On a specified domain, a function f has an inverse function, or is invertible, if each output value of f is mapped from a unique input value. A graph test for this is the horizontal line test: if any horizontal line crosses the graph more than once, some output has two inputs.
Example. f(x) = x² is not invertible on all real numbers, because f(2) = 4 and f(−2) = 4. But the domain of a function may be restricted in many ways to make it invertible. On x ≥ 0, f(x) = x² is invertible, and so is it on x ≤ 0.
A function is invertible when each output value comes from a unique input value. Slide the horizontal line and count how many inputs land on it. If any line hits the graph twice, the function is not invertible on that domain.
An inverse function, written f⁻¹, maps the output values of f on its invertible domain back to their corresponding input values. If f(a) = b, then f⁻¹(b) = a. On the invertible domain, if f consists of input-output pairs (a, b), then f⁻¹ consists of the pairs (b, a).
Example. If f(3) = 1, then f⁻¹(1) = 3. To find f⁻¹(9) from a table, look for the 9 in the output row of f and read its input.
An inverse function is a reverse mapping. If f(a) = b, then f⁻¹(b) = a. So the input-output pairs (a, b) of f become the pairs (b, a) of the inverse. Pick a table and a pair.
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| f(x) | 4 | 7 | 1 | 9 | 6 |
| input | 1 | 4 | 6 | 7 | 9 |
|---|---|---|---|---|---|
| reverse | 3 | 1 | 5 | 2 | 4 |
The composition of a function f and its inverse f⁻¹ is the identity function, in either order:
This is the check you can use on any proposed inverse. Just as additive inverses add to 0 and multiplicative inverses multiply to 1, function inverses compose to give the identity function I(x) = x from Lesson 2.7.
Example. f(x) = 2x + 1 and g(x) = (x − 1)/2. Then f(g(x)) = 2·(x − 1)/2 + 1 = x and g(f(x)) = ((2x + 1) − 1)/2 = x. So g = f⁻¹. But h(x) = (x + 1)/2 gives f(h(x)) = x + 2, which is not x, so h is not the inverse.
A function and its inverse undo each other, so composing them in either order returns the input: f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. That is the identity function from Lesson 2.7. Test several pairs, including one that is not an inverse pair.
On a function's invertible domain, the range of f is the domain of f⁻¹, and the domain of f is the range of f⁻¹. The inverse of a table of values of y = f(x) is found by reversing the input-output pairs: (a, b) corresponds to (b, a).
Example. Let f(x) = √(x − 1) + 2 with domain x ≥ 1 and range y ≥ 2. Then f⁻¹ has domain x ≥ 2 and range y ≥ 1. Since f(5) = 4, f⁻¹(4) = 5.
On an invertible domain, the range and domain of f are the domain and range of f⁻¹. If (a, b) is on the graph of f, then (b, a) is on the graph of f⁻¹, so the two graphs mirror each other across the dashed line y = x. Blue is f and gold is f⁻¹.
To determine the inverse of a function on an invertible domain, describe what f does to its input as a list of steps, then undo the steps in reverse order. The last operation of f is the first one you undo.
Example. f(x) = 2x + 3 multiplies by 2, then adds 3. To undo it, subtract 3, then divide by 2. So f⁻¹(y) = (y − 3)/2. Check: f(f⁻¹(y)) = 2·(y − 3)/2 + 3 = y ✓.
To find an inverse, describe what f does to its input step by step, then undo the steps in reverse order using the opposite operations. The last step of f is the first step you undo.
In a real situation, an inverse answers the reverse question. If f turns Celsius into Fahrenheit, f⁻¹ turns Fahrenheit into Celsius. The meaning and units of the input and output swap, so always say what f⁻¹(b) = a means in words.
Example. A taxi costs C(m) = 3 + 2m dollars for m miles. C(10) = 23 means a 10-mile ride costs $23. The inverse answers a different question: C⁻¹(23) = 10 means that $23 buys a 10-mile ride.
In a context, the inverse swaps the meaning and the units of the input and output. If f(a) = b, then f⁻¹(b) = a, and the units follow the swap.