The big idea: The inverse function f⁻¹ undoes what f does.__LINEBREAK__If f takes x to y, then f⁻¹ takes y back to x.__LINEBREAK__For example, if f(3) = 7, then f⁻¹(7) = 3.
- the inverse of f — reads "f inverse"
- applying f then f⁻¹ brings you back to the start
Real-world inverse: temperature conversion
The formula F = 1.8C + 32 converts Celsius (C) to Fahrenheit (F). Find the inverse formula that converts Fahrenheit to Celsius.
Step by step
- Write the original function. F is the output, C is the input.
- Solve for C (the original input) in terms of F (the original output).
- Divide by 1.8.
Final answer
The inverse is C = (F − 32) / 1.8. This is exactly the type of inverse question IB asks.
Critical: f⁻¹(x) is NOT 1/f(x): The notation f⁻¹ looks like an exponent of −1, but it is NOT a reciprocal.__LINEBREAK__If f(x) = 3x − 5, then: ✅ f⁻¹(x) = (x + 5) / 3 (the inverse function) ❌ 1/f(x) = 1/(3x − 5) (the reciprocal — a completely different thing)__LINEBREAK__This is the most common confusion with inverse notation.
- inverse function f⁻¹
- The function that reverses f: if f(a) = b, then f⁻¹(b) = a.
- one-to-one function
- A function where every output comes from exactly one input — required for the inverse to also be a function.
The three-step method: 1. Write y = f(x) 2. Swap x and y (swap the variable roles) 3. Solve for y__LINEBREAK__The result y = ... is f⁻¹(x) — rename it properly.
Finding the inverse of a linear function
Find f⁻¹(x) for f(x) = 3x − 5.
Step by step
- Write y = f(x).
- Swap x and y.
- Solve for y: add 5 to both sides.
- Divide by 3.
- Write the inverse using correct notation.
Final answer
f⁻¹(x) = (x + 5) / 3
Finding the inverse of a fractional function
Find f⁻¹(x) for f(x) = (x + 2) / 4.
Step by step
- Write y = f(x).
- Swap x and y.
- Multiply both sides by 4.
- Solve for y.
- Write using inverse notation.
Final answer
f⁻¹(x) = 4x − 2
Always write f⁻¹(x) = ...: Do not just write the expression without the correct notation.__LINEBREAK__IB awards a mark for correct notation: f⁻¹(x) = (x + 5)/3.__LINEBREAK__Writing just "(x + 5)/3" is incomplete — you haven't told IB what you found.
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The big idea: For an inverse to exist as a function, the original function must be one-to-one — every output comes from exactly one input.__LINEBREAK__A parabola like f(x) = x² is NOT one-to-one: f(3) = 9 and f(−3) = 9. Two inputs give the same output.__LINEBREAK__Fix: restrict the domain (e.g. x ≥ 0) so each output comes from exactly one input.
Restricting the domain for f(x) = x²
f(x) = x² with domain restricted to x ≥ 0. Find f⁻¹(x).
Step by step
- Write y = x² (with domain x ≥ 0, so y ≥ 0).
- Swap x and y.
- Solve for y — take the positive square root (y ≥ 0).
- Write the inverse.
Final answer
f⁻¹(x) = √x, domain x ≥ 0. Without the restriction, f⁻¹ would not be a function.
| Property | f(x) | f⁻¹(x) |
|---|---|---|
| Domain | restricted domain of f | becomes the range of f⁻¹ |
| Range | restricted range of f | becomes the domain of f⁻¹ |
| Rule | domain of f → range of f | range of f → domain of f (reversed) |
[Diagram: math-inverse-function-diagram] - Available in full study mode
IB gives you the restriction — use it: IB will usually tell you the restricted domain in the question.__LINEBREAK__If the question says "f(x) = x² for x ≥ 0", the restriction is already given.__LINEBREAK__Your job: find f⁻¹(x) AND state its domain (which equals the range of f).
The big idea: The graph of f⁻¹ is the reflection of the graph of f over the line y = x.__LINEBREAK__Every point (a, b) on f maps to the point (b, a) on f⁻¹.__LINEBREAK__This is because f and f⁻¹ swap the x and y roles.
[Diagram: math-inverse-function-diagram] - Available in full study mode
Reading the inverse graph from f
f passes through (1, 3), (2, 5), and (4, 9). State three points on f⁻¹.
Step by step
- Each point (a, b) on f gives point (b, a) on f⁻¹.
Final answer
f⁻¹ passes through (3, 1), (5, 2), and (9, 4).
| Property | Link between f and f⁻¹ |
|---|---|
| Graph relationship | f⁻¹ is the reflection of f over y = x |
| Point relationship | if (a, b) is on f, then (b, a) is on f⁻¹ |
| Domain / Range | domain of f = range of f⁻¹, and vice versa |
| Intersection | f and f⁻¹ intersect on the line y = x |
Sketching f⁻¹ in an exam: If IB gives you the graph of f and asks you to sketch f⁻¹:__LINEBREAK__1. Draw the line y = x (lightly, as a guide). 2. Reflect each key point of f over that line. 3. Join the reflected points with a smooth curve.__LINEBREAK__Always label the reflected curve as y = f⁻¹(x).