Inverse Functions

Undoing a function machine, swap and solve, x on the top and bottom, and squared functions with a domain.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel IGCSE Maths 4MA1 · 76 real past-paper questions below

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What an Inverse Does

A function is a machine: a number goes in, a new number comes out. The inverse machine runs it backwards, from the output to the input.

Worked example

1
$x \to \times 3 \to + 2 \to 20$ — f is a two-step machine: times 3, then add 2. We want the input that gives 20.
2
$20 \to - 2 \to \div 3 \to 6$ — Go backwards and do the opposite of each step, last one first: undo +2 with −2, then undo ×3 with ÷3.
3
$f^{- 1} (20) = 6$ — Check by running f forwards: 6 × 3 + 2 = 20.

Common mistake

Reading $f^{- 1} (x)$ as $\frac{1}{f ( x )}$ . The −1 means “undo”, not “one over”.

Key idea

The Roman writer Suetonius tells us Julius Caesar hid messages by moving each letter 3 places on: A became D. To read them, you need the inverse: move each letter 3 places back. Try it: FDW means CAT.

Swap and Solve

To find $f^{- 1} (x)$ : write y for f(x), make x the subject, then swap the letters.

Worked example

1
$y = \frac{2 x + 7}{5}$ — f(x) = (2x + 7) ÷ 5
2
$5 y = 2 x + 7$ — multiply both sides by 5
3
$5 y - 7 = 2 x$ — take 7 from both sides
4
$x = \frac{5 y - 7}{2}$ — divide by 2
5
$f^{- 1} (x) = \frac{5 x - 7}{2}$ — swap y back to x

Key idea

Check with a number: f(4) = 3, so $f^{- 1} (3)$ must come out as 4. It does: $(5 \times 3 - 7) \div 2 = 4$ .

x on Top and Bottom

When x is on the top and the bottom of a fraction, gather the x terms on one side and factorise.

Worked example

1
$y = \frac{3 x}{x - 2}$ — Write y for f(x). The letter x is on the top and the bottom.
2
$y (x - 2) = 3 x$ — Multiply both sides by (x − 2) to clear the bottom of the fraction.
3
$y x - 2 y = 3 x$ — Expand the bracket: y times x, and y times −2.
4
$y x - 2 y - 3 x = 0$ — Take 3x from both sides, so the x terms are together.
5
$y x - 3 x = 2 y$ — Add 2y to both sides, so the terms without x are on the right.
6
$x (y - 3) = 2 y$ — Take x out as a common factor, so x is written only once.
7
$x = \frac{2 y}{y - 3}$ — Divide both sides by (y − 3).
8
$f^{- 1} (x) = \frac{2 x}{x - 3}$ — Swap y for x.

Common mistake

Stopping one line early. While x is in two places, it is not the subject. Factorise to get one x.

A Squared Function (Higher)

A square has two roots, so the question gives a domain like $x \geq 3$ . The domain tells you which sign to keep.

Worked example

1
$y = (x - 3)^{2} + 2, x \geq 3$ — the function
2
$y - 2 = (x - 3)^{2}$ — take 2 from both sides
3
$\pm y - 2 = x - 3$ — square root both sides
4
$x = 3 + y - 2$ — x ≥ 3, so keep the plus
5
$f^{- 1} (x) = 3 + x - 2$ — swap y to x

Key idea

The inputs of f are the outputs of the inverse. Here f only gives numbers from 2 up, so the inverse only takes x ≥ 2.

Inverse of a Quadratic (Higher)

Exams give you $x^{2} + b x + c$ with a domain. Write it as a squared bracket first. Then swap and solve, and let the domain choose the sign.

Worked example

1
$y = 2 x^{2} + 8 x + 3, x \geq - 2$ — Write y for f(x). The x terms are in two places, so complete the square.
2
$y = 2 (x^{2} + 4 x) + 3$ — Take out the 2 from the x terms, so that x² has no number in front.
3
$y = 2 ((x + 2)^{2} - 4) + 3$ — Half of 4 is 2. The bracket (x + 2)² has an extra 4, so take 4 away.
4
$y = 2 (x + 2)^{2} - 8 + 3$ — Multiply the −4 by the 2 outside the bracket.
5
$y = 2 (x + 2)^{2} - 5$ — Work out −8 + 3.
6
$y + 5 = 2 (x + 2)^{2}$ — Add 5 to both sides.
7
$\frac{y + 5}{2} = (x + 2)^{2}$ — Divide both sides by 2.
8
$\pm \frac{y + 5}{2} = x + 2$ — Square root both sides. A square has two roots.
9
$x = - 2 + \frac{y + 5}{2}$ — The domain x ≥ −2 means x + 2 is 0 or more, so keep the plus. Then take 2 from both sides.
10
$f^{- 1} (x) = - 2 + \frac{x + 5}{2}$ — Swap y for x.

Key idea

The lowest value of f is $f (- 2) = - 5$ , so the inverse takes $x \geq - 5$ . That is the same as saying the number under the root cannot be negative.

A Function and Its Inverse

A function followed by its own inverse changes nothing. Exams use this to simplify, and to spot a function that is its own inverse.

Formula

f (f^{- 1} (x)) = x f^{- 1} (f (x)) = x

Rule

The answer is just $x$ . It is not 1 and it is not 0. Do the function and then undo it, and you are back where you began.

Worked example

1
$f (x) = 9 - x$ — Test whether doing f twice gives x back. If it does, f undoes itself.
2
$f f (x) = f (9 - x) = 9 - (9 - x)$ — Put the whole of f(x) into f, inside a bracket.
3
$f f (x) = 9 - 9 + x$ — Remove the bracket. The minus changes the sign of every term inside, so −(9 − x) = −9 + x.
4
$f f (x) = x$ — 9 − 9 = 0, so only x is left.
5
$f^{- 1} (x) = f (x)$ — Doing f twice gets x back, so f is its own inverse.

Worked example

1
$h (x) = \frac{1}{x}, x \neq = 0$ — Another test: is h its own inverse?
2
$hh (x) = \frac{1}{1/ x}$ — Put 1/x into h. The bottom of the fraction is now 1/x.
3
$hh (x) = x$ — 1 divided by 1/x flips back to x, so h is its own inverse.

Key idea

Any function $\frac{k}{x}$ is its own inverse. From $y = \frac{k}{x}$ you get $x = \frac{k}{y}$ , which is the same rule.

Which Method? Read the Question

Look at where x appears.

You see	You do
x once	Undo each step, last first
x top and bottom	Gather x terms, factorise
A square and a domain	Square root, keep one sign
“Find $f^{- 1} (3)$ ”	Solve f(x) = 3
“Solve f(x) = f⁻¹(x)”	Solve f(x) = x instead

Key idea

$f^{- 1} (3) = 4$ means $f (4) = 3$ . You do not always need the whole inverse.

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