In short

A function is bounded if all its output values stay within some finite interval — there exist real numbers m and M such that m \le f(x) \le M for every x in the domain. The supremum is the smallest such upper bound; the infimum is the largest such lower bound. A function that is not bounded is unbounded.

A temperature sensor sits on a rooftop in Jaipur. Over one full year, it records a temperature every hour — 8,760 readings. The lowest reading is 4°\text{C} (a cold January night). The highest is 48°\text{C} (a brutal May afternoon). No reading ever goes below 4 or above 48.

You could say: "this sensor's output is trapped between 4 and 48." That is the core idea of a bounded function — its outputs never escape a finite strip.

But here is a subtler question. Suppose someone tells you "the temperature never exceeds 50°\text{C}." That is also true — 50 is an upper bound. So is 100, and so is 10{,}000. There are infinitely many upper bounds. Among all of them, the tightest one — the smallest number that still sits above every reading — is the one that matters. That tightest upper bound has a name: the supremum. And the tightest lower bound is the infimum.

Upper and lower bounds

Take a function f defined on some domain D.

A number M is an upper bound for f if f(x) \le M for every x \in D. The function's outputs never exceed M.

A number m is a lower bound for f if f(x) \ge m for every x \in D. The function's outputs never dip below m.

Consider f(x) = x^2 on D = [-3, 3]. The outputs range from 0 (at x = 0) to 9 (at x = \pm 3). So M = 9 is an upper bound, and so is M = 10, M = 100, or any number \ge 9. Similarly, m = 0 is a lower bound, and so is m = -1, m = -50, or any number \le 0.

A parabola on a closed interval trapped inside a horizontal stripThe parabola y equals x squared is drawn from x equals negative 3 to x equals 3. Two horizontal dashed lines at y equals 0 and y equals 9 form a strip. The curve stays entirely within this strip. The strip is lightly shaded. x y m = 0 M = 9 3 −3 1 −1 (0, 0) (3, 9)
The parabola $y = x^2$ on $[-3, 3]$, trapped inside a horizontal strip between $m = 0$ and $M = 9$. Every point on the curve lies within the shaded region. The function is bounded.

Now consider g(x) = x^2 on D = \mathbb{R} (all real numbers). The lower bound is still 0, but there is no upper bound — you can always pick a larger x and get a larger output. The function is bounded below but not bounded above, so it is unbounded.

Bounded vs unbounded

Bounded function

A function f: D \to \mathbb{R} is bounded if there exist real numbers m and M such that

m \le f(x) \le M \quad \text{for all } x \in D.

Equivalently, f is bounded if there exists a number K > 0 such that |f(x)| \le K for all x \in D.

If only the upper-bound condition holds (f(x) \le M for all x), f is bounded above. If only the lower-bound condition holds (f(x) \ge m for all x), f is bounded below.

A function that is not bounded is unbounded.

The equivalence in the definition is quick to see. If m \le f(x) \le M, then |f(x)| \le K where K = \max(|m|, |M|). Conversely, if |f(x)| \le K, set m = -K and M = K.

Here are some examples to build your intuition:

Function Domain Bounded? Why
f(x) = \sin x \mathbb{R} Yes -1 \le \sin x \le 1 always
f(x) = x^2 \mathbb{R} No No upper bound
f(x) = \frac{1}{1+x^2} \mathbb{R} Yes 0 < f(x) \le 1 always
f(x) = x^3 \mathbb{R} No No upper or lower bound
f(x) = x^2 [-5, 5] Yes 0 \le f(x) \le 25

The last row shows something important: the domain matters. The same formula x^2 is unbounded on \mathbb{R} but bounded on [-5, 5]. Boundedness is a property of the function together with its domain, not of the formula alone.

Comparison of bounded and unbounded functionsTwo small graphs side by side. Left: y equals sine x oscillates between negative 1 and 1, bounded. Right: y equals x cubed goes to positive and negative infinity, unbounded. sin x — bounded 1 −1 x³ — unbounded
Left: $\sin x$ oscillates forever between $-1$ and $1$ — it is bounded. Right: $x^3$ grows without limit in both directions — it is unbounded.

Maximum and minimum values

When a function achieves its best possible bound — when some input x_0 actually produces the bound value — that bound has a sharper name.

If there exists x_0 \in D such that f(x_0) \ge f(x) for all x \in D, then f(x_0) is the maximum value of f on D.

If there exists x_0 \in D such that f(x_0) \le f(x) for all x \in D, then f(x_0) is the minimum value of f on D.

Take f(x) = x^2 on [-3, 3]. The minimum value is f(0) = 0 — actually achieved at x = 0. The maximum value is f(3) = 9 — achieved at x = 3 (and also at x = -3).

But now take g(x) = \frac{1}{x} on (0, 1]. The maximum value is g(1) = 1, achieved at x = 1. What about the minimum? As x approaches 0 from the right, g(x) grows without limit. There is no minimum value — the function is unbounded above. And it has no largest value either, in this case the function is unbounded above rather than below.

Now consider h(x) = \frac{x}{x+1} on [0, \infty). As x grows, h(x) approaches 1 but never reaches it. The minimum value is h(0) = 0, achieved at x = 0. But the value 1 is never achieved — so h has no maximum, even though it is bounded above by 1.

This is the crucial distinction: a bounded function might not achieve its bounds.

A function that achieves its bounds versus one that does notTwo graphs. Left: y equals x squared on negative 3 to 3, with filled dots at the maximum (3,9) and minimum (0,0). Right: y equals x over x plus 1 on 0 to infinity, approaching y equals 1 as a dashed asymptote, with a filled dot at the minimum (0,0) but no dot at the supremum. max and min exist min = 0 max = 9 no maximum y = 1 min = 0 sup = 1 (never reached)
Left: $f(x) = x^2$ on $[-3, 3]$ achieves both its minimum ($0$) and maximum ($9$). Right: $h(x) = \frac{x}{x+1}$ on $[0, \infty)$ achieves its minimum ($0$) but only approaches $1$ — the supremum exists, but the maximum does not.

Supremum and infimum

This is where the language gets precise. If a function is bounded above, there may be many upper bounds. The smallest one has a special name.

Supremum and Infimum

Let f: D \to \mathbb{R} be bounded above. The supremum (or least upper bound) of f on D is the smallest real number M such that f(x) \le M for all x \in D. It is written \sup_{x \in D} f(x).

Similarly, if f is bounded below, the infimum (or greatest lower bound) is the largest real number m such that f(x) \ge m for all x \in D. It is written \inf_{x \in D} f(x).

The supremum has two defining properties:

  1. It is an upper bound: f(x) \le \sup f for every x in the domain.
  2. Nothing smaller works: for any \epsilon > 0, there exists some x_0 in the domain with f(x_0) > \sup f - \epsilon.

Property 2 is the key. It says that the supremum is tight — you can always find a function value that comes within \epsilon of it, no matter how small \epsilon is.

When the supremum is actually achieved — when some x_0 gives f(x_0) = \sup f — the supremum equals the maximum. When it is not achieved, the supremum exists but the maximum does not.

Here is a concrete comparison:

Number line showing infimum and supremum of a bounded functionA number line with a shaded region representing the set of output values of a bounded function. The left endpoint is labelled inf f and the right endpoint is labelled sup f. Various output values are marked as dots inside the shaded region. y inf f sup f output values of f any number below inf f is a lower bound any number above sup f is an upper bound
The output values of a bounded function sit inside a finite interval on the $y$-axis. The infimum is the left edge of that interval — the tightest lower bound. The supremum is the right edge — the tightest upper bound. Every number to the left of $\inf f$ is also a lower bound, but none of them is as tight.

Why the domain matters (again)

Consider f(x) = \frac{1}{x} on different domains:

The function 1/x on two different domains showing bounded vs unboundedTwo graphs of y equals 1 over x. Left: on the domain 1 to infinity, the curve stays between 0 and 1, bounded. Right: on the domain 0 to 1, the curve shoots upward without bound. D = [1, ∞): bounded 1 0 D = (0, 1]: unbounded
Same formula, different behaviour. On $[1, \infty)$, $\frac{1}{x}$ is bounded and stays in $(0, 1]$. On $(0, 1]$, the same formula is unbounded above — the outputs grow without limit as $x$ approaches $0$.

Two worked examples

Example 1: Show that $f(x) = \frac{x}{x^2 + 1}$ is bounded on $\mathbb{R}$, and find $\sup f$ and $\inf f$

Step 1. Check that the function is well-defined on all of \mathbb{R}. The denominator is x^2 + 1, which is always \ge 1, so there is no division by zero.

Why: before analysing bounds, confirm the function actually works on the stated domain.

Step 2. Find the critical points. Set f'(x) = 0. Using the quotient rule:

f'(x) = \frac{(x^2+1) \cdot 1 - x \cdot 2x}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}

This equals zero when 1 - x^2 = 0, i.e., x = \pm 1.

Why: the extreme values of a differentiable function on \mathbb{R} occur at critical points — where the derivative is zero.

Step 3. Evaluate f at the critical points.

f(1) = \frac{1}{1+1} = \frac{1}{2}, \qquad f(-1) = \frac{-1}{1+1} = -\frac{1}{2}

Why: these are the candidates for the global maximum and minimum.

Step 4. Check the behaviour as x \to \pm\infty. As x grows large, f(x) = \frac{x}{x^2+1} \approx \frac{1}{x} \to 0. So the function approaches 0 at both ends, confirming that the critical-point values are the extremes.

Why: if the function grew without bound as x \to \infty, the critical points would not give global extremes.

Result: f is bounded on \mathbb{R}. \sup f = \frac{1}{2} (achieved at x = 1). \inf f = -\frac{1}{2} (achieved at x = -1). Since both are achieved, \max f = \frac{1}{2} and \min f = -\frac{1}{2}.

Graph of f(x) = x/(x^2+1) showing maximum and minimumThe graph of f(x) equals x over x squared plus 1. The curve passes through the origin, rises to a maximum of 1/2 at x equals 1, then decreases back toward 0. On the left side it dips to a minimum of negative 1/2 at x equals negative 1. Horizontal dashed lines at y equals 1/2 and y equals negative 1/2 mark the bounds. x y 1 2 −1 −2 ½ −½ max = ½ at x = 1 min = −½ at x = −1
The curve $f(x) = \frac{x}{x^2+1}$ is trapped between $y = -\frac{1}{2}$ and $y = \frac{1}{2}$. Both bounds are achieved — the function touches the dashed lines at $x = \pm 1$.

The graph confirms the algebra. The curve rises to \frac{1}{2} at x = 1 and dips to -\frac{1}{2} at x = -1, then tapers toward the x-axis in both directions. The entire curve lives inside the horizontal strip [-\frac{1}{2}, \frac{1}{2}].

Example 2: Show that $g(x) = \frac{1}{x}$ on $D = (0, \infty)$ is bounded below but not bounded above, and find $\inf g$

Step 1. Check for a lower bound. For all x > 0, g(x) = \frac{1}{x} > 0. So m = 0 is a lower bound.

Why: a positive number divided by a positive number is always positive.

Step 2. Show that 0 is the greatest lower bound. For any \epsilon > 0, choose x = \frac{1}{\epsilon}. Then g\!\left(\frac{1}{\epsilon}\right) = \epsilon. This output is within \epsilon of 0.

Why: property 2 of the infimum — for any distance \epsilon from the proposed infimum, you must exhibit an output that close. Choosing x = 1/\epsilon does the job.

Step 3. Check whether g achieves this infimum. Is there any x > 0 with g(x) = 0? That would require \frac{1}{x} = 0, which has no solution. The infimum is not achieved.

Why: \frac{1}{x} can be made as small as desired, but it never actually reaches 0. So \inf g = 0 but \min g does not exist.

Step 4. Show the function is unbounded above. For any proposed upper bound M > 0, choose x = \frac{1}{M+1}. Then g(x) = M + 1 > M. No finite M works as an upper bound.

Why: this is a direct proof that no upper bound exists — for each candidate, you produce an output that exceeds it.

Result: g(x) = \frac{1}{x} on (0, \infty) is bounded below with \inf g = 0 (not achieved). It is unbounded above — \sup g does not exist.

Graph of 1/x on (0, infinity) showing no upper boundThe graph of y equals 1 over x for positive x. The curve drops steeply near x equals 0 and flattens out approaching the x-axis as x grows. A dashed line at y equals 0 shows the infimum. An upward arrow near x equals 0 indicates the function grows without bound. x y 1 2 3 4 2 3 unbounded above inf = 0 (never reached) (1, 1)
The curve $y = \frac{1}{x}$ for $x > 0$. As $x \to 0^+$, the output shoots upward without limit — no upper bound exists. As $x \to \infty$, the output approaches $0$ but never reaches it — the infimum is $0$, not achieved.

The picture matches. The curve hugs the x-axis from above (approaching the infimum 0) while climbing without limit near the y-axis (no upper bound).

Common confusions

Interactive: explore boundedness

Interactive graph of sin(x)/x with draggable pointThe curve y equals sine x plotted over the interval negative 5 to 5. A draggable point moves along the curve. Readouts show the x and y coordinates, demonstrating that y stays between negative 1 and 1. x y 1 2 −1 −2 1 −1 sup = 1 inf = −1 drag the red point
Drag the point along $y = \sin x$. No matter where you place it, $y$ stays between $-1$ and $1$. The function is bounded, with $\sup = 1$ and $\inf = -1$ — both achieved.

The material above covers all of the bounded-functions content needed for class 11. The following extends the ideas to a more rigorous setting.

The completeness property and why supremum exists

A natural question: how do you know the supremum always exists? Could there be a set of real numbers that is bounded above but has no least upper bound?

The answer is no — and this is not a theorem you can prove from simpler facts. It is an axiom of the real numbers, called the completeness property (or the least upper bound property):

Every non-empty subset of \mathbb{R} that is bounded above has a supremum in \mathbb{R}.

This is one of the properties that distinguishes \mathbb{R} from \mathbb{Q}. Consider the set S = \{q \in \mathbb{Q} : q^2 < 2\} — all rational numbers whose square is less than 2. This set is bounded above (by 2, for instance), but its least upper bound in \mathbb{R} is \sqrt{2}, which is irrational. If you tried to find the supremum within \mathbb{Q} alone, you would fail — there is no rational number that serves as the least upper bound of S within \mathbb{Q}. The completeness of \mathbb{R} fills this gap.

Bounded functions and continuous functions on closed intervals

There is a deep result that connects boundedness to continuity:

Extreme Value Theorem. If f is continuous on a closed interval [a, b], then f is bounded on [a, b] and achieves both its maximum and minimum values.

This is why many of the "nice" functions you encounter in class 11 — polynomials, trigonometric functions on closed intervals, absolute-value functions — always have a maximum and minimum on a closed interval. The theorem fails if the interval is open (the function might shoot off to infinity near the endpoint) or if the function is not continuous (it might "jump" past its bound).

Sup and inf as functions of the domain

If f is bounded on D and E \subseteq D, then the range of f on E is a subset of the range on D. This means:

\inf_{x \in D} f(x) \le \inf_{x \in E} f(x) \le \sup_{x \in E} f(x) \le \sup_{x \in D} f(x)

Restricting the domain can only tighten the bounds (or keep them the same), never loosen them. Enlarging the domain can only loosen the bounds (or keep them the same).

Where this leads next

You now know what it means for a function to be bounded, how to identify upper and lower bounds, and how the supremum and infimum give the tightest possible bounds. Here is where these ideas connect.