1.02PureCore

Composite Functions

Q: Which function goes first in fg?

g. Read fg(x) as f( g(x)) — the function closest to x is applied first.

Q: How does this connect to the chain rule?

The chain rule differentiates a composite: d dx f( g(x)) = f^ ( g(x)), g^ (x). Decomposing correctly is the first step.

Q: What does f^3(x) mean?

fff(x) — apply f three times in succession.

A thermostat reads a temperature, converts it to a voltage, and the voltage sets a fan speed. Each step is a function; the whole chain is a $composite function$ . In notation, $fg (x)$ means "apply $g$ first, then $f$ " — and the order is the thing students most often get backwards.

35 min Functions

The big picture

Composition is how complicated functions are built from simple ones, and recognising the pieces is a skill you will use constantly. The chain rule in differentiation is nothing more than a rule for differentiating composites; integration by substitution undoes one; and graph transformations like $f (2 x)$ or $f (x) + 3$ are compositions with very simple functions. Two technical points make or break exam answers: $fg$ is evaluated from right to left, and the composite only exists where the range of the inner function lies inside the domain of the outer one.

What you'll be able to do

Evaluate composite functions $fg (x)$ numerically and algebraically
Recognise that $fg$ and $gf$ are usually different
Determine the domain and range of a composite function
Solve equations involving composite functions and decompose functions into composites

Reading and evaluating fg

$fg (x)$ means $f (g (x))$ : work out $g (x)$ first, then use that result as the input to $f$ . The function written nearest $x$ acts first.

Some texts write $f \circ g$ ; OCR generally writes $fg$ . Repeated application, $ff (x)$ , is often written $f^{2} (x)$ — which does $not$ mean $(f (x))^{2}$ .

Numerically, evaluate inside out. With $f (x) = 2 x + 1$ and $g (x) = x^{2}$ : $fg (3) = f (9) = 19$ , while $gf (3) = g (7) = 49$ .

fg (x) = f (g (x))

Apply g first, then f.

Why this works

Composition is not commutative. Putting socks on then shoes differs from shoes then socks, and $fg$ differs from $gf$ for almost every pair of functions you will meet.

Algebraic composites

To find $fg (x)$ as an expression, substitute the whole of $g (x)$ , in brackets, wherever $x$ appears in $f$ .

With $f (x) = \frac{3}{x - 1}$ and $g (x) = x^{2} + 2$ : $fg (x) = \frac{3}{( x ^{2} + 2 ) - 1} = \frac{3}{x ^{2} + 1}$ , and $gf (x) = (\frac{3}{x - 1})^{2} + 2$ .

fg (x) = (2 x + 5)^{2} - 3

2Set equal to

46

(2 x + 5)^{2} = 49

3So

2 x + 5 = 7

2 x + 5 = - 7

Answer

x = 1

x = - 6

Tip — When the composite contains a perfect square, take square roots rather than expanding — and remember the negative root.

Domain and range of a composite

For $fg$ to exist, every output of $g$ must be an acceptable input to $f$ . In other words, the $range of g$ must lie within the $domain of f$ .

The domain of $fg$ is the domain of $g$ (restricted further if necessary). To find the range of $fg$ , take the range of $g$ and ask what $f$ does to that set.

1The range of

g

g (x) \geq 1

2Every value

\geq 1

is positive, so it lies in the domain of

f

; hence

fg

exists.

ln

is increasing, so on inputs

\geq 1

it takes values

\geq ln 1 = 0

Answer

fg (x) = ln (x^{2} + 1)

with range

fg (x) \geq 0

Why this works

Think of the range of the inner function as the only doorway into the outer function. The composite range is whatever the outer function produces from that doorway — often much smaller than the outer function’s full range.

Decomposing a function

Going the other way — spotting how a function is built — is just as important. $h (x) = 3 x - 2$ is $fg (x)$ with $g (x) = 3 x - 2$ (inside) and $f (x) = x$ (outside).

There is usually more than one valid decomposition. The natural one takes the innermost operation as $g$ and the last operation you would press on a calculator as $f$ .

This is precisely the recognition step for the chain rule: $\frac{d}{d x} 3 x - 2$ differentiates the outer square root and multiplies by the derivative of the inner $3 x - 2$ .

1Inner operation: add 4, so

g (x) = x + 4

2Then cube and take the reciprocal:

f (x) = \frac{1}{x ^{3}}

Answer

h = fg

with

g (x) = x + 4

f (x) = x^{- 3}

Think like an examiner

Show the intermediate step, such as $fg (3) = f (9)$ , before the final value.
Substitute the inner function in brackets when forming $fg (x)$ .
Justify existence of a composite by comparing the range of the inner with the domain of the outer.
For $f^{2} (x)$ , apply $f$ twice — do not square.

Common misconceptions

Applying

f

first in

fg (x)

The function next to

x

— here

g

— acts first.

Assuming

fg = gf

Check both; they are usually different.

Interpreting

f^{2} (x)

[f (x)]^{2}

f^{2} (x) = ff (x)

Giving the range of

fg

as the full range of

f

Apply

f

only to the range of

g

Composition

fg (x) = f (g (x))

Right to left.

f^{2} (x) = ff (x)

Repeated application.

range of g \subseteq domain of f

Condition for fg to exist.

Stretch yourself

The functions $f$ and $g$ are defined for all real $x$ by $f (x) = x^{2} - 2 x$ and $g (x) = x + k$ , where $k$ is a non-zero constant. Show that the equation $fg (x) = gf (x)$ has exactly one solution, and find the value of $k$ for which that solution is $x = - 1$ .

Hint — Form both composites and subtract. The $x^{2}$ terms cancel, leaving a linear equation in $x$ .

Questions students ask

Which function goes first in fg?

g

. Read

fg (x)

f (g (x))

— the function closest to

x

is applied first.

Can fg exist when gf does not?

Yes. Existence depends on whether the inner function’s range fits the outer function’s domain, and that can hold one way but not the other.

How does this connect to the chain rule?

The chain rule differentiates a composite:

\frac{d}{d x} f (g (x)) = f^{'} (g (x)) g^{'} (x)

. Decomposing correctly is the first step.

What does

f^{3} (x)

mean?

fff (x)

— apply

f

three times in succession.

Key takeaways

$fg (x) = f (g (x))$ : apply $g$ first.
In general $fg \neq = gf$ .
fg exists when the range of $g$ lies within the domain of $f$ .
The range of fg is found by applying $f$ to the range of $g$ .
Decomposing functions into composites prepares you for the chain rule.

How this fits the course

Build on

Domain and Range

Derivatives of Functions

Leads to

Inverse Functions

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Real past-paper questions on Composite Functions, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

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Composite Functions

Reading and evaluating fg

Algebraic composites

Domain and range of a composite

Decomposing a function

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Related topics

Composite Functions

Reading and evaluating fg

Algebraic composites

Domain and range of a composite

Decomposing a function

Practise this topic

Related topics