Running a function backwards

Inverses from a table, from a graph, and by swapping x and y in the equation.

A currency converter turns pounds into euros. Standing at the other counter you want the opposite: given the euros, how many pounds was that? The second machine undoes the first, and it is built from the same information.

Start with a function stored as a table and reverse the lookup by hand.

Every pair got turned round. Where the original said 3 goes to 11, the reversed one says 11 goes to 3. Sending a value forwards and then backwards returns it unchanged.

The reversed function is called the inverse of the original, and it is written $f^{-1}$, said as f inverse. Its rule is exactly the promise you watched: it takes an output of $f$ and returns the input that produced it.

On a graph: reflect over the line y = x

If $(a, b)$ is on the graph of $f$, then $(b, a)$ is on the graph of $f^{-1}$. Swapping the two coordinates of a point is exactly what reflecting it over the line $y = x$ does. Run it and check the distances.

The midpoint of a point and its partner always has equal coordinates, so it lies on $y = x$. That is what makes the reflection exact rather than approximate: the line $y = x$ is the mirror.

From the equation: swap and solve

The algebra follows the same swap. Since the pairs $(a,b)$ turn into $(b,a)$, exchange $x$ and $y$ in the equation and then rearrange to get $y$ on its own.

One worked example in full. Find the inverse of $f(x) = 3x - 4$.

1. Write it with $y$: $y = 3x - 4$.
2. Swap the letters: $x = 3y - 4$.
3. Add 4 to both sides: $x + 4 = 3y$.
4. Divide both sides by 3: $y = \frac{x + 4}{3}$.
5. Write it in the proper notation: $f^{-1}(x) = \frac{x + 4}{3}$.

Test it on a number. $f(5) = 15 - 4 = 11$, and $f^{-1}(11) = \frac{11 + 4}{3} = \frac{15}{3} = 5$. The 5 came back.

There is a second way to think about the same work, which some people find faster: list what $f$ does to the input in order, then undo each step in the opposite order. For $3x - 4$ the steps are multiply by 3, then subtract 4. Undone backwards: add 4, then divide by 3. That is the rule derived above.

One more consequence of swapping the pairs, and it is examined often: the domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. The inputs and outputs have traded places, so their sets trade places too.

An example makes that concrete. If $f$ has domain $[0, \infty)$ and range $[2, \infty)$, then $f^{-1}$ has domain $[2, \infty)$ and range $[0, \infty)$. Handing $f^{-1}$ the number 1 is meaningless, because no input to $f$ ever produced 1 in the first place.

The undoing-the-steps view also explains why the steps come back in reverse. If getting dressed is socks and then shoes, undressing is shoes and then socks. Since $f$ multiplies by 3 and then subtracts 4, its inverse adds 4 and then divides by 3, and doing those two in the other order gives a different rule that does not undo anything.