Inverse Functions

Undoing what a function does

Start with a rule simple enough to see straight through: $f(x) = x + 3$, the rule “add three.” Feed it $5$ and it returns $8$; feed it $10$ and it returns $13$. To send those outputs back where they came from, you do the obvious thing and subtract three: $8$ goes back to $5$, and $13$ goes back to $10$. The rule “subtract three,” written $g(x) = x - 3$, reverses everything “add three” did. That reversing rule is the inverse of $f$.

Look at what happens to a single input-output pair. The function $f$ ties the input $5$ to the output $8$, a pairing we can write as $(5, 8)$. Its inverse ties $8$ back to $5$, the pair $(8, 5)$. The inverse takes every pair of $f$ and swaps its two entries: whatever was the input becomes the output, and whatever was the output becomes the input. That one sentence, swap the input and the output, is the whole idea of an inverse, and everything else in this lesson follows from it.

The inverse of $f$ has its own symbol, $f^{-1}$, read aloud as ”$f$ inverse.” It is defined to undo $f$, and “undo” can be stated precisely in two directions. If you run $f$ and then run $f^{-1}$, you return to the input you started with:

Reading from the inside out, $f$ turns $x$ into $f(x)$, and $f^{-1}$ turns that back into $x$. The reverse order must work too. If you run $f^{-1}$ first and then $f$, you again come back to the start:

Both equations say the same thing from opposite ends: $f$ and $f^{-1}$ cancel each other, whichever one you apply first. For $f(x) = x + 3$ and $f^{-1}(x) = x - 3$, check both directions:

Adding three and then subtracting three lands you back where you began, and so does subtracting first and then adding.

The superscript that is not a reciprocal

The symbol $f^{-1}$ carries a trap, and it is worth defusing before you use it. The raised $-1$ everywhere else in algebra makes a reciprocal, so $f^{-1}(x)$ looks like it should mean $\dfrac{1}{f(x)}$. It does not. Here $f^{-1}$ is the name of the inverse function, and it has nothing to do with dividing $1$ by anything.

Why reuse a symbol that invites the confusion? The choice is a deliberate analogy, and understanding it keeps the two ideas apart for good. For a nonzero number, its multiplicative inverse (its reciprocal) is the number that multiplies with it to give $1$: the reciprocal of $5$ is $\dfrac{1}{5}$, and $5 \cdot \dfrac{1}{5} = 1$. The number $1$ is special for multiplication because multiplying by it changes nothing. Functions tell the same story with composition in place of multiplication. The inverse $f^{-1}$ is the function that composes with $f$ to give the rule that changes nothing, the identity that returns its input unchanged. That requirement is exactly the statement $f^{-1}(f(x)) = x$. So the raised $-1$ marks “the inverse under the operation in play,” multiplication for a number, composition for a function. Because the two operations are different, the two inverses are different objects.

Make the difference concrete. For $f(x) = x + 3$, the inverse function is $f^{-1}(x) = x - 3$, while the reciprocal is $\dfrac{1}{f(x)} = \dfrac{1}{x + 3}$. These are not close. At $x = 5$,

One returns the input that produced the output $8$; the other is a small fraction with nothing to do with undoing $f$. Whenever you see $f^{-1}$, read “the function that reverses $f$,” never “one over $f$.”

Finding an inverse: swap, then solve

Guessing the inverse works for “add three,” but you need a method that works every time. The swap principle hands you one directly. The pairs of $f$ are (input, output), and the inverse’s pairs are (output, input). So to build the inverse you interchange the input and the output and then rearrange. In symbols, write $y = f(x)$, swap $x$ and $y$, and solve the new equation for $y$. The recipe is not a ritual; each step is the input-output swap written in algebra.

Watch it work on $f(x) = 2x + 3$. Write the output as $y$:

Here $x$ is the input and $y$ is the output. Swap them, so that the letter standing for the output now plays the role of the input we feed in. The letter standing for the input is then the value we want back:

Solve this for $y$ by subtracting three and dividing by two:

Rename the result, and the inverse is $f^{-1}(x) = \dfrac{x - 3}{2}$. The rule “double, then add three” is undone by “subtract three, then halve,” with the steps reversed and each one turned around, exactly as an undo should be.

Domain and range trade places

The forbidden-input swap you just saw is one instance of a rule that always holds: an inverse turns the domain and range of $f$ around. It follows straight from swapping input and output. Every pair of $f$ has the form (input from the domain, output from the range). The inverse swaps each pair into (output, input). So the numbers that were outputs are now the inputs of $f^{-1}$, and the numbers that were inputs are now its outputs. Therefore

• the domain of $f^{-1}$ is the range of $f$ (the old outputs are the new inputs), and
• the range of $f^{-1}$ is the domain of $f$ (the old inputs are the new outputs).

Return to $f(x) = \dfrac{1}{x - 3}$ from the last example. Its domain is every number except $3$, and its range is every number except $0$, because the reciprocal of a nonzero quantity is never $0$. Its inverse $f^{-1}(x) = \dfrac{1}{x} + 3$ has domain every number except $0$ and range every number except $3$. Line them up and the trade is exact: the domain of one is the range of the other, both ways.

When a function has no inverse

Every example so far has had a clean inverse, but some functions have none, and it is important to see why. The trouble starts when two different inputs produce the same output. Consider the squaring rule on all numbers, $f(x) = x^2$. It sends both $3$ and $-3$ to $9$:

Now ask the inverse to do its job. It must send the output $9$ back to the input it came from, but here $9$ came from two inputs. Should $f^{-1}(9)$ be $3$ or $-3$? There is no honest answer. A function is allowed only one output per input, and this would force the single input $9$ to have two outputs. So squaring on all numbers has no inverse function.

The property that avoids this trap has a name. A function is one-to-one when different inputs always give different outputs, that is, when no two inputs share an output. Squaring fails to be one-to-one because $3$ and $-3$ collide. A line like $f(x) = 4x - 7$ is one-to-one, because two different inputs, stretched and shifted the same way, stay different.

The horizontal line test

One-to-one has a picture, and it is the mirror image of a test you already know. In the lesson on functions you met the vertical line test: a graph represents a function when no vertical line crosses it more than once. The reason is that a vertical line gathers all the points with a single input, and a function may not give that input two outputs. One-to-one asks the reverse question, so it uses the reverse line.

A horizontal line gathers all the points that share a single output, since every point on the line $y = c$ sits at the same height $c$. If some horizontal line meets the graph in two points, those two points have the same output but different inputs, and the function is not one-to-one. So a graph belongs to a one-to-one function exactly when no horizontal line crosses it more than once. The vertical line test checks “one output per input,” which makes the graph a function. The horizontal line test checks “one input per output,” which makes that function invertible.

The rising line on the left passes: every height is reached exactly once, so it is one-to-one and has an inverse. The parabola $y = x^2$ on the right fails: the horizontal line hits it twice, at $x = 3$ and $x = -3$ for the height $9$, the same collision that blocked its inverse. (Swapping every point $(a, b)$ for $(b, a)$ also reflects a graph across the line $y = x$. That reflection is how the inverse is drawn in a later lesson; here the only question is whether the inverse exists.)

A function that fails the test can often be repaired by shrinking its domain until it is one-to-one. Squaring collides only because it accepts both a positive and a negative input for each output. Throw away the negatives, keeping $f(x) = x^2$ for $x \ge 0$, and every output now comes from a single nonnegative input. On that restricted domain squaring is one-to-one, and it has the inverse you already met, the square root.

Verifying an inverse by composition

You now have a way to find an inverse. But how do you check that a proposed answer is right, or test whether two functions handed to you are really inverses? The definition already said it: $f$ and $g$ are inverses exactly when they undo each other in both directions,

These are compositions, the operation from the last lesson, so verifying an inverse is just building the two composite functions and checking that each one collapses to $x$. Both are required, and the reason goes back to one-to-one. A single direction can hold while the other fails, and when it does the two functions are not inverses.

Running a process backward

An inverse earns its keep whenever a function models a process and you need to run that process in reverse. If a rule turns a starting value into a result, its inverse turns the result back into the starting value, which answers a question the forward rule cannot.

Common mistakes