Domain and range

Lesson 1 put numbers and expressions into functions without asking whether every input is allowed. Some are not: you can’t divide by zero, for example. This lesson names the allowed inputs and the outputs, and shows why the two matter for inverses.

The domain is the set of inputs the function accepts.
The range is the set of outputs it actually gives.

From domain to rangeEach input has exactly one output

A function must give one output for each input. Two inputs may share an output, as $-2$ and $2$ both go to 4 here.

Finding the domain: three things to avoid

For a function on the real numbers, the domain is every $x$ except those that break one of these rules:

Division by zero. Leave out the $x$ values that make a denominator zero.
Square roots of negatives. What is under $\sqrt{\ }$ must be $\ge 0$ .
Logarithms of zero or negatives. What is inside $\log$ must be $> 0$ .

So a fraction is undefined exactly where its bottom is zero. Factorise the bottom and make each bracket zero.

More: where a function is undefined

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q1

Given $f(y) = \log_2\left(\dfrac{y + 1}{y + 3}\right)$ :

find the domain of $f(y)$ ;

solve $f(y) = 1$ .

The rule for logs
- A logarithm is defined only for positive numbers.
- So ${\frac{y + 1}{y + 3} > 0}$ .
Think first. What must be true of the number inside the log?
When is a fraction positive?
- Both positive: ${y + 1 > 0}$ and ${y + 3 > 0}$ , which means ${y > -1}$ .
- Both negative: ${y + 1 < 0}$ and ${y + 3 < 0}$ , which means ${y < -3}$ .
- So the domain is ${\{y : y < -3 \text{ or } y > -1\}}$ .
Think first. The top and bottom must have the same sign. When are both positive? Both negative?
Solve f(y) = 1
- ${\log_2\frac{y + 1}{y + 3} = 1}$ means ${\frac{y + 1}{y + 3} = 2^1 = 2}$ .
- Multiply by $y + 3$ : ${y + 1 = 2y + 6}$ .
- Take $y + 6$ from both sides: ${y = -5}$ .
Think first. Write log₂(…) = 1 without the log.
Check against the domain
- ${-5 < -3}$ , so $y = -5$ is in the domain.
- And ${\frac{-4}{-2} = 2}$ ✓.

Finding the range

The range is every value the function actually reaches. Picture the graph, or look at how the rule is built:

A square is never negative, so $(x - 2)^2 + 5$ is never less than 5. Its range is $y \ge 5$ .
$x^2 + 1$ is at least 1, so $\dfrac{4}{x^2 + 1}$ is at most $\frac41 = 4$ . It is always positive, but gets as close to 0 as you like.

Reading a rangeHighest value 4; never reaches 0: 0 < y ≤ 4

Pick a function, then slide $x$ into and out of its domain:

Domain and rangePick a function, slide x

x ≥ 2domain (x-axis)y ≥ 0range (y-axis)1y at x = 3

x = 3

x = 3 is in the domain, and gives y = 1, which is in the range. The domain is x ≥ 2; the range is y ≥ 0.

The domain of an inverse

The inverse runs the function backwards, so its inputs are the function’s outputs:

\text{domain of } f^{-1} = \text{range of } f

Only a function that is one-to-one (each output comes from one input) has a proper inverse. $x^2$ is not: $-2$ and $2$ both give 4, so the inverse can’t know which to return. Questions then either restrict the domain (say, to $x \ge 0$ ) or accept the answer with $\pm$ .

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q3

Two functions $f$ and $g$ are defined on the set of real numbers, $\mathbb{R}$ , by $f : x \to x^2 + 2$ and $g : x \to \dfrac{1}{x + 2}$ , $x \ne -2$ . Find the domain of $(g \circ f)^{-1}$ .

The composite
- $f$ acts first: ${g \circ f(x) = g(x^2 + 2)}$ .
- Replace $x$ in $g$ : ${\frac{1}{(x^2 + 2) + 2} = \frac{1}{x^2 + 4}}$ .
Think first. g∘f: f first. What is g(x² + 2)?
Make x the subject
- Write ${y = \frac{1}{x^2 + 4}}$ , so ${x^2 + 4 = \frac1y}$ .
- Take 4 from both sides: ${x^2 = \frac1y - 4 = \frac{1 - 4y}{y}}$ .
- So ${x = \pm\sqrt{\frac{1 - 4y}{y}}}$ , and ${(g \circ f)^{-1}(x) = \pm\sqrt{\frac{1 - 4x}{x}}}$ .
Think first. Write y = 1/(x² + 4) and solve for x.
Its domain
- The bottom can’t be zero: ${x \ne 0}$ .
- What is under the root must be at least 0: ${\frac{1 - 4x}{x} \ge 0}$ .
- Both parts positive: ${0 < x \le \frac14}$ (both negative is impossible).
- So the domain is ${\{x : 0 < x \le \frac14\}}$ .
Think first. What must be true for the root to exist?
Check with the range
- ${x^2 + 4 \ge 4}$ , so ${\frac{1}{x^2 + 4}}$ is at most ${\frac14}$ and always positive.
- The range of $g \circ f$ is ${0 < y \le \frac14}$ : the same set ✓.

Equations built from functions

When a question sets two functions equal, as in “where do $y = f(x)$ and $y = g(x)$ meet?”, write $f(x) = g(x)$ and solve. A substitution, such as $u = 2x - 5$ , often turns it into a simple equation.

More: equations with functions

Your turn

WAEC 2023 · Paper 2 · Q9 (a)

Two functions $g$ and $h$ are defined on the set of real numbers, $\mathbb R$ , by $g: x \to \dfrac{2 + x}{x^2 + x - 20}$ and $h: x \to x + 3$ .

(a)(i)
Find $g \circ h(x)$ ;
(a)(ii)
the values of $x$ for which $g \circ h(x)$ is undefined.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

$g \circ h(x) = g(x + 3)$ : replace every $x$ in $g$ by $x + 3$ .
The top: $2 + (x + 3) = x + 5$ .
The bottom: $(x + 3)^2 + (x + 3) - 20 = x^2 + 6x + 9 + x + 3 - 20$
$= x^2 + 7x - 8$ .
So $g \circ h(x) = \dfrac{x + 5}{x^2 + 7x - 8}$ .

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