Inverse Functions

The inverse of a function undoes what the function does: it maps each output back to the original input. Inverse functions exist precisely when a function is one-to-one, and their graphs are reflections of each other across the line $y = x$ .

Quick Reference

Concept	Formula / Rule
Definition	$x = f^{- 1} (y) \Leftrightarrow y = f (x)$
Domain/range swap	$Dom (f) = Rng (f^{- 1})$ ; $Rng (f) = Dom (f^{- 1})$
Cancellation equations	$f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$
When an inverse exists	If and only if $f$ is one-to-one
Graph reflection	Graph of $f^{- 1}$ is the reflection of graph of $f$ across $y = x$
Warning	$f^{- 1} (y) \neq \frac{1}{f (y)}$

The Idea of Inversion

Consider $f (x) = x^{3}$ . Given $x = 2$ , we get $f (2) = 8$ . Now suppose we are told that $f (x) = 27$ and asked to find $x$ . We solve $x^{3} = 27$ , giving $x = \sqrt[3]{27} = 3$ .

In general, for any $y$ in the range of $f$ , the value of $x$ satisfying $y = f (x)$ is given by $x = \sqrt[3]{y}$ . This defines $x$ as a function of $y$ , which we call $g (y) = \sqrt[3]{y}$ . The function $g$ is the inverse of $f$ , since it undoes the effect of $f$ :

g (f (x)) = \sqrt[3]{x^{3}} = x and f (g (y)) = {(\sqrt[3]{y})}^{3} = y .

Definition

Suppose $f$ is a one-to-one function with domain $A$ and range $B$ . The inverse function $f^{- 1}$ is defined by

x = f^{- 1} (y) means y = f (x)

for every $y$ in $B$ .

Diagram showing the relationship between a function f and its inverse g, reversing the mapping between x and y. — The function $f$ and its inverse $f^{- 1}$ undo the effects of each other.

The domain and range of $f$ and $f^{- 1}$ simply swap:

Dom (f) = Rng (f^{- 1}), Rng (f) = Dom (f^{- 1}) .

Important notation warning: The $- 1$ in $f^{- 1}$ denotes an inverse function, not an exponent. In particular, $f^{- 1} (y) \neq \frac{1}{f (y)}$ .

Given that $f$ has an inverse and $f (1) = 3$ , $f (2) = - 4$ , and $f (5) = - 1$ , find $f^{- 1} (3)$ , $f^{- 1} (- 4)$ , and $f^{- 1} (- 1)$ .

Solution

By the definition of inverse function:

f^{- 1} (3) = 1 because f (1) = 3,

f^{- 1} (- 4) = 2 because f (2) = - 4,

f^{- 1} (- 1) = 5 because f (5) = - 1.

Arrow diagram showing specific points mapping from domain to range via f. — Arrow diagram for $f$ : maps 1 to 3, 2 to −4, 5 to −1.

Arrow diagram showing specific points mapping from range to domain via f⁻¹. — Arrow diagram for $f^{- 1}$ : the mapping is exactly reversed.

Verifying an Inverse Function

Let $f$ be a one-to-one function with domain $A$ and range $B$ , and let $g$ be a function with domain $B$ and range $A$ . Then $g = f^{- 1}$ if and only if both of the following hold:

g (f (x)) = x for every x \in A,

f (g (y)) = y for every y \in B .

Proof

Suppose

g = f^{- 1}

. By definition,

x = g (y) \Leftrightarrow y = f (x)

. Substituting

f (x)

for

y

gives

x = g (f (x))

. Substituting

g (y)

for

x

gives

y = f (g (y))

. Both conditions follow. Conversely, suppose (1) and (2) hold. If

x = g (y)

, then by (2),

f (g (y)) = y

, i.e.,

f (x) = y

. If

y = f (x)

, then by (1),

g (f (x)) = x

, i.e.,

g (y) = x

. So

x = g (y) \Leftrightarrow y = f (x)

, which is the definition of

g = f^{- 1}

Writing the independent variable of $f^{- 1}$ as $x$ (rather than $y$ ), the cancellation equations become:

f^{- 1} (f (x)) = x for every x in the domain of f,

f (f^{- 1} (x)) = x for every x in the domain of f^{- 1} .

A function has an inverse function if and only if it is one-to-one. Equivalently, every increasing or decreasing (monotonic) function has an inverse function.

Verify that $f (x) = 5 x^{3} + 2$ and $g (x) = \sqrt[3]{\frac{x - 2}{5}}$ are inverses of each other.

Solution

Both functions have domain

ℝ

. We verify both cancellation equations:

f (g (x)) = f (\sqrt[3]{\frac{x - 2}{5}}) = 5 {(\sqrt[3]{\frac{x - 2}{5}})}^{3} + 2 = 5 \cdot \frac{x - 2}{5} + 2 = x . ✓

g (f (x)) = g (5 x^{3} + 2) = \sqrt[3]{\frac{(5 x^{3} + 2) - 2}{5}} = \sqrt[3]{\frac{5 x^{3}}{5}} = \sqrt[3]{x^{3}} = x . ✓

Determine if each function has an inverse. (a) $f (x) = x^{3} + 1$ . (b) $h (x) = x^{3} - x$ .

Solution

(a) We show

f

is one-to-one. If

f (x_{1}) = f (x_{2})

x_{1}^{3} + 1 = x_{2}^{3} + 1 ⟹ x_{1}^{3} = x_{2}^{3} ⟹ x_{1} = x_{2} .

f

is one-to-one and has an inverse. The graph of

f

is an upward shift of

y = x^{3}

and passes the horizontal line test.

Graph of y = x³ + 1 which is one-to-one. — $f (x) = x^{3} + 1$ passes the horizontal line test.

(b) Note that

h (x) = x (x^{2} - 1)

, so

h (- 1) = h (0) = h (1) = 0

. Three distinct inputs give the same output, so

h

is not one-to-one and has no inverse.

Graph of y = x³ - x which is not one-to-one. — $h (x) = x^{3} - x$ fails the horizontal line test.

How to Find the Inverse Function

Steps to find $f^{- 1}$ :

Write down the equation $y = f (x)$ .
Solve for $x$ in terms of $y$ : $x = g (y)$ .
The formula $f^{- 1} (y) = g (y)$ defines the inverse, with $y$ restricted to the range of $f$ .
If you prefer, replace every $y$ with $x$ to write $f^{- 1} (x) = g (x)$ .

Given $f (x) = 2 x^{5} + 3$ , find its inverse function.

Solution

Set

y = f (x)

and solve for

x

2 x^{5} + 3 = y ⟹ 2 x^{5} = y - 3 ⟹ x^{5} = \frac{y - 3}{2} ⟹ x = {(\frac{y - 3}{2})}^{1 / 5} .

Therefore

f^{- 1} (x) = {(\frac{x - 3}{2})}^{1 / 5}

Given $g (x) = 2 x - 1$ , find $g^{- 1}$ .

Solution

Set

y = 2 x - 1

and solve:

x = \frac{y + 1}{2}

. Replacing

y

with

x

g^{- 1} (x) = \frac{x + 1}{2} .

Verification:

g (g^{- 1} (x)) = 2 \cdot \frac{x + 1}{2} - 1 = x

. ✓

Given $h (x) = \sqrt{2 x + 3}$ , find $h^{- 1}$ .

Solution

Set

y = \sqrt{2 x + 3}

and solve:

y^{2} = 2 x + 3

, so

x = \frac{y^{2} - 3}{2}

. The range of

h

[0, \infty)

, so:

h^{- 1} (x) = \frac{x^{2} - 3}{2}, x \geq 0.

Given $u (x) = x^{2} + 1$ with domain $x \leq 0$ , find $u^{- 1}$ .

Solution

Set

y = x^{2} + 1

and solve:

x = \pm \sqrt{y - 1}

. Since

x \leq 0

, we choose

x = - \sqrt{y - 1}

. The range is

[1, \infty)

u^{- 1} (x) = - \sqrt{x - 1}, x \geq 1.

Graph of the Inverse Function

If $(a, b)$ is a point on the graph of $f$ , then $(b, a)$ is on the graph of $f^{- 1}$ . The point $(b, a)$ is the reflection of $(a, b)$ across the line $y = x$ .

The graphs of a function and its inverse are reflections of one another across the line $y = x$ .

Graph of the polynomial function f(x)=2x⁵ + 3 and its inverse reflected across the line y = x. — (a) $f (x) = 2 x^{5} + 3$ and $f^{- 1} (x) = {(\frac{x - 3}{2})}^{1 / 5}$ , reflected across $y = x$ .

Graph of the linear function g(x) = 2x - 1 and its inverse reflected across the line y = x. — (b) $g (x) = 2 x - 1$ and $g^{- 1} (x) = \frac{x + 1}{2}$ , reflected across $y = x$ .

Graph of the function h(x)=√(2x+3) and its inverse reflected across the line y = x. — (c) $h (x) = \sqrt{2 x + 3}$ and $h^{- 1} (x) = \frac{1}{2} (x^{2} - 3)$ , $x \geq 0$ , reflected across $y = x$ .

Given the graph of $f (x)$ below, sketch the graph of $f^{- 1} (x)$ .

Graph of a function with specific points labeled (2, 4), (3, 2), (4, 1), and (5, 0.5). — Graph of $f$ with points $(2,4)$, $(3,2)$, $(4,1)$, $(5, 0.5)$ labeled.

Solution

Reflect across

y = x

by swapping coordinates:

(2, 4) \to (4, 2)

(3, 2) \to (2, 3)

(4, 1) \to (1, 4)

(5, 0.5) \to (0.5, 5)

Solution graph showing the original function and its inverse plotted with reflected points. — The graphs of $f$ and $f^{- 1}$ are reflections of one another across $y = x$ .

Frequently Asked Questions

f^{- 1} (x)

the same as

1 / f (x)

No. The notation

f^{- 1}

denotes the inverse function, not the reciprocal. For example, if

f (x) = 2 x + 1

, then

f^{- 1} (x) = (x - 1) / 2

, which is very different from

1 / (2 x + 1)

Does every function have an inverse?

No. A function has an inverse if and only if it is one-to-one. For instance,

f (x) = x^{2}

does not have an inverse on all of

ℝ

because

f (2) = f (- 2) = 4

What does it mean that

f

and

f^{- 1}

are inverses of each other?

The relationship is symmetric: if

g = f^{- 1}

, then

g^{- 1} = f

. In other words,

f

and

f^{- 1}

undo each other in both directions.

Why does the graph of f⁻¹ look like a reflection of f across y = x?

Because the roles of

x

and

y

are swapped:

(a, b)

is on the graph of

f

exactly when

(b, a)

is on the graph of

f^{- 1}

. Swapping coordinates geometrically corresponds to reflecting across the diagonal line

y = x

How does restricting the domain allow us to define an inverse?

If a function is not one-to-one on its full domain, we can select a subdomain on which it is one-to-one. For example,

u (x) = x^{2} + 1

restricted to

x \leq 0

is strictly decreasing and hence one-to-one, so it has an inverse on that restricted domain.

Quick Reference

The Idea of Inversion

Definition

Verifying an Inverse Function

How to Find the Inverse Function

Graph of the Inverse Function

Frequently Asked Questions

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