The inverse of a function reverses it: it takes each output of f back to the input that produced it. To find f⁻¹(x) you write y = f(x), rearrange to make x the subject, then rename the variable. To finish properly you also state the domain of f⁻¹.
This lesson builds on composite functions, because f⁻¹ is the function for which f⁻¹f(x) = x. It leads into the question of when an inverse does not exist.
How do you find the inverse?
- Write y = f(x).
- Rearrange to make x the subject. Collect all the x terms on one side and factorise if x appears more than once.
- Replace y by x to write f⁻¹(x).
- Find the range of f. This gives the domain of f⁻¹.
- Check with a number: f⁻¹(f(a)) should return a.
Worked example
The function f is defined by f(x) = (2x + 3)/(x − 1) for x ≠ 1. Find f⁻¹(x) and state its domain.
Step 1: y = (2x + 3)/(x − 1).
Step 2, clear the fraction: y(x − 1) = 2x + 3, so yx − y = 2x + 3.
Step 3, collect x terms: yx − 2x = y + 3, so x(y − 2) = y + 3.
Step 4, make x the subject: x = (y + 3)/(y − 2).
Step 5, rename: f⁻¹(x) = (x + 3)/(x − 2).
Domain: write f(x) = (2(x − 1) + 5)/(x − 1) = 2 + 5/(x − 1). The term 5/(x − 1) is never zero, so f(x) never equals 2. The range of f is all real values except 2, so the domain of f⁻¹ is x ≠ 2. This matches the denominator of f⁻¹, which is zero at x = 2.
Check: f(3) = 9/2 = 4.5. Then f⁻¹(4.5) = 7.5/2.5 = 3. It returns to 3.
The mistake to watch for
A common slip is to treat f⁻¹ as the reciprocal of f.
Mistaken answer: f⁻¹(x) = 1/f(x) = (x − 1)/(2x + 3).
Test it: f(3) = 4.5, so a true inverse must send 4.5 back to 3. This rule gives (4.5 − 1)/(2 × 4.5 + 3) = 3.5/12 ≈ 0.29, not 3.
The correction is to remember that the −1 is a label for “inverse” and not an index. The inverse comes from rearranging for x, not from flipping the fraction. A single numerical check, as in the worked example, exposes the confusion quickly.
Check yourself
Find f⁻¹(x) and its domain where required, then open the answer.
1. f(x) = 3x − 5
Show answer
y = 3x − 5, so 3x = y + 5, so x = (y + 5)/3. f⁻¹(x) = (x + 5)/3, defined for all real x.
2. f(x) = x² + 1 for x ≥ 0
Show answer
y = x² + 1 gives x² = y − 1, and since x ≥ 0 we take the positive root: x = √(y − 1). f⁻¹(x) = √(x − 1). The range of f is f(x) ≥ 1, so the domain of f⁻¹ is x ≥ 1.
3. f(x) = 5/(x + 2) for x ≠ −2
Show answer
y(x + 2) = 5, so x + 2 = 5/y, so x = 5/y − 2. f⁻¹(x) = 5/x − 2, which is (5 − 2x)/x. The output of f is never 0, so the domain is x ≠ 0. Check: f(1) = 5/3, and f⁻¹(5/3) = 3 − 2 = 1.
Where this leads next
The natural next lesson is explaining a many-to-one mapping that has no unrestricted inverse, which shows why the domain matters even more. Practise the whole module in the functions and restrictions practice set, and use the function composition and inverse explorer to see the graph of f and f⁻¹ reflect in y = x.
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