1The idea in one paragraph

Functions can be joined end to end. Put x into g, then put the result into f, and you have a new function, the composite f ∘ g, written (f ∘ g)(x) = f(g(x)). The function nearest the x acts first, so f ∘ g and g ∘ f are usually different. The identity function is the one that does nothing, x ↦ x. A function and its inverse cancel out to the identity: f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. That is the definition of the inverse, and it gives the method for finding one: undo the steps of f in reverse order, which in practice means writing y = f(x), swapping x and y, and making y the subject again. The inverse only exists when f is one-to-one, its domain is the range of f, and its graph is the reflection of f in the line y = x.

2Composite functions

A composite function is two functions applied one after the other. Figure 1 draws it as two machines in a row, with f(x) = 2x + 1 and g(x) = x².

(f ∘ g)(x) = f(g(x)): g acts first, then f. Read from the inside out, or from right to left.

The notation looks backwards until you read it as a bracket: f(g(x)) is "f of g of x", and you always work out the inside bracket first.

At a number. Work from the inside out, one function at a time.

As an expression. Replace every x in the outer function with the whole inner function, in brackets.

Check the expression against the number: (f ∘ g)(3) = 2(9) + 1 = 19. It agrees.

Order matters. Figure 2 runs the same input through both orders. (f ∘ g)(3) = 19 but (g ∘ f)(3) = 7² = 49. Squaring then doubling is not the same as doubling then squaring. In general f ∘ g ≠ g ∘ f, and a question that asks for one will not accept the other.

A function can also be composed with itself. (f ∘ f)(x) = f(f(x)) = 2(2x + 1) + 1 = 4x + 3.

Solving an equation with a composite. Build the composite first, then solve as usual.

Both answers are valid, because g squares away the sign. Check one: (f ∘ g)(−3) = 2(9) + 1 = 19.

3When the inner output must fit the outer domain

The output of g becomes the input of f, so it has to be something f is allowed to take. With f(x) = √x (domain x ≥ 0) and g(x) = x − 5:

• (f ∘ g)(x) = √(x − 5). The inner result x − 5 must be at least 0, so the composite only works for x ≥ 5. The domain of f ∘ g is smaller than the domain of g.
• (g ∘ f)(x) = √x − 5. Here f acts first, so x ≥ 0 is needed, and after that g accepts anything. Domain x ≥ 0.

So (f ∘ g)(2) does not exist: g(2) = −3, and f cannot take −3. When a question asks you to explain why a composite is undefined at some value, this is the explanation: the inner output falls outside the domain of the outer function.

Taking a composite apart. Often you need to see a function as a composite. h(x) = (3x − 1)⁴ is "take 3x − 1, then raise it to the fourth power", so h = f ∘ g with g(x) = 3x − 1 and f(x) = x⁴. That skill is exactly what the chain rule in calculus needs, so practise naming the inside and the outside now.

In context. Composites describe chains of conversions. Suppose, as an illustration, that one US dollar buys 0.9 euros and one euro buys 160 yen. Then g(x) = 0.9x converts dollars to euros and f(x) = 160x converts euros to yen. The composite (f ∘ g)(x) = 160(0.9x) = 144x converts dollars straight to yen: $50 is 7200 yen. The order is forced by the units: g must act first, because f expects euros.

4The identity function, and what an inverse really is

The identity function leaves every input unchanged: I(x) = x. Composing with it changes nothing: f(I(x)) = f(x) and I(f(x)) = f(x). Its graph is the line y = x, the mirror line from 2.2.

The inverse of f is the function that brings you back to where you started, whichever order you apply the two in.

(f ∘ f⁻¹)(x) = (f⁻¹ ∘ f)(x) = x

This is the whole meaning of an inverse, written as algebra. It also gives you a way to check any inverse you find: compose the two and simplify. If you do not get x, the inverse is wrong.

Take the temperature conversion from 2.2: f(x) = 1.8x + 32 turns Celsius into Fahrenheit, and f⁻¹(x) = (x − 32) ÷ 1.8 turns it back.

5Which functions have an inverse

An inverse has to send each output back to exactly one input. So f has an inverse only if it is one-to-one: no two inputs share an output. On a graph, every horizontal line meets it at most once (the horizontal line test from 2.2).

When a function is not one-to-one, restrict its domain until it is. For f(x) = (x − 3)² + 1 on all real numbers, f(2) = f(4) = 2, so there is no inverse. Its vertex is at x = 3, and on either side of the vertex it is one-to-one. The usual choice is the right-hand half, x ≥ 3, which is also the answer to "find the smallest value of k such that f has an inverse on x ≥ k": k = 3, the x-coordinate of the vertex.

6Finding the inverse by algebra

The inverse undoes the steps of f in reverse order, as Figure 3 shows for f(x) = 4x − 7: f multiplies by 4 and then subtracts 7, so f⁻¹ adds 7 and then divides by 4.

For anything longer than two steps, use the algebraic method. It works for every function on the syllabus.

To find f⁻¹(x): write y = f(x), swap x and y, then rearrange to make y the subject.

Swapping x and y is the algebra of reflecting in y = x: every point (a, b) becomes (b, a). You may also rearrange first and swap last; the answer is the same. What you must not do is stop before the swap, because then your formula has y in terms of x the wrong way round.

Example 1 · a linear function. f(x) = 4x − 7.

Example 2 · a rational function. f(x) = (2x + 3) ÷ (x − 1), x ≠ 1. The move to learn is collecting every y term on one side and factorising y out.

The domain of f⁻¹ is the range of f. The range of f is every real number except 2, its horizontal asymptote, so the domain of f⁻¹ is x ≠ 2, which agrees with the denominator. Figure 4 shows what the reflection does to the asymptotes: f has x = 1 and y = 2, and f⁻¹ has x = 2 and y = 1. The asymptotes swap, as every other feature does.

Check with the identity. Substituting f⁻¹ into f:

Example 3 · a restricted quadratic. f(x) = (x − 3)² + 1, for x ≥ 3. Here the square root brings a choice of sign, and the domain makes it.

The range of f is f(x) ≥ 1 (the vertex is (3, 1)), so the domain of f⁻¹ is x ≥ 1. The domain of f is x ≥ 3, so the range of f⁻¹ is y ≥ 3, and that is why the positive root is the right one: 3 − √(x − 1) would give values below 3. Figure 5 draws both. They meet on the line y = x, at the point where f(x) = x:

For an increasing function like this one, the graphs of f and f⁻¹ can only meet on y = x, so solving f(x) = x is the fast way to find where f and f⁻¹ intersect.

Example 4 · an exponential. If you have met logarithms: f(x) = 2ˣ − 1. Swap to x = 2ʸ − 1, so 2ʸ = x + 1 and y = log₂(x + 1). The range of f is f(x) > −1, so f⁻¹(x) = log₂(x + 1), x > −1. Exponentials and logarithms are inverses of each other, which is the whole story of 2.9.

Some functions are their own inverse. For f(x) = (3x + 1) ÷ (x − 3), swapping and rearranging gives f⁻¹(x) = (3x + 1) ÷ (x − 3) again, so f(f(x)) = x. Its graph is symmetric in y = x. It is a curiosity, but a question that asks you to "find f⁻¹ and comment" wants you to notice it.

7Using inverses without finding the formula

Two shortcuts save time on Paper 1.

f⁻¹ at one value. f⁻¹(k) is the input that gives output k, so solve f(x) = k. For f(x) = (2x + 3) ÷ (x − 1), f⁻¹(7) solves (2x + 3) = 7(x − 1), so 5x = 10 and f⁻¹(7) = 2. Check: f(2) = 7 ÷ 1 = 7. Using the formula from Example 2, (7 + 3) ÷ (7 − 2) = 2, the same.

Composites with inverses. (f⁻¹ ∘ g)(x) means g first, then f⁻¹. Work from the inside out exactly as before: find g(x) at your number, then apply f⁻¹ to the result.

And a composite chain can be undone too. To undo "g then f", you undo f first and then undo g, just as you take off your shoes before your socks. In the currency example, (f ∘ g)(x) = 144x changes dollars to yen, and its inverse x ÷ 144 changes yen back to dollars: 7200 yen is $50.