side:

$\qquad f(x)=0$

Then ask:

Is $f$ continuous?

Does $f$ change sign on some interval?

Is $f$ increasing or decreasing?

How many turning points can $f$ have?

Can one root be transformed into a root of another equation?

---

Existence of a Root

📐 Intermediate Value Theorem

If $f$ is continuous on $[a,b]$ and

$\qquad f(a)\cdot f(b)<0$

then there exists at least one $c\in(a,b)$ such that

$\qquad f(c)=0$

❗ What IVT Gives

The Intermediate Value Theorem gives existence of at least one root.

It does not by itself give uniqueness.

---

Uniqueness of a Root

📐 Monotonicity Test

If $f$ is differentiable on an interval and

$\qquad f'(x)>0$ for all $x$ there, then $f$ is strictly increasing

$\qquad f'(x)<0$ for all $x$ there, then $f$ is strictly decreasing

A strictly monotone function can cross the

x

-axis at most once.

So if you already know a root exists and

f

is strictly monotone, then the root is unique.

❗ Most Common Pattern

To prove a unique solution in an interval:

show existence using IVT or direct evaluation

show $f'(x)$ keeps one sign there

---

Counting Roots with Derivatives

📐 Critical Points Control Root Behaviour

If $f'(x)=0$ only at a few points, then $f$ has only a few turning points.

This helps in estimating how many times the graph can cross the $x$ -axis.

💡 Very Useful Idea

For a cubic:

if the local maximum is positive

and the local minimum is negative

then the graph usually has three distinct real roots.

---

Rolle-Type Logic

📐 If Two Roots Exist, Derivative Must Vanish Somewhere

If $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and

$\qquad f(a)=f(b)=0$

then there exists some $c\in(a,b)$ such that

$\qquad f'(c)=0$

This is often used in reverse:

if $f'$ cannot vanish enough times, then $f$ cannot have too many roots.

::: ---

Standard Method for Root-Location

💡 Five-Step Method

Rewrite the equation as $f(x)=0$ .

Identify the natural domain.

Check continuity on that domain.

Evaluate $f$ at convenient points to get sign information.

Use derivative sign to prove uniqueness or to bound the number of roots.

---

Minimal Worked Examples

Example 1: Existence and uniqueness Solve the root-location problem for

\qquad x+\ln x=2,\qquad x>0

Define

\qquad f(x)=x+\ln x-2

Then

f

is continuous on

(0,\infty)

and

\qquad f'(x)=1+\dfrac{1}{x}>0 \quad \text{for } x>0

f

is strictly increasing on

(0,\infty)

. Now check values:

\qquad f(1)=1+0-2=-1

\qquad f(2)=2+\ln 2-2=\ln 2>0

So there is a root in

(1,2)

by IVT, and it is unique because

f

is strictly increasing. --- Example 2: Counting roots Consider

\qquad f(x)=x^3-3x+1

Then

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

So the critical points are

x=-1

and

x=1

. Now evaluate:

\qquad f(-1)=-1+3+1=3>0

\qquad f(1)=1-3+1=-1<0

Also,

as $x\to-\infty$ , $f(x)\to-\infty$

as $x\to\infty$ , $f(x)\to\infty$

So:

one root lies in $(-\infty,-1)$

one root lies in $(-1,1)$

one root lies in $(1,\infty)$

Hence

f(x)=0

has exactly three real roots. ---

Relating Roots of Two Equations

❗ Transformation Trick

Sometimes two equations are secretly linked.

Example:
suppose $a$ satisfies

$\qquad \log_{2021} a = 2022-a$

Then

$\qquad a = 2021^{\,2022-a}$

Now let

$\qquad b=2022-a$

Then

$\qquad 2021^b = a = 2022-b$

So $b$ becomes a root of the related equation

$\qquad 2021^b = 2022-b$

This is a powerful CMI-style idea:

solve uniqueness separately

then relate the roots by substitution

---

Common Patterns

📐 Patterns That Repeat

$f(x)=g(x)$ rewritten as $f(x)-g(x)=0$

logarithmic equation on $(0,\infty)$ with derivative always positive

exponential equation with monotone behaviour

cubic or quartic where derivative reveals turning points

two equations connected by the same transformed function

---

Common Mistakes

⚠️ Avoid These Errors

❌ Using IVT without checking continuity

❌ Claiming uniqueness from sign change alone

❌ Forgetting the natural domain of logarithms or square roots

❌ Counting roots without checking critical points

❌ Missing the possibility that two equations are transform-related

---

CMI Strategy

💡 How to Think in CMI Problems

Turn the equation into $f(x)=0$ immediately.

Use the easiest interval checks first.

Use $f'(x)$ to prove monotonicity and uniqueness.

For multiple-root questions, inspect turning points through $f'$ .

If two equations appear together, look for a substitution linking one root to the other.

---

Practice Questions

:::question type="MCQ" question="The equation

x+\ln x=2

has" options=["no real solution","exactly one real solution in

(0,\infty)

","exactly two real solutions in

(0,\infty)

","infinitely many real solutions"] answer="B" hint="Use

f(x)=x+\ln x-2

and study

f'(x)

." solution="Let

\qquad f(x)=x+\ln x-2

(0,\infty)

. Then

\qquad f'(x)=1+\dfrac{1}{x}>0

for all

x>0

, so

f

is strictly increasing. Also,

\qquad f(1)=-1<0,\qquad f(2)=\ln 2>0

So by IVT there is a root in

(1,2)

, and since

f

is strictly increasing, the root is unique. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Find the number of real roots of the equation

x^3-3x+1=0

." answer="3" hint="Check the derivative and the values at the critical points." solution="Let

\qquad f(x)=x^3-3x+1

Then

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

So the critical points are

x=-1

and

x=1

. Now,

\qquad f(-1)=3>0,\qquad f(1)=-1<0

Also,

\qquad f(x)\to-\infty

x\to-\infty

and

\qquad f(x)\to\infty

x\to\infty

Therefore there is one root in each of the intervals

\qquad (-\infty,-1),\ (-1,1),\ (1,\infty)

So the equation has exactly

\boxed{3}

real roots." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If a continuous function changes sign on an interval, then it has a root in that interval","If

f'(x)>0

on an interval, then

f

can have at most one root there","If

f(a)=0

and

f(b)=0

with

a<b

, then

f'(x)

must be positive somewhere in

(a,b)

","If a differentiable function has two distinct roots, then its derivative vanishes at some point between them"] answer="A,B,D" hint="Use IVT, monotonicity, and Rolle's theorem." solution="1. True, by the Intermediate Value Theorem.

True, because

f'(x)>0

implies

f

is strictly increasing.

False. The correct guaranteed conclusion is that

f'(c)=0

for some

c\in(a,b)

, not necessarily positive.

True, by Rolle's theorem.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Suppose

a

is a solution of

\log_{m} a = k-a

, where

m>0

and

m\ne1

. Show that

b=k-a

is a solution of

m^b=k-b

." answer="The transformed value

b=k-a

satisfies

m^b=k-b

." hint="Exponentiate the logarithmic equation." solution="Given

\qquad \log_m a = k-a

Exponentiating base

m

, we get

\qquad a = m^{\,k-a}

Now define

\qquad b=k-a

Then the above equation becomes

\qquad a=m^b

Also,

\qquad k-b = k-(k-a)=a

Therefore,

\qquad m^b = a = k-b

b=k-a

is indeed a solution of

\qquad m^b = k-b

Hence the required statement is proved." ::: ---

Summary

❗ Key Takeaways for CMI

Root-location begins by rewriting the equation as $f(x)=0$ .

IVT gives existence; monotonicity gives uniqueness.

Derivatives help count roots by controlling turning points.

Rolle's theorem helps restrict how many roots are possible.

Some paired equations are linked by a substitution between their roots.

---

Chapter Summary

❗ Applications of derivatives — Key Points

The first derivative $f'(x)$ determines the monotonicity of $f(x)$ : $f'(x) > 0$ implies increasing, $f'(x) < 0$ implies decreasing. Critical points occur where $f'(x)=0$ or is undefined.
The equation of the tangent to $y=f(x)$ at $(x_0, y_0)$ is $y-y_0 = f'(x_0)(x-x_0)$ . The normal line has slope $-1/f'(x_0)$ .
Local extrema (maxima or minima) occur at critical points. The First Derivative Test examines the sign change of $f'(x)$ , while the Second Derivative Test uses the sign of $f''(x)$ .
Global extrema on a closed interval $[a, b]$ are found by comparing $f(x)$ values at critical points within $(a, b)$ and at the endpoints $a$ and $b$ .
Two curves $y=f(x)$ and $y=g(x)$ are tangent at $x=c$ if and only if $f(c)=g(c)$ and $f'(c)=g'(c)$ .
Rolle's Theorem and the Mean Value Theorem are fundamental for root-location arguments, guaranteeing the existence of specific points where the derivative takes certain values.
* Optimization problems involve formulating a function representing the quantity to be optimized and finding its extrema using derivative tests.

---

Chapter Review Questions

:::question type="MCQ" question="For what values of $x$ is the function $f(x) = x^3 - 6x^2 + 9x + 5$ strictly decreasing?" options=[" $(-\infty, 1) \cup (3, \infty)$ "," $(1, 3)$ "," $(-\infty, 3)$ "," $(1, \infty)$ "] answer=" $(1, 3)$ " hint="Determine the sign of the first derivative." solution="The first derivative is $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$ . For $f(x)$ to be strictly decreasing, $f'(x) < 0$ . This inequality holds when $1 < x < 3$ . Thus, the function is strictly decreasing on the interval $(1, 3)$ ."
:::

:::question type="NAT" question="If the tangent to the curve $y = x^3 - 3x + 5$ at a point $(x_0, y_0)$ is parallel to the line $y = 9x - 2$ , what is the sum of all possible values of $x_0$ ?" answer="0" hint="The slope of the tangent at $(x_0, y_0)$ is $f'(x_0)$ . Parallel lines have equal slopes." solution="The derivative of the curve is $y' = 3x^2 - 3$ . The slope of the tangent at $(x_0, y_0)$ is $3x_0^2 - 3$ . The given line $y = 9x - 2$ has a slope of 9. Since the tangent is parallel to this line, their slopes must be equal:

3x_0^2 - 3 = 9

3x_0^2 = 12

x_0^2 = 4

x_0 = \pm 2

The possible values for

x_0

are

2

and

-2

. The sum of these values is

2 + (-2) = 0

."
:::

:::question type="MCQ" question="A rectangular box with a square base and an open top is to have a volume of $32 \operatorname{m}^3$ . What is the minimum possible surface area of the box?" options=[" $64 \operatorname{m}^2$ "," $48 \operatorname{m}^2$ "," $32 \operatorname{m}^2$ "," $16 \operatorname{m}^2$ "] answer=" $48 \operatorname{m}^2$ " hint="Express the surface area as a function of one variable, then use calculus to find its minimum." solution="Let the side length of the square base be $x$ and the height of the box be $h$ .
The volume is $V = x^2 h = 32$ , so $h = \frac{32}{x^2}$ .
The surface area $A$ (open top) is $A = x^2 + 4xh$ .
Substitute $h$ : $A(x) = x^2 + 4x \left(\frac{32}{x^2}\right) = x^2 + \frac{128}{x}$ .
To find the minimum surface area, we take the derivative of $A(x)$ with respect to $x$ and set it to zero:

A'(x) = 2x - \frac{128}{x^2}

Set

A'(x) = 0

2x - \frac{128}{x^2} = 0

2x = \frac{128}{x^2}

2x^3 = 128

x^3 = 64

x = 4

Now find

h

h = \frac{32}{4^2} = \frac{32}{16} = 2

.
The minimum surface area is

A(4) = 4^2 + 4(4)(2) = 16 + 32 = 48 \operatorname{m}^2

."
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Having thoroughly explored the applications of derivatives, your understanding of rates of change, optimization, and curve behavior is significantly enhanced. These foundational concepts naturally lead to the inverse operation: Integration. The ability to reverse differentiation allows for the calculation of areas, volumes, and the modeling of accumulation, forming the basis for solving Differential Equations and paving the way for advanced topics in Multivariable Calculus.