In short

The domain of a function is the set of all inputs for which the function produces a valid output. The range is the set of all outputs the function actually produces. For real-valued functions, finding the domain means locating values that cause division by zero, square roots of negatives, or logarithms of non-positives — and excluding them. Finding the range means determining which output values are actually achieved. When no domain is stated, the natural (implied) domain is the largest subset of \mathbb{R} where the formula works.

A railway ticket counter sells tickets from Delhi to Jaipur. The counter accepts cash in denominations of ₹10, ₹20, ₹50, ₹100, ₹200, ₹500, and ₹2000. For each denomination you hand over, the counter gives back a specific change amount and a ticket. But hand over a ₹5 coin and the counter refuses — that denomination is not in the system.

The set of denominations the counter accepts is its domain. The set of change amounts it actually returns is its range. A ₹2000 note and a ₹500 note might both produce the same change amount (if the ticket costs ₹480, both produce valid change), but the counter will never accept a denomination it was not programmed for.

Functions work the same way. The formula f(x) = \frac{1}{x-3} accepts every real number except 3 — feed in 3 and the machine breaks (division by zero). The domain is \mathbb{R} \setminus \{3\}. The outputs that actually come out — that is the range. Finding both is the core skill of this article.

What domain and range mean

From the article on functions, you know that a function f: A \to B assigns each element of A to exactly one element of B. The set A is the domain, and the range is the subset of B that actually gets hit:

\text{Range}(f) = \{f(x) \mid x \in A\}

When A and B are subsets of \mathbb{R}, the domain is an interval or a union of intervals on the x-axis, and the range is an interval or union of intervals on the y-axis. The graph of the function sits inside the rectangle \text{domain} \times \text{codomain}, and its shadow on the y-axis is the range.

Graph showing domain as shadow on x-axis and range as shadow on y-axisA curve is drawn on an x-y plane. Its horizontal shadow on the x-axis is shaded and labelled domain. Its vertical shadow on the y-axis is shaded and labelled range. x y domain range
The domain is the "shadow" of the curve on the $x$-axis — every $x$-value where the curve exists. The range is the shadow on the $y$-axis — every $y$-value the curve actually reaches.

Natural domain (implied domain)

When a problem says "find the domain of f(x) = \sqrt{x - 2}" without specifying a set A, it is asking for the natural domain — the largest subset of \mathbb{R} for which the formula produces a real output. This is also called the implied domain.

Three operations create restrictions on the natural domain:

  1. Division by zero. If the formula has a denominator, exclude every x that makes the denominator 0.
  2. Even roots of negatives. If the formula contains \sqrt[n]{\text{expression}} with n even, require the expression inside to be \ge 0.
  3. Logarithms of non-positives. If the formula contains \log(\text{expression}), require the expression inside to be > 0.

These three rules cover virtually every domain question in class 11. Odd roots (\sqrt[3]{x}) and polynomials (x^3 - 5x + 2) are defined for all real numbers, so they impose no restrictions.

Three domain restrictions: division, even root, logarithmThree rows. First row: 1 over expression, exclude where expression equals 0. Second row: square root of expression, need expression at least 0. Third row: log of expression, need expression greater than 0. 1 / g(x) Exclude x where g(x) = 0 root of g √g(x) Need g(x) ≥ 0 root of g domain log g(x) Need g(x) > 0 root of g domain
The three common domain restrictions. For division: exclude the zeros of the denominator (hollow dot). For square roots: include the zero and everything to the right where the expression is non-negative (solid dot). For logarithms: exclude the zero, include everything to the right where the expression is positive (hollow dot).

Finding the domain: the systematic method

Here is a clean procedure for any formula-based function.

Step 1. Identify every sub-expression that creates a restriction: denominators, even roots, logarithms.

Step 2. For each restriction, write an inequality or an exclusion.

Step 3. Intersect all the conditions. The intersection is the domain.

Take f(x) = \frac{\sqrt{x+1}}{x-2}. Two restrictions appear:

Intersect: x \ge -1 and x \ne 2. In interval notation: [-1, 2) \cup (2, \infty).

Number line showing the domain of sqrt(x+1) over (x-2)A number line from negative 3 to 5. A solid dot at negative 1 begins a shaded region that extends to the right, with a hollow dot at 2 indicating that 2 is excluded, and then continues to the right beyond 5. −3 −1 1 2 4 domain = [−1, 2) ∪ (2, ∞)
The domain of $f(x) = \frac{\sqrt{x+1}}{x-2}$. The square root starts at $x = -1$ (solid dot, included). The point $x = 2$ is punched out (hollow dot, excluded) because it makes the denominator zero.

Finding the range: algebraic method

Finding the range is harder than finding the domain because you need to determine which output values are actually achieved. The standard algebraic method works like this:

Step 1. Set y = f(x).

Step 2. Rearrange to express x in terms of y (or to get a polynomial in x with coefficients involving y).

Step 3. Determine which values of y allow at least one real solution for x that lies in the domain.

The values of y that survive are the range.

For a quadratic function f(x) = ax^2 + bx + c with a > 0, the parabola opens upward. Its minimum value is at the vertex: y_{\min} = c - \frac{b^2}{4a}. The range is \left[c - \frac{b^2}{4a}, \infty\right).

For a rational function like f(x) = \frac{1}{x}, you set y = \frac{1}{x}, which gives x = \frac{1}{y}. This has a solution for every y \ne 0, so the range is \mathbb{R} \setminus \{0\}.

Parabola showing minimum value as start of rangeA parabola y equals x squared minus 2x minus 3 with vertex at (1, negative 4). A horizontal dashed line at y equals negative 4 marks the minimum. The range is negative 4 to infinity, shown as a shaded strip on the y-axis. x y −4 4 8 1 vertex (1, −4) range = [−4, ∞)
The parabola $y = x^2 - 2x - 3 = (x-1)^2 - 4$. The vertex at $(1, -4)$ is the lowest point, so the range is $[-4, \infty)$. The shaded strip on the $y$-axis marks the range.

Graphical method: reading domain and range from a graph

When you have the graph of a function, the domain and range are visible.

This is the "shadow" idea from the first figure. Project the curve onto the x-axis for the domain, and onto the y-axis for the range.

Interactive parabola with draggable point showing domain and rangeThe parabola y equals x squared is drawn. A red draggable point sits on the curve. As the point is dragged, dashed lines project to both axes, illustrating how x maps to y. Readouts show the current x and y coordinates. x y 1 2 −1 −2 1 4 drag the red point
Drag the point along $y = x^2$. Every real $x$ is in the domain, but the $y$-values never go below $0$ — the range is $[0, \infty)$.

Two worked examples

Example 1: Find the domain and range of $f(x) = \sqrt{4 - x^2}$

Step 1. Identify the restriction. The square root requires 4 - x^2 \ge 0.

Why: you cannot take the square root of a negative number in \mathbb{R}, so the expression under the root must be non-negative.

Step 2. Solve the inequality.

4 - x^2 \ge 0 \implies x^2 \le 4 \implies -2 \le x \le 2

Why: rearranging gives x^2 \le 4. Since x^2 is non-negative, this holds exactly when |x| \le 2, i.e., x \in [-2, 2].

Step 3. State the domain.

\text{Domain} = [-2, 2]

Step 4. Find the range. The expression under the root, 4 - x^2, takes its maximum when x = 0: \sqrt{4 - 0} = 2. It takes its minimum when x = \pm 2: \sqrt{4 - 4} = 0. As x moves from 0 to \pm 2, the value \sqrt{4 - x^2} decreases continuously from 2 to 0. Every value between 0 and 2 is achieved.

Why: the square root function is continuous, so by the intermediate value theorem, every y between 0 and 2 is hit.

Step 5. State the range.

\text{Range} = [0, 2]

Result: Domain = [-2, 2]. Range = [0, 2].

Graph of f(x) = sqrt(4 - x squared), a semicircleA semicircle of radius 2 centred at the origin, in the upper half-plane. The domain is from negative 2 to 2 on the x-axis, and the range is from 0 to 2 on the y-axis. Both are marked with shaded regions on the respective axes. x y 2 −2 1 −1 2 1 (0, 2)
The graph of $f(x) = \sqrt{4 - x^2}$ is the upper half of a circle of radius $2$. The domain is the interval $[-2, 2]$ (shaded on the $x$-axis). The range is $[0, 2]$ (shaded on the $y$-axis). The highest point $(0, 2)$ is where $x^2 = 0$, giving the maximum output.

The graph is the upper semicircle of x^2 + y^2 = 4. You can see that x cannot go beyond \pm 2 (that is the domain) and y never exceeds 2 or goes below 0 (that is the range).

Example 2: Find the domain and range of $g(x) = \frac{x+1}{x-3}$

Step 1. Find the domain. The denominator x - 3 must not be zero, so x \ne 3.

Why: division by zero is undefined. Every other real number produces a valid output.

Step 2. State the domain.

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

Step 3. Find the range. Set y = \frac{x+1}{x-3} and solve for x.

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

Why: by isolating x, you determine which y-values produce a valid x. This expression is defined when y - 1 \ne 0, i.e., y \ne 1.

Step 4. Check that the solution x lies in the domain. You need x \ne 3. Substitute: \frac{3y+1}{y-1} = 3 gives 3y + 1 = 3y - 3, i.e., 1 = -3, which is impossible. So no value of y forces x = 3. Every y \ne 1 is achieved.

Why: the impossible equation confirms that x = 3 never arises as a solution, so no additional values of y need to be excluded beyond y = 1.

Result: Domain = \mathbb{R} \setminus \{3\}. Range = \mathbb{R} \setminus \{1\}.

Graph of g(x) = (x+1)/(x-3) showing asymptotesThe graph of g of x equals x plus 1 over x minus 3. A vertical dashed line at x equals 3 shows the vertical asymptote. A horizontal dashed line at y equals 1 shows the horizontal asymptote. The curve approaches but never touches either asymptote. x y 3 1 y = 1 x = 3 (0, −1/3)
The graph of $g(x) = \frac{x+1}{x-3}$. The vertical asymptote at $x = 3$ (dashed) shows the point excluded from the domain. The horizontal asymptote at $y = 1$ (dashed) shows the value excluded from the range — the curve approaches $y = 1$ but never reaches it.

The horizontal asymptote tells you the missing value in the range. As x \to \pm\infty, the function approaches 1 but never equals it — and the algebra confirms this: y = 1 has no solution for x.

Domain and range of common functions

Here is a quick reference for functions you will see repeatedly.

Function Domain Range
f(x) = c (constant) \mathbb{R} \{c\}
f(x) = x (identity) \mathbb{R} \mathbb{R}
f(x) = x^2 \mathbb{R} [0, \infty)
f(x) = x^3 \mathbb{R} \mathbb{R}
f(x) = |x| \mathbb{R} [0, \infty)
f(x) = \sqrt{x} [0, \infty) [0, \infty)
f(x) = \frac{1}{x} \mathbb{R} \setminus \{0\} \mathbb{R} \setminus \{0\}

Each row encodes a pair of facts: what you can feed in, and what comes out. The graphs of basic functions article draws all of these curves, so you can verify each row visually.

Domain and range comparison for three common functionsThree small graphs in a row: y equals x squared (parabola, range 0 to infinity), y equals square root of x (half parabola sideways, domain 0 to infinity, range 0 to infinity), and y equals 1 over x (hyperbola, domain and range exclude 0). y = x² ℝ [0,∞) y = √x [0,∞) [0,∞) y = 1/x ℝ\{0} ℝ\{0}
Three common functions and their ranges. The parabola covers all $y \ge 0$. The square root also covers $y \ge 0$ but starts at $x = 0$. The reciprocal function covers all $y$ except $0$.

Common confusions

Going deeper

If you are comfortable finding domains and ranges for class 11 functions, the material above is sufficient. The rest of this section covers two subtler points.

Piecewise functions and domain by cases

A piecewise function defines different rules on different intervals. Each piece has its own natural domain, and the overall domain is the union of the intervals where each piece is defined.

Take:

f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ \sqrt{x} & \text{if } x \ge 0 \end{cases}

The first piece is x^2, defined for all x < 0. The second piece is \sqrt{x}, defined for all x \ge 0. The union is (-\infty, 0) \cup [0, \infty) = \mathbb{R}. For the range: the first piece gives (0, \infty) (since x^2 > 0 for x < 0). The second piece gives [0, \infty). The union is [0, \infty).

Domain of composite functions

When you have h(x) = f(g(x)), the domain of h is the set of all x such that:

  1. x is in the domain of g (so g(x) is defined), and
  2. g(x) is in the domain of f (so f(g(x)) is defined).

For example, h(x) = \sqrt{\frac{1}{x}}. Think of this as f(g(x)) where g(x) = \frac{1}{x} and f(t) = \sqrt{t}. Condition 1: x \ne 0. Condition 2: \frac{1}{x} \ge 0, which means x > 0. Intersecting: x > 0, i.e., the domain is (0, \infty).

This becomes important in function composition, which you will study in a later chapter.

Where this leads next

You now know how to find the domain and range of real-valued functions using algebraic methods, graphical inspection, and the three key restriction rules. Here is where this knowledge connects.