What is an inverse function?
A function takes an input x and gives one output f(x). The inverse function, written f⁻¹, goes the other way. It takes the output and gives back the input.
If f(a) = b, then f⁻¹(b) = a.
Doing one and then the other brings you back to the start: f⁻¹(f(x)) = x and f(f⁻¹(x)) = x.
Careful: f⁻¹ does not mean 1 ÷ f(x). The small −1 here is a name, not a power.
The domain (allowed inputs) of f⁻¹ is the range (outputs) of f, and the range of f⁻¹ is the domain of f.
How to find the inverse of a function
Four steps:
- Write y = f(x).
- Swap x and y.
- Solve the new equation for y.
- Write the answer as f⁻¹(x) = … and state its domain.
Example: f(x) = 2x + 1. Write y = 2x + 1. Swap: x = 2y + 1. Solve: y = (x − 1)/2. So f⁻¹(x) = (x − 1)/2.
Check: f(3) = 7 and f⁻¹(7) = 3. It works.
Quick method for simple chains: list the steps of f and undo them in reverse order. f = “×2 then +1”, so f⁻¹ = “−1 then ÷2”.
Graph of an inverse function
Every point (a, b) on f becomes (b, a) on f⁻¹. Swapping coordinates is the same as reflecting in the line y = x.
So to sketch f⁻¹, draw y = x as a mirror and flip the graph of f across it.
- Where f meets y = x, f⁻¹ meets it at the same point.
- If f crosses the y-axis at (0, c), then f⁻¹ crosses the x-axis at (c, 0).
- Famous pairs: 2ˣ and log₂ x; x³ and ∛x; x² (x ≥ 0) and √x; eˣ and ln x.
One-to-one functions and restricting the domain
The undo machine must know which input to give back. So each output must come from only one input. Such a function is called one-to-one (injective).
Horizontal line test: if any flat line cuts the graph more than once, the function is not one-to-one and has no inverse.
x² fails: 2² = 4 and (−2)² = 4. Fix it by keeping only part of the graph. With domain x ≥ 0, f(x) = x² is one-to-one and its inverse is the square root function f⁻¹(x) = √x, with domain x ≥ 0.
Strictly increasing or strictly decreasing functions always pass the test.
Try it: the undo game
With a friend: one person secretly picks a number, doubles it, adds 5 and says only the result. You must find the secret number. Write the rule as f(x) = 2x + 5 and use f⁻¹(x) = (x − 5)/2. In the 3D, choose a function, move the slider and check that the blue and orange points always mirror each other.
Key formulas and definitions
- If f(a) = b then f⁻¹(b) = a
- f⁻¹(f(x)) = x and f(f⁻¹(x)) = x
- Domain of f⁻¹ = range of f; range of f⁻¹ = domain of f
- Graph of f⁻¹ = reflection of graph of f in y = x
- f(x) = ax + b ⇒ f⁻¹(x) = (x − b)/a
- x² (x ≥ 0) ↔ √x; aˣ ↔ logₐ x
Worked examples
1. Find the inverse of f(x) = 3x − 6.
y = 3x − 6. Swap: x = 3y − 6. Add 6: x + 6 = 3y. Divide by 3: y = (x + 6)/3. So f⁻¹(x) = (x + 6)/3. Check: f(4) = 6, f⁻¹(6) = 12/3 = 4.
2. f(x) = 5x + 2. Find f⁻¹(17) without finding the formula.
f⁻¹(17) is the x for which f(x) = 17. 5x + 2 = 17 ⇒ 5x = 15 ⇒ x = 3. So f⁻¹(17) = 3.
3. Find the inverse of f(x) = (x + 4)/2 and verify f(f⁻¹(x)) = x.
y = (x + 4)/2. Swap: x = (y + 4)/2. 2x = y + 4 ⇒ y = 2x − 4. f⁻¹(x) = 2x − 4. Check: f(2x − 4) = (2x − 4 + 4)/2 = x. ✔
4. Find the inverse of f(x) = x³ + 1.
y = x³ + 1. Swap: x = y³ + 1. y³ = x − 1. y = ∛(x − 1). So f⁻¹(x) = ∛(x − 1), defined for all real x.
5. Find the inverse of f(x) = (x − 2)², x ≥ 2, and state its domain.
Range of f is y ≥ 0. y = (x − 2)². Swap: x = (y − 2)². Take the root (y ≥ 2 so y − 2 ≥ 0): √x = y − 2. f⁻¹(x) = √x + 2, domain x ≥ 0, range ≥ 2.
6. Find the inverse of f(x) = (2x + 1)/(x − 3), x ≠ 3.
y = (2x + 1)/(x − 3). Swap: x = (2y + 1)/(y − 3). x(y − 3) = 2y + 1 ⇒ xy − 3x = 2y + 1 ⇒ xy − 2y = 3x + 1 ⇒ y(x − 2) = 3x + 1. f⁻¹(x) = (3x + 1)/(x − 2), x ≠ 2.
Common mistakes
- Thinking f⁻¹(x) means 1/f(x). It does not: for f(x) = 2x, f⁻¹(x) = x/2 but 1/f(x) = 1/(2x).
- Forgetting to swap x and y, and just rearranging y = f(x) for x — then naming the answer wrongly.
- Undoing the steps in the same order instead of the reverse order ("−1 then ÷2", not "÷2 then −1").
- Giving x² an inverse without limiting the domain, or forgetting that √x needs x ≥ 0.