Composite functions — functions inside functions — Mathematics

Before this

What is a function
Expanding double brackets
Percentage increase and decrease · coming soon

This leads to

Inverse functions
Combined transformations — order matters · coming soon
The chain rule · coming soon

Two discounts, two different prices

A shop is selling a jacket for $80. This week it takes 25% off every price. You also have a coupon worth $10. The shop will apply both. The only question is the order.

The cashier, Mei Ling, takes the 25% off first. A quarter of 80 is 20, so the price falls to $60. Then she takes off the $10 coupon. You pay $50.

At the next till, Kwame uses the coupon first. The price falls to $70. Then he takes 25% off. A quarter of 70 is 17.50, so you pay $52.50.

Same jacket. Same discount. Same coupon. Two different prices. Nobody made an arithmetic mistake. The only thing that changed was the order of the two steps.

Now try a different jacket, at $120. Discount first: $90, then $80. Coupon first: $110, then $82.50. The gap is $2.50 again. It is $2.50 for every price. By the end of this lecture you will be able to prove that, in two lines. You will also know exactly why the order matters, and how to write it so that nobody can confuse the two.

Feeding one machine into another

The last lecture ended with a promise. Because a function gives each input exactly one output, you can feed that output straight into a second function. This lecture keeps that promise.

Each step at the shop is a function. Call the discount $d$ and the coupon $c$ :

d (x) = 0.75 x c (x) = x - 10

Here $x$ is the price going in. The discount keeps three quarters of the price. The coupon subtracts 10.

Mei Ling put the price into $d$ first. Then she put the answer into $c$ . In function notation, the output $d (x)$ became the input of $c$ :

c (d (x))

Read this from the inside out. The innermost brackets hold $x$ . The function closest to $x$ is $d$ , so $d$ acts first. Its output then sits inside the brackets of $c$ , so $c$ acts second.

A function built like this, by feeding the output of one function into another, is called a composite function. The two steps are joined, or composed, into one.

The same two functions,

f (x) = 2 x + 1

and

g (x) = x^{2}

, in two orders. Top: the input 3 goes through

g

first, giving 9, and then through

f

, giving 19. Bottom: 3 goes through

f

first, giving 7, and then through

g

, giving 49.

The notation: fg(x) means "do g first"

Writing two sets of brackets every time is slow. So we shorten $f (g (x))$ to $f g (x)$ . We read it as "f of g of x", or "f g of x".

Composite function

$f g (x) = f (g (x))$ . The function $g$ acts on $x$ first. Then $f$ acts on the output of $g$ .

In the same way, $g f (x) = g (f (x))$ : $f$ first, then $g$ .

This is the hard part of the notation, so spend time here. In $f g (x)$ , the letter $f$ is written first, but it acts second. We read from left to right, but the work happens from right to left. That feels backwards, and it causes more mistakes in this topic than anything else.

The way to make it feel natural is to see where the short form comes from. Write the long form, $f (g (x))$ . Now remove the inner brackets. What is left is $f g (x)$ . So the letter next to $x$ is the function that touches $x$ first. The letter further away waits for its turn. Do not learn "right to left" as a rule. Look at which function is next to the input.

Some books write $(f \circ g) (x)$ , read "f circle g of x". It means exactly the same as $f g (x)$ . We will use $f g (x)$ in these lectures.

Let us work out the two pipelines in the figure, using $f (x) = 2 x + 1$ and $g (x) = x^{2}$ .

f g (3) = f (g (3)) = f (9) = 2 (9) + 1 = 19

g f (3) = g (f (3)) = g (7) = 7^{2} = 49

The two answers are very different, as they were at the shop.

Stop and think

With the same $f$ and $g$ , a student says: " $f g (3)$ means $f (3)$ times $g (3)$ , so it is $7 \times 9 = 63$ ." What has gone wrong?

Show the answer

Writing two function names side by side does not mean multiplying them. In $f g (3)$ , the letters $f$ and $g$ are names of machines, not numbers. Nothing is multiplied. The 3 goes into $g$ , and $g$ 's output goes into $f$ . So $f g (3) = f (9) = 19$ , not 63. This is the same idea as in the last lecture: $f (3)$ is not "f times 3". If a question wants the product, it writes $f (x) \times g (x)$ , with the times sign.

From numbers to a formula

With one number, you can work step by step. But you can also find a formula for the whole pipeline. Then any input can go in at once.

Use the method from the last lecture: put the input into the function in brackets. For $f g (x)$ , the input of $f$ is $g (x)$ , which is $x^{2}$ :

f g (x) = f (x^{2}) = 2 (x^{2}) + 1 = 2 x^{2} + 1

For $g f (x)$ , the input of $g$ is $f (x)$ , which is $2 x + 1$ . The function $g$ squares its input, so the whole of $2 x + 1$ is squared:

g f (x) = g (2 x + 1) = (2 x + 1)^{2} = 4 x^{2} + 4 x + 1

Check both with $x = 3$ . First, $2 (9) + 1 = 19$ . Second, $4 (9) + 12 + 1 = 49$ . Both agree with the pipelines.

The formulas show clearly that $f g$ and $g f$ are two different functions. One is $2 x^{2} + 1$ . The other is $4 x^{2} + 4 x + 1$ . So in general:

f g (x) \neq = g f (x)

Now return to the shop. Discount first is $c d (x) = c (0.75 x) = 0.75 x - 10$ . Coupon first is $d c (x) = d (x - 10) = 0.75 (x - 10) = 0.75 x - 7.5$ . Compare the two formulas. For every price $x$ , the second is 2.5 more than the first. That is the proof promised at the start. It also shows why. When the coupon goes first, the discount then shrinks the coupon too. You only get three quarters of its value, $7.50 instead of $10. So the best order for the customer is discount first, then coupon.

Try many pairs of functions below and look for the rare pairs where the order does not matter.

Explore — does the order matter?

Choose a function for f and a function for g. Then move the input. The top pipeline runs g first; the bottom one runs f first. Look at the outputs and the two formulas. Can you find a pair where the two orders always agree?

f(x) = g(x) =

Input: x = 3

Five pairs of functions, each composed in both orders.

f(x)	g(x)	fg(x)	gf(x)	Same?
2x + 1	x²	2x² + 1	4x² + 4x + 1	No
2x	x + 3	2x + 6	2x + 3	No
x²	x + 3	x² + 6x + 9	x² + 3	No
x + 3	x − 4	x − 1	x − 1	Yes
2x	3x	6x	6x	Yes

Two "add" functions always agree in either order, and so do two "multiply" functions. Mixing kinds of step usually breaks the agreement. With f(x) = x² and g(x) = 2x + 1, the orders give the same output at x = 0 and x = −2, but different outputs everywhere else.

What should you have seen? For most pairs, the two pipelines give different outputs and different formulas. A few pairs always agree: two functions that both add, such as $x + 3$ and $x - 4$ , or two that both multiply, such as $2 x$ and $3 x$ . Adding 3 and then subtracting 4 is the same as subtracting 4 and then adding 3. When the two kinds of step are mixed, the agreement breaks, as it did at the shop. You may also have found an input where two different formulas happen to give the same number, such as $x = 0$ for $x^{2}$ and $2 x + 1$ . One matching number does not make two functions the same. The formulas must agree for every input.

A function inside itself

Nothing stops you from feeding a function into itself. The output of $f$ goes back into $f$ . We write this as $ff (x)$ :

ff (x) = f (f (x))

With $f (x) = 2 x + 1$ , the input of the outer $f$ is $2 x + 1$ . So

ff (x) = 2 (2 x + 1) + 1 = 4 x + 3

Some books write $ff (x)$ as $f^{2} (x)$ . Be careful with this. Here the small 2 means "apply $f$ twice". It does not mean "square the output", which is $(f (x))^{2} = (2 x + 1)^{2}$ . The two are very different: at $x = 1$ , $ff (1) = 7$ , but $(f (1))^{2} = 9$ . In these lectures we write $ff (x)$ , so there is no doubt.

When the pipeline jams

There is one more thing to check. A composite only works if the output of the first function is an allowed input of the second. Look at these two functions:

f (x) = x - 4 g (x) = x

The function $g$ only accepts inputs that are 0 or more, because a negative number has no real square root.

Now try $g f (13)$ . First $f (13) = 9$ . Then $g (9) = 9 = 3$ . That works.

Now try $g f (2)$ . First $f (2) = - 2$ . Then $g$ receives $- 2$ , and $- 2$ is not a real number. The pipeline jams. So $g f (x) = x - 4$ only works when $x - 4 \geq 0$ , which means $x \geq 4$ .

The other order is different. $f g (2) = f (2) = 2 - 4 \approx - 2.59$ . This works, because $f$ accepts any number. So $f g (x) = x - 4$ works whenever $x \geq 0$ .

The rule behind this is short. The outputs of the inner function must be allowed inputs of the outer function. The lecture on domain and range, later in this topic, uses this rule to find exactly which inputs a composite can accept.

Worked examples

Worked Example 1 — Both orders as formulas

$f (x) = 3 x - 2$ and $g (x) = x + 5$ . Find $f g (x)$ and $g f (x)$ .

For $f g (x)$ : $g$ is next to $x$ , so $g$ acts first. The input of $f$ is $x + 5$ , in brackets: $f g (x) = f (x + 5) = 3 (x + 5) - 2 = 3 x + 15 - 2 = 3 x + 13$ .

For $g f (x)$ : $f$ acts first. The input of $g$ is $3 x - 2$ : $g f (x) = g (3 x - 2) = (3 x - 2) + 5 = 3 x + 3$ .

The two differ by 10 for every $x$ . This is the shop again. In $f g$ , the 5 is added before multiplying by 3, so it becomes 15. In $g f$ , it is added after, so it stays 5.

Check with $x = 1$ : $g (1) = 6$ and $f (6) = 16$ , and $3 (1) + 13 = 16$ . ✓ Also $f (1) = 1$ and $g (1) = 6$ , and $3 (1) + 3 = 6$ . ✓

Worked Example 2 — A value, step by step

$h (x) = x^{2} + 2$ and $k (x) = 3 x$ . Find $hk (3)$ .

When the input is a number, the safest method is to work with the number. $k$ is next to the input, so it acts first: $k (3) = 9$ . Then $h (9) = 81 + 2 = 83$ . So $hk (3) = 83$ .

Check with the formula: $hk (x) = h (3 x) = (3 x)^{2} + 2 = 9 x^{2} + 2$ . At $x = 3$ , this is $81 + 2 = 83$ . ✓ Notice the brackets around $3 x$ . Without them you might write $3 x^{2} + 2$ , which gives 29.

For comparison, $kh (3) = k (11) = 33$ . The order matters again.

Worked Example 3 — Solving an equation with a composite

$f (x) = 2 x + 1$ . Solve $ff (x) = 23$ .

First find the formula. We found it above: $ff (x) = 2 (2 x + 1) + 1 = 4 x + 3$ .

Now solve $4 x + 3 = 23$ . Then $4 x = 20$ , so $x = 5$ .

Check by running the pipeline: $f (5) = 11$ , and $f (11) = 23$ . ✓

There is a second way to think about it, working backwards through the pipeline. The last step gave 23, so its input was the number that $f$ turns into 23. That number is 11. The first step gave 11, so its input was 5. Running a pipeline backwards is the idea of the next lecture.

Worked Example 4 — Taking a function apart

$h (x) = (3 x - 1)^{5}$ . Write $h$ as a composite of two simpler functions.

Ask: if I put a number into $h$ , what do I do first? First I work out $3 x - 1$ . Then I raise the result to the power 5. So there are two steps. Let the first step be $g (x) = 3 x - 1$ and the second $f (x) = x^{5}$ .

The first step is next to $x$ , so $h (x) = f g (x)$ .

Check: $f g (x) = f (3 x - 1) = (3 x - 1)^{5}$ . ✓ With $x = 1$ : $g (1) = 2$ and $f (2) = 32$ , and $(3 - 1)^{5} = 32$ . ✓

Taking functions apart like this looks like a puzzle for its own sake. It is not. In calculus, the chain rule finds the rate of change of a composite from the rates of change of its parts. You must first see the parts.

Below is a piece of working with one mistake in it. Find it.

Spot the error — finding gf(x)

A student was asked to find gf(x), where f(x) = x + 3 and g(x) = x² − 1. Tap the line where the working first goes wrong.

A student's working for gf(x), where f(x) = x + 3 and g(x) = x² − 1, with the wrong line marked.

gf(x) = g(f(x)), so f acts first. (Correct.)
f(x) = x + 3, so the input of g is x + 3. (Correct.)
The error: g(x + 3) = x² + 3 − 1. The whole input x + 3 must replace x, in brackets: g(x + 3) = (x + 3)² − 1.
= x² + 2 (the arithmetic is right, but it follows from the wrong line).
So gf(x) = x² + 2.

The correct answer is gf(x) = (x + 3)² − 1 = x² + 6x + 8. Check with x = 1: f(1) = 4 and g(4) = 15, and 1 + 6 + 8 = 15. The wrong answer, x² + 2, is in fact fg(x): the student squared x first and added 3 last.

The mistake is worth a second look. Forgetting the brackets did not just give a wrong answer. It quietly swapped the order of the two steps. The student squared $x$ first and added 3 last, which is $f g (x)$ , not $g f (x)$ . Brackets are how the notation keeps the order. Lose them, and you lose the order.

Three classic mistakes

Trap 1 — Doing the steps in reading order

In $f g (x)$ , the letter $f$ comes first on the page, but $g$ acts first. Look at which function is next to $x$ . If you are unsure, write the long form $f (g (x))$ and work from the innermost brackets outwards.

Trap 2 — Reading fg(x) as f(x) × g(x)

Writing two function names side by side means "one after the other", not "multiply". With $f (x) = 2 x + 1$ and $g (x) = x^{2}$ , $f g (3) = 19$ , while $f (3) \times g (3) = 63$ .

Trap 3 — Losing the brackets around the inner function

The whole output of the inner function goes into the outer one. For $g (x) = x^{2}$ and $f (x) = 2 x + 1$ , $g f (x) = (2 x + 1)^{2}$ , not $2 x^{2} + 1$ . The same care applies to $ff (x)$ , which is not $(f (x))^{2}$ .

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

$f (x) = x^{2} - 3$ and $g (x) = 2 x + 5$ . Find $f g (- 1)$ and $g f (- 1)$ .

Hint 1
For $f g (- 1)$ , the function next to the input acts first. Which one is it?

Hint 2
$g (- 1) = 2 (- 1) + 5 = 3$ . Now put 3 into $f$ .

Full answer
$f g (- 1)$ : $g (- 1) = 3$ , then $f (3) = 9 - 3 = 6$ . So $f g (- 1) = 6$ . $g f (- 1)$ : $f (- 1) = (- 1)^{2} - 3 = - 2$ , then $g (- 2) = - 4 + 5 = 1$ . So $g f (- 1) = 1$ . Check with formulas: $f g (x) = (2 x + 5)^{2} - 3 = 4 x^{2} + 20 x + 22$ , and at $- 1$ this is $4 - 20 + 22 = 6$ . ✓ $g f (x) = 2 (x^{2} - 3) + 5 = 2 x^{2} - 1$ , and at $- 1$ this is $2 - 1 = 1$ . ✓
$f (x) = x^{2}$ and $g (x) = x - 3$ . Find $f g (x)$ and $g f (x)$ . Then find the input for which $f g (x) = g f (x)$ .

Hint 1
$f g (x) = f (x - 3)$ . The whole of $x - 3$ is squared.

Hint 2
Set $x^{2} - 6 x + 9 = x^{2} - 3$ . The $x^{2}$ terms cancel.

Full answer
$f g (x) = (x - 3)^{2} = x^{2} - 6 x + 9$ . $g f (x) = x^{2} - 3$ . Setting them equal: $- 6 x + 9 = - 3$ , so $- 6 x = - 12$ and $x = 2$ . Check: $f g (2) = f (- 1) = 1$ and $g f (2) = g (4) = 1$ . ✓ The two functions agree at one input only, so they are still different functions.
$h (x) = (2 x - 7)^{3}$ . Find two functions $f$ and $g$ such that $h (x) = f g (x)$ . Then find $g f (x)$ for the same two functions, and show that $g f (2) \neq = f g (2)$ .

Hint 1
What do you do first to a number when you work out $h$ ? That step is the inner function.

Hint 2
Inner: $g (x) = 2 x - 7$ . Outer: $f (x) = x^{3}$ . For $g f (x)$ , cube first and then double and subtract 7.

Full answer
$g (x) = 2 x - 7$ and $f (x) = x^{3}$ , so $f g (x) = (2 x - 7)^{3} = h (x)$ . In the other order, $g f (x) = g (x^{3}) = 2 x^{3} - 7$ . At $x = 2$ : $f g (2) = (4 - 7)^{3} = (- 3)^{3} = - 27$ , but $g f (2) = 2 (8) - 7 = 9$ . ✓ They are different, as expected.
A science thermometer reads in kelvin (K). The function $C (k) = k - 273.15$ changes kelvin into degrees Celsius. The function $F (c) = 1.8 c + 32$ changes degrees Celsius into degrees Fahrenheit. Find the composite that changes kelvin straight into Fahrenheit, and use it to convert 300 K. Which order, $FC$ or $CF$ , makes sense here, and why?

Hint 1
The kelvin reading must go into $C$ first. Which composite has $C$ next to the input?

Hint 2
$FC (k) = F (k - 273.15) = 1.8 (k - 273.15) + 32$ . Expand the bracket.

Full answer
$FC (k) = 1.8 (k - 273.15) + 32 = 1.8 k - 491.67 + 32 = 1.8 k - 459.67$ . For 300 K: $1.8 (300) - 459.67 = 540 - 459.67 = 80.33$ , so about 80.3 °F. Check step by step: $C (300) = 26.85$ °C, and $F (26.85) = 48.33 + 32 = 80.33$ . ✓ Only $FC$ makes sense. $C$ expects a temperature in kelvin, and $F$ expects one in Celsius. In $CF$ , the kelvin reading would go into $F$ , which treats it as Celsius, and then the answer in Fahrenheit would go into $C$ , which treats it as kelvin. The arithmetic would work, but the meaning would not. A composite needs each output to be the right kind of input for the next step. As a final check, $FC (0) = - 459.67$ : absolute zero is −459.67 °F.

The bigger idea

Composition is how mathematics builds complicated things out of simple ones. Almost every formula you meet is a chain of simple steps: add, multiply, square, take a root. Seeing the chain lets you take any formula apart and put it back together.

The order of the steps matters in many places, not only in algebra. Put on socks and then shoes, and you are ready to go out. Do it the other way round, and you are not. In the topic on graph transformations, a stretch followed by a shift gives a different graph from a shift followed by a stretch. In calculus, the chain rule is the method for composites, and it starts by asking which function is inside.

Sometimes the second function exactly undoes the first. Then the pipeline sends every input back to where it started: $f g (x) = x$ . That pair is special enough to have its own lecture, and it is the next one.

Key concepts in this lesson

A composite function feeds the output of one function into another function.
$f g (x)$ means $f (g (x))$ , so the function next to $x$ , here $g$ , acts first.
In general $f g (x) \neq = g f (x)$ , because the order of the steps changes the result.
To find a composite formula, put the whole inner function into the outer one, in brackets.
$ff (x)$ means applying $f$ twice, which is different from squaring $f (x)$ .
A composite works only when the outputs of the inner function are allowed inputs of the outer function.