The domain of a function is the complete set of all possible independent input values (typically x) for which the function is mathematically defined and outputs a real number (f(x)∈ℝf(x) \in \mathbb{R}).

If you consider a function as an algebraic processing machine:

Input x∈Domain⟶ Function f⟶Output f(x)∈Range\text{Input } x \in \text{Domain} \longrightarrow \fbox{ Function } f \longrightarrow \text{Output } f(x) \in \text{Range}

If an input value produces an impossible mathematical operation—such as division by zero, taking the even root of a negative number, or evaluating the logarithm of a non-positive number—that value cannot belong to the domain and must be excluded.

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Mathematical Domain Notation

UK exam boards accept three standard conventions for expressing domains:

  1. Inequality Notation: Direct algebraic inequalities.
    • Example:x>2x > 2, or −3≤x≤5-3 \le x \le 5.
  2. Set-Builder Notation: Formal set notation stating membership in the real numbers ℝ\mathbb{R}.
    • Example:{x∈ℝ:x≠4}\{x \in \mathbb{R} : x \neq 4\} ("the set of all real numbers x such that x does not equal 4").
    • Example:{x∈ℝ:x≥0}\{x \in \mathbb{R} : x \ge 0\}.
  3. Interval Notation: Bracket notation representing continuous spans on the real number line.
    • Round parentheses (a, b) indicate non-inclusive boundaries (< or >).
    • Square brackets [a, b] indicate inclusive boundaries (≤\le or ≥\ge).
    • Example:[3,∞)[3, \infty) represents x≥3x \ge 3.
    • Example:(−∞,1)∪(1,∞)(-\infty, 1) \cup (1, \infty) represents x≠1x \neq 1.

The Three Fundamental Domain Restrictions

Unless a function involves one of three specific algebraic operations, its natural domain is simply all real numbers (x∈ℝx \in \mathbb{R}).

Whenever you analyse a function, check for these three critical conditions:

Restriction 1: Division by Zero (Rational Functions)

Division by zero is undefined in mathematics. For any rational function:

f(x)=P(x)Q(x)f(x) = \dfrac{P(x)}{Q(x)}

The domain requires that the denominator cannot equal zero:

Q(x)≠0Q(x) \neq 0

Values of x that make Q(x) = 0 represent vertical asymptotes or points of discontinuity (holes) and must be excluded from the domain.

Restriction 2: Even Roots of Negative Numbers (Radical Functions)

In real analysis, you cannot take an even root (square root (x)\sqrt(x), fourth root (4x)\sqrt[4](x), etc.) of a negative number:

f(x)=g(x)⟹g(x)≥0f(x) = \sqrt{g(x)} \implies g(x) \ge 0

Odd Roots vs. Even Roots:

Odd roots (such as cube roots x3\sqrt[3]{x} or fifth roots x5\sqrt[5]{x}) can accept negative inputs because (−2)3=−8⟹−83=−2(-2)^3 = -8 \implies \sqrt[3]{-8} = -2.

Therefore, for f(x)=g(x)3f(x) = \sqrt[3]{g(x)}, the radical itself imposes no domain restriction (x∈ℝx \in \mathbb{R}, provided g(x)g(x) is defined).

If an even root appears in a denominator, it cannot equal zero either, turning the weak inequality into a strict inequality:

f(x)=1g(x)⟹g(x)>0f(x) = \dfrac{1}{\sqrt{g(x)}} \implies g(x) > 0

Restriction 3: Logarithms of Non-Positive Numbers

The logarithm of zero or a negative number is undefined over the set of real numbers. For any base a>0,a≠1a > 0, a \neq 1:

f(x)=ln⁡(g(x))orf(x)=loga⁡(g(x))⟹g(x)>0f(x) = \ln(g(x)) \quad \text{or} \quad f(x) = \log_a(g(x)) \implies g(x) > 0

The argument of the logarithm must be strictly positive.

Worked Example

Polynomial Functions

Find the domain of each of the following polynomial functions:

(a) f(x)=5x3−4x2+11f(x) = 5x^3 - 4x^2 + 11

(b) g(x)=1−14xg(x) = 1 - \dfrac{1}{4}x

Step 1: Check for algebraic restrictions

  • Are there any denominators containing variables? No.
  • Are there any even roots? No.
  • Are there any logarithms or inverse trigonometric expressions? No.

Step 2: Formulate the domain statement

Both functions are polynomials defined for every real number.

Domain of f(x):x∈ℝor(−∞,∞)\text{Domain of } f(x): x \in \mathbb{R} \quad \text{or} \quad (-\infty, \infty)

Domain of g(x):x∈ℝor(−∞,∞)\text{Domain of } g(x): x \in \mathbb{R} \quad \text{or} \quad (-\infty, \infty)

Practice Questions & Solutions

1

Find the domain of the function:

Express your answer in set-builder notation.

Solution

Set the denominator equal to zero to find the excluded input value:

The numerator is a polynomial defined for all real numbers.

The only restriction is that

.

In set-builder notation:

2

Determine the domain of the function:

Express your answer using inequality notation.

Solution

The expression inside the square root is in the denominator. Therefore, the radicand must be strictly greater than zero:

18 - 6x > 0

Rearrange the inequality:

18 > 6x

Divide by 6:

The numerator is defined for all real numbers.

Therefore, the domain is:

x <3

3

Find the domain of the function:

Solution

For the natural logarithm to be defined, its argument must be strictly positive:

A quotient is strictly positive when either:

Both numerator and denominator are positive:

and

Intersection:

Both numerator and denominator are negative:

and

Intersection:

(Alternatively, note that the critical values are and . Testing values in the intervals , , and reveals the quotient is positive when or ).

Domain:

4

The function k(x) is defined by:

Find the set of values of x for which k(x) is defined, expressing your answer in interval notation.

Solution

The function k(x) is the sum of two square root terms. Both must be defined simultaneously:

First radical condition:

Second radical condition:

Find the intersection of both intervals:

In interval notation:

5

Find all values of in the interval that must be excluded from the domain of:

Solution

The function is undefined wherever the denominator is zero:

Within the specified interval

:

First quadrant solution:


Second quadrant solution:


Excluded values:

Summarise with AI:

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Gianpiero Placidi

UK-based Chemistry graduate with a passion for education, providing clear explanations and thoughtful guidance to inspire student success.