The inverse function f⁻¹ undoes f: if f sends a to b, then f⁻¹ sends b back to a. Two things can go wrong, the rearrangement and the domain, and this lesson covers both.
This lesson follows composite functions in the right order, because the check at the end is a composite. The chapter overview is at Functions.
How do I find an inverse?
Use four steps. Write y = f(x), solve for x in terms of y, swap the letters so the answer reads f⁻¹(x), then state the domain.
Take f(x) = (2x + 3) ÷ (x − 1), with x ≠ 1.
- y = (2x + 3) ÷ (x − 1), so y(x − 1) = 2x + 3.
- Expand: yx − y = 2x + 3, so yx − 2x = y + 3.
- Factorise: x(y − 2) = y + 3, so x = (y + 3) ÷ (y − 2).
- Swap: f⁻¹(x) = (x + 3) ÷ (x − 2), x ≠ 2.
The domain restriction x ≠ 2 is the value f can never output. Check: f(3) = 9 ÷ 2 = 4.5, and f⁻¹(4.5) = 7.5 ÷ 2.5 = 3. It returns the input.
Worked example: same rule, two domains
Let f(x) = (x − 2)². On all real numbers this has no inverse, because f(1) = f(3) = 1.
Domain x ≥ 2. Set y = (x − 2)². Then x − 2 = ±√y. Because x ≥ 2, x − 2 is not negative, so x − 2 = +√y and x = 2 + √y.
The range of f is y ≥ 0, so f⁻¹(x) = 2 + √x, x ≥ 0.
Domain x ≤ 2. Now x − 2 is not positive, so x − 2 = −√y and x = 2 − √y.
The inverse is f⁻¹(x) = 2 − √x, x ≥ 0. The two inverses differ only in the sign, and the domain of f decides which one is right.
Check the first one: f⁻¹(9) = 2 + 3 = 5, and f(5) = 3² = 9. It works.
The mistake that costs marks
The usual slip is to write ± in the final answer, or to choose the sign without reading the domain. An inverse is a function, so it gives one output for each input, and ± gives two.
The domain of f is what removes the ambiguity. Read it before you take the square root.
Check yourself
f(x) = x² + 1 with domain x ≥ 0. Find f⁻¹(x) and f⁻¹(10).
Answer
y = x² + 1, so x² = y − 1. The domain x ≥ 0 means x is not negative, so x = √(y − 1).
The range of f is y ≥ 1, so f⁻¹(x) = √(x − 1), x ≥ 1.
f⁻¹(10) = √9 = 3. Check: f(3) = 9 + 1 = 10.
What to study next
Move on to solving equations involving an unknown function, where inverses and composites appear together. Try your own pairs in the composite and inverse function explorer.
If restrictions still slip, online one-to-one Additional Mathematics tuition gives you a teacher reading your working line by line.