Find the unknown function inside a composite

In 3 of 20 recent exams
Paper 1, no calculator
3 to 7 marks

How it is set in the exam

Paper 1 (no calculator), 3–7 marks, often straight after "find $f^{-1}(x)$ ", whose answer it uses. Typical (3 of 20 exam versions, all on Paper 1 (no calculator)).

Spot it in the exam

Exam phrases to spot (3)

“given that $(f \circ g)(x) = \ldots$ , find $g(x)$ ”
“ $g(x) = ax + b$ , where $a$ and $b$ are constants … given that $(f \circ g)(x) = \ldots$ for all $x$ , find the possible values of $a$ and $b$ ”
“the function $g$ is such that $f(g(x)) = \ldots$ ; find $g(x)$ ”

In disguise (2)

the input is scaled: $f(g(3x)) = \ldots$ , or $f\left(g\left(\frac{x}{2}\right)\right) = \ldots$ . Undo the outer function, then replace $x$ .
the outer function is an inverse: $f^{-1}(g(x)) = h(x)$ , so $g(x) = f(h(x))$

Variants

the outer function linear, rational or quadratic; one answer, or two when the outer function is a quadratic (paired by the $x$ -coefficient).

Looks like it, but isn't (3)

the composite equals $x$ for all $x$ , with $x$ itself as the input, $h(k(x)) = x$ : the unknown is the inverse, Find the inverse function, with its domain. A scaled input stays here: $f\left(g\left(\frac{x}{2}\right)\right) = x$ gives $g\left(\frac{x}{2}\right) = f^{-1}(x)$ , so $g(x) = f^{-1}(2x)$ , not $f^{-1}(x)$
only the composite at one point is known, such as $(f \circ g)(2) = 7$ , and a constant is asked for: Form a composite function and use it. Coefficients can be matched only when the composite is given for every $x$ .
one value of the inner function, when only the composite at one point is known: $g(a) = f^{-1}\left((f \circ g)(a)\right)$ , Values of f, f∘g and f⁻¹ from a graph or a formula

Worked example

Undo a rational outer function; the input is scaled

Paper 1 (no calculator)

Question

Let $f(x) = \dfrac{3x + 1}{x + 1}$ , for $x \ne -1$ .
(a) Find $f^{-1}(x)$ . [3]
(b) The function $g$ is such that $f(g(2x)) = x + 3$ , for $x \ne 0$ . Find $g(x)$ . [4]

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Which method?

one-to-one with an inverse you can find (linear, rational, exponential, log)
Method
apply $f^{-1}$ to both sides: $g(x) = f^{-1}(\text{the given composite})$
Why
undoing the outer function leaves the inner one
a quadratic (its inner function is then given as $ax + b$ ), or any outer function when the inner one is given as $ax + b$
Method
compose with the letters, expand, equate the coefficients, solve, then pair the answers; for a quadratic, completing the square on both sides and keeping both signs also works (Example 3.2, other route)
Why
a quadratic has no inverse on $\mathbb{R}$ ; the given form $ax + b$ is what narrows $g$ down to two functions

The outer function is…	Method	Why
one-to-one with an inverse you can find (linear, rational, exponential, log)	apply $f^{-1}$ to both sides: $g(x) = f^{-1}(\text{the given composite})$	undoing the outer function leaves the inner one
a quadratic (its inner function is then given as $ax + b$ ), or any outer function when the inner one is given as $ax + b$	compose with the letters, expand, equate the coefficients, solve, then pair the answers; for a quadratic, completing the square on both sides and keeping both signs also works (Example 3.2, other route)	a quadratic has no inverse on $\mathbb{R}$ ; the given form $ax + b$ is what narrows $g$ down to two functions

Either way, if the input is scaled, as in $g(2x)$ , find that first and then replace $x$ by $\frac{x}{2}$ . Markschemes accept both methods whenever both work.

Steps

Undo the outer function
if it is one-to-one: apply $f^{-1}$ to both sides, so $g(x) = f^{-1}(\ldots)$ (if the outer function is $f^{-1}$ , apply $f$ ). Otherwise write $f(ax + b)$ with the letters and expand.
Equating coefficients:
match the $x^2$ , $x$ and constant terms. Each gives an equation.
Pair the answers:
$a^2 = 9$ gives two values of $a$ , and the constant term may give two of $b$ . The $x$ -coefficient equation says which $b$ goes with which $a$ .
A scaled input:
after step 1 you have $g(2x)$ (say); replace $x$ by $\frac{x}{2}$ to get $g(x)$ .
Check
by composing forwards: $f(g(x))$ must give the composite in the question.

Common mistakes

Composing in the wrong order
$g(f(x))$ instead of $f(g(x))$ earns method marks only.
Applying $f$ instead of $f^{-1}$
to strip $f$ off $f(g(x))$ , use its inverse.
Stopping at $g(2x)$
when $g(x)$ is asked for.

More mistakes (3)

Pairing $a$ and $b$ wrongly,
or giving all four combinations: the $x$ -coefficient equation (or the completed-square route) pairs them, and one is given for each correct function.
Trying to undo a quadratic with an inverse
it has none on $\mathbb{R}$ . Equate coefficients, or complete the square and keep both signs.
$g = f^{-1}$ when the input is scaled
$f\left(g\left(\frac{x}{2}\right)\right) = x$ gives $g(x) = f^{-1}(2x)$ ; $g = f^{-1}$ would give $f\left(f^{-1}\left(\frac{x}{2}\right)\right) = \frac{x}{2}$ , not $x$ .

Easy to confuse with

Next typeThe graphs of f and f⁻¹

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