Function behaviour

This chapter rigorously examines the behaviour of functions, focusing on their increasing and decreasing intervals, and the intuitive principles of graph sketching. A thorough understanding of these concepts is crucial for determining the number of roots and solving complex problems that integrate algebraic and calculus techniques, forming a cornerstone of CMI examination success.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Increasing-decreasing behaviour | | 2 | Graph sketch intuition | | 3 | Number of roots from graph | | 4 | Combined algebra-calculus problems |

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We begin with Increasing-decreasing behaviour.

Part 1: Increasing-decreasing behaviour

Increasing-Decreasing Behaviour

Overview

Increasing-decreasing behaviour is not just about checking whether a derivative is positive or negative. In stronger exam questions, this topic connects monotonicity with boundedness, limits, uniqueness of roots, and even construction of unusual examples. The PYQs for this topic clearly show that CMI is interested not only in routine sign charts, but also in proof-based reasoning and counterexample construction. ---

Learning Objectives

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Core Definitions

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Derivative Tests

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Mean Value Theorem Link

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Sign Chart Method

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Important Subtleties

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Bounded Monotone Functions

This is extremely useful in reasoning-based questions. ---

A Key PYQ-Type Result

Reason: Since $f$ is increasing and differentiable, we must have $f'(x)\ge0$, so $L\ge0$. If $L>0$, then for all sufficiently large $x$ we would have $\qquad f'(x)\ge \dfrac{L}{2}$ Then by the mean value theorem, $f$ would increase at least linearly on the tail, contradicting boundedness. Hence the only possible limit is $\boxed{0}$. ---

Construction Ideas

Typical construction template

Take a continuous nonnegative “bump train” $h(x)$:

• each bump has small area

• total area is finite

• bump heights do not settle to one limit

Then define $\qquad f(x)=\int_0^x h(t)\,dt$ Now:

• $f'(x)=h(x)\ge0$, so $f$ is weakly increasing

• $\int_0^\infty h(t)\,dt<\infty$, so $f$ is bounded

• if $h$ oscillates forever without approaching a limit, then $\lim_{x\to\infty} f'(x)$ does not exist

To make the function strictly increasing everywhere, add a small positive continuous integrable term such as $e^{-x^2}$ to the bump train. ---

Minimal Worked Example

Example Determine the increasing-decreasing behaviour of $\qquad f(x)=x^3-3x^2$ Differentiate: $\qquad f'(x)=3x^2-6x=3x(x-2)$ Now:

• if $x<0$, then $f'(x)>0$

• if $0<x<2$, then $f'(x)<0$

• if $x>2$, then $f'(x)>0$

So:

• increasing on $\qquad (-\infty,0)$ and $(2,\infty)$

• decreasing on $\qquad (0,2)$

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Root and Uniqueness Consequences

This is often how monotonicity is used in proof-based problems. ---

Common Patterns in Questions

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Suppose $f$ is differentiable on $\mathbb{R}$ and $f'(x)\ge0$ for all $x$. Which statement must be true?" options=["$f$ is strictly increasing","$f$ is weakly increasing","$f$ is bounded","$f'(x)>0$ for all $x$"] answer="B" hint="Use the mean value theorem." solution="If $f'(x)\ge0$ everywhere, then by the mean value theorem $f(b)-f(a)\ge0$ whenever $a<b$. Hence $f$ is weakly increasing. Strict increase need not hold, because the function may be constant on some interval. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="If $f$ is differentiable, increasing, bounded, and $\lim_{x\to\infty} f'(x)$ exists, then what must that limit be?" answer="0" hint="If the limit were positive, the function would eventually grow too fast." solution="Since $f$ is increasing and differentiable, we have $f'(x)\ge0$, so the limit $L$ must satisfy $L\ge0$. If $L>0$, then for sufficiently large $x$ we would have $f'(x)\ge L/2$. By the mean value theorem this would force $f$ to grow at least linearly on the tail, contradicting boundedness. Hence the only possible value is $\boxed{0}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["A differentiable increasing function must satisfy $f'(x)\ge0$ everywhere","A strictly increasing differentiable function can have $f'(a)=0$ at some point","If $f'(x)>0$ on an interval, then $f$ is strictly increasing there","A bounded increasing differentiable function must satisfy $\lim_{x\to\infty} f'(x)=0$"] answer="A,B,C" hint="Think separately about derivative sign, strict increase, and existence of derivative limits." solution="1. True. This is the converse derivative fact for differentiable increasing functions.

True. Example: $f(x)=x^3$ is strictly increasing and $f'(0)=0$.

True. Positive derivative on an interval implies strict increase.

False. The limit, if it exists, must be $0$, but it need not exist at all.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Prove that if $f$ is differentiable on an interval $I$ and $f'(x)\ge0$ for all $x\in I$, then $f$ is increasing on $I$." answer="Use the mean value theorem." hint="Apply the mean value theorem to $[a,b]$ with $a<b$." solution="Let $a<b$ be points in $I$. Since $f$ is differentiable on $I$, it is continuous on $[a,b]$ and differentiable on $(a,b)$. By the mean value theorem, there exists $c\in(a,b)$ such that $\qquad f(b)-f(a)=f'(c)(b-a)$. Now $f'(c)\ge0$ by hypothesis, and $b-a>0$. Hence $\qquad f(b)-f(a)\ge0$. So $f(a)\le f(b)$ whenever $a<b$. Therefore $f$ is increasing on $I$." ::: ---

Summary

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Part 2: Graph sketch intuition

Graph Sketch Intuition

Overview

Graph sketch intuition is the art of predicting the shape of a graph from a small amount of information: sign of the function, derivative, second derivative, symmetry, asymptotes, and end behaviour. In CMI-style questions, you are often not asked to draw a perfect graph; instead, you are asked to infer one decisive feature such as whether a function bends upward or downward, whether a midpoint inequality holds, or whether a turning point is a maximum or a minimum. ---

Learning Objectives

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Core Idea

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First Derivative and Shape

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Second Derivative and Curvature

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The Midpoint / Chord Intuition

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Tangent Intuition

If $f'(a)>0$, the graph rises there. If $f'(a)<0$, the graph falls there. If $f'(a)=0$, the tangent is horizontal. ---

Symmetry Clues

These are powerful sketching shortcuts. ---

End Behaviour and Boundedness

Examples:

• $\arctan x$ is bounded above by $\dfrac{\pi}{2}$ and below by $-\dfrac{\pi}{2}$

• $\dfrac{1}{x}$ has two branches and vertical/horizontal asymptotes

• $e^{-x}$ decreases but stays positive and tends to $0$ as $x\to\infty$

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A Key Example: $\arctan x$

This means that on the positive side, midpoint comparison gives: $\qquad \arctan(a-h)+\arctan(a+h) < 2\arctan(a)$ for suitable positive $a,h$. ---

Minimal Worked Examples

Example 1 Suppose $f'(x)>0$ and $f''(x)<0$ on an interval. Then the graph is:

• increasing

• concave down

So the function rises, but the slope keeps decreasing. --- Example 2 For $f(x)=x^3-3x$, $\qquad f'(x)=3x^2-3=3(x-1)(x+1)$ So the stationary points are at $x=\pm1$. Now check sign of $f'$:

• positive for $x<-1$

• negative for $-1<x<1$

• positive for $x>1$

Therefore:

• local maximum at $x=-1$

• local minimum at $x=1$

This is the kind of structure graph-sketch problems often test. ---

What Information Is Not Enough?

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If a differentiable function satisfies $f'(x)>0$ and $f''(x)<0$ on an interval, then on that interval the graph is" options=["decreasing and concave up","increasing and concave down","increasing and concave up","decreasing and concave down"] answer="B" hint="Use the meanings of first and second derivatives." solution="Since $f'(x)>0$, the function is increasing. Since $f''(x)<0$, the graph is concave down. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="For $f(x)=x^3-3x$, how many stationary points does the graph have?" answer="2" hint="Solve $f'(x)=0$." solution="We have $\qquad f'(x)=3x^2-3=3(x^2-1)=3(x-1)(x+1)$ So $f'(x)=0$ at $\qquad x=1,\ -1$ Hence the graph has $\boxed{2}$ stationary points." ::: :::question type="MSQ" question="Which of the following statements about $f(x)=\arctan x$ are true?" options=["$f'(x)>0$ for all real $x$","$f''(x)<0$ for all $x>0$","$f$ has an inflection point at $x=0$","$f$ is unbounded above"] answer="A,B,C" hint="Differentiate once and twice." solution="For $f(x)=\arctan x$, $\qquad f'(x)=\dfrac{1}{1+x^2}>0$ for all real $x$, so statement 1 is true. Also, $\qquad f''(x)=\dfrac{-2x}{(1+x^2)^2}$ which is negative for $x>0$, so statement 2 is true. Since $f''$ changes sign at $x=0$, statement 3 is true. Finally, $\arctan x$ is bounded above by $\dfrac{\pi}{2}$, so statement 4 is false. Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Explain why a concave-down graph on an interval tends to satisfy $f(a-h)+f(a+h)<2f(a)$ for nearby points." answer="Because a concave-down graph lies above its chord, so the midpoint value is larger than the average of nearby endpoint values." hint="Think of the point at the middle compared with the chord joining the two side points." solution="If the graph is concave down, then between two nearby points it lies above the line segment joining them. So the value of the function at the midpoint is greater than the average of the values at the side points. Therefore, $\qquad f(a)>\dfrac{f(a-h)+f(a+h)}{2}$ which gives $\qquad f(a-h)+f(a+h)<2f(a)$. This is the required reason." ::: ---

Summary

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Part 3: Number of roots from graph

Number of Roots from Graph

Overview

Many CMI-style questions do not ask you to find roots explicitly. Instead, they ask you to count them using the graph of a function or using graphical information such as maxima, minima, sign, monotonicity, and intersections with lines. The central idea is simple: a root is an intersection with the $x$-axis, and more general equations are counted by intersections of graphs. ---

Learning Objectives

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Core Meaning of a Root

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Main Graph Rules

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Crossing vs Touching

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Role of Continuity

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Horizontal Line Method

This is one of the most common ways graph-based root questions are asked. ---

End Behaviour + Turning Points

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Standard Root-Counting Patterns

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Symmetry and Absolute Value Tricks

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Minimal Worked Examples

Example 1 A continuous graph has local maximum $3$, local minimum $-2$, and tends to $+\infty$ as $x\to\pm\infty$. How many roots does $f(x)=0$ have? Since the graph is:

• high on the far left,

• goes above $0$ at the local maximum,

• below $0$ at the local minimum,

• high again on the far right,

the graph must cut the $x$-axis three times. So the number of roots is $\boxed{3}$. --- Example 2 If the graph of $y=f(x)$ intersects the line $y=2$ exactly at four points, then the equation $f(x)=2$ has $\boxed{4}$ roots. ---

CMI Strategy

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Practice Questions

:::question type="MCQ" question="A continuous function $f$ has local maximum value $4$, local minimum value $-1$, and satisfies $f(x)\to +\infty$ as $x\to \pm\infty$. How many real roots does the equation $f(x)=0$ have?" options=["$1$","$2$","$3$","$4$"] answer="C" hint="Track the graph from left to right using end behaviour and the extreme values." solution="Since $f(x)\to +\infty$ as $x\to \pm\infty$, the graph is high on both ends. The local maximum is $4>0$ and the local minimum is $-1<0$. So while moving from left to right, the graph must:

• cross the $x$-axis once before reaching the positive local maximum,

• cross once while descending from positive values to the negative minimum,

• cross once again while rising from negative values to positive values on the right.

Hence the equation $f(x)=0$ has exactly $\boxed{3}$ real roots. So the correct option is $\boxed{C}$." ::: :::question type="NAT" question="The graph of $y=f(x)$ intersects the line $y=2$ at exactly three distinct points and the line $y=-2$ at exactly two distinct points. How many real roots does the equation $|f(x)|=2$ have?" answer="5" hint="Split $|f(x)|=2$ into two equations." solution="We use the identity $\qquad |f(x)|=2 \iff f(x)=2 \text{ or } f(x)=-2$ The graph intersects:

• $y=2$ at exactly $3$ points

• $y=-2$ at exactly $2$ points

Therefore the total number of roots is $\qquad 3+2=5$ Hence the answer is $\boxed{5}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["Roots of $f(x)=0$ are the intersection points of the graph of $y=f(x)$ with the $x$-axis","If a graph touches the $x$-axis and turns back, that point is still a root","The number of roots of $f(x)=k$ equals the number of intersections of $y=f(x)$ with $y=k$","A graph must cross the $x$-axis at every root"] answer="A,B,C" hint="Focus on what a root means graphically." solution="1. True. That is exactly the graphical meaning of $f(x)=0$.

True. Touching the axis also means the function value is zero there.

True. Solving $f(x)=k$ means counting intersections with the horizontal line $y=k$.

False. A graph may only touch the axis, as in $y=x^2$ at $x=0$.

Hence the correct statements are $\boxed{A,B,C}$." ::: :::question type="SUB" question="A continuous function $f$ satisfies $f(x)\to -\infty$ as $x\to -\infty$, $f(x)\to +\infty$ as $x\to +\infty$, has local maximum value $5$, and local minimum value $-1$. Determine the number of real roots of the equation $f(x)=2$." answer="3" hint="Compare the line $y=2$ with the local maximum and local minimum." solution="We count intersections of the graph of $y=f(x)$ with the horizontal line $y=2$. Given:

• left end goes to $-\infty$

• right end goes to $+\infty$

• local maximum is $5$, which is above $2$

• local minimum is $-1$, which is below $2$

So:

As the graph rises from $-\infty$ to the local maximum $5$, it must cross $y=2$ once.

As it falls from $5$ to $-1$, it must cross $y=2$ once again.

As it rises from $-1$ to $+\infty$, it must cross $y=2$ once more.

Therefore the equation $f(x)=2$ has exactly $\boxed{3}$ real roots." ::: ---

Summary

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Part 4: Combined algebra-calculus problems

Combined Algebra-Calculus Problems

Overview

This topic is not a single formula topic. It is a reasoning topic where algebraic structure controls calculus behaviour. In CMI-style problems, you are often asked to combine several ideas at once:

• domain restrictions from logarithms, roots, or denominators

• slope analysis using derivatives or the Fundamental Theorem of Calculus

• asymptotic growth comparisons such as $\ln x$ versus powers

• tangency interpreted through repeated roots

• inverse functions mixed with differentiation

• continuous functional equations such as $f(x^2)=f(x)^2$

• parity and symmetry inside definite integrals

So the correct mindset is: simplify the algebra first, then apply the calculus cleanly. ---

Learning Objectives

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Lens 1: Domain Before Calculus

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Lens 2: Integral-Defined Functions and Slope

This is one of the most important bridges between algebra and calculus. If a problem asks where $g(x)$ has:

• maximum slope, maximize $g'(x)=h(x)$

• minimum slope, minimize $g'(x)=h(x)$

So the slope question becomes an algebraic/extremum question about the integrand. ---

Lens 3: Growth Comparison

That means:

• $\dfrac{\ln x}{x^a}\to 0$

• $\dfrac{x^a}{b^x}\to 0$

• $\dfrac{\ln x}{1/x^n} = x^n \ln x \to \infty$

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Lens 4: Tangency and Repeated Roots

This is the cleanest way to decide whether a single line can be tangent at many points. ---

Lens 5: Inverse Functions and Derivatives

If a problem imposes a relation such as $\qquad f'(x)=f^{-1}(x)$, then algebra and calculus are tied together very tightly. This shows how trial families reduce a hard functional-differential problem to algebra. ---

Lens 6: Functional Equations with Continuity

This is extremely useful when combined with continuity. For example, on $[0,1]$, $\qquad x^{2^n}\to 0$ So if $f$ is continuous, $\qquad f(x)^{2^n} = f(x^{2^n}) \to f(0)$ This often forces very strong restrictions on $f$. ---

Lens 7: Parity and Symmetry

Since

• $\cos(nx)$ is even

• $\sin(nx)$ is odd

we get quick conclusions such as:

• $x^{2023}\cos(nx)$ is odd

• $x^4\sin x$ is odd

So both integrate to $0$ over $[-a,a]$. ---

Minimal Worked Examples

Example 1 Let $\qquad g(x)=\int_1^x \ln(t^2+1)\,dt$ Then $\qquad g'(x)=\ln(x^2+1)$ On $[1,3]$, since $x^2+1$ increases, $\ln(x^2+1)$ also increases. Therefore the slope of $g$ is largest at $\boxed{x=3}$. --- Example 2 Suppose a line is tangent to a polynomial at exactly $3$ distinct points. Then the difference polynomial has at least $6$ roots counting multiplicity. So the polynomial must have degree at least $\boxed{6}$. ---

Common Patterns in This Topic

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $g(x)=\int_0^x \ln(t^2+1)\,dt$, then on $[0,2]$ the maximum slope of $g$ occurs at" options=["$0$","$1$","$2$","No maximum exists"] answer="C" hint="Use $g'(x)$." solution="By the Fundamental Theorem of Calculus, $\qquad g'(x)=\ln(x^2+1)$. Since $x^2+1$ increases on $[0,2]$, $\ln(x^2+1)$ also increases. Therefore the maximum slope occurs at $\boxed{2}$, so the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find the smallest possible degree of a polynomial that admits one line tangent to its graph at exactly $4$ distinct real points." answer="8" hint="Use repeated roots." solution="If one line is tangent at exactly $4$ distinct real points, then the difference between the polynomial and the line has at least $2\cdot 4=8$ roots counting multiplicity. Hence the degree must be at least $8$. This bound is achievable by taking $\qquad p(x)=L(x)+\prod_{i=1}^4 (x-a_i)^2$. So the smallest possible degree is $\boxed{8}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\dfrac{\ln x}{x^{1/10}}\to 0$ as $x\to\infty$","$\dfrac{x^{10}}{e^x}\to 0$ as $x\to\infty$","If $F(x)$ is odd, then $\int_{-a}^{a}F(x)\,dx=0$","If $f(x^2)=f(x)^2$, then necessarily $f(x)=x^2$"] answer="A,B,C" hint="Use growth hierarchy and parity." solution="1. True, because logarithms grow slower than positive powers.

True, because exponentials dominate polynomials.

True, by symmetry of odd functions on $[-a,a]$.

False. Many functions satisfy the equation, for example $f(x)=x^r$ for any real $r$, and also the zero function on suitable domains.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Suppose $f(x)=ax^b$ on $\mathbb{R}_+$. Find the equation that $b$ must satisfy if $f'(x)=f^{-1}(x)$ for all $x>0$." answer="$b^2-b-1=0$" hint="Match exponents of $x$ on both sides." solution="We have $\qquad f(x)=ax^b$ so $\qquad f'(x)=abx^{b-1}$ Also, $\qquad f^{-1}(x)=\left(\dfrac{x}{a}\right)^{1/b}=a^{-1/b}x^{1/b}$ If $f'(x)=f^{-1}(x)$ for all $x>0$, then the exponents of $x$ must match: $\qquad b-1=\dfrac{1}{b}$ Multiplying by $b$, $\qquad b^2-b-1=0$ So the required equation is $\boxed{b^2-b-1=0}$." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="For the function $f(x) = x^3 - 6x^2 + 9x + 1$, which of the following statements is true?" options=["$f(x)$ is increasing on $(1, 3)$.", "$f(x)$ has a local minimum at $x=1$.", "$f(x)$ has a local maximum at $x=1$.", "$f(x)$ is decreasing on $(-\infty, 1) \cup (3, \infty)$."] answer="$f(x)$ has a local maximum at $x=1$." hint="Compute the first derivative $f'(x)$ and analyse its sign changes." solution="The first derivative is $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$.
Critical points are at $x=1$ and $x=3$.

• For $x < 1$, $f'(x) > 0$, so $f(x)$ is increasing.


• For $1 < x < 3$, $f'(x) < 0$, so $f(x)$ is decreasing.


• For $x > 3$, $f'(x) > 0$, so $f(x)$ is increasing.

Since $f'(x)$ changes from positive to negative at $x=1$, there is a local maximum at $x=1$.
Since $f'(x)$ changes from negative to positive at $x=3$, there is a local minimum at $x=3$.
Therefore, the statement '$f(x)$ has a local maximum at $x=1$' is true."
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:::question type="NAT" question="Determine the number of distinct real roots for the equation $x^3 - 3x^2 + 6 = 0$." answer="1" hint="Define $f(x) = x^3 - 3x^2 + 6$ and analyse its local extrema using the first derivative." solution="Let $f(x) = x^3 - 3x^2 + 6$.
Then $f'(x) = 3x^2 - 6x = 3x(x-2)$.
Setting $f'(x)=0$ gives critical points at $x=0$ and $x=2$.
Evaluate $f(x)$ at these critical points:

• $f(0) = (0)^3 - 3(0)^2 + 6 = 6$. This is a local maximum (since $f'(x)$ changes from positive to negative at $x=0$).


• $f(2) = (2)^3 - 3(2)^2 + 6 = 8 - 12 + 6 = 2$. This is a local minimum (since $f'(x)$ changes from negative to positive at $x=2$).

Also, consider the limits:

• As $x \to -\infty$, $f(x) \to -\infty$.


• As $x \to \infty$, $f(x) \to \infty$.

Sketching the graph: $f(x)$ starts from $-\infty$, increases to a local maximum of $6$ at $x=0$, then decreases to a local minimum of $2$ at $x=2$, and then increases to $\infty$.
Since the local maximum value ($6$) and the local minimum value ($2$) are both positive, the graph crosses the x-axis only once, for $x<0$.
Thus, there is only 1 distinct real root."
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:::question type="MCQ" question="For the function $g(x) = x^4 - 4x^3$, which of the following describes its inflection points?" options=["Only at $x=0$.", "Only at $x=2$.", "At $x=0$ and $x=2$.", "There are no inflection points."] answer="At $x=0$ and $x=2$." hint="Compute the second derivative $g''(x)$ and identify points where its sign changes." solution="The first derivative is $g'(x) = 4x^3 - 12x^2$.
The second derivative is $g''(x) = 12x^2 - 24x = 12x(x-2)$.
Setting $g''(x)=0$ gives potential inflection points at $x=0$ and $x=2$.
Now, we check the sign of $g''(x)$ around these points:

• For $x < 0$, $g''(x) = 12x(x-2) = (negative)(negative) > 0$ (concave up).


• For $0 < x < 2$, $g''(x) = 12x(x-2) = (positive)(negative) < 0$ (concave down).


• For $x > 2$, $g''(x) = 12x(x-2) = (positive)(positive) > 0$ (concave up).

Since $g''(x)$ changes sign at both $x=0$ and $x=2$, both are inflection points.
Therefore, the statement 'At $x=0$ and $x=2$' is true."
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:::question type="NAT" question="For the function $f(x) = \frac{x^2}{x-1}$, what is the value of its local maximum?" answer="0" hint="Calculate $f'(x)$ and find the critical points. Then apply the First Derivative Test or Second Derivative Test to classify them." solution="To find local extrema, we first compute the first derivative:

Setting $f'(x)=0$ gives $x(x-2)=0$, so critical points are at $x=0$ and $x=2$. Note that $x=1$ is a vertical asymptote and not in the domain of $f(x)$.
Now, we analyse the sign of $f'(x)$ around the critical points:

• For $x < 0$, $f'(x) = \frac{(-)( -)}{(+)} > 0$, so $f(x)$ is increasing.


• For $0 < x < 1$, $f'(x) = \frac{(+)( -)}{(+)} < 0$, so $f(x)$ is decreasing.


• For $1 < x < 2$, $f'(x) = \frac{(+)( -)}{(+)} < 0$, so $f(x)$ is decreasing.


• For $x > 2$, $f'(x) = \frac{(+)( +)}{(+)} > 0$, so $f(x)$ is increasing.

At $x=0$, $f'(x)$ changes from positive to negative, indicating a local maximum.
The value of the function at $x=0$ is $f(0) = \frac{0^2}{0-1} = 0$.
At $x=2$, $f'(x)$ changes from negative to positive, indicating a local minimum.
The value of the function at $x=2$ is $f(2) = \frac{2^2}{2-1} = \frac{4}{1} = 4$.
The local maximum value is 0."
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What's Next?