2.2 Functions, domain and range

Introduction

A function is a rule that turns each input into exactly one output. Converting a temperature from Celsius to Fahrenheit, working out the price of an order from the number of items, and predicting the height of a thrown ball from the time since it was released are all functions: give the rule an input, and it returns one definite answer. Functions are the main tool for describing how one quantity depends on another, and every model in topic 2 is a function.

Each function comes with a set of allowed inputs (its domain) and a set of outputs it produces (its range). In a real context the domain carries meaning: time cannot be negative, a number of people must be a whole number, and no temperature is colder than absolute zero. Many functions can also be run backwards. An inverse function takes an output and returns the input that produced it, answering questions such as "what temperature gives a speed of sound of 340 m s⁻¹?".

What a function is

Mappings between sets

A relationship between two sets of numbers can be shown by a set of arrows from each input to the output it is paired with. Each input sits in one set, each output in another, and an arrow joins them.

Not every mapping is a function. For a function, each input must have exactly one output, so exactly one arrow leaves each input. Several inputs may share an output, but no input may have two.

The three diagrams show the possible patterns. The first follows the rule $x \mapsto 2x + 1$ and pairs each input with its own output. The second, "square the number", sends both $-2$ and $2$ to 4: still a function, because each input has one output. The third, "the numbers whose square is the input", sends 4 to both $-2$ and $2$, so an input has two outputs and the mapping is not a function.

The function $x \mapsto 2x + 1$ is one-to-one: two different inputs can never give the same output, because doubling and adding 1 keeps different numbers apart. A one-to-one function can be run backwards without ambiguity, which matters for inverse functions.

Four ways to represent a function

The same function can be given as a rule in words, a formula, a table of values, a mapping diagram or a graph. A shop that charges $4.50 per T-shirt plus a $15 set-up fee could describe its prices in any of these ways: the formula $C = 4.5n + 15$, a price list, or a graph of cost against the number of shirts.

The vertical line test

On a graph, each input $x$ is a vertical line, and the outputs paired with it are the points where the graph meets that line. A graph represents a function exactly when no input has two outputs, which gives a quick visual test.

The sideways parabola fails the test because the input $x = 1.2$ is paired with two outputs, $y = \sqrt{3.2}$ and $y = -\sqrt{3.2}$. A circle fails for the same reason. A vertical straight line $x = k$ fails completely: one input is paired with every value of $y$, which is why vertical lines have no equation of the form $y = mx + c$ (subtopic 2.1).

Function notation

A function is usually named by a letter, and its output for an input $x$ is written $f(x)$, read "$f$ of $x$". The notation puts the input in brackets after the name of the function. It is not a multiplication: $f(3)$ means "the output of $f$ when the input is 3".

The letters can be chosen to fit the context, which makes a model easier to read. A velocity that depends on time is written $v(t)$; the cost of $n$ items is $C(n)$; a height after $t$ seconds is $h(t)$. The letter in the brackets names the input variable, and every domain or answer about the input is then written with that letter.

The output depends on the input, so it is the other variable of the pair.

Function notation separates two different questions. "Find $f(5)$" gives the input and asks for the output: substitute. "Solve $f(x) = 5$" gives the output and asks for the input: solve an equation. Mixing up the two is behind many errors with inverse functions.

Domain and range

The domain

Every function has a set of inputs it accepts. Some inputs make a formula meaningless, and some make no sense in the situation being described.

When no domain is stated, a function's domain is taken to be the largest possible domain: every real number for which the formula gives a real output. For most functions met in practice this is all real numbers, $x \in \mathbb{R}$. Two operations restrict it:

• division by zero: $\frac{5}{x - 3}$ is undefined at $x = 3$, so the domain is $x \in \mathbb{R},\ x \ne 3$;

• square roots of negative numbers: $\sqrt{2 - x}$ needs $2 - x \ge 0$, so the domain is $x \le 2$

The logarithms of subtopic 1.5 add a third restriction, since $\log x$ and $\ln x$ need $x > 0$. Functions that are not standard models, such as $f(x) = x + \frac{12}{x^2}$, follow the same rules: find where the formula breaks down, and exclude those values. Here $x = 0$ is excluded, so the domain is $x \in \mathbb{R},\ x \ne 0$.

The range

Once the inputs are fixed, the outputs they produce form the range.

The range is easiest to see on a graph: it is the set of heights the graph reaches, read on the vertical axis. For $f(x) = \sqrt{2 - x}$, a square root is never negative, and as $x$ decreases from 2 the output grows without limit, so the range is $f(x) \ge 0$. The graph of $f$, which ends at $(2, 0)$ and climbs steadily to the left, shows both facts.

The range depends on the domain. Over all real numbers, $g(x) = x^2$ has range $g(x) \ge 0$; restricted to $1 \le x \le 3$, its range is only $1 \le g(x) \le 9$.

Finding a range from a graph

For a function on an interval, the smallest and largest outputs are not always at the ends of the interval. A curve can turn inside the domain, so its lowest or highest point may be a local minimum or local maximum in between. The safe method is to graph the function over its domain and look for every point where the graph is highest or lowest.

For $f(x) = 0.5x^2 - 4x + 10$ with $0 \le x \le 10$, the end values are $f(0) = 10$ and $f(10) = 20$. Taking only these would suggest a range of $10 \le f(x) \le 20$, which is wrong: the graph dips to a minimum at $(4, 2)$. The range is $2 \le f(x) \le 20$.

Finding the range with the GDC

1. Enter the function with its domain. On a TI-84 Plus CE one way is to divide by the domain condition, $Y_1 = (0.5X^2 - 4X + 10)/(X \ge 0 \text{ and } X \le 10)$; on a Casio fx-CG50, append the interval after a comma, $Y_1 = 0.5X^2 - 4X + 10, [0, 10]$.

2. Choose a viewing window slightly wider than the domain, and use the table of values to choose the $y$ range so that the whole curve is visible.

3. Use the minimum and maximum tools for any turning points inside the domain, and evaluate the function at both endpoints.

4. The range runs from the smallest of these values to the largest.

On paper, write the endpoint values and any turning point found, then the range as an inequality: "$f(0) = 10$, $f(10) = 20$, minimum $f(4) = 2$, so $2 \le f(x) \le 20$".

The graph of $g$ shows the gap in the range: no horizontal line between $y = 2$ and $y = 5$ meets either piece. Such a gap is hard to spot from the formula alone, so a GDC graph is the usual first step in finding a range.

Writing domains and ranges

A domain is written in the input variable and a range in the output, either with inequalities or in interval notation. The IB uses square brackets that point inwards for included endpoints and outwards for excluded ones.

Infinity is never included, because it is not a number: its bracket always points outwards. Both styles are correct, and inequalities are the more common choice. The choice between $\le$ and $<$ matters. A value that the function actually reaches is included ($\le$); a value it approaches but never reaches is excluded ($<$).

The Khan Academy video finds the domain and range of functions given by formulas. It starts from the inputs that make each formula undefined, then reasons about which outputs can occur.

Functions as mathematical models

A function used to describe a real situation is a model. The independent variable is the quantity that is set or that passes (time, number of items, temperature), and the dependent variable is the quantity that responds. Once a situation is written as a function, its questions become questions about the function: a value, a maximum, a solution of an equation.

A model is a simplification. It keeps the features that matter and ignores others, so its predictions are approximate and only valid over a limited set of inputs. Choosing, fitting and testing models is the subject of subtopics 2.5 and 2.6, which develop the full modelling cycle.

Domains that make sense in context

The domain of a model comes from the situation, not only from the formula. The formula $K = C + 273.15$, which converts a Celsius temperature into kelvin, accepts any real number. Physically, though, no temperature is colder than absolute zero, $-273.15$ °C, which is defined as 0 K. The model's domain is therefore $C \ge -273.15$, and its range is $K \ge 0$.

Source: Wikipedia, "Absolute zero", accessed 2025.

The appropriate domain of a model (also called a reasonable domain) is the set of inputs for which the model describes the real situation. Typical restrictions are:

• time from the start of the situation ($t \ge 0$) to its end (landing, emptying, the end of a ride);

• whole numbers for things that are counted: people, items, trips ($n \in \mathbb{Z}$, often $n \ge 1$);

• non-negative lengths, masses and amounts;

• limits set in the question (a maximum order, a speed limit, the capacity of a tank).

Discrete domains and cost functions

Some inputs can only be whole numbers. A function whose domain is a set of integers has a graph made of separate points, not a continuous curve, and its answers must be whole numbers rounded in the direction that makes sense.

For the T-shirt printer, the cost of $n$ shirts is $C(n) = 4.5n + 15$, with domain $n \in \mathbb{Z},\ n \ge 1$. The fixed cost of $15 is paid once; each extra shirt adds $4.50, the gradient of the points. The graph is a row of points on a straight line, because 2.5 shirts cannot be ordered.

Inverse functions

Undoing a function

An inverse function reverses the effect of a function: it takes each output back to the input it came from. The conversion from Celsius to Fahrenheit, built in subtopic 2.1, is $F(C) = 1.8C + 32$: multiply by 1.8, then add 32. To convert back, the steps are undone in the reverse order: subtract 32, then divide by 1.8.

Each function undoes the other. A temperature of 20 °C converts to $1.8 \times 20 + 32 = 68$ °F, and $\frac{68 - 32}{1.8} = 20$ returns the original 20 °C.

The notation $f^{-1}(x)$ is read "$f$ inverse of $x$". The $-1$ is part of the name and does not mean a power, so $f^{-1}(x)$ and $\frac{1}{f(x)}$ are different things.

Finding a value of an inverse

Since $f^{-1}$ sends each output back to its input, finding $f^{-1}(10)$ means finding the input that gives output 10. That is the same as solving the equation $f(x) = 10$.

No formula for $f^{-1}$ is needed. The equation can be solved by hand when it is simple, or with the GDC: graph $y = f(x)$ and $y = 10$ and find the point of intersection, or use the equation solver. The answer is the $x$-coordinate.

The graph of an inverse: reflection in y = x

If $f$ sends $a$ to $b$, the point $(a, b)$ is on the graph of $f$. The inverse sends $b$ back to $a$, so $(b, a)$ is on the graph of $f^{-1}$. Swapping the coordinates of every point reflects the graph in the line $y = x$: the points $(a, b)$ and $(b, a)$ are mirror images in that line.

The point $(20, 68)$ on the conversion line becomes $(68, 20)$ on its inverse. The two lines meet on $y = x$ at $(-40, -40)$: $-40$ °C is the same temperature as $-40$ °F, the only reading that is identical on both scales. A point on the line of reflection is unchanged by the reflection.

Source: Wikipedia, "Fahrenheit", accessed 2025.

Where the graphs of a function and its inverse meet on the line $y = x$, the coordinates satisfy $f(x) = x$. Such points lie on both graphs and are invariant under the reflection. For a function that decreases, the two graphs can also cross at pairs of points away from $y = x$.

Sketching the graph of an inverse

To sketch $y = f^{-1}(x)$ from the graph of $y = f(x)$:

1. Draw the line $y = x$ (with equal scales on both axes, so that the reflection is not distorted).

2. Reflect the key points of $f$ by swapping their coordinates: endpoints, intercepts and any labelled points. An intercept $(0, c)$ becomes $(c, 0)$.

3. Mark any invariant points where the graph of $f$ crosses $y = x$; the inverse passes through them too.

4. Join the reflected points with a curve of the mirror-image shape, over the domain of $f^{-1}$ only.

The graph of $f(x) = 0.1x^3 + 1$, $0 \le x \le 4$, crosses $y = x$ twice, at about $x = 1.15$ and $x = 2.42$ (the solutions of $0.1x^3 + 1 = x$). Its inverse passes through the same two points, and every other point is swapped: $(0, 1)$ becomes $(1, 0)$, and $(3, 3.7)$ becomes $(3.7, 3)$.

When an inverse exists

A function can be reversed only if every output came from exactly one input. If two inputs share an output, the inverse would have to send that output back to two places at once, and then it would not be a function. Only one-to-one functions have inverses. The graph gives a test for this, using horizontal lines in the way the vertical line test uses vertical ones.

The function $f(x) = x^2$, $x \in \mathbb{R}$, fails the test: the line $y = 9$ meets the graph at $x = 3$ and $x = -3$, so "$f^{-1}(9)$" would have two values. Its reflection in $y = x$ is the sideways parabola that failed the vertical line test, which is the same failure seen from the other side. Every non-horizontal straight line, every exponential function and every function that only rises (or only falls) over its domain passes the test.

Domain and range of an inverse

Because the inverse swaps inputs and outputs, it swaps the domain and the range.

For the stopping-distance model $d(v) = 0.2v + 0.006v^2$, $0 \le v \le 130$, the function only increases, so its range runs from $d(0) = 0$ to $d(130) = 127.4$: $0 \le d \le 127.4$. The inverse $d^{-1}$ accepts stopping distances from 0 to 127.4 m (its domain) and returns speeds from 0 to 130 km/h (its range). Asking for $d^{-1}(150)$ makes no sense: no speed in the model's domain gives a stopping distance of 150 m.

The Khan Academy video introduces inverses as functions that map each output back to its input. If two inputs shared an output, that output would have to map back to both, which is the horizontal line test in words.

Exam focus

Functions, domains and ranges are examined on Paper 1 as short questions of 4–8 marks and as the first parts of Paper 2 modelling questions. The command terms are write down (a value or a feature read off directly), state (a domain or range without working), find (with working: substitution, a GDC graph or an equation), interpret (say what a value means in context) and explain. The recurring shapes are: the range of a function on a given domain; a value of an inverse such as $f^{-1}(70)$, found by solving $f(x) = 70$, followed by its interpretation in context; the domain or range of an inverse; and an appropriate domain for a model such as a projectile from launch to landing, or a ride of a given length. Questions can use functions that are not among the standard models (a sum of a power and a fraction, for example); the domain is then the largest possible one unless the question states it, and a GDC graph then gives the range and other features.

A range or domain usually earns two marks: one for the correct endpoints and one for the correct inequality signs or interval notation, in the right variable. Values of an inverse earn a method mark for the equation $f(x) = \text{value}$ and an accuracy mark for the answer.

The model is the standard linear approximation for the speed of sound in dry air near 0 °C (source: Wikipedia, "Speed of sound", accessed 2025).

Part (e) catches many candidates, who copy their range of $v$ from part (b) instead of using the domain of $v$. Part (d) needs both quantities named with units: a bare "14.4" earns nothing.

Finding a value of an inverse is quicker with the GDC than with algebra. Writing $f(x) = 10$ on the page, then the solution from the graph or solver, earns both marks; attempting a general formula for $f^{-1}$ wastes time and, for many functions, is not possible by hand.

Practice questions

1. The function $f$ is defined by $f(x) = 3x^2 - 5$. Write down the values of $f(2)$ and $f(-1)$. [2]

2. Write down the largest possible domain of $g(x) = \frac{6}{x + 4}$. [1]

3. Find the domain and range of $f(x) = \sqrt{x - 3} + 2$. [3]

4. State, with a reason, whether each of these represents $y$ as a function of $x$: (a) $\{(2, 3), (4, 3), (6, 5)\}$; (b) $\{(1, 2), (1, 5), (3, 4)\}$

5. A taxi fare, in dollars, for a journey of $d$ km is modelled by $C(d) = 3.5 + 1.8d$. (a) Find $C(12)$. (b) Find $C^{-1}(30)$

6. A ball's height in metres $t$ seconds after it is thrown is $h(t) = -5t^2 + 12t + 2$. (a) Find when it hits the ground. (b) State an appropriate domain for $t$

7. The graph of a one-to-one function $f$ passes through $(1, 4)$ and $(3, 7)$. Write down two points on the graph of $f^{-1}$, and the value of $f^{-1}(7)$

8. The function $h$ is defined by $h(x) = \frac{120}{x + 1} + 4$ for $0 \le x \le 9$

9. Explain why the function $f(x) = x^2 + 1$, $x \in \mathbb{R}$, does not have an inverse function. [2]

10. A water tank is emptied through a valve. The volume of water in the tank, in litres, $t$ minutes after the valve is opened is $V(t) = 2000 - 25t$. (a) State an appropriate domain for $t$ and the range of $V$. (b) Find $V^{-1}(1200)$

Summary

• A function assigns to each input exactly one output; a graph is a function of $x$ if it passes the vertical line test.

• Function notation $f(x)$ names the output for input $x$; "find $f(5)$" means substitute, "solve $f(x) = 5$" means find the input.

• The domain is the set of allowed inputs (largest possible unless stated, usually $\mathbb{R}$, excluding division by zero and square roots of negatives); the range is the set of outputs.

• Find a range from the graph over the domain: check both endpoints and every turning point, using the GDC's minimum and maximum tools.

• Write domains in the input variable and ranges in $f(x)$ or $y$, with inequalities or interval notation ($[a, b]$, $]a, b[$); $\le$

• In context, the domain comes from the situation: $t \ge 0$ up to an end event, whole numbers for counted items, physical limits such as absolute zero.

• An inverse function $f^{-1}$ undoes $f$: if $f(a) = b$ then $f^{-1}(b) = a$

• The graph of $f^{-1}$ is the reflection of the graph of $f$ in $y = x$; points on $y = x$

• Only one-to-one functions (passing the horizontal line test) have inverses; the domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

Self-check

• To do:

  I can decide whether a mapping, a set of pairs or a graph represents a function, and justify my answer.

• To do:

  I can tell one-to-one and many-to-one functions apart from a mapping diagram or a graph.

• To do:

  I can evaluate $f(a)$ for a number or an expression, keeping negative inputs in brackets.

• To do:

  I can identify the independent and dependent variables of a model.

• To do:

  I can find the largest possible domain of a function with a fraction or a square root.

• To do:

  I can find the range of a function on a given domain using a GDC graph, checking turning points as well as endpoints.

• To do:

  I can write a domain or range with the correct variable, the correct order and the correct strict or weak signs.

• To do:

  I can convert between inequalities and interval notation, including ranges in two pieces.

• To do:

  I can state an appropriate domain for a model in context and reject roots that make no sense.

• To do:

  I can round answers on a discrete domain in the direction the context requires.

• To do:

  I can find $f^{-1}(a)$ by solving $f(x) = a$, by hand or with my GDC, and interpret the answer with units.

• To do:

  I can sketch the graph of an inverse by reflecting key points in $y = x$ and marking invariant points.

• To do:

  I can explain, using the horizontal line test, why a function does or does not have an inverse.

• To do:

  I can write down the domain and range of an inverse function from those of the function.

1. (a)

   Write down the largest value of $k$ for which $g$ has an inverse.

   [1]1 mark

   Markscheme (a)

   $k = 3$ A1

   [1 mark]

2. (b)

   Using this value of $k$, find $g^{-1}(x)$, stating its domain.

   [4]4 marks

   Markscheme (b)

   $g(x) = (x - 3)^2 + 2$ (A1)

   $x = (y - 3)^2 + 2 \Rightarrow y = 3 \pm \sqrt{x - 2}$

3. (c)

   Find $g^{-1}(18)$.

   [1]1 mark

   Markscheme (c)

   $g^{-1}(18) = 3 - \sqrt{16} = -1$

(b) The input is the whole expression $a + 1$.

(c) Set the output equal to 13 and solve.

So $x = 3$ or $x = -1.5$. With a GDC, the polynomial root finder (coefficients 2, −3, −9) or the intersection of $y = 2x^2 - 3x + 4$ with $y = 13$ gives the same two values. Part (a) already showed $f(3) = 13$: two inputs give the output 13, so $f$ is many-to-one.

On the left piece, $g(-6) = 3 - \frac{8}{-4} = 5$, and the graph rises without limit as $x$ approaches $-2$ from the left.

On the right piece, the graph comes up from very large negative values just to the right of $x = -2$ and reaches $g(6) = 3 - \frac{8}{8} = 2$.

The range is $g(x) \le 2$ or $g(x) \ge 5$. No output lies strictly between 2 and 5.

(b) The denominator is zero at $x = 3$: domain $x \in \mathbb{R},\ x \ne 3$. A fraction with numerator 5 can be any non-zero number but never 0: range $g(x) \ne 0$ (that is, $g(x) \in \mathbb{R},\ g(x) \ne 0$).

(c) There is no restriction: domain $x \in \mathbb{R}$. Since $x^2 \ge 0$, the largest output is $7 - 0 = 7$: range $h(x) \le 7$.

A quick graph of each on the GDC confirms the ranges.

The GDC's polynomial root finder (coefficients −4.9, 18, 1.5) gives $t = 3.7549\ldots$ and $t = -0.0815\ldots$. Time after the throw cannot be negative, so the negative root is rejected. The ball lands after 3.75 s.

(c) The model applies from the throw to the landing: $0 \le t \le 3.75$, or $t \in [0, 3.75]$.

(d) The GDC's maximum tool on the graph of $y = h(t)$ gives the highest point at $t = 1.84$, $h = 18.03\ldots$. The lowest height is 0, at landing. The range is $0 \le h(t) \le 18.0$.

The number of shirts must be a whole number no greater than 52.2…, so the club can buy at most 52 shirts. Rounding up to 53 would cost $4.5 \times 53 + 15 = 253.5$, that is $253.50, over the budget.

With the GDC, the intersection of $y = 0.2x + 0.006x^2$ and $y = 40$ for $0 \le x \le 130$ is at $x = 66.666\ldots$ (the root finder also gives $-100$, outside the domain). So $d^{-1}(40) = 66.7$.

(c) A car that needs 40 m to stop was travelling at about 66.7 km/h.

The range is $7 \le k(x) \le 22$.

(b) Solve $k(x) = 8$.

So $k^{-1}(8) = 10$. Check: $k(10) = 6 + 2 = 8$. ✓

(c) The domain of $k^{-1}$ is the range of $k$: $7 \le x \le 22$. The range of $k^{-1}$ is the domain of $k$: $3 \le k^{-1}(x) \le 12$.

Note: Award (M1) for substituting either endpoint of the domain.

$307 \le v(T) \le 356$ ($307.06 \le v(T) \le 355.54$) A1A1

Note: Award A1 for both correct endpoints, A1 for a correct inequality with weak signs in the correct order. Accept $[307, 356]$.

[3 marks]

(c) $331.3 + 0.606T = 340$ (M1)

$T = 14.4$ (14.3564…) A1

[2 marks]

(d) When the speed of sound is 340 m s⁻¹, the air temperature is 14.4 °C. A1

Note: The answer must refer to both the speed and the temperature, with units.

[1 mark]

(e) $-40 \le v^{-1}(x) \le 40$ A1

Note: Accept $-40 \le T \le 40$ or $[-40, 40]$. Do not award for the answer to part (b).

[1 mark]

(f) $-300$ °C is below absolute zero ($-273.15$ °C), so no such temperature exists (it is also outside the domain of the model). R1

[1 mark]

Total [10 marks]

$g^{-1}(x) = 3 - \sqrt{x - 2}$ A1

$x \ge 2$ A1

Note: Award A0 for the third mark if $3 + \sqrt{x - 2}$ or $\pm$ is given.

[4 marks]

Note: Follow through from part (b) only if their answer lies in the domain $x \le 3$.

[1 mark]

[1]1 mark

[2]2 marks