Is ^-1 x the same as 1 x?

No. ^-1x is the inverse of sine on a restricted domain; 1 x is cosec x.

1.02PureStretch

Inverse Functions

Q: Where do a function and its inverse intersect?

For an increasing function, always on the line y = x, so solve f(x) = x. Decreasing functions can also meet off that line.

If $\mathrm{f}$ converts Celsius to Fahrenheit, its $\textbf{inverse}$ $\mathrm{f}^{-1}$ converts back. Finding an inverse is mostly careful rearrangement, but the real content of this lesson is the conditions: an inverse only exists for a one-to-one function, and its domain and range are the original’s range and domain, swapped.

40 min Functions

The big picture

An inverse function reverses a mapping, so applying a function and then its inverse returns you exactly where you started: $\mathrm{f}^{-1}\mathrm{f}(x) = x$ . That requirement is why many-to-one functions have no inverse — if two inputs share an output, there is no single way back. The fix, restricting the domain, is how $\sqrt{x}$ , $\ln x$ and the inverse trigonometric functions are defined at all. Graphically, swapping inputs and outputs is a reflection in $y = x$ , which gives a quick check on any inverse you find and explains why the natural log and exponential graphs are mirror images.

What you'll be able to do

Explain why only one-to-one functions have inverses
Find $\mathrm{f}^{-1}(x)$ algebraically and state its domain and range
Sketch $y = \mathrm{f}^{-1}(x)$ as the reflection of $y = \mathrm{f}(x)$ in $y = x$
Use $\mathrm{ff}^{-1}(x) = \mathrm{f}^{-1}\mathrm{f}(x) = x$ and restrict domains to create inverses

When an inverse exists

The inverse $\mathrm{f}^{-1}$ must send each output of $\mathrm{f}$ back to the input that produced it. If $\mathrm{f}$ is many-to-one, some output came from two inputs, and $\mathrm{f}^{-1}$ would have to give two answers — so it would not be a function.

Therefore $\mathrm{f}^{-1}$ exists $\textbf{if and only if f is one-to-one}$ on its domain.

Notation warning: $\mathrm{f}^{-1}(x)$ means the inverse function, not $\dfrac{1}{\mathrm{f}(x)}$ .

\mathrm{f}^{-1}\mathrm{f}(x) = x \quad\text{and}\quad \mathrm{ff}^{-1}(x) = x

An inverse undoes the function, in either order.

Why this works

The swap also exchanges the sets: the domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$ , and the range of $\mathrm{f}^{-1}$ is the domain of $\mathrm{f}$ . Many marks on inverse questions are for stating these, not for the algebra.

Finding an inverse algebraically

Write $y = \mathrm{f}(x)$ , rearrange to make $x$ the subject, then swap the letters to write $\mathrm{f}^{-1}(x)$ in terms of $x$ .

Then state the domain of $\mathrm{f}^{-1}$ — which is the range of $\mathrm{f}$ , so you usually need to find that first.

Example 1

\mathrm{f}(x) = \dfrac{2x + 3}{x - 1}

x > 1

. Find

\mathrm{f}^{-1}(x)

and state its domain.

1Let

y = \dfrac{2x + 3}{x - 1}

, so

y(x - 1) = 2x + 3

2Expand and collect

x

terms:

xy - 2x = y + 3

, so

x(y - 2) = y + 3

3Hence

x = \dfrac{y + 3}{y - 2}

; swap letters.

4Range of

\mathrm{f}

: writing

\mathrm{f}(x) = 2 + \dfrac{5}{x-1}

, for

x > 1

the fraction takes every positive value, so

\mathrm{f}(x) > 2

Answer

\mathrm{f}^{-1}(x) = \dfrac{x + 3}{x - 2}

, domain

x > 2

Tip — Whenever $x$ appears more than once, multiply out, gather every $x$ term on one side, and factorise $x$ out. That single technique handles almost every rational inverse.

The graphs of f and its inverse

If $(a, b)$ lies on $y = \mathrm{f}(x)$ , then $\mathrm{f}(a) = b$ , so $\mathrm{f}^{-1}(b) = a$ and $(b, a)$ lies on $y = \mathrm{f}^{-1}(x)$ . Swapping coordinates is a reflection in the line $y = x$ .

So the graph of $\mathrm{f}^{-1}$ is the reflection of the graph of $\mathrm{f}$ in $y = x$ . Intercepts swap axes and asymptotes swap orientation: a horizontal asymptote $y = 2$ becomes a vertical asymptote $x = 2$ .

Draw axes to equal scales when sketching this, or the reflection will look wrong.

Example 1

\mathrm{f}(x) = \mathrm{e}^{x} + 2

x \in \mathbb{R}

. Find

\mathrm{f}^{-1}(x)

and describe how the graphs are related.

y = \mathrm{e}^{x} + 2 \Rightarrow \mathrm{e}^{x} = y - 2 \Rightarrow x = \ln(y - 2)

2Range of

\mathrm{f}

\mathrm{f}(x) > 2

, so the domain of the inverse is

x > 2

3The asymptote

y = 2

\mathrm{f}

becomes the asymptote

x = 2

\mathrm{f}^{-1}

Answer

\mathrm{f}^{-1}(x) = \ln(x - 2)

x > 2

— the reflection of

y = \mathrm{e}^{x} + 2

y = x

Restricting the domain

A many-to-one function can be given an inverse by restricting its domain to a part on which it is one-to-one.

$\mathrm{f}(x) = (x - 3)^{2} + 1$ on $\mathbb{R}$ is many-to-one, with vertex at $(3, 1)$ . On $x \ge 3$ it is increasing and one-to-one.

Solving $y = (x - 3)^{2} + 1$ gives $x = 3 \pm \sqrt{y - 1}$ . The restriction $x \ge 3$ selects the positive root, so $\mathrm{f}^{-1}(x) = 3 + \sqrt{x - 1}$ , $x \ge 1$ .

Points where $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ always lie on $y = x$ when $\mathrm{f}$ is increasing, so they can be found by solving the easier equation $\mathrm{f}(x) = x$ .

Example 1For

\mathrm{f}(x) = (x - 3)^{2} + 1

x \ge 3

, solve

\mathrm{f}(x) = \mathrm{f}^{-1}(x)

\mathrm{f}

is increasing on its domain, so solve

\mathrm{f}(x) = x

(x-3)^{2} + 1 = x

x^{2} - 7x + 10 = 0

, so

(x - 2)(x - 5) = 0

3Only

x = 5

is in the domain

x \ge 3

Answer

x = 5

Tip — When you take a square root in an inverse, the domain restriction decides the sign. State the reason for your choice in words.

Think like an examiner

Justify existence by stating that the function is one-to-one on its domain.
Always give the domain of $\mathrm{f}^{-1}$ — it is the range of $\mathrm{f}$ .
Choose the sign of a square root using the original domain, and say so.
Sketch inverse graphs as reflections in $y = x$ , with intercepts and asymptotes swapped.

Common misconceptions

Writing

\mathrm{f}^{-1}(x) = \dfrac{1}{\mathrm{f}(x)}

The

-1

denotes an inverse function, not a reciprocal.

Giving the domain of

\mathrm{f}^{-1}

as the domain of

\mathrm{f}

The domain of

\mathrm{f}^{-1}

is the range of

\mathrm{f}

Keeping

\pm

in an inverse involving a square root.

Choose the one sign consistent with the restricted domain.

Reflecting in the

y

-axis instead of

y = x

Swapping coordinates is a reflection in

y = x

Inverse functions

\mathrm{f}^{-1}\mathrm{f}(x) = \mathrm{ff}^{-1}(x) = x

Defining property.

\text{domain of } \mathrm{f}^{-1} = \text{range of } \mathrm{f}

Sets swap.

(a, b) \mapsto (b, a)

Reflection in

y = x

Stretch yourself

The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = \dfrac{ax + 1}{x - a}$ , $x \neq a$ , where $a$ is a constant. Show that $\mathrm{f}$ is self-inverse — that is, $\mathrm{f}^{-1}(x) = \mathrm{f}(x)$ — and explain what this means for its graph.

Hint — Rearrange $y = \dfrac{ax + 1}{x - a}$ for $x$ in the usual way and compare the result with $\mathrm{f}$ .

Reveal solution

SolutionLet

y = \dfrac{ax+1}{x-a}

. Then

xy - ay = ax + 1

, so

xy - ax = ay + 1

and

x(y - a) = ay + 1

, giving

x = \dfrac{ay + 1}{y - a}

. Swapping letters,

\mathrm{f}^{-1}(x) = \dfrac{ax + 1}{x - a} = \mathrm{f}(x)

. Equivalently,

\mathrm{ff}(x) = x

. Graphically, reflecting

y = \mathrm{f}(x)

y = x

gives the same curve, so the graph is symmetric about

y = x

: its vertical asymptote

x = a

and horizontal asymptote

y = a

are reflections of each other, which is consistent with that symmetry. For example, with

a = 2

\mathrm{f}(3) = 7

and

\mathrm{f}(7) = \dfrac{15}{5} = 3

✓.

\blacksquare

Questions students ask

Why does a many-to-one function have no inverse?

Two inputs share an output, so reversing the mapping would send that output to two places — which is not allowed for a function.

How do I find the domain of the inverse?

Find the range of the original function. That set is exactly the domain of the inverse.

\sin^{-1} x

the same as

\dfrac{1}{\sin x}

No.

\sin^{-1}x

is the inverse of sine on a restricted domain;

\dfrac{1}{\sin x}

\operatorname{cosec} x

Where do a function and its inverse intersect?

For an increasing function, always on the line

y = x

, so solve

\mathrm{f}(x) = x

. Decreasing functions can also meet off that line.

Key takeaways

An inverse exists only for a one-to-one function.
Find $\mathrm{f}^{-1}$ by rearranging $y = \mathrm{f}(x)$ for $x$ and swapping letters.
Domain and range swap between $\mathrm{f}$ and $\mathrm{f}^{-1}$ .
The graph of $\mathrm{f}^{-1}$ is the reflection of $\mathrm{f}$ in $y = x$ .
Restricting the domain can make a function one-to-one and invertible.