Inverse Function

Reversing an Input and Output Pair

You often translate words or sentences from English to Indonesian, for example, when watching movies or reading news. This translation process is similar to how a function works: there is an input (English word) and an output (Indonesian word).

Real language translation can be ambiguous. Here we use a deliberately small dictionary in which each selected word has exactly one paired translation.

Here, the "Translation Machine" acts like a function that transforms "Mathematics" (input) into "Matematika" (output).

The reverse mapping sends "Matematika" back to "Mathematics". This mapping is an inverse function.

Definition of Inverse Function

An inverse function is a function that "reverses" the operation of an initial function. If function $f$ maps element $x$ from domain $A$ to element $y$ in codomain $B$, then its inverse function, denoted as $f^{-1}$ (read "f inverse"), maps element $y$ from $B$ back to element $x$ in $A$.

Mathematically:

In other words, if $f$ changes $x$ to $y$, then $f^{-1}$ changes $y$ back to $x$. The inverse function "undoes" the effect of the original function.

The notation $f^{-1}(x)$ denotes the inverse function. The multiplicative inverse or reciprocal is written $\frac{1}{f(x)}$.

Condition for a Function to Have an Inverse

For a function $f:A\to B$ to have a two-sided inverse $f^{-1}:B\to A$, the function $f$ must be bijective. A bijective function is both injective (one-to-one) and surjective (onto).

• Injective (One-to-one): Every distinct element in the domain maps to a distinct element in the codomain. No two different inputs produce the same output.
• Surjective (Onto): Every element in the codomain is the result of mapping from at least one element in the domain. All possible outputs occur.

The specified domain and codomain matter. If $f$ is injective but does not reach all of $B$, we can replace its codomain by its actual range. Writing that range as $\operatorname{Im}(f)$, the resulting function $f:A\to \operatorname{Im}(f)$ is bijective and therefore has an inverse $f^{-1}:\operatorname{Im}(f)\to A$. If $f$ is not injective, reversing its pairs sends one output to more than one input, so the reversed relation is not a function unless the original domain is restricted.

For example, $q(x)=x^2$ is not injective on $\mathbb{R}$ because $q(2)=q(-2)=4$. Restricting its domain and codomain gives the bijection

This function has the inverse $q^{-1}(x)=\sqrt{x}$.

Determining the Formula for an Inverse Function

To find the formula for the inverse function $f^{-1}(x)$ from a function $f(x)$, you can follow these steps:

1. Replace $f(x)$ with $y$.
2. Swap the positions of the variables $x$ and $y$ in the equation.
3. Solve the equation for $y$ in terms of $x$.
4. Replace $y$ with $f^{-1}(x)$ to get the inverse function formula.

Example:

Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x) = 2x + 3$. Find its inverse.

1. Replace $f(x)$ with $y$:

   $$y = 2x + 3$$

2. Swap $x$ and $y$:

   $$x = 2y + 3$$

3. Solve for $y$:

   $$x - 3 = 2y$$

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

4. Replace $y$ with $f^{-1}(x)$:

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

So, the inverse function of $f(x) = 2x + 3$ is $f^{-1}(x) = \frac{x - 3}{2}$.

Graph of a Function and Its Inverse

The graph of the inverse function $f^{-1}(x)$ is a reflection of the graph of the original function $f(x)$ across the line $y = x$.

For example, look at the graph of $f(x) = 2x + 3$, its inverse $f^{-1}(x) = \frac{x - 3}{2}$, and the identity line $y=x$.

Check Your Understanding

Let $r:\mathbb{R}\to\mathbb{R}$ be defined by $r(x)=3x-5$. Work through the three steps below for this function. Check both compositions at the end, because a correct inverse undoes the original in both orders.

1. Find $r^{-1}(x)$.
2. Verify both compositions $r^{-1}(r(x))$ and $r(r^{-1}(x))$.
3. State the domain and range of both functions.

Solution

Starting with $y=3x-5$, swap the variables and solve for $y$:

Now verify both directions:

Both $r$ and $r^{-1}$ have domain $\mathbb{R}$ and range $\mathbb{R}$. The two identities confirm that the functions undo each other on their complete domains.