Inverse functions

An inverse function of a function $f$ is one that reverses the operation of $f$ and is usually denoted as $f^{-1}$

Definition

Let $f$ be some function with domain $\mathbf{X}$ with codomain $\mathbf{Y}$. If there exists a function $g$ whose domain is $\mathbf{Y}$ and codomain is $\mathbf{X}$ then $f$ is an invertible function where $g$ is the inverse function.

\begin{align} y &= f(g(y)) & \text{ for all } , y \in \mathbf{Y} \ x &= g(f(x)) = f^{-1}(y) &\text{ for all } , x \in \mathbf{X} \ \end{align}

We show below a list of common functions and their inverses. For each function $y=f(x)$ we show its inverse $x=f^{-1}(y)$

Inverse of Addition function

$y=f(x)=x+a$ $\text{Inverse function is subtraction}$ $x=f^{-1}(y)=y-a$

Inverse of Subtraction function

$y=f(x)=x-a$

$\text{Inverse function is addition}$ $x=f^{-1}(y)=y+a$

Inverse of Multiplication function

$y=f(x)=mx$

$\text{Inverse function is division}$ $x=f^{-1}(y)=\frac{y}{m}$ where $m\ne 0$

Inverse of Division function

$y=f(x)=\frac{x}{a}$

$\text{Inverse function is multiplication}$ $x=f^{-1}(y)=y\cdot a$ where $m\ne 0$

Inverse of Reciprocal function

$y=f(x)=\frac{1}{x}\equiv x^{-1}$

$\text{Inverse function is itself}$ $x=f^{-1}(y)=\frac{1}{y}\equiv y^{-1}$ where $x,y\ne 0$

Inverse of Power function

$y=f(x)=x^n$

$\text{Inverse function is the}$ [[Root|${\mathbf{n}}^{\mathbf{th}}\text{root}$]] $x=f^{-1}(y)=\sqrt[n]{y}=y^{1/n}$ where:

\begin{align} n & \in \mathbb{N} \ y & \in \left{ \begin{aligned} &\ \mathbb{R+} \text{ if } n \text{ is even} \ &\ \mathbb{R} \text{ if } n \text{ is odd } \end{aligned} \right. \end{align}

basically we restrict the domain and codomain to only positive numbers $\mathbf{R+}$ otherwise each element in the domain would correspond to two elements in the codomain since a square root has +ve and -ve solutions.

Inverse of Exponential function

$y=f(x)=b^x$

$\text{Inverse function is}$ [[Logarithm|$\text{logarithm}$]] $x=f^{-1}(y)={\log }_{b}y$ where:

\begin{align} y & \in \mathbb{R} \ b & \in \mathbb{R+}; \ b \neq 1 \end{align}

Notice since Exponentiation is not commutative ($x^y\ne y^x$), we have two different inverse functions.