Composite functions

AS · P1 · 11 min

A composite function applies one function and then another to the result. If $f$ doubles a number and $g$ squares it, then "do $f$ , then $g$ " is a new function, written $gf$ . Paper 1 asks you to form composites, evaluate them, solve equations involving them, and decide whether a composite can be formed at all, which depends on matching the range of one function to the domain of the other.

The idea

Think of each function as a machine. The output of the first machine is fed straight into the second.

x \ \xrightarrow{\ \ f\ \ }\ f(x) \ \xrightarrow{\ \ g\ \ }\ g\big(f(x)\big)

Definition

The composite function $gf$ is defined by $gf(x) = g\big(f(x)\big)$ . It means: apply $f$ first, then apply $g$ to the result.

$ff$ (also written $f^2$ ) means apply $f$ twice: $ff(x) = f\big(f(x)\big)$ .

The order is the most important thing to get right. $gf$ means $f$ first, because $f$ is next to the $x$ . Read it from right to left, as you would read $g\big(f(x)\big)$ from the inside out.

In general $gf \neq fg$ . If $f(x) = x + 1$ and $g(x) = x^2$ :

$gf(x) = g(x + 1) = (x + 1)^2$ (add one, then square);
$fg(x) = f\left(x^2\right) = x^2 + 1$ (square, then add one).

Forming a composite function

Identify which function acts first: in $gf(x)$ it is $f$ .
Write $g(\ \ )$ with an empty bracket, using the rule for $g$ with every $x$ replaced by a bracket.
Put the whole expression $f(x)$ into each bracket.
Simplify.
For a numerical value such as $gf(3)$ , it is often quicker to work out $f(3)$ first and then apply $g$ to that number.

Forming composites in both orders

The functions $f$ and $g$ are defined for $x \in \mathbb{R}$ by $f(x) = 2x + 1$ and $g(x) = x^2 - 3$ . Find (a) $fg(x)$ , (b) $gf(x)$ , (c) $gf(2)$ .

Solution

(a) $g$ first, then $f$ . The rule for $f$ is "double and add one":

fg(x) = f\left(x^2 - 3\right) = 2\left(x^2 - 3\right) + 1 = 2x^2 - 5

(b) $f$ first, then $g$ . The rule for $g$ is "square and subtract three":

gf(x) = g(2x + 1) = (2x + 1)^2 - 3 = 4x^2 + 4x - 2

Applying a function twice

The function $f$ is defined by $f(x) = 3x - 2$ for $x \in \mathbb{R}$ . Find $ff(x)$ and solve $ff(x) = x$ .

Solution

ff(x) = f(3x - 2) = 3(3x - 2) - 2 = 9x - 8

$ff(x) = x$ gives $9x - 8 = x$ , so $8x = 8$ and $x = 1$ . (Indeed $f(1) = 1$ , so $ff(1) = 1$ .)

Solving an equation with a composite

With $f(x) = 2x + 1$ and $g(x) = x^2 - 3$ for $x \in \mathbb{R}$ , solve $gf(x) = 6$ .

Solution

From the first example, $gf(x) = 4x^2 + 4x - 2$ . So

4x^2 + 4x - 2 = 6 \quad\Rightarrow\quad 4x^2 + 4x - 8 = 0 \quad\Rightarrow\quad x^2 + x - 2 = 0 \quad\Rightarrow\quad (x + 2)(x - 1) = 0

So $x = 1$ or $x = -2$ .

Alternatively, work backwards: $g(f(x)) = 6$ means $\big(f(x)\big)^2 - 3 = 6$ , so $f(x) = \pm 3$ , giving $2x + 1 = 3$ or $2x + 1 = -3$ , i.e. $x = 1$ or $x = -2$ .

When can a composite be formed?

$gf(x)$ feeds the outputs of $f$ into $g$ . Every one of those outputs must be an allowed input for $g$ .

Key result

The composite $gf$ can be formed only if

\text{the range of } f \text{ is contained in the domain of } g

The domain of $gf$ is the domain of $f$ .

If even one output of $f$ is outside the domain of $g$ , then $gf$ does not exist as a function on that domain.

Deciding whether a composite exists

The functions $f$ and $g$ are defined by $f(x) = x - 4$ for $x \in \mathbb{R}$ and $g(x) = \sqrt{x}$ for $x \ge 0$ . Explain why $gf$ cannot be formed, and find $fg(x)$ .

Solution

The range of $f$ is all real numbers, $\mathbb{R}$ . The domain of $g$ is $x \ge 0$ . The range of $f$ is not contained in the domain of $g$ : for example $f(0) = -4$ , and $g(-4) = \sqrt{-4}$ is undefined. So $gf$ cannot be formed.

For $fg$ : the range of $g$ is $g(x) \ge 0$ , which is contained in the domain of $f$ ( $\mathbb{R}$ ). So $fg$ exists:

fg(x) = f\left(\sqrt{x}\right) = \sqrt{x} - 4, \quad \text{for } x \ge 0

The range of a composite

The functions $f$ and $g$ are defined by $f(x) = 3 - 2x$ for $-1 \le x \le 2$ and $g(x) = x^2$ for $x \in \mathbb{R}$ . Find the range of $gf$ .

Solution

First find the range of $f$ , the set of values fed into $g$ . $f$ is decreasing, with $f(-1) = 5$ and $f(2) = -1$ , so the range of $f$ is $-1 \le f(x) \le 5$ .

Now apply $g$ (squaring) to every number from $-1$ to $5$ . The squares of numbers in this interval run from $0$ (at $0$ , which is in the interval) up to $25$ (at $5$ ).

Range of $gf$ : $0 \le gf(x) \le 25$ .

Squaring the end values alone would give $1$ and $25$ , but $0$ lies between $-1$ and $5$ , and $0^2 = 0$ is smaller than both.

Finding constants from a composite

The function $f$ is defined by $f(x) = ax + b$ for $x \in \mathbb{R}$ , where $a$ and $b$ are constants. Given that $ff(x) = 9x - 8$ , find the possible pairs of values of $a$ and $b$ .

Solution

ff(x) = a(ax + b) + b = a^2x + ab + b

Compare coefficients with $9x - 8$ :

$x$ : $a^2 = 9$ , so $a = 3$ or $a = -3$ ;
constant: $ab + b = -8$ , i.e. $b(a + 1) = -8$ .

If $a = 3$ : $4b = -8$ , $b = -2$ . If $a = -3$ : $-2b = -8$ , $b = 4$ .

So $a = 3,\ b = -2$ or $a = -3,\ b = 4$ . (Check the second: $f(x) = -3x + 4$ , $ff(x) = -3(-3x + 4) + 4 = 9x - 8$ .)

The least constant for which a composite exists

The functions $f$ and $g$ are defined by $f(x) = x^2 + 2x + k$ for $x \in \mathbb{R}$ and $g(x) = \sqrt{x}$ for $x \ge 0$ , where $k$ is a constant. Find the least value of $k$ for which $gf$ can be formed.

Solution

Complete the square: $f(x) = (x + 1)^2 + k - 1$ . The range of $f$ is $f(x) \ge k - 1$ .

For $gf$ to exist, every value of $f$ must be in the domain of $g$ , i.e. $\ge 0$ . So we need

k - 1 \ge 0 \quad\Rightarrow\quad k \ge 1

The least value of $k$ is $1$ .

Watch out

Wrong order. $fg(x)$ means $g$ first. Students who do $f$ first lose every mark in the part. Say it aloud: " $fg$ of $x$ is $f$ of $g$ of $x$ ".

Multiplying instead of composing. $fg(x)$ is not $f(x) \times g(x)$ . For $f(x) = 2x + 1$ and $g(x) = x^2 - 3$ , $f(x)g(x) = (2x + 1)\left(x^2 - 3\right)$ , which is something else.

Reading $f^2(x)$ as a square. In Cambridge function notation $f^2(x)$ means $ff(x)$ , not $\big(f(x)\big)^2$ .

Range of a composite from end points. As in the squaring example, the composite may reach values between the images of the end points. Track the whole interval.

Ignoring the domain condition. Before using $gf$ , check that the range of $f$ lies inside the domain of $g$ .

Exam tip

Questions on composites are usually short (2 or 3 marks), but they set up later parts, such as solving $fg(x) = 0$ or finding an inverse. Simplify the composite fully.
To "explain why $gf$ cannot be formed", compare the range of $f$ with the domain of $g$ and say which values cause the problem. A numerical example (like $f(0) = -4$ above) makes the explanation complete.
For $gf(a)$ with a number $a$ , working out $f(a)$ first is quicker and less error-prone than forming $gf(x)$ .
Keep brackets around substituted expressions until the last line.

Summary

$gf(x) = g\big(f(x)\big)$ : apply $f$ first.
In general $fg \neq gf$ .
$ff(x) = f^2(x)$ means apply $f$ twice.
To form $gf$ , replace every $x$ in $g$ by the expression $f(x)$ , in brackets.
$gf$ exists only if the range of $f$ lies within the domain of $g$ .
The range of $gf$ : find the range of $f$ , then apply $g$ to that whole set.
Unknown constants: expand the composite and compare coefficients.

Practice questions

Question

$f(x) = 3x - 1$ and $g(x) = x^2 + 2$ for $x \in \mathbb{R}$ . Find $fg(x)$ , $gf(x)$ and $gf(1)$ .
$f(x) = 2x + 3$ for $x \in \mathbb{R}$ . Find $ff(x)$ and solve $ff(x) = 21$ .
$f(x) = \dfrac{1}{x + 1}$ for $x > -1$ . Show that $ff(x) = \dfrac{x + 1}{x + 2}$ .
$f(x) = x^2$ and $g(x) = x - 3$ for $x \in \mathbb{R}$ . Solve $fg(x) = gf(x)$ .
$f(x) = x + 5$ for $x \in \mathbb{R}$ and $g(x) = \sqrt{x}$ for $x \ge 0$ . Explain why $gf$ cannot be formed. Find $fg(x)$ and state its range.
$f(x) = 2x - 1$ for $0 \le x \le 3$ and $g(x) = x^2 - 2x$ for $x \in \mathbb{R}$ . Find the range of $gf$ .
$f(x) = ax + b$ for $x \in \mathbb{R}$ . Given that $f(2) = 7$ and $ff(1) = 13$ , find the possible values of $a$ and $b$ .
$f(x) = 2x + k$ for $x \ge 0$ and $g(x) = \sqrt{x - 3}$ for $x \ge 3$ . (a) Find the set of values of $k$ for which $gf$ can be formed. (b) For $k = 7$ , solve $gf(x) = 4$ .
$g(x) = 4x - 1$ for $x \in \mathbb{R}$ , and $f$ is a function such that $fg(x) = 16x^2 - 8x + 3$ . Find $f(x)$ .

Answers

$fg(x) = 3\left(x^2 + 2\right) - 1 = 3x^2 + 5$ . $gf(x) = (3x - 1)^2 + 2 = 9x^2 - 6x + 3$ . $gf(1) = g(2) = 6$ .
$ff(x) = 2(2x + 3) + 3 = 4x + 9$ . $4x + 9 = 21$ gives $x = 3$ .
$ff(x) = \dfrac{1}{\frac{1}{x + 1} + 1} = \dfrac{1}{\frac{1 + (x + 1)}{x + 1}} = \dfrac{x + 1}{x + 2}$ . (The range of $f$ is $f(x) > 0$ , which is inside the domain $x > -1$ , so $ff$ exists.)
$(x - 3)^2 = x^2 - 3$ gives $x^2 - 6x + 9 = x^2 - 3$ , so $6x = 12$ and $x = 2$ .
The range of $f$ is $\mathbb{R}$ , which includes negative numbers (e.g. $f(-6) = -1$ ), and these are not in the domain of $g$ . $fg(x) = \sqrt{x} + 5$ for $x \ge 0$ , with range $fg(x) \ge 5$ .
The range of $f$ is $-1 \le f(x) \le 5$ . Now $g(t) = (t - 1)^2 - 1$ for $-1 \le t \le 5$ : least value $-1$ at $t = 1$ ; $g(-1) = 3$ , $g(5) = 15$ . Range of $gf$ : $-1 \le gf(x) \le 15$ .
$2a + b = 7$ and $ff(1) = a(a + b) + b = 13$ . Substituting $b = 7 - 2a$ : $a(7 - a) + 7 - 2a = 13$ , so $a^2 - 5a + 6 = 0$ , giving $a = 2$ or $a = 3$ . Then $a = 2,\ b = 3$ or $a = 3,\ b = 1$ .
(a) The range of $f$ is $f(x) \ge k$ ; this must lie in $x \ge 3$ , so $k \ge 3$ . (b) $gf(x) = \sqrt{2x + 7 - 3} = \sqrt{2x + 4}$ . $\sqrt{2x + 4} = 4$ gives $2x + 4 = 16$ , so $x = 6$ .
Let $t = 4x - 1$ , so $4x = t + 1$ . Then $16x^2 - 8x + 3 = (4x)^2 - 2(4x) + 3 = (t + 1)^2 - 2(t + 1) + 3 = t^2 + 2$ . So $f(t) = t^2 + 2$ , i.e. $f(x) = x^2 + 2$ . Check: $f(4x - 1) = 16x^2 - 8x + 1 + 2$ .

The idea

When can a composite be formed?

Practice questions

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