MathsISC Class 11 · Relations and Functions

Real Functions: Domain, Range, Standard Graphs and the Algebra of Functions

Updated 17 October 2026 · Fact-checked

A function f : A → B assigns to each element of A exactly one element of B. For a real function the domain is the set of real x for which f(x) is defined (denominators non-zero, square-root arguments non-negative), and the range is the set of values taken. Functions can be added, multiplied and divided on the intersection of their domains, with the quotient excluding zeros of the denominator.

Understand Functions: Domain, Range, Graphs and Algebra

Function. A relation in which every element of the domain has exactly one image. The vertical line test checks a graph: any vertical line must cut it at most once. Notation f(x), with x the independent variable.

Finding the domain. (1) For a polynomial, the domain is ℝ. (2) For a rational function p(x)/q(x), exclude the zeros of q. (3) For $g (x)$ , require g(x) ≥ 0. (4) For $1/ g (x)$ , require g(x) > 0. (5) For a combination, intersect the individual domains.

Finding the range. Write y = f(x), solve for x in terms of y, and find which y give real x in the domain. For $9 - x^{2}$ the values lie between 0 and 3.

Standard functions and graphs. *Constant*: f(x) = c, range {c}. *Identity*: f(x) = x. *Modulus*: f(x) = |x|, domain ℝ, range [0, ∞), graph is a V. *Greatest integer*: f(x) = ⌊x⌋, the greatest integer not exceeding x, domain ℝ, range ℤ, a step graph. *Fractional part*: {x} = x − ⌊x⌋, range [0, 1). *Signum*: f(x) = 1 for x > 0, 0 for x = 0, −1 for x < 0; domain ℝ, range {−1, 0, 1}. *Reciprocal*: f(x) = 1/x, domain ℝ − {0}, range ℝ − {0}. *Square root*: domain [0, ∞), range [0, ∞).

Algebra of functions. For f and g with domains D_f and D_g: (f + g)(x) = f(x) + g(x), (f − g)(x), (fg)(x) = f(x)g(x), on D_f ∩ D_g; (f/g)(x) = f(x)/g(x) on D_f ∩ D_g with g(x) ≠ 0; (cf)(x) = c f(x).

Functions: Domain, Range, Graphs and Algebra: formulas to remember

Domain rules: $Rational: denominator ≠ 0 | $\sqrt{g}$: $g ≥ 0$ | $1/\sqrt{g}$: $g > 0$$; Intersect for combinations.
Standard ranges: $$|x|$: $[0, \infty)$ | $⌊x⌋$: ℤ | $\{x\}$: $[0, 1)$ | $1/x$: ℝ − {0}$; Domains: ℝ, ℝ, ℝ, ℝ − {0}.
Algebra: $$(f \pm g)(x) = f(x) \pm g(x)$ | $(fg)(x) = f(x)g(x)$ | $(f/g)(x) = f(x)/g(x)$, $g(x) \neq 0$$; Domain D_f ∩ D_g.
Modulus: $$|x - a| = k$ gives $x = a \pm k$ (k \geq 0)$; Two solutions for k > 0.

Functions: Domain, Range, Graphs and Algebra in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A pricing rule with rounding

Suppose a parking fee is Rs 10 for each hour or part of an hour, modelled as $C (t) = 10 ⌈ t ⌉$ for t > 0 hours.

Find the fee for 2.3 hours, exactly 3 hours and 0.1 hour.

Try it first, then show the solutionSolution

$⌈ 2.3 ⌉ = 3$ , so the fee is Rs 30.
$⌈ 3 ⌉ = 3$ , so the fee is Rs 30.
$⌈ 0.1 ⌉ = 1$ , so the fee is Rs 10.

Answer: Rs 30, Rs 30 and Rs 10.

Memory hook: A step function: the value stays constant between jumps.

Tricks to remember Functions: Domain, Range, Graphs and Algebra

Short memory aids. Say each one aloud once, then use it on a question the same day.

Denominator and root
Never zero in a denominator; never negative under a square root.
Use it when: Finding domains.
Vertical line test
A graph is a function if no vertical line meets it twice.
Use it when: Identifying functions from graphs.

Functions: Domain, Range, Graphs and Algebra: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Domain of a root function · 1 mark · MCQBoard level

The domain of $f (x) = x - 2$ is:
(a) $(2, \infty)$
(b) $[2, \infty)$
(c) $(- \infty, 2]$
(d) ℝ

Try it first, then show the solutionSolution

We need $x - 2 \geq 0$ , so $x \geq 2$ .

Answer: (b) $[2, \infty)$

Example 2: Domain and range · 3 marks · Short answerBoard level

Find the domain and the range of (i) $f (x) = x^{2} + 1$ and (ii) $g (x) = 1/ (x - 3)$ .

Try it first, then show the solutionSolution

(i) Domain ℝ since f is a polynomial. (0.75 mark)
Since $x^{2} \geq 0$ , $f (x) \geq 1$ : range $[1, \infty)$ . (0.75 mark)
(ii) $x \neq = 3$ : domain ℝ − {3}. (0.75 mark)
For $y = 1/ (x - 3)$ , $x = 3 + 1/ y$ , so y can be any non-zero real: range ℝ − {0}. (0.75 mark)

Answer: (i) ℝ, [1, ∞); (ii) ℝ − {3}, ℝ − {0}.

Example 3: A semicircle function · 3 marks · Short answerHarder board level

Find the domain and range of $f (x) = 9 - x^{2}$ , and the domain of $h (x) = 9 - x^{2} / (x - 1)$ .

Try it first, then show the solutionSolution

$9 - x^{2} \geq 0$ gives $- 3 \leq x \leq 3$ : domain $[- 3, 3]$ . (1 mark)
The expression is between 0 (at $x = \pm 3$ ) and 3 (at $x = 0$ ): range $[0, 3]$ . (1 mark)
For h also $x \neq = 1$ : domain $[- 3, 1) \cup (1, 3]$ . (1 mark)

Answer: Domain [−3, 3], range [0, 3]; h: [−3, 1) ∪ (1, 3].

Example 4: Algebra of functions · 4 marks · Short answerHarder board level

f(x) = √(x − 1) and g(x) = √(5 − x). Find the domains of f + g, fg and f/g and the value of (f/g)(2).

Try it first, then show the solutionSolution

D_f = [1, ∞) and D_g = (−∞, 5], so D_f ∩ D_g = [1, 5]. (1 mark)
f + g and fg are defined on [1, 5]. (1 mark)
f/g needs g(x) ≠ 0, so x ≠ 5: domain [1, 5). (1 mark)
$(f / g) (2) = 1 / 3 = 1/ 3 \approx 0.577$ . (1 mark)

Answer: [1, 5]; [1, 5]; [1, 5); 1/√3.

Example 5: Modulus and greatest integer functions · 5 marks · Long answerHOTS / competency

Evaluate ⌊−2.3⌋, ⌊3.99⌋ and ⌊5⌋. Solve ⌊x⌋ = 2 and |x − 2| = 3. State the range of $f (x) = x - ⌊ x ⌋$ and sketch its behaviour on [0, 3).

Try it first, then show the solutionSolution

⌊−2.3⌋ = −3, ⌊3.99⌋ = 3, ⌊5⌋ = 5. (1 mark)
⌊x⌋ = 2 gives $2 \leq x < 3$ . (1 mark)
|x − 2| = 3 gives $x - 2 = 3$ or $x - 2 = - 3$ , so x = 5 or x = −1. (1 mark)
f(x) = x − ⌊x⌋ is the fractional part; range [0, 1). (1 mark)
On [0, 3) it rises from 0 to nearly 1 on each unit interval and drops back to 0 at each integer: a saw-tooth. (1 mark)

Answer: −3, 3, 5; x ∈ [2, 3); x = 5 or −1; range [0, 1) saw-tooth.

Where students lose marks in Functions: Domain, Range, Graphs and Algebra

The slips examiners see most often, and what to write instead.

Mistake: Forgetting to exclude zeros of a denominator
Giving ℝ as the domain of a rational function.
Do this: Remove the points where the denominator vanishes.
Usually costs: 1 to 2 marks.
Mistake: Mixing up the range with the domain
Quoting the interval of x as the range.
Do this: Range is the set of y values.
Usually costs: 1 mark.
Mistake: Wrong value of ⌊x⌋ for negatives
Taking ⌊−2.3⌋ = −2.
Do this: It is the greatest integer not exceeding x: −3.
Usually costs: 1 to 2 marks.

Exam tips

Write the conditions for the domain as inequalities.
Solve y = f(x) for x to find the range.
Sketch standard graphs to support the answer.
For combinations, state D_f ∩ D_g clearly.

Practice questions on Functions: Domain, Range, Graphs and Algebra

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The range of $f (x) = ∣ x ∣$ is:
(a) ℝ
(b) $(0, \infty)$
(c) $[0, \infty)$
(d) $(- \infty, 0]$
Hint
The graph is a V above the x-axis.
Answer
(c) $[0, \infty)$
Q2 · 2 marks · Short answerBoard level
Find the domain of $f (x) = (x + 2) / (x^{2} - 4)$ .
Hint
Factor the denominator.
Answer
$x^{2} - 4 = (x - 2) (x + 2) \neq = 0$ , so the domain is ℝ − {−2, 2}.
Q3 · 3 marks · Short answerHarder board level
Find the domain of $f (x) = 1/ x^{2} - 5 x + 6$ .
Hint
Strictly positive under the root.
Answer
$x^{2} - 5 x + 6 = (x - 2) (x - 3) > 0$ for $x < 2$ or $x > 3$ : domain $(- \infty, 2) \cup (3, \infty)$ .
Q4 · 3 marks · Short answerHarder board level
Find the range of $f (x) = (x - 1) / (x + 2)$ , x ≠ −2.
Hint
Solve for x in terms of y.
Answer
$y (x + 2) = x - 1$ gives $x = (- 2 y - 1) / (y - 1)$ , defined for $y \neq = 1$ . Range: ℝ − {1}.
Q5 · 3 marks · Short answerHarder board level
If f(x) = x² and g(x) = 2x + 1, find f(g(2)), (f + g)(2) and (fg)(2).
Hint
Evaluate step by step.
Answer
$g (2) = 5$ , so $f (g (2)) = 25$ . $(f + g) (2) = 4 + 5 = 9$ ; $(f g) (2) = 4 \times 5 = 20$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a taxi fare is Rs 50 for the first kilometre and Rs 12 for every additional kilometre or part of it, modelled by $F (d) = 50 + 12 ⌈ d - 1 ⌉$ for d ≥ 1.
(i) Find F(1). (1 mark)
(ii) Find F(3.2). (1 mark)
(iii) State the range of F over d ≥ 1. (1 mark)
(iv) State whether the graph is continuous. (1 mark)
Hint
The ceiling function creates steps.
Answer
(i) $F (1) = 50 + 12 \times 0 = 50$ . (ii) $⌈ 2.2 ⌉ = 3$ , so $F = 50 + 36 = 86$ . (iii) The values 50, 62, 74, ... : ${50 + 12 k : k = 0, 1, 2, ...}$ . (iv) No: it is a step function with jumps.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Domain of $1/ x$ ?Show answer
ℝ − {0}.
2.Range of ⌊x⌋?Show answer
The integers ℤ.
3.Domain of f/g?Show answer
D_f ∩ D_g with g(x) ≠ 0.
4.Range of $x - ⌊ x ⌋$ ?Show answer
[0, 1).

Functions: Domain, Range, Graphs and Algebra: frequently asked questions

Is every relation a function?

No. A function needs exactly one image for every element of the domain.

How do I find the range quickly?

Express x in terms of y, then find the values of y for which a real x in the domain exists; or use bounds such as x^2 ≥ 0.

What is the difference between a function and its graph?

The graph is a picture of the function: the set of points (x, f(x)).