2.2 Functions, domain, range and inverse

The graph of a function is one-to-one, and so the function has an inverse, when every horizontal line meets it at most once.

A function is a rule that turns each input into exactly one output. The idea is at the centre of the course: sequences, graphs, models, trigonometry and calculus are all built on functions. A function can be given as a formula, a graph, a table of values or a description in words, and moving between these representations is one of the main skills in mathematics.

Every function comes with a set of allowed inputs and a set of outputs it produces. Knowing these sets matters as much as knowing the formula, especially when a function models a real situation, where only some inputs make sense.

What a function is

A relation pairs inputs with outputs. A function is a relation with one extra condition: every input gives exactly one output. Two different inputs may share an output, but one input may never produce two different outputs.

A mapping diagram shows a relation as arrows from inputs to outputs. In a function, exactly one arrow leaves each input.

The first mapping is $x \mapsto 2x + 1$ and the second is $x \mapsto x^2$; both are functions. The third sends each number to both of its square roots, so the input $4$ has two outputs, $-2$ and $2$, and the relation is not a function. The symbol $\sqrt{x}$ avoids this problem by meaning only the non-negative square root, so $\sqrt{4} = 2$.

Function notation

A function is usually named with a letter, and $f(x)$, read "$f$ of $x$", is the output when the input is $x$. The equation $f(x) = 3x - 1$ and the mapping notation $f: x \mapsto 3x - 1$ describe the same function. Here $x$ is the independent variable (the input chosen freely) and $f(x)$, or $y$ when the function is graphed, is the dependent variable.

Other letters are used when they suit the context: $v(t)$ for a velocity at time $t$, $C(n)$ for the cost of $n$ items, $A(r)$ for the area of a circle of radius $r$. The letter in brackets is a placeholder, so $f(x) = x^2$ and $f(t) = t^2$ define the same function.

To evaluate a function, replace every occurrence of the variable by the input, in brackets.

Graphs of functions and the vertical line test

The graph of a function $f$ is the set of all points $(x, f(x))$, so its equation is $y = f(x)$. Since each $x$ in the domain has exactly one output, each vertical line meets the graph at most once. This gives a quick visual test.

The circle $x^2 + y^2 = 9$ fails the test. The vertical line $x = 1$ meets it at two points, $(1, \sqrt{8})$ and $(1, -\sqrt{8})$, so the input $1$ would have two outputs. Each half of the circle on its own passes the test: the upper semicircle is the graph of the function $y = \sqrt{9 - x^2}$, and the lower one is the graph of $y = -\sqrt{9 - x^2}$.

Domain and range

The domain of a function is its set of inputs and the range is the set of outputs it actually produces. On a graph, the domain is the spread of the graph along the $x$-axis and the range is its spread along the $y$-axis.

A function is not fully defined until its domain is known. The same formula on different domains gives different functions, with different ranges and graphs. When a domain is stated, it is written after the formula: $f(x) = x^2$, $x \in \mathbb{R}$, $-2 \le x \le 3$.

Interval and set notation

Domains and ranges are sets of real numbers, usually intervals. They can be written as inequalities, in interval notation or in set notation.

The symbol $\cup$ (union) joins two intervals. The short form "$x \in \mathbb{R}$, $x \neq 2$" is the most common way to write the last domain. A domain can also be a set of separate values, such as $n \in \{1, 2, 3, 4\}$ or $n \in \mathbb{N}$, when only whole numbers make sense.

Finding the range

The range depends on the domain and on the shape of the graph. The safest method is to sketch the graph over the given domain and read off the lowest and highest values of $y$. Checking only the values at the endpoints of the domain is not enough when the graph turns in between.

The natural domain

When no domain is given, the domain is taken to be the largest set of real numbers for which the formula gives a real output. This is the natural domain. Three operations restrict it.

When a formula contains more than one of these, every condition must hold at the same time.

Knowing the natural domain also helps with the range. Since $\sqrt{x - 3}$ is never negative, the range of $f(x) = \sqrt{x - 3}$ is $f(x) \ge 0$. Since $\frac{1}{x + 2}$ can never equal $0$, the range of $g(x) = \frac{1}{x + 2}$ is $g(x) \neq 0$ (this is explored in 2.8). The range of $\ln x$ is all real numbers (2.9).

Functions as mathematical models

A function used to describe a real situation is a mathematical model. The variables stand for real quantities with units, and the formula captures how one depends on the other. The key concept of modelling runs through the whole course, and it starts here: a model is useful only within the set of inputs where it makes sense.

Contexts restrict a domain in several ways. A time, a length or a number of items cannot be negative. A quantity may run out, such as fuel in a tank. A number of people or items must be a whole number, which makes the domain a set of separate values and the graph a set of points rather than a continuous curve.

The same ideas apply with other notation. A car that brakes from 24 m s⁻¹ with constant deceleration has velocity $v(t) = 24 - 6t$, where $t$ is the time in seconds; the model applies only for $0 \le t \le 4$, after which the car has stopped and stays at rest. A cinema that charges €12 per ticket and a €5 booking fee per order has cost $C(n) = 12n + 5$ for $n$ tickets, with $n \in \{1, 2, 3, \ldots\}$, so its graph is a set of separate points.

One-to-one and many-to-one functions

Functions fall into two types according to whether outputs can repeat. The type decides whether the function can be reversed.

The function $x \mapsto 2x + 1$ is one-to-one. The function $x \mapsto x^2$ on $\mathbb{R}$ is many-to-one, since $(-3)^2 = 3^2 = 9$. On a graph, a repeated output appears as a horizontal line meeting the graph more than once.

The horizontal line test and the vertical line test answer different questions. The vertical line test decides whether a graph is a function at all; the horizontal line test decides whether that function is one-to-one.

Inverse functions

An inverse function reverses, or undoes, a function. If $f$ takes the input $a$ to the output $b$, the inverse takes $b$ back to $a$. For example, $f(x) = 2x + 3$ means "multiply by 2, then add 3". To undo it, reverse the steps in the opposite order: "subtract 3, then divide by 2". So

and, for instance, $f(4) = 11$ while $f^{-1}(11) = 4$. A systematic algebraic method for finding $f^{-1}(x)$ is developed in 2.5.

Why only one-to-one functions have inverses

If $f$ is many-to-one, two inputs share an output. For $f(x) = x^2$ on $\mathbb{R}$, both $-3$ and $3$ give $9$, so reversing $f$ would send $9$ back to two values. That reversed relation is not a function. A many-to-one function can be given an inverse by restricting its domain until it is one-to-one: on the domain $x \ge 0$, $f(x) = x^2$ is one-to-one and its inverse is $f^{-1}(x) = \sqrt{x}$.

Values of an inverse without its formula

A value of $f^{-1}$ can be found without a formula for $f^{-1}$. Since $f^{-1}(b) = a$ means $f(a) = b$, finding $f^{-1}(b)$ means solving the equation $f(x) = b$.

A table of values of $f$ is also a table of values of $f^{-1}$, read from the bottom row to the top row.

The graph of an inverse function

If the point $(a, b)$ lies on the graph of $y = f(x)$, then $f(a) = b$, so $f^{-1}(b) = a$ and the point $(b, a)$ lies on the graph of $y = f^{-1}(x)$. Swapping the coordinates of every point is the same as reflecting in the line $y = x$. So the graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$.

The graph shows how the domain and range swap. The function $f(x) = x^2 + 1$, $x \ge 0$, has range $f(x) \ge 1$; its inverse $f^{-1}(x) = \sqrt{x - 1}$ has domain $x \ge 1$ and range $f^{-1}(x) \ge 0$.

Exam-style questions on functions and inverses

Questions often combine the domain, the range and an inverse with an equation that has more than one solution, where the domain decides which solution is valid.

Self-check

• To do:

  I can decide whether a relation is a function from a mapping, a list of pairs or a graph.

• To do:

  I can evaluate $f(x)$ for numbers and expressions, and solve $f(x) = k$.

• To do:

  I can write domains and ranges as inequalities, in interval notation and in set notation.

• To do:

  I can find the natural domain of a function with square roots, fractions or logarithms.

• To do:

  I can find the range of a function on a given domain from a sketch of its graph.

• To do:

  I can choose a sensible domain for a function that models a real situation.

• To do:

  I can tell whether a function is one-to-one or many-to-one and explain why only one-to-one functions have inverses.

• To do:

  I can find values of $f^{-1}$, state its domain and range, and sketch its graph by reflection in $y = x$.

1. (a)

   Show that $\cos\mathrm{C\hat{A}B} = \dfrac{5x - 80}{3x}$.

   [4]4 marks

   Markscheme (a)

   $\mathrm{BC} = 50 - x$ (A1)

   $\cos\mathrm{C\hat{A}B} = \frac{30^2 + x^2 - (50 - x)^2}{2(30)(x)}$

2. (b)

   Find the value of $x$ for which $\mathrm{C\hat{A}B} = 90°$.

   [2]2 marks

   Markscheme (b)

   $5x - 80 = 0$

3. (c)

   Find the set of possible values of $x$.

   [2]2 marks

   Markscheme (c)

   use of the triangle inequality, e.g. $x + 30 > 50 - x$ and $(50 - x) + 30 > x$

4. (d)

   Show that the area of the enclosure, $E$ m$^2$, is given by $E = 20\sqrt{-x^2 + 50x - 400}$

5. (e)

   Find the maximum area of the enclosure and the value of $x$ for which it occurs.

   [3]3 marks

   Markscheme (e)

   attempt to find the maximum of $E$ (graph, or vertex of $-x^2 + 50x - 400$) (M1)

6. (f)

   Find the values of $x$ for which the area of the enclosure is $200$ m$^2$.

   [2]2 marks

   Markscheme (f)

   $20\sqrt{-x^2 + 50x - 400} = 200$

(c) A function. Each student has exactly one date of birth, even though several students may share one.

(d) Not a function, in general. Two students born on the same day would make one input give two outputs.

(b) Replace $x$ by $(a + 1)$:

(c) Here the output is given and the input is unknown:

so $x = 4$ or $x = -1$. Both inputs give the output $4$, so $f$ is many-to-one.

(c) A function. Every input gives the output $7$; the graph is a horizontal line, which meets each vertical line once. A function whose output never changes is called a constant function.

(d) Not a function. The single input $7$ is paired with every value of $y$; the graph is a vertical line.

The range is $-3 \le f(x) < 5$, or $[-3, 5)$. The endpoint $5$ is excluded because $x = 3$ is not in the domain.

(b) Since $x^2 \ge 0$ for every real $x$, $x^2 + 1 \ge 1$, with equality at $x = 0$. Every value above $1$ is also reached. The range is $g(x) \ge 1$, or $[1, \infty)$.

(c) Sketch the graph on the domain. The endpoints give $h(-2) = 4$ and $h(3) = 9$, but the graph falls to $0$ at $x = 0$ in between. The range is $0 \le h(x) \le 9$, or $[0, 9]$.

(c) The square root needs $2x + 6 \ge 0$, but it is also in the denominator, so it must not be zero. Together, $2x + 6 > 0$, which gives $x > -3$. The domain is $(-3, \infty)$.

(d) The logarithm needs $x^2 > 0$, which is true for every real $x$ except $0$. The domain is $x \in \mathbb{R}$, $x \neq 0$.

So $f^{-1}(11) = 2$.

(b) Solve $x^3 + 3 = -5$:

So $f^{-1}(-5) = -2$.

(c) $f(1) = 5$, and $f^{-1}(5) = 1$. So $f^{-1}\left(f(1)\right) = 1$.

Parts (b) and (c) show the defining property: applying a function and then its inverse, in either order, returns the starting value.

(b) The domain of $f^{-1}$ is the range of $f$: $x \ge 1$. The range of $f^{-1}$ is the domain of $f$: $f^{-1}(x) \ge 3$.

(c) Plot a few points of $f$: $(3, 1)$, $(4, 2)$, $(7, 3)$. Swap the coordinates to get points of $f^{-1}$: $(1, 3)$, $(2, 4)$, $(3, 7)$. Reflect the curve in $y = x$.

(c) Solve $f(x) = 4$:

So $f^{-1}(4) = 10$.

(d) Domain of $f^{-1}$: $x \ge 0$ (the range of $f$). Range of $f^{-1}$: $f^{-1}(x) \ge 2$ (the domain of $f$).

(e) The graph of $f$ starts at $(2, 0)$ on the $x$-axis and passes through $(4, 2)$ and $(10, 4)$. Reflecting in $y = x$, the graph of $f^{-1}$ starts at $(0, 2)$ on the $y$-axis and passes through $(2, 4)$ and $(4, 10)$.

A sketch earns its marks for the correct shape, the labelled endpoints on the axes, the restricted domains (no curve to the left of $x = 2$ for $f$, none to the left of $x = 0$ for $f^{-1}$) and the visible symmetry in $y = x$.

(c) On the domain $x \ge 3$, $f$ is increasing, so every horizontal line meets the graph at most once. Hence $f$ is one-to-one and has an inverse.

(d) Solve $f(x) = 5$:

so $x = 1$ or $x = 5$. Since the domain of $f$ is $x \ge 3$, reject $x = 1$. Hence $f^{-1}(5) = 5$.

(e) Part (d) shows that $g(1) = 5$ and $g(5) = 5$. Two different inputs give the same output, so $g$ is many-to-one and has no inverse.

$= \frac{900 + x^2 - 2500 + 100x - x^2}{60x}$ A1

$= \frac{100x - 1600}{60x}$ A1

$= \frac{5x - 80}{3x}$ AG

$x = 16$ A1

$10 < x < 40$ A1

Note: Award (M1)A0 for non-strict inequalities.

[4]4 marks

maximum area $= 300$ m$^2$ A1

$x = 13.8$ or $x = 36.2$ $(13.8196\ldots, 36.1803\ldots)$ A1

Note: Accept exact values $25 \pm 5\sqrt{5}$.

[3]3 marks