Inverse Functions

Undoing a function

The function f(x) = 3x + 1 does two things to its input, in order: it multiplies by 3, then it adds 1. For example, f(3) = 3 × 3 + 1 = 10.

The inverse function of f, written $f^{-1}$ and read "f inverse", undoes f. It sends each output of f back to the input it came from: since f(3) = 10, $f^{-1}\left(10\right) = 3$. The −1 is part of the name, not a power: $f^{-1}\left(x\right)$ does not mean 1 divided by f(x).

Reverse the order, and reverse each step

Think of putting on socks and then shoes. To undo that, the shoes come off first and then the socks: the last thing done is the first thing undone. A function is undone the same way.

The last step of f was "add 1", so undo it first, by subtracting 1. The step before was "multiply by 3", so undo it next, by dividing by 3. Starting from 10: 10 − 1 = 9, then 9 ÷ 3 = 3, which is the input that f turned into 10. As a formula, $f^{-1}\left(x\right) = \frac{{x - 1}}{3}$.

Both changes are needed. Keeping the original order, dividing by 3 and then subtracting 1, turns 10 into $\frac{10}{3} - 1 = \frac{7}{3}$, not 3. Reversing the order without reversing the steps, adding 1 and then multiplying by 3, turns 10 into 33.

$f^{-1}$ undoes f = 3x + 1 backward: it subtracts 1 first, then divides by 3, and takes 10 back to 3.

The same steps, by algebra

The inverse can also be found by solving. Write y = 3x + 1 and make x the subject. Subtract 1 from both sides: y − 1 = 3x. Divide both sides by 3: $x = \frac{{y - 1}}{3}$. The algebra undoes the steps in the same reversed order, subtract 1 and then divide by 3.

This says which input x gives the output y. The inverse is a function in its own right, so write it with x as its input: $f^{-1}\left(x\right) = \frac{{x - 1}}{3}$.

Check by composing. $f^{-1}f\left(x\right) = f^{-1}\left(3x + 1\right) = \frac{{\left(3x + 1\right) - 1}}{3} = \frac{3x}{3} = x$, and ff⁻¹$\left(x\right) = f\left(\frac{{x - 1}}{3}\right) = 3 \times \frac{{x - 1}}{3} + 1 = x - 1 + 1 = x$. Each function undoes the other, whichever acts first.

The area of a circle comes from its radius in two steps: square r, then multiply by $\pi$. Drag the radius to change the circle, and choose the order of the undoing steps. $\surd$ first and then $\div \pi$ lands on a different number. $\div \pi$ first and then $\surd$, the reverse order, brings back r for every radius.

Why the graph is a reflection in y = x

Every point of the graph y = f(x) is an input with its output. f(1) = 3 × 1 + 1 = 4, so (1, 4) is on the graph of f. Then $f^{-1}\left(4\right) = 1$, so (4, 1) is on the graph of $f^{-1}$. The inverse swaps input and output, so every point (a, b) on the graph of f becomes the point (b, a) on the graph of $f^{-1}$.

Swapping the coordinates of a point reflects it in the line y = x. To reflect a point in a line is to move it straight across the line, at a right angle, to the same distance on the other side. Check that with (1, 4) and (4, 1). The segment joining them has gradient (1 − 4) ÷ (4 − 1) = −3 ÷ 3 = −1, and y = x has gradient 1. Since −1 × 1 = −1, the segment crosses y = x at a right angle. Its midpoint is ((1 + 4) ÷ 2, (4 + 1) ÷ 2) = (2.5, 2.5), which is on y = x. So (4, 1) is the same distance from the line as (1, 4), on the other side.

The same holds for every point, so the whole graph of $f^{-1}$ is the graph of f reflected in the line y = x. A point that lies on y = x is its own reflection, so wherever the graph of f crosses the line y = x, the graph of $f^{-1}$ crosses it at the same point.

The gold line through the origin is y = x. The steep gold line is y = 3x + 1 and the plain line is its inverse, $y = \frac{{x - 1}}{3}$. The marked points (1, 4) and (4, 1) are reflections of each other in y = x.

The gold curve is $y = x^{3}$ and the green dashed curve is its reflection in y = x, which is the graph of the inverse, the cube root. Drag P along $y = x^{3}$ to (2, 8): its mirror P' is at (8, 2), because f(2) = 8 means $f^{-1}\left(8\right) = 2$. Then choose $f\left(x\right) = x^{2}$: the point at the same height on the other side of the y-axis reflects onto the same vertical line as P', so the reflected curve is not the graph of a function.

Only a one-to-one function has an inverse

Take $f\left(x\right) = x^{2}$. Since $3^{2} = 9$ and $\left(-3\right)^{2} = 9$, two inputs give the output 9. An inverse would have to send 9 back to 3 and also to −3, which is two outputs for one input, and a function gives each input exactly one output. So $x^{2}$, with every real number allowed in, has no inverse.

A function has an inverse only when each of its outputs comes from exactly one input. Such a function is called one-to-one. On a graph, this means no horizontal line meets the graph more than once. The horizontal line y = 9 meets $y = x^{2}$ twice, at (−3, 9) and (3, 9). Reflected in y = x, it becomes the vertical line x = 9, which would meet the reflected curve twice, at (9, −3) and (9, 3). A vertical line meeting a graph twice means that graph is not a function.

The fix is to cut the domain. With the domain $x \ge 0$, the function $f\left(x\right) = x^{2}$ is one-to-one: each output 0 or more comes from exactly one input that is 0 or more. Its inverse is $f^{-1}\left(x\right) = \sqrt{x}$, because $\surd$ means the square root that is not negative. Check: f(3) = 9 and $f^{-1}\left(9\right) = \sqrt{9} = 3$.

The domain and range change places. $f\left(x\right) = x^{2}$ with $x \ge 0$ has the range $f\left(x\right) \ge 0$, and that is exactly the domain of $\sqrt{x}$. In general, the inputs of $f^{-1}$ are the outputs of f, and the outputs of $f^{-1}$ are the inputs of f.

The plain curve is $y = x^{2}$ with its domain cut to $x \ge 0$. The gold curve is its inverse, $y = \sqrt{x}$, and the straight gold line is y = x. The points (2, 4) and (4, 2) are reflections of each other: $2^{2} = 4$ and $\sqrt{4} = 2$.

The usual mistakes

Undoing the steps in the original order. For f(x) = 3x + 1, dividing by 3 and then subtracting 1 gives $\frac{x}{3} - 1$, which sends 10 to $\frac{7}{3}$, not back to 3. The last step, adding 1, is undone first.

Reversing only one step. 3x − 1 turns the add into a subtract but still multiplies by 3. Every step is replaced by its opposite, so the multiply becomes a divide as well.

Reading $f^{-1}$ as a reciprocal. $f^{-1}\left(x\right) = \frac{{x - 1}}{3}$, not $\frac{1}{{3x + 1}}$.

Giving $x^{2}$ an inverse without cutting its domain. Only with $x \ge 0$, or only with $x \le 0$, is $x^{2}$ one-to-one.