Find the inverse function, with its domain

In 10 of 20 recent exams
1% of the marks
Paper 1 or 2
1 to 4 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), 1–4 marks per part. Rational functions are the commonest (6 of its 14 parts), then logs (3), roots (2), lines (2) and an exponential (1). Very common (10 of 20 exam versions, 4 of them on Paper 2 (calculator)). On Paper 2 (calculator) it is still algebra, often part (a) of a question whose later parts use the GDC.

Spot it in the exam

Exam phrases to spot (6)

“find $f^{-1}(x)$ ”
“find the inverse function of $f$ ”
“show that $f^{-1}(x) = \ldots$ ”
“find $f^{-1}(x)$ and state its domain”
“show that $f^{-1}(x) = f(x)$ ”
“given that $g^{-1}(x) = g(x)$ , find the value of $c$ ”

In disguise (4)

"write down the inverse of $a^x$ ": recall $\log_a x$ , and the other way round (Exponentials and logs are inverses)
a composite with an unknown function that equals $x$ for all $x$ : $h(k(x)) = x$ means $k = h^{-1}$ (if the input is scaled first, as in $h\left(k\left(\frac{x}{2}\right)\right) = x$ , then $k(x) = h^{-1}(2x)$ : Find the unknown function inside a composite)
"show that $f(f(x)) = x$ ": the function is its own inverse
the equation of the reflection of the graph of $f$ in $y = x$ : that curve is $y = f^{-1}(x)$ (sketching it, or where it meets the graph of $f$ , is The graphs of f and f⁻¹)

Variants

rational (the denominator may be written with its constant first); a log with a scaled argument; a log plus a constant; an exponential (a show-that); a root (the inverse, its domain and its range); linear; self-inverse, either shown or used to find a constant.

Looks like it, but isn't (3)

one value, $f^{-1}(b)$ , with no formula asked for: Values of f, f∘g and f⁻¹ from a graph or a formula
a sketch of the graph of $f^{-1}$ , or where the graphs of $f$ and $f^{-1}$ meet: The graphs of f and f⁻¹: reflect in y = x, find where they meet
choosing a domain yourself so that $f^{-1}$ exists ("the largest value of $k$ for which …"): HL (AHL 2.14), not SL

Worked example

A rational function: the inverse and its domain

Paper 1 (no calculator)

Question

Let $f(x) = \dfrac{3x + 4}{5 - x}$ , for $x \ne 5$ .
(a) Find $f^{-1}(x)$ . [4]
(b) Find the domain of $f^{-1}$ . [2]

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Which method?

"find $f^{-1}(x)$ "
Route
swap $x$ and $y$ , then make $y$ the subject by the family's technique (Finding an inverse)
Also accepted
make $x$ the subject first, then swap
"show that $f^{-1}(x) = \ldots$ "
Route
derive it forwards from $y = f(x)$ , ending on the given formula
Also accepted
verify that $f(f^{-1}(x)) = x$ , using the given formula
"show that $f^{-1}(x) = f(x)$ "
Route
find $f^{-1}$ and see that it is $f$
Also accepted
show that $f(f(x)) = x$
"given that $g^{-1}(x) = g(x)$ , find $c$ "
Route
find $g^{-1}$ in terms of $c$ and compare it with $g$
Also accepted
a rational $g$ : set its asymptotes equal, $x = k$ and $y = k$

The question says	Route	Also accepted
"find $f^{-1}(x)$ "	swap $x$ and $y$ , then make $y$ the subject by the family's technique (Finding an inverse)	make $x$ the subject first, then swap
"show that $f^{-1}(x) = \ldots$ "	derive it forwards from $y = f(x)$ , ending on the given formula	verify that $f(f^{-1}(x)) = x$ , using the given formula
"show that $f^{-1}(x) = f(x)$ "	find $f^{-1}$ and see that it is $f$	show that $f(f(x)) = x$
"given that $g^{-1}(x) = g(x)$ , find $c$ "	find $g^{-1}$ in terms of $c$ and compare it with $g$	a rational $g$ : set its asymptotes equal, $x = k$ and $y = k$

Steps

Write $y = f(x)$ and swap $x$ and $y$ .
Make $y$ the subject
, using the family's technique (Finding an inverse): collect and factor for a rational function; isolate, then square for a root; isolate the power, then take logs for an exponential; write a log as a power.
Write $f^{-1}(x) = \ldots$
(" $y = \ldots$ " is accepted), in any equivalent form unless a form is asked for.
The domain of $f^{-1}$ is the range of $f$ :
on its largest domain, $\frac{ax + b}{cx + d}$ takes every value except its horizontal asymptote $y = \frac{a}{c}$ ; a root is $\ge 0$ ; $e^x > 0$ ; a log takes every value. If the question states a smaller domain, find the range from that domain's ends (a sketch helps). The range of $f^{-1}$ is the domain of $f$ .
Check
one point ( $f(a) = b$ , so $f^{-1}(b) = a$ ), or $f(f^{-1}(x)) = x$ . On , graph $f$ , your $f^{-1}$ and $y = x$ : the two curves should be reflections in the line (graph a function on a given domain).
Self-inverse:
compare $f^{-1}$ with $f$ , or show $f(f(x)) = x$ with a general $x$ . For a constant, compare coefficients (or set the two asymptotes equal).

Common mistakes

Not finishing the swap
an answer in terms of $y$ , or " $x = \ldots$ ", is not $f^{-1}(x)$ .
Sign slips when collecting the $y$ terms
move one term at a time, and check with one point at the end.
$f^{-1}(x) = \frac{1}{f(x)}$
the $-1$ is not a power.

More mistakes (5)

The domain of $f^{-1}$ from its formula alone
$\frac{(4 - x)^2 - 6}{2}$ looks defined everywhere, but the domain is the range of $f$ , $x \le 4$ .
The wrong horizontal asymptote
when the denominator starts with its constant: $\frac{3x + 4}{5 - x}$ has $y = -3$ , not $y = 3$ .
Taking $\ln$ before isolating the exponential
$\ln\left(4e^{2y} + 1\right)$ is not $\ln 4 + 2y + \ln 1$ .
Scaling the wrong way
from $x = \log_3\frac{y}{9}$ , $\frac{y}{9} = 3^x$ , so $y = 9 \times 3^x$ , not $\frac{3^x}{9}$ .
"Showing" self-inverse with one number
use a general $x$ .

Easy to confuse with

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