# Inverse Function

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https://algebrica.org/inverse-function/
Fetched from algebrica.org post 14110; source modified 2025-11-22T19:07:43.

## What is an inverse function

In the introduction to [functions](../functions.md), we saw that a function \\( f: X \\to Y \\) is called bijective if it is both injective and surjective, that is, for every \\( y \\in Y \\), there exists a unique \\( x \\in X \\) such that \\( f(x) = y \\).

-   \\( X \\) is the domain.
-   \\( Y \\) is the codomain.
-   A function is called injective if for every \\( x\_1, x\_2 \\in X \\), with \\( x\_1 \\ne x\_2 \\), we have \\( f(x\_1) \\ne f(x\_2) \\). In other words, for every \\( y \\in Y \\), there exists at most one \\( x \\in X \\) such that \\( f(x) = y \\).
-   A function is called surjective if for every \\( y \\in Y \\), there exists at least one \\( x \\in X \\) such that \\( f(x) = y \\).

* * *

A function \\( f : X \\to Y \\) is bijective if and only if there exists a function \\( g : Y \\to X \\) such that:

-   \\( (g \\circ f)(x) = g(f(x)) = x \\) for every \\( x \\in X \\)
-   \\( (f \\circ g)(y) = f(g(y)) = y \\) for every \\( y \\in Y \\)

In this case, the function \\( g \\) is unique and is called the **inverse function** of \\( f \\), denoted by:

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

##### \\( (g \\circ f)(x) = g(f(x)) \\) is called the composite function, which means applying \\( f \\) to \\( x \\) first, and then applying \\( g \\) to the result.

## Making a function invertible by restricting its domain

Consider the function \\( f(x) = x^2 \\), defined on \\( \\mathbb{R} \\). This is a quadratic function, represented by a [parabola](../parabola.md) with its vertex at the origin of the Cartesian plane. On its full [domain](../determining-the-domain-of-a-function.md) \\( \\mathbb{R} \\), the function is not invertible, since it is not injective: distinct inputs can produce the same output, for example \\( f(-2) = f(2) \\).

However, if we restrict the domain to \\( \[0, +\\infty) \\), the function becomes bijective and therefore invertible. In this case, the inverse function is:

\\[
f(x) = x^2 \\rightarrow f^{-1}(x) = \\sqrt{x} \\quad \\text{for} \\; x \\geq 0
\\]

![](https://algebrica.org/wp-content/uploads/resources/images/inverse-function-1.png)

The graph of a function and that of its inverse are symmetric with respect to the line \\(y = x\\), which is the diagonal bisecting the first and third quadrants of the Cartesian plane.

If a function \\( f \\) is [composed](../composite-functions.md) with its inverse \\( f^{-1} \\), the result is the **identity function**, which maps each element of a set to itself:

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

## How to find the inverse of a general function

-   Check whether the function is bijective, or determine a restriction of its domain that makes it bijective.

-   Replace \\( f(x) \\) with \\( y \\), so that you work with the equation \\( y = f(x) \\).

-   Swap the variables \\( x \\) and \\( y \\): write \\( x = f(y) \\). This reflects the idea of inverting input and output.

-   Solve the equation for \\( y \\), isolating it explicitly.

-   Rewrite the result as \\( f^{-1}(x) = \\ldots \\), using \\( x \\) as the input variable for the inverse.


## Example

We want to find its inverse of the function \\( f(x) = \\dfrac{2x - 1}{x + 3} \\).

##### The function \\( f \\) is bijective on its domain \\( \\mathbb{R} \\setminus {-3} \\), because it is strictly increasing: its derivative is always positive. This ensures that \\( f \\) is injective, and since the image of \\( f \\) covers all real numbers except a single point, it is also surjective onto its codomain.

* * *

Write the function as an equation:

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

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

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

* * *

Solve for \\( y \\). Multiply both sides by \\( y + 3 \\):

\\[
x(y + 3) = 2y - 1
\\]

Distribute the left-hand side:

\\[
xy + 3x = 2y - 1
\\]

* * *

Bring all terms to one side and factor \\( y \\) on the left-hand side:

\\[
\\begin{align} &xy - 2y = -1 - 3x\\[0.5em] &y(x - 2) = -1 - 3x \\end{align}
\\]

Solve for \\( y \\):

\\[
y = \\dfrac{-1 - 3x}{x - 2}
\\]

The inverse function is:

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

## Inverse function theorem

A useful result from basic analysis is the one–dimensional version of the inverse function theorem. The idea is quite intuitive: if a function behaves regularly on an interval, then it can be inverted without difficulty. More precisely, suppose a function \\(f\\) is continuous and differentiable on an interval \\(I\\), and its [derivative](../derivatives.md) never vanishes:

\\[
f’(x) \\neq 0 \\quad \\forall \\, x \\in I
\\]

Under these conditions, the function is strictly monotonic on \\(I\\), which guarantees that it is invertible on that interval. As a consequence, an inverse function \\(f^{-1}\\) exists on \\(f(I)\\). This inverse is not only continuous but also differentiable, and its derivative is given by the relation:

\\[
\\bigl(f^{-1}\\bigr)'(y) = \\frac{1}{f’\\!\\bigl(f^{-1}(y)\\bigr)}
\\]

This result shows how a local condition (the derivative never becomes zero) ensures a global property such as invertibility.