The Central Idea

Consider

\[ f(x)=2x+3. \]

If

\[ f(4)=11, \]

then the inverse should reverse that assignment:

\[ f^{-1}(11)=4. \]

The original function sends

\[ 4\longmapsto11. \]

The inverse sends

\[ 11\longmapsto4. \]

An inverse function reverses the arrows of a function.

What Is an Inverse Function?

If

\[ f:A\to B \]

is bijective, then its inverse is a function

\[ \boxed{f^{-1}:B\to A} \]

such that

\[ f^{-1}(f(x))=x \]

for every \(x\in A\), and

\[ f(f^{-1}(y))=y \]

for every \(y\in B\).

The Inverse Reverses the Assignment

If

\[ f(a)=b, \]

then

\[ \boxed{f^{-1}(b)=a}. \]

This is the defining reversal.

Inverse Notation Does Not Mean Reciprocal

The notation

\[ f^{-1}(x) \]

does not mean

\[ \frac{1}{f(x)}. \]

For example, if

\[ f(x)=2x, \]

then

\[ f^{-1}(x)=\frac{x}{2}, \]

while

\[ \frac1{f(x)}=\frac1{2x}. \]

These are different functions.

Why Bijectivity Matters

For a full inverse function from codomain back to domain, the original function must be bijective.

That means it must be both:

Why Injectivity Is Necessary

Suppose two distinct inputs satisfy

\[ f(a)=f(b)=c. \]

If we reverse the assignment, the inverse would need to send

\[ c \]

to both \(a\) and \(b\).

That would give one input two outputs, so the inverse relation would not be a function.

Why Surjectivity Is Necessary

Suppose the codomain contains an element \(y\) that is never produced by \(f\).

Then \(f^{-1}(y)\) would have no value.

So a full inverse defined on the entire original codomain requires surjectivity.

Inverse Relation Versus Inverse Function

Every relation can have its ordered pairs reversed.

If the original relation contains

\[ (a,b), \]

the inverse relation contains

\[ (b,a). \]

But the inverse relation is a function only when the reversal still assigns exactly one output to each input.

Ordered-Pair Example

Suppose

\[ f=\{(1,4),(2,5),(3,6)\}. \]

Then

\[ f^{-1} = \{(4,1),(5,2),(6,3)\}. \]

This inverse relation is also a function because the original outputs were all distinct.

Noninvertible Ordered-Pair Example

Suppose

\[ f=\{(1,4),(2,4),(3,6)\}. \]

Then the inverse relation would contain

\[ (4,1) \]

and

\[ (4,2). \]

So the inverse relation is not a function.

Horizontal-Line Test

A real-valued function has an inverse function on its domain if it is one-to-one there.

Graphically, this is tested with the horizontal-line test.

If every horizontal line meets the graph at most once, the function is injective and can be reversed on its range.

Why the Horizontal-Line Test Works

A horizontal line corresponds to one output value.

If that line intersects the graph more than once, the same output came from several inputs.

Reversing those assignments would fail the function rule.

Domain and Range Swap

If

\[ f:A\to B \]

is bijective, then

\[ f^{-1}:B\to A. \]

Thus:

\[ \boxed{\operatorname{dom}(f^{-1})=\operatorname{range}(f)} \]

and

\[ \boxed{\operatorname{range}(f^{-1})=\operatorname{dom}(f)}. \]

Graph of an Inverse

If

\[ (a,b) \]

lies on the graph of \(f\), then

\[ (b,a) \]

lies on the graph of \(f^{-1}\).

So the graphs of a function and its inverse are reflections across the line

\[ \boxed{y=x}. \]

Why Reflection Across \(y=x\) Appears

Reversing an ordered pair swaps its coordinates:

\[ (x,y)\longmapsto(y,x). \]

Geometrically, swapping coordinates reflects a point across the line

\[ y=x. \]

Finding an Inverse Algebraically

For a function given by a formula, a common workflow is:

  1. write \(y=f(x)\);
  2. swap \(x\) and \(y\);
  3. solve for \(y\);
  4. rename the result \(f^{-1}(x)\);
  5. state the domain and range.

Linear Inverse Example

Let

\[ f(x)=3x-5. \]

Write

\[ y=3x-5. \]

Swap \(x\) and \(y\):

\[ x=3y-5. \]

Solve for \(y\):

\[ x+5=3y, \] \[ y=\frac{x+5}{3}. \]

Therefore,

\[ \boxed{f^{-1}(x)=\frac{x+5}{3}}. \]

Verify by Composition

For

\[ f(x)=3x-5 \]

and

\[ f^{-1}(x)=\frac{x+5}{3}, \]

check:

\[ f^{-1}(f(x)) = \frac{(3x-5)+5}{3} = x. \]

Also,

\[ f(f^{-1}(x)) = 3\left(\frac{x+5}{3}\right)-5 = x. \]

Both compositions return the identity.

Identity Function Connection

The identity function is

\[ I(x)=x. \]

For inverse functions,

\[ \boxed{f^{-1}\circ f=I} \]

and

\[ \boxed{f\circ f^{-1}=I}. \]

Why Composition Is the Strongest Check

Finding a formula that seems to reverse another function is not enough.

Composition verifies that the two functions actually undo each other on the correct domains.

The Problem with \(f(x)=x^2\)

Consider

\[ f(x)=x^2 \]

on

\[ \mathbb{R}. \]

The function is not injective because

\[ f(2)=4 \]

and

\[ f(-2)=4. \]

So it has no inverse function on all of \(\mathbb{R}\).

Algebra Reveals the Problem

Write

\[ y=x^2. \]

Swap:

\[ x=y^2. \]

Then

\[ y=\pm\sqrt{x}. \]

The symbol

\[ \pm \]

reveals two possible outputs for many inputs.

That is an inverse relation, not one inverse function.

Restricted Domains

To create an inverse function, we can restrict the original domain to a region where the function is one-to-one.

For

\[ f(x)=x^2, \]

a common restriction is

\[ \boxed{x\ge0}. \]

Square Function with Restricted Domain

Define

\[ f:[0,\infty)\to[0,\infty), \qquad f(x)=x^2. \]

This function is strictly increasing, injective, and surjective onto its codomain.

Therefore, it is bijective.

Inverse of the Restricted Square Function

Now the inverse is

\[ \boxed{f^{-1}(x)=\sqrt{x}}. \]

The principal square root corresponds precisely to the nonnegative domain restriction on the original square function.

A Different Restriction Gives a Different Inverse

Instead, restrict

\[ f(x)=x^2 \]

to

\[ (-\infty,0]. \]

Then the inverse becomes

\[ \boxed{f^{-1}(x)=-\sqrt{x}}. \]

The chosen branch reflects the chosen original domain.

Restricted Domain Is Part of the Function

The formulas

\[ f(x)=x^2,\qquad x\ge0 \]

and

\[ g(x)=x^2,\qquad x\le0 \]

are different functions because their domains differ.

They therefore have different inverses.

Strict Monotonicity and Inverses

A strictly increasing or strictly decreasing function is injective on that interval.

Therefore, strict monotonicity is a powerful way to identify intervals on which an inverse function can exist.

Quadratic Domain Restriction

Consider

\[ f(x)=(x-3)^2+2. \]

The vertex occurs at

\[ x=3. \]

The function is decreasing on

\[ (-\infty,3] \]

and increasing on

\[ [3,\infty). \]

Either interval can be used as a one-to-one restriction.

Inverse on the Right Branch

Restrict

\[ f(x)=(x-3)^2+2 \]

to

\[ x\ge3. \]

Write

\[ y=(x-3)^2+2. \]

Then

\[ y-2=(x-3)^2. \]

Because \(x\ge3\),

\[ x-3=\sqrt{y-2}. \]

So

\[ x=3+\sqrt{y-2}. \]

Therefore,

\[ \boxed{f^{-1}(x)=3+\sqrt{x-2}}. \]

Inverse on the Left Branch

If instead the original domain is

\[ x\le3, \]

then

\[ x-3=-\sqrt{y-2}. \]

So the inverse is

\[ \boxed{f^{-1}(x)=3-\sqrt{x-2}}. \]

The Original Range Becomes the Inverse Domain

For

\[ f(x)=(x-3)^2+2 \]

on either one-to-one branch, the range is

\[ [2,\infty). \]

Therefore, the inverse domain is

\[ \boxed{[2,\infty)}. \]

Absolute Value and Restricted Domains

The absolute-value function

\[ f(x)=|x| \]

is not injective on \(\mathbb{R}\) because

\[ f(a)=f(-a). \]

But on

\[ [0,\infty), \]

it becomes

\[ f(x)=x, \]

which is bijective onto

\[ [0,\infty). \]

Exponential and Logarithmic Functions

The exponential function

\[ f(x)=b^x, \qquad b>0,\quad b\ne1, \]

is strictly monotonic and has range

\[ (0,\infty). \]

Its inverse is the logarithmic function

\[ \boxed{f^{-1}(x)=\log_b x}. \]

Domain and Range Swap for Exponentials

For

\[ f(x)=b^x, \]

we have

\[ \operatorname{dom}(f)=\mathbb{R}, \] \[ \operatorname{range}(f)=(0,\infty). \]

Therefore,

\[ \operatorname{dom}(f^{-1})=(0,\infty), \] \[ \operatorname{range}(f^{-1})=\mathbb{R}. \]

Cube and Cube Root

Consider

\[ f(x)=x^3. \]

This function is strictly increasing on all real numbers and has range \(\mathbb{R}\).

Therefore, it is bijective from \(\mathbb{R}\) to \(\mathbb{R}\).

Its inverse is

\[ \boxed{f^{-1}(x)=\sqrt[3]{x}}. \]

Reciprocal Function

Consider

\[ f(x)=\frac1x, \qquad x\ne0. \]

Its range also excludes zero.

If

\[ f(x)=y, \]

then

\[ y=\frac1x \]

implies

\[ x=\frac1y. \]

So

\[ \boxed{f^{-1}(x)=\frac1x}. \]

The reciprocal function is its own inverse.

Self-Inverse Functions

A function satisfying

\[ f^{-1}=f \]

is called self-inverse or an involution.

Such a function satisfies

\[ f(f(x))=x. \]

Reflection Across \(y=x\) for Self-Inverse Functions

If a function is its own inverse, its graph is symmetric about

\[ y=x. \]

The reciprocal function is a familiar example.

Inverse of a Linear Function

For

\[ f(x)=ax+b, \qquad a\ne0, \]

solve

\[ y=ax+b. \]

Then

\[ x=ay+b. \]

So

\[ y=\frac{x-b}{a}. \]

Therefore,

\[ \boxed{f^{-1}(x)=\frac{x-b}{a}}. \]

Constant Functions Have No Inverse Function

If

\[ f(x)=c, \]

then every input shares the same output.

Such a function is never injective when the domain has more than one element.

So it cannot have an inverse function on that domain.

Piecewise Functions Can Be Invertible

A piecewise function may have an inverse if the entire function is one-to-one and reaches its intended codomain.

Its inverse may itself require piecewise notation.

Piecewise Inverse Example

Suppose

\[ f(x)= \begin{cases} x-2, & x<0,\\ x+2, & x\ge0. \end{cases} \]

The first branch has range

\[ (-\infty,-2), \]

and the second has range

\[ [2,\infty). \]

These ranges do not overlap, so the function is injective.

Inverse of the Piecewise Example

Reverse each branch:

\[ y=x-2 \Rightarrow x=y+2, \]

and

\[ y=x+2 \Rightarrow x=y-2. \]

Thus,

\[ f^{-1}(x)= \begin{cases} x+2, & x<-2,\\ x-2, & x\ge2. \end{cases} \]

The inverse domain is the original range.

Inverse Functions and Composition

If \(f\) and \(g\) are invertible, then

\[ (g\circ f)^{-1} = f^{-1}\circ g^{-1}. \]

The order reverses.

Why the Order Reverses

If the original process applies

\[ f \]

first and then

\[ g, \]

the inverse process must undo \(g\) first and then undo \(f\).

This mirrors the reverse-order logic used in solving equations.

Inverse Functions and Solving Equations

Suppose

\[ f(x)=y. \]

If \(f\) is invertible, apply \(f^{-1}\) to both sides:

\[ f^{-1}(f(x)) = f^{-1}(y). \]

Then

\[ \boxed{x=f^{-1}(y)}. \]

Inverse functions provide a general framework for undoing operations.

Inverse Operations and Inverse Functions

Solving

\[ 2x+3=11 \]

uses subtraction and division to undo the original steps.

The inverse function

\[ f^{-1}(x)=\frac{x-3}{2} \]

packages those reverse operations into one function.

Inverse Functions Are Directional

The function

\[ f:A\to B \]

and its inverse

\[ f^{-1}:B\to A \]

reverse roles:

Codomain Choice Still Matters

Suppose

\[ f:\mathbb{R}\to\mathbb{R}, \qquad f(x)=e^x. \]

The function is injective but not surjective onto \(\mathbb{R}\).

So it is not bijective under that codomain declaration.

Choose the Natural Codomain for Invertibility

Instead define

\[ f:\mathbb{R}\to(0,\infty), \qquad f(x)=e^x. \]

Now the range equals the codomain.

The function is bijective and has inverse

\[ \ln x. \]

Inverse Function Versus Solving for \(x\)

Solving one equation such as

\[ f(x)=7 \]

asks for preimages of one output.

Finding \(f^{-1}\) constructs a new function that solves the entire family of equations

\[ f(x)=y \]

at once.

Inverse Functions and Fixed Points

If

\[ f(a)=a, \]

then \(a\) is a fixed point.

For an invertible function,

\[ f^{-1}(a)=a \]

as well.

Fixed points remain fixed under both directions of the mapping.

Inverse Graphs Share Fixed Points

A function and its inverse are reflected across

\[ y=x. \]

Points lying on that reflection line remain unchanged.

So fixed points are shared by a function and its inverse.

Parametric View of Inverses

If a graph is parametrized by

\[ x=t, \qquad y=f(t), \]

then the inverse relation swaps the coordinates:

\[ x=f(t), \qquad y=t. \]

This gives another way to see the reflection across \(y=x\).

Worked Example: Find and Verify an Inverse

Let

\[ f(x)=5x+2. \]

Write

\[ y=5x+2. \]

Swap:

\[ x=5y+2. \]

Solve:

\[ x-2=5y, \] \[ y=\frac{x-2}{5}. \]

Thus,

\[ \boxed{f^{-1}(x)=\frac{x-2}{5}}. \]

Check:

\[ f^{-1}(f(x)) = \frac{5x+2-2}{5} = x. \]

Worked Example: Restricted Quadratic

Let

\[ f(x)=x^2+1 \]

with domain

\[ x\ge0. \]

Then

\[ y=x^2+1. \]

Swap:

\[ x=y^2+1. \]

Solve:

\[ y^2=x-1, \] \[ y=\sqrt{x-1}. \]

Therefore,

\[ \boxed{f^{-1}(x)=\sqrt{x-1}}. \]

The inverse domain is

\[ [1,\infty). \]

Worked Example: Determine Whether an Inverse Exists

Let

\[ f(x)=x^2-4 \]

on \(\mathbb{R}\).

Because

\[ f(2)=0 \]

and

\[ f(-2)=0, \]

the function is not injective.

So it has no inverse function on all real numbers.

Worked Example: Choose a Restricted Domain

For

\[ f(x)=x^2-4, \]

restrict to

\[ x\ge0. \]

Then

\[ y=x^2-4. \]

Swap:

\[ x=y^2-4. \]

So

\[ y^2=x+4. \]

Using the nonnegative branch,

\[ \boxed{f^{-1}(x)=\sqrt{x+4}}. \]

The inverse domain is

\[ [-4,\infty). \]

A Reliable Inverse-Function Workflow

  1. State the original domain and codomain.
  2. Check injectivity using algebra, monotonicity, or the horizontal-line test.
  3. Check whether the range matches the intended codomain.
  4. If necessary, restrict the domain to a one-to-one interval.
  5. Write \(y=f(x)\).
  6. Swap \(x\) and \(y\).
  7. Solve for \(y\).
  8. Rename the result \(f^{-1}(x)\).
  9. State the inverse domain and range.
  10. Verify by composition.

A Reliable Domain-Restriction Workflow

  1. Identify where the function fails to be one-to-one.
  2. Use turning points or monotonic intervals to locate one-to-one branches.
  3. Choose a branch appropriate to the problem or convention.
  4. Record the restricted domain explicitly.
  5. Determine the corresponding range.
  6. Use that range as the domain of the inverse.

A Reliable Graph Workflow

  1. Apply the horizontal-line test.
  2. If the graph fails, identify a one-to-one branch.
  3. Restrict the domain to that branch.
  4. Reflect the restricted graph across \(y=x\).
  5. Check that original domain and range have swapped.

Common Errors

Confusing \(f^{-1}(x)\) with \(1/f(x)\)

Inverse functions reverse mappings; reciprocals divide \(1\) by an output.

Finding an Algebraic Inverse Without Checking Injectivity

A formula that produces \(\pm\) branches signals that the original function may need a restricted domain.

Forgetting Domain and Range

The inverse domain is the original range, and the inverse range is the original domain.

Ignoring Codomain

A function may be injective but not bijective under an overly large codomain.

Choosing Both Square-Root Branches

An inverse function must return exactly one output. The original restricted domain determines which branch is correct.

Assuming Every Function Has an Inverse Function

Only one-to-one functions can be reversed as functions on their ranges.

Failing to Verify by Composition

The strongest check is that both compositions reduce to the identity on the appropriate domains.

Practice

1. Linear Inverse

Find the inverse of

\[ f(x)=4x+7. \]
Check Answer \[ y=4x+7. \]

Swap:

\[ x=4y+7. \]

Solve:

\[ y=\frac{x-7}{4}. \]

Therefore,

\[ \boxed{f^{-1}(x)=\frac{x-7}{4}}. \]

2. Does an Inverse Exist?

Does

\[ f(x)=x^2 \]

have an inverse function on \(\mathbb{R}\)?

Check Answer

No. It is not injective because

\[ f(a)=f(-a). \]

3. Restrict the Square Function

Restrict

\[ f(x)=x^2 \]

to \(x\ge0\). Find the inverse.

Check Answer \[ \boxed{f^{-1}(x)=\sqrt{x}}. \]

4. Domain and Range Swap

If

\[ \operatorname{dom}(f)=[2,\infty) \]

and

\[ \operatorname{range}(f)=(-1,\infty), \]

what are the domain and range of \(f^{-1}\)?

Check Answer \[ \operatorname{dom}(f^{-1})=(-1,\infty), \] \[ \operatorname{range}(f^{-1})=[2,\infty). \]

5. Composition Check

Let

\[ f(x)=2x-3 \]

and

\[ g(x)=\frac{x+3}{2}. \]

Compute \(g(f(x))\).

Check Answer \[ g(f(x)) = \frac{(2x-3)+3}{2} = x. \]

So \(g=f^{-1}\).

6. Quadratic Restriction

Let

\[ f(x)=(x-2)^2+1 \]

with domain \(x\ge2\). Find the inverse.

Check Answer \[ y=(x-2)^2+1. \] \[ y-1=(x-2)^2. \]

Because \(x\ge2\),

\[ x-2=\sqrt{y-1}. \]

So

\[ \boxed{f^{-1}(x)=2+\sqrt{x-1}}. \]

7. Reciprocal Function

What is the inverse of

\[ f(x)=\frac1x, \qquad x\ne0? \]
Check Answer \[ \boxed{f^{-1}(x)=\frac1x}. \]

The function is self-inverse.

8. Graphical Relationship

How are the graphs of \(f\) and \(f^{-1}\) related?

Check Answer

They are reflections across

\[ \boxed{y=x}. \]

Key Ideas

Where This Leads

Inverse functions complete the basic structural picture of functions by showing when an input-output relationship can be reversed without ambiguity.

These ideas lead directly into logarithms, inverse trigonometric functions, coordinate transformations, inverse matrices, solving equations with inverse operations, inverse modeling problems, implicit relations, change of variables, and higher mathematics where reversible mappings and structural equivalence are central.