Domain and range of functions

Every calculator, spreadsheet formula and physics equation is quietly asking two questions before it does anything else: what inputs am I allowed to feed this rule, and what outputs can it possibly hand back? A function is a relation with one extra guarantee — every permitted input produces exactly one output — and the domain and range are simply the honest bookkeeping of those allowed inputs and achievable outputs. This article stays inside the ISC Class 11 "Relations and Functions" scope: defining a function as a special relation $y = f(x)$ , and finding the domain and range of the nine standard real-valued functions — constant, identity, polynomial, rational, modulus, signum, exponential, logarithmic, and greatest-integer. Formal classification (one-one, onto, bijective) and composition or inverse of functions belong to Class XII and are not covered here. By the end you should be able to spot instantly why $\frac{1}{x-3}$ excludes $x=3$ , why $[-1.3]=-2$ and not $-1$ , and how to hunt for a range by solving for $x$ in terms of $y$ .

The core insight

A function assigns each input in its domain exactly one output, and finding domain/range means locating every input where the rule breaks down (domain) and every output value the rule can actually produce (range).

The core idea

A function $f$ from set $A$ to set $B$ is a relation $f \subseteq A \times B$ in which every element of $A$ appears as the first entry of exactly one ordered pair. Writing $y = f(x)$ means: for each input $x$ in the domain, $f$ assigns a unique output $y$ . Two definitions do all the work in this topic:

Domain = the set of all permissible inputs $x$ , i.e. the values for which $f(x)$ is actually defined.
Range = the set of all outputs $f(x)$ actually attained as $x$ ranges over the domain — not just "wherever the formula could land," but every value it truly reaches.

That distinction between domain and range mirrors two different kinds of question. Domain-hunting is about spotting forbidden inputs: division by zero, a negative number under an even root, a non-positive argument inside a logarithm. Range-hunting is about tracing achievable outputs, usually by solving the equation $y = f(x)$ for $x$ and asking which $y$ values leave a valid $x$ behind.

\text{Domain}(f) = \{x \in \mathbb{R} : f(x) \text{ is defined}\}, \qquad \text{Range}(f) = \{f(x) : x \in \text{Domain}(f)\}

where: $f(x)$ is the rule assigning an output to each input $x$ ; "defined" means the arithmetic in the rule does not break (no division by zero, no even root of a negative, no logarithm of a non-positive number). The nine standard real-valued functions on the syllabus, with their domain and range:

Function	Rule	Domain	Range
Constant	$f(x) = c$	$\mathbb{R}$	$\{c\}$
Identity	$f(x) = x$	$\mathbb{R}$	$\mathbb{R}$
Polynomial	$f(x) = a_nx^n + \dots + a_0$	$\mathbb{R}$	depends on degree
Rational	$f(x) = \dfrac{p(x)}{q(x)}$	$\mathbb{R}$ except where $q(x)=0$	depends
Modulus	$f(x) = \|x\|$	$\mathbb{R}$	$[0,\infty)$
Signum	$f(x) = \begin{cases}1,& x>0\\0,& x=0\\-1,& x<0\end{cases}$	$\mathbb{R}$	$\{-1,0,1\}$
Exponential	$f(x) = a^x,\ a>0, a\ne 1$	$\mathbb{R}$	$(0,\infty)$
Logarithmic	$f(x) = \log_a x,\ a>0,a\ne1$	$(0,\infty)$	$\mathbb{R}$
Greatest integer	$f(x) = [x]$ , largest integer $\le x$	$\mathbb{R}$	$\mathbb{Z}$

For a rational function the domain excludes zeros of the denominator; the range then depends on whether the equation $y = p(x)/q(x)$ can be solved back for $x$ at every $y$ . For modulus and greatest-integer functions the domain is all of $\mathbb{R}$ , but the outputs are restricted in a distinctive shape. The greatest integer function always "rounds down" toward $-\infty$ , never toward zero: $[2.7] = 2$ , but $[-1.3] = -2$ (not $-1$ ), since the largest integer that is still $\le -1.3$ is $-2$ .

Figure

Number line picture of the domain of one over x minus three showing every real number allowed except the single excluded point three

Derivation AFinding the range by solving for x in terms of y

For a function given by an algebraic formula, the range is found not by guessing but by treating $y = f(x)$ as an equation, solving it for $x$ in terms of $y$ , and asking for which values of $y$ that expression for $x$ is itself a valid real number lying in the original domain — every such $y$ is in the range, and every $y$ that breaks the solved expression is excluded.

Step 1. Start from $y = f(x)$ and rearrange algebraically until $x$ is isolated as an expression in $y$ , exactly as you would solve any equation for one variable given the other.

Step 2. Examine the resulting expression $x = g(y)$ for restrictions: division by an expression in $y$ that could be zero, a square root of an expression in $y$ that could be negative, or any other operation that fails for particular $y$ values.

Step 3. Every value of $y$ for which $g(y)$ is defined and gives an $x$ actually inside the domain of $f$ belongs to the range; every value of $y$ that makes $g(y)$ undefined is excluded from the range.

Step 4. Write the range as the set of all such permissible $y$ , usually as $\mathbb{R}$ minus a finite set of excluded points, or as an interval, matching the shape of the restriction found in Step 2.

Worked example

Problem. Find the domain and range of $f(x) = \dfrac{1}{x-3}$ , and then find the domain and range of $g(x) = |x-2| - 3$ .

Step 1. For $f(x) = \dfrac{1}{x-3}$ , this is a rational function, undefined wherever the denominator is zero. Setting $x - 3 = 0$ gives $x = 3$ , so

\text{Domain}(f) = \mathbb{R} - \{3\} = (-\infty, 3) \cup (3, \infty)

Step 2. To find the range of $f$ , let $y = \dfrac{1}{x-3}$ and solve for $x$ :

y(x-3) = 1 \implies x - 3 = \frac{1}{y} \implies x = 3 + \frac{1}{y}

This expression for $x$ is defined for every real $y$ except $y=0$ (division by $y$ ), and no value of $x$ ever makes $y=0$ since $\frac{1}{x-3}=0$ has no solution. So $\text{Range}(f) = \mathbb{R} - \{0\}$ .

Step 3. For $g(x) = |x-2| - 3$ , the modulus $|x-2|$ is defined for every real $x$ , and subtracting the constant $3$ does not restrict this further, so $\text{Domain}(g) = \mathbb{R}$ . Since $|x-2| \ge 0$ for all $x$ , with the minimum value $0$ at $x=2$ , $g(x) \ge 0-3 = -3$ , and as $x$ moves away from $2$ in either direction $|x-2|$ grows without bound, so $g$ takes every value from $-3$ upward.

Step 4. Hence $\text{Range}(g) = [-3, \infty)$ . Checking one point, $g(-1) = |-1-2| - 3 = |-3| - 3 = 3 - 3 = 0$ .

Final answer: $\text{Domain}(f) = \mathbb{R}-\{3\}$ , $\text{Range}(f) = \mathbb{R}-\{0\}$ ; $\text{Domain}(g) = \mathbb{R}$ , $\text{Range}(g) = [-3,\infty)$ , with $g(-1)=0$ .

Competitive stretch

For $h(x) = \dfrac{1}{\sqrt{x-3}}$ , the domain tightens further: the denominator must be nonzero and the quantity under the square root must be nonnegative, so $x - 3 > 0$ (strict, since $x=3$ still gives division by zero), giving $\text{Domain}(h) = (3, \infty)$ — combining a rational restriction with a square-root restriction in a single expression.

Common traps

Forgetting to exclude denominator zeros when finding the domain of a rational function — always factor and solve $q(x) = 0$ first, then remove those points from $\mathbb{R}$ .
Rounding the greatest integer function toward zero instead of toward $-\infty$ : $[-1.3] = -2$ , not $-1$ , because $[x]$ is the largest integer that is less than or equal to $x$ .
Assuming the domain of $\log_a x$ allows zero or negative inputs — $\log_a x$ is defined only for $x>0$ ; $\log_a 0$ and $\log_a(\text{negative})$ are undefined, not zero or negative outputs.
Confusing signum's three-value range $\{-1,0,1\}$ with a continuous interval — signum only ever outputs one of these three exact numbers, never anything between them.

Ask the tutor: Ask the AI tutor to walk through finding the domain and range of two or three more mixed functions — like a rational function combined with a square root, or a modulus function combined with a greatest-integer function — using the solve-for-x technique step by step.

Why does the greatest integer function round down instead of to the nearest integer

Because $[x]$ is defined as the largest integer that is less than or equal to $x$ , not the closest one. For a positive number like $2.7$ , rounding down and rounding to nearest happen to agree at first glance, but for a negative number like $-1.3$ the largest integer still $\le -1.3$ is $-2$ , since $-1$ is actually greater than $-1.3$ . The rule always moves toward $-\infty$ , which is why it is also called the "floor function."

How do I know when to use the solve-for-x method versus just reasoning about the shape of the function

Use the solve-for-x method whenever the formula is algebraic and rearrangeable, like a rational function — it gives an exact, provable range. For functions built from modulus, signum, or greatest-integer, algebraic rearrangement is often clumsy, so it is faster to reason about the shape directly: modulus expressions have a known minimum, signum only ever takes three fixed values, and greatest-integer only ever outputs integers. Both methods must agree when applicable — the algebra is just the rigorous backup.

Is every relation a function, and how can I quickly check

No — a relation is a function only if every input has exactly one output. The quickest check on a graph is the vertical line test: if any vertical line crosses the graph more than once, some input has two outputs, so it fails to be a function. Algebraically, watch for rules like $y^2 = x$ , which give two values of $y$ (positive and negative square root) for most positive $x$ , so it is a relation but not a function.

Domain and range of functions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads