Bounded Functions

A function is bounded on a set if its output values are confined between two horizontal lines. This concept captures the idea that the function cannot grow without limit on that set.

Quick Reference

Concept	Meaning
$f$ bounded on $E$	$\exists M > 0$ such that $< / t d >< / t r >< t r >< t d >$ f$ unbounded on $E$	No single $M$ can bound $\lvert f(x) \rvert $f o r a l l$ x \in E$
Geometric interpretation	Graph of $f$ above $E$ lies between the lines $y = M $a n d$ y = -M$
"Bounded" without qualifier	Refers to boundedness on the entire domain

Definition

If $E$ is a subset of the domain of $f $, w e s a y$ f$ is bounded on $E$ (by $M$) if there exists a positive number $M$ such that

| f (x) | < M for all x \in E .

Otherwise, $f$ is said to be unbounded on $E$.

Geometrically, $f$ is bounded on $E$ when the portion of the graph above $E$ lies entirely between the horizontal lines $y = M $a n d$ y = -M$ for some $M > 0$.

Examples

The floor function $f(x) = \lfloor x \rfloor$ (which outputs the greatest integer less than or equal to $x$) is bounded on the interval $E = (-1, 1)$.

Solution

For all $x \in (-1, 1)$, the value $\lfloor x \rfloor$ is either $0$ (when $0 \leq x < 1

) o r

-1$ (when $-1 < x < 0$). In either case:

| f (x) | \leq 1.

So $f$ is bounded on $(-1, 1)$ with $M = 1

(o r a n y

M > 1$). However, $f(x) = \lfloor x \rfloor$ is unbounded on all of $\mathbb{R}$: for any candidate bound $M$, there exist integers $n > M

, s o

|f(n)| = n > M

. < f i g u r e c l a s s = " i m a g e i m a g e_{r} e s i z e d " s t y l e = " m a x - w i d t h : 480 p x; " >< i m g s r c = " h t t p s : / / a d a p t i v e b o o k s . o r g / b o o k - i m a g e s / p r e c a l c u l u s / c h 5 - b o u n d e d - 1. p n g " a l t = " G r a p h o f t h e f l o o r f u n c t i o n y = ⌊ x ⌋ s h o w i n g s t e p i n t e r v a l s a n d h o r i z o n t a l b o u n d s . " >< f i g c a p t i o n > G r a p h o f

f(x) = \lfloor x \rfloor$: bounded on $(-1,1)$, unbounded on $\mathbb{R}$.

Let $g(x) = \dfrac{1}{x}$. Determine whether $g$ is bounded on $[1, \infty)$ and on its full domain $\mathbb{R} - \{0\}$.

Solution

On $[1, \infty)$: Since $x \geq 1$, we have $0 < g(x) = 1/x \leq 1$. Therefore:

| g (x) | \leq 1 for all x \in [1, \infty) .

So $g$ is bounded on $[1, \infty)$ with $M = 1

. < f i g u r e c l a s s = " i m a g e i m a g e_{r} e s i z e d " s t y l e = " m a x - w i d t h : 400 p x; " >< i m g s r c = " h t t p s : / / a d a p t i v e b o o k s . o r g / b o o k - i m a g e s / p r e c a l c u l u s / c h 5 - b o u n d e d - 2 a . p n g " a l t = " G r a p h o f y = 1 / x o n t h e i n t e r v a l [1, \infty) s h o w i n g i t i s b o u n d e d b y y = 1 a n d y = - 1. " >< f i g c a p t i o n >

g(x) = 1/x$ is bounded on $[1, \infty)$: the graph stays between $y = -1

a n d

y = 1$. On $\mathbb{R} - \{0\}$: As $x \to 0^+$, $g(x) \to +\infty

, a n d a s

x \to 0^-$, $g(x) \to -\infty$. No single value of $M$ can bound $|g(x)|

f o r a l l

x \neq 0

. S o

i s < s t r o n g > u n b o u n d e d < / s t r o n g > o n i t s f u l l d o m a i n . < f i g u r e c l a s s = " i m a g e i m a g e_{r} e s i z e d " s t y l e = " m a x - w i d t h : 400 p x; " >< i m g s r c = " h t t p s : / / a d a p t i v e b o o k s . o r g / b o o k - i m a g e s / p r e c a l c u l u s / c h 5 - b o u n d e d - 2 b . p n g " a l t = " G r a p h o f y = 1 / x o n i t s f u l l d o m a i n, s h o w i n g i t g o e s t o p o s i t i v e a n d n e g a t i v e i n f i n i t y n e a r z e r o . " >< f i g c a p t i o n >

g(x) = 1/x$ is unbounded on $\mathbb{R} - \{0\}$: the graph grows without limit near $x = 0$.

Let $h(x) = \dfrac{1}{x^2 + 1}$. Show that $h$ is bounded on all of $\mathbb{R}$.

Solution

For all $x \in \mathbb{R}$:

x^{2} + 1 \geq 1 (with equality at x = 0),

0 < h (x) = \frac{1}{x^{2} + 1} \leq 1.

Therefore $|h(x)| \leq 1

f o r a l l

x \in \mathbb{R}

, a n d

h$ is bounded on its entire domain with $M = 1

. A s

|x| \to \infty$, the denominator grows without bound, causing $h(x) \to 0$. The maximum value of $h

i s

h(0) = 1

. < f i g u r e c l a s s = " i m a g e i m a g e_{r} e s i z e d " s t y l e = " m a x - w i d t h : 400 p x; " >< i m g s r c = " h t t p s : / / a d a p t i v e b o o k s . o r g / b o o k - i m a g e s / p r e c a l c u l u s / c h 5 - b o u n d e d - 3. p n g " a l t = " G r a p h o f y = 1 / (x ² + 1) s h o w i n g t h e b e l l s h a p e b o u n d e d b e t w e e n y = 0 a n d y = 1. " >< f i g c a p t i o n >

h(x) = 1/(x^2 + 1)$ is bounded on $\mathbb{R}$: the entire graph lies between $y = 0

a n d

y = 1$.

Remark: When we say a function is "bounded" or "unbounded" without specifying a set, we refer to its behavior on its entire domain. For instance, $g(x) = 1/x$ is unbounded and $h(x) = 1/(x^2+1)$ is bounded.

Frequently Asked Questions

What does "bounded by M" mean exactly?

It means that $|f(x)| < M

f o r a l l

x$ in the specified set. Geometrically, the entire graph above that set lies in the horizontal strip between $y = -M

a n d

y = M$.

Can a function be bounded on a small set but unbounded on a larger one?

Yes. For example, $g(x) = 1/x$ is bounded on $[1, \infty)$ but unbounded on $\mathbb{R} - \{0\}$. Boundedness always depends on the choice of set $E$.

Does a bounded function have to attain its maximum and minimum values?

Not necessarily. For instance, $h(x) = 1/(x^2+1)$ is bounded by $M = 1$ and attains the maximum value $1

a t

x = 0$, but it never reaches $0$. More extreme cases exist where bounded functions approach their bounds without ever reaching them.

Is every constant function bounded?

Yes. If $f(x) = c

f o r a l l

x$, then $|f(x)| = |c|$ everywhere. Choosing $M = |c| + 1$ gives $|f(x)| < M

f o r a l l

x$.

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Definition

Examples

Frequently Asked Questions

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