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 $f^{-1}$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, $f^{-1}\left(10\right)=3$f⁻¹(10) = 3. The −1 is part of the name, not a power: $f^{-1}\left(x\right)$f⁻¹(x) 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, $f^{-1}\left(x\right)=\frac{\left(x-1\right)}{3}$f⁻¹(x) = (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}$10/3 − 1 = 7/3, not 3. Reversing the order without reversing the steps, adding 1 and then multiplying by 3, turns 10 into 33.

$f^{-1}$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: $x=\frac{\left(y-1\right)}{3}$x = (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: $f^{-1}\left(x\right)=\frac{\left(x-1\right)}{3}$f⁻¹(x) = (x − 1)/3.

Check by composing. $f^{-1}f\left(x\right)=f^{-1}\left(3x+1\right)=\frac{\left(\left(3x+1\right)-1\right)}{3}=\frac{3x}{3}=x$f⁻¹f(x) = f⁻¹(3x + 1) = ((3x + 1) − 1)/3 = 3x/3 = x, and ff⁻¹$\left(x\right)=f\left(\frac{\left(x-1\right)}{3}\right)=3\times \frac{\left(x-1\right)}{3}+1=x-1+1=x$(x) = f((x − 1)/3) = 3 × (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 $\pi$π. Drag the radius to change the circle, and choose the order of the undoing steps. $\surd$√ first and then $÷\pi$÷ π lands on a different number. $÷\pi$÷ π 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 $f^{-1}\left(4\right)=1$f⁻¹(4) = 1, so (4, 1) is on the graph of $f^{-1}$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 $f^{-1}$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 $f^{-1}$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 $f^{-1}$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 white line is its inverse, $y=\frac{\left(x-1\right)}{3}$y = (x − 1)/3. The marked points (1, 4) and (4, 1) are reflections of each other in y = x.

The gold curve is $y=x^3$y = x³ and the green dashed curve is its reflection in y = x, which is the graph of the inverse, the cube root. Drag P along $y=x^3$y = x³ to (2, 8): its mirror P' is at (8, 2), because f(2) = 8 means $f^{-1}\left(8\right)=2$f⁻¹(8) = 2. Then choose $f\left(x\right)=x^2$f(x) = x²: 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 $f\left(x\right)=x^2$f(x) = x². Since $3^2=9$3² = 9 and ${\left(-3\right)}^2=9$(−3)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 $x^2$x², 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 $y=x^2$y = x² 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 $x\ge 0$x ≥ 0, the function $f\left(x\right)=x^2$f(x) = x² is one-to-one: each output 0 or more comes from exactly one input that is 0 or more. Its inverse is $f^{-1}\left(x\right)=\sqrt{x}$f⁻¹(x) = √x, because $\surd$√ means the square root that is not negative. Check: f(3) = 9 and $f^{-1}\left(9\right)=\sqrt{9}=3$f⁻¹(9) = √9 = 3.

The domain and range change places. $f\left(x\right)=x^2$f(x) = x² with $x\ge 0$x ≥ 0 has the range $f\left(x\right)\ge 0$f(x) ≥ 0, and that is exactly the domain of $\sqrt{x}$√x. In general, the inputs of $f^{-1}$f⁻¹ are the outputs of f, and the outputs of $f^{-1}$f⁻¹ are the inputs of f.

The white curve is $y=x^2$y = x² with its domain cut to $x\ge 0$x ≥ 0. The gold curve is its inverse, $y=\sqrt{x}$y = √x, and the straight gold line is y = x. The points (2, 4) and (4, 2) are reflections of each other: $2^2=4$2² = 4 and $\sqrt{4}=2$√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 $\frac{x}{3}-1$x/3 − 1, which sends 10 to $\frac{7}{3}$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 $f^{-1}$f⁻¹ as a reciprocal. $f^{-1}\left(x\right)=\frac{\left(x-1\right)}{3}$f⁻¹(x) = (x − 1)/3, not $\frac{1}{\left(3x+1\right)}$1/(3x + 1).

Giving $x^2$x² an inverse without cutting its domain. Only with $x\ge 0$x ≥ 0, or only with $x\le 0$x ≤ 0, is $x^2$x² one-to-one.