What is an inverse function
In the introduction to functions, we saw that a function $f : X \rightarrow Y$ 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 $x_{1} \neq x_{2}$, we have $f ( x_{1} ) \neq f ( x_{2} )$. 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 $f : X \rightarrow Y$ is bijective if and only if there exists a function $g : Y \rightarrow X$ 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 with its vertex at the origin of the Cartesian plane. On its full domain $\mathbb{R}$, 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:
\[f ( x ) = x^{2} \rightarrow f^{- 1} ( x ) = \sqrt{x} \text{for} x \geq 0\]
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 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 $f ( x ) = \frac{2 x - 1}{x + 3}$.
The function $f$ is bijective on its domain $\mathbb{R} \backslash - 3$, 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: \(y = \frac{2 x - 1}{x + 3}\)
Swap $x$ and $y$ \(x = \frac{2 y - 1}{y + 3}\)
Solve for $y$. Multiply both sides by $y + 3$: \(x ( y + 3 ) = 2 y - 1\)
Distribute the left-hand side: \(x y + 3 x = 2 y - 1\)
Bring all terms to one side and factor $y$ on the left-hand side: \(& x y - 2 y = - 1 - 3 x \\ & y ( x - 2 ) = - 1 - 3 x\)
Solve for $y$: \(y = \frac{- 1 - 3 x}{x - 2}\)
The inverse function is: \(f^{- 1} ( x ) = \frac{- 1 - 3 x}{x - 2}\)
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 never vanishes:
\[f^{'} ( x ) \neq 0 \forall x \in I\]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:
\[( f^{- 1} )^{'} ( y ) = \frac{1}{f^{'} ( f^{- 1} ( y ) )}\]This result shows how a local condition (the derivative never becomes zero) ensures a global property such as invertibility.