Real Analysis

Properties of continuous functions

Properties of continuous functions form the bedrock of real analysis, encompassing crucial principles like the Intermediate Value Theorem, Boundedness Theorem, and Extreme Value Theorem. This topic covers these structural properties.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Intermediate Value Theorem (IVT) & Extreme Value Theorem (EVT)

Let $f: [a, b] \to \mathbb{R}$ be a continuous function. \\ 1. (EVT): There exist $c, d \in [a, b]$ such that $f(c) = \inf_{x \in [a, b]} f(x)$ and $f(d) = \sup_{x \in [a, b]} f(x)$. \\ 2. (IVT): If $u$ is any real number strictly between $f(a)$ and $f(b)$, then there exists at least one $c \in (a, b)$ such that $f(c) = u$.

📐 Geometric Interpretation

Geometrically, EVT ensures that a continuous curve defined on a compact interval attains both a global maximum and a global minimum. IVT guarantees that any horizontal line $y = u$ lying between the endpoint height values $f(a)$ and $f(b)$ must intersect the graph of the function at least once.

⚙️ Algebraic & Structural Meaning

Algebraically, these properties link topological compactness and connectedness of the domain to the range. Specifically, the continuous image of a compact set is compact, and the continuous image of a connected set is connected.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Continuity on Domain	`[a, b] (Closed and bounded interval)`	Strictly Mandatory / Necessary	If continuity fails (e.g., jump discontinuity), the function may skip intermediate values, violating IVT.
Compactness of Domain	`[a, b] (Closed interval)`	Strictly Mandatory for EVT	On an open interval like $(0, 1)$, a continuous function like $f(x) = \frac{1}{x}$ or $f(x) = x$ fails to attain bounds or extremes.
Sign change at endpoints (for Bolzano's Theorem)	`Endpoints $a$ and $b$`	Essential for root location	If $f(a) \cdot f(b) > 0$, root existence is not guaranteed by IVT alone, although roots may still exist.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Bolzano's Root Location Test

$$ f(a) \cdot f(b) < 0 \implies \exists c \in (a, b) \text{ s.t. } f(c) = 0 $$

Rule: Substitute boundary points into polynomial or transcendental expressions. If signs differ, at least one real root exists in $(a, b)$.

Application: Directly tested in UP PGT to determine the exact interval containing a root of cubic equations.

Fixed Point Theorem for Continuous Mappings

$$ f: [0, 1] \to [0, 1] \text{ continuous} \implies \exists c \in [0, 1] \text{ s.t. } f(c) = c $$

Rule: Construct $g(x) = f(x) - x$. If $g(0) \ge 0$ and $g(1) \le 0$, a fixed point exists instantly.

Polynomial Root Count on Odd vs Even Degree

$$ \lim_{x \to \pm \infty} f(x) = \pm \infty \text{ (Odd degree polynomial)} \implies \text{At least one real root} $$

Rule: Every polynomial of odd degree with real coefficients has at least one real root by IVT.

Range of Continuous Functions on Compact Sets

$$ f([a, b]) = [m, M] \text{ where } m = \min f, M = \max f $$

Rule: The image of a closed, bounded interval under a continuous function is always a closed, bounded interval.

Intermediate Value Property for Derivatives (Darboux's Theorem)

$$ f'(a) = k_1, \, f'(b) = k_2 \implies \forall k \in [k_1, k_2], \, \exists c \in (a, b) \text{ s.t. } f'(c) = k $$

Rule: Derivatives possess the intermediate value property even if they are not continuous.

Application: Asked in high-level PGT screening tests.

Rapid Elimination via Boundedness

$$ f \in C([a, b]) \implies f \text{ is bounded} $$

Rule: Immediately eliminate any option stating an unbounded range for a continuous function on a closed interval.

Application: Quick sanity check for matching list questions.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Assuming Open Intervals Attain Extrema

Common Pitfall: Candidates apply EVT to open intervals like $(0, 1)$ for functions like $f(x) = x$.

Counterexample: $f(x) = x \text{ on } (0, 1)$

Why it fails: The function is continuous, but the range $(0, 1)$ has neither a maximum nor a minimum element within the set.

🛡️ Exam Defense: Always check if the domain is closed and bounded before asserting maximum/minimum attainment.

⚠️ COMMON ERROR #02

Converse of IVT Fallacy

Common Pitfall: Believing that if a function takes all intermediate values between $f(a)$ and $f(b)$, the function must be continuous.

Counterexample: $f(x) = \sin(1/x) \text{ for } x \neq 0 \text{ and } f(0) = 0$

Why it fails: Darboux functions can have intermediate value property without being continuous everywhere.

🛡️ Exam Defense: Remember that IVT is a necessary condition for continuous functions, not a sufficient condition for continuity.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Extreme Value Theorem	$f: [a, b] \to \mathbb{R}$ continuous	$\exists c, d \in [a, b]$ s.t. $f(c) \le f(x) \le f(d)$	Graph has highest and lowest points	Keywords: 'maximum', 'minimum', 'attains bounds'
Intermediate Value Theorem	$f: [a, b] \to \mathbb{R}$ continuous	$f(a) < u < f(b) \implies f(c) = u$	Graph crosses every horizontal strip between endpoints	Keywords: 'root exists', 'takes value', 'equation has solution'
Bolzano's Theorem	$f: [a, b] \to \mathbb{R}$ continuous, $f(a)f(b) < 0$	$\exists c \in (a, b)$ s.t. $f(c) = 0$	Graph intersects the x-axis	Keywords: 'at least one real root in interval'
Boundedness Theorem	$f: [a, b] \to \mathbb{R}$ continuous	$\|f(x)\| \le M$ for all $x \in [a, b]$	Graph stays within horizontal bounds	Keywords: 'bounded', 'finite supremum'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $f: [0, 1] \to [0, 1]$ is a continuous function, then the equation $f(x) = x^2$ has:

(A) No solution in $(0, 1)$(B) At least one solution in $[0, 1]$(C) Exactly two solutions(D) Infinitely many solutions

📚 Method A: Formal Textbook Derivation

Step 1: Define auxiliary function $g(x) = f(x) - x^2$. Step 2: Since $f$ maps $[0, 1]$ to $[0, 1]$, $g(0) = f(0) \ge 0$ and $g(1) = f(1) - 1 \le 0$. Step 3: By IVT, since $g$ is continuous on $[0, 1]$, there exists $c \in [0, 1]$ such that $g(c) = 0 \implies f(c) = c^2$.

⚡ Method B: 25-Second Exam Shortcut

Set $g(x) = f(x) - x^2$, check boundary signs ($g(0) \ge 0, g(1) \le 0$), and instantly apply Bolzano's theorem.

💡 Tip: Whenever a problem links $f(x)$ with $x^2$ or $x$, define an auxiliary difference function and check boundary conditions.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(1) = 2$ and $f(3) = 10$. Which of the following is definitely true?

(A) $f(x) = 6$ has no solution in $(1, 3)$(B) $f(x) = 6$ has at least one solution in $(1, 3)$(C) $f(x) = 12$ has a solution in $(1, 3)$(D) $f(x)$ is strictly increasing

📚 Method A: Formal Textbook Derivation

Step 1: Identify that $f$ is continuous on $[1, 3]$. Step 2: Note that $6$ lies strictly between $f(1) = 2$ and $f(3) = 10$ ($2 < 6 < 10$). Step 3: By the Intermediate Value Theorem, there exists $c \in (1, 3)$ such that $f(c) = 6$.

⚡ Method B: 25-Second Exam Shortcut

Check if target value lies between endpoint values ($2 < 6 < 10$). Yes $\implies$ IVT guarantees solution.

💡 Tip: Direct application of IVT. Look for values lying strictly inside the range of endpoint values.

Practice questions

Choose an answer to check it and see the solution:

If $f:[a, b] \to \mathbb{R}$ is continuous and $f(a) f(b) < 0$, then:

A The equation $f(x) = 0$ has no root in $(a, b)$

B The equation $f(x) = 0$ has at least one root in $(a, b)$

C The function $f$ is monotonic on $(a, b)$

D The function $f$ is unbounded on $(a, b)$

Derivation & Analysis:

By Bolzano's theorem (a direct corollary of IVT), if a continuous function has opposite signs at endpoints, it must cross the x-axis at least once.

Which of the following functions attains both its supremum and infimum on the given interval?

A $f(x) = \frac{1}{x}$ on $(0, 1)$

B $f(x) = \tan x$ on $(-\frac{\pi}{2}, \frac{\pi}{2})$

C $f(x) = x^3 - 3x$ on $[-2, 2]$

D $f(x) = e^x$ on $(-\infty, \infty)$

Derivation & Analysis:

By the Extreme Value Theorem, a continuous function on a closed and bounded interval $[-2, 2]$ must attain its bounds. The other intervals are open or unbounded.

Every polynomial of degree $n$ where $n$ is odd has:

A No real roots

B At least one real root

C Exactly $n$ distinct real roots

D Only complex roots

Derivation & Analysis:

By limits at infinity and IVT, an odd-degree polynomial changes sign from $-\infty$ to $+\infty$ (or vice versa), guaranteeing at least one real root.

If $f: [0, 1] \to [0, 1]$ is continuous, then the fixed point theorem guarantees the existence of $c \in [0, 1]$ such that:

A $f(c) = 0$

B $f(c) = 1$

C $f(c) = c$

D $f'(c) = 0$

Derivation & Analysis:

Brouwer's Fixed Point Theorem in 1D states that any continuous mapping from a closed interval to itself has at least one fixed point $f(c) = c$.

The range of the continuous function $f(x) = \sin x$ on $[0, 2\pi]$ is:

A $(-1, 1)$

B $[0, 1]$

C $[-1, 1]$

D $[0, 2\pi]$

Derivation & Analysis:

By EVT and connectedness, the continuous image of $[0, 2\pi]$ under sine is the closed interval $[-1, 1]$.

If $f$ is continuous on $[a, b]$ and $f(x) \neq 0$ for all $x \in [a, b]$, then:

A $f(x) > 0$ everywhere or $f(x) < 0$ everywhere on $[a, b]$

B $f(a) = f(b)$

C $f$ attains zero at some interior point

D $f$ is constant

Derivation & Analysis:

If $f$ took values of opposite signs, IVT would force $f(c) = 0$ for some $c$, contradicting $f(x) \neq 0$. Thus, $f$ maintains a constant sign throughout.

Let $f: [0, 1] \to \mathbb{R}$ be continuous with $f(0) = 1$ and $f(1) = 0$. Which point is guaranteed to be in the range of $f$?

A $1.5$

B $-0.5$

C $0.5$

D $2$

Derivation & Analysis:

By IVT, the range includes all values between $f(1) = 0$ and $f(0) = 1$. Since $0.5 \in [0, 1]$, it is guaranteed.

Which of the following properties is NOT necessarily preserved under continuous mappings?

A Compactness

B Connectedness

C Boundedness

D Differentiability

Derivation & Analysis:

Continuity does not imply differentiability; for example, $f(x) = |x|$ is continuous but not differentiable at $x = 0$.

If $f(x) = x^5 + 5x + 1$, how many real roots does $f(x) = 0$ have?

A 5

B 3

C 1

D 0

Derivation & Analysis:

Since $f'(x) = 5x^4 + 5 > 0$ everywhere, $f$ is strictly increasing. With $\lim_{x \to -\infty} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = \infty$, IVT and monotonicity guarantee exactly one real root.

Q10

Let $f$ be a continuous function on $[0, 2]$ such that $f(0) = f(2)$. Then:

A There exists $c \in (0, 2)$ such that $f(c) = 1$

B There exists $c \in (0, 2)$ such that $f'(c) = 0$ (assuming differentiability)

C $f$ must be a constant function

D $f$ cannot cross the x-axis

Derivation & Analysis:

This is Rolle's Theorem hypothesis: equal endpoint values for a continuous and differentiable function guarantee a horizontal tangent in the interior.

Common questions

Yes. Darboux's theorem shows that derivatives of differentiable functions can have the intermediate value property even if the derivative function itself is discontinuous.

Failing to check whether the domain is closed and bounded when applying the Extreme Value Theorem, leading to errors with open intervals where bounds are not attained.

Mastering Bolzano's sign test and auxiliary function construction ($g(x) = f(x) - x$) allows candidates to identify root existence and fixed points within 15 to 20 seconds.

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