IB Maths AA HLTopic 2 — FunctionsPaper 1 & 2~9 min read
Inverse Functions
An inverse function f−1 is a function that undoes what f does. Feed an output back through f−1 and you get the original input. Practically that means inverses are useful for solving equations — if you can build f−1, then f(x) = k becomes x = f−1(k) immediately. The mechanics are short: swap x and y, rearrange. The catches are about whether the inverse exists at all and getting the domain right.
📘 What you need to know
Inverse undoes the original: f(f−1(x)) = f−1(f(x)) = x — applying both in either order gives you back what you started with (the identity function).
Notation warning: f−1(x) is the inverse function, NOT the reciprocal 1/f(x). Despite the −1.
How to find f−1: write y = f(x), swap x and y, then rearrange to make y the subject. The result is f−1(x).
Domain and range swap: domain of f = range of f−1; range of f = domain of f−1.
Inverse exists only if f is one-to-one. Many-to-one functions have no inverse unless you restrict the domain to make them one-to-one.
Graph of f−1 is the reflection of f in the line y = x. Every point (a, b) on f becomes (b, a) on f−1.
Solving with inverses: f(x) = k ⟺ x = f−1(k).
What an inverse actually does
Inverse propertyf(f−1(x)) = x and f−1(f(x)) = x
Quick demo. If f(x) = 4x, then f−1(x) = x/4. Check: f(f−1(20)) = f(5) = 20. The inverse “undid” the original. The composition f ∘ f−1 just gives back the input — that’s called the identity function, id(x) = x.
Important warning: f−1(x) is the inverse function. The reciprocal is 1/f(x) = [f(x)]−1. They look similar but are entirely different concepts. Don’t confuse them.
What an inverse looks like on a graph
Reflecting a graph in the line y = x swaps the x and y coordinates of every point. So if (a, b) is on y = f(x), then (b, a) is on y = f−1(x).
f and its inverse — reflections in y = x
If y = f(x) crosses the line y = x, the inverse curve passes through the same point — fixed points of f are also fixed points of f−1.
How to find f−1(x) — the swap method
🧭 Recipe — finding an inverse algebraically
Writey = f(x).
Swapx and y everywhere.
Rearrange to make y the subject.
The result isf−1(x).
State its domain (this equals the range of the original f).
Domain and range swap
Domain of f ⟶ Range of f−1
inputs become outputs
e.g. domain of f is x ≥ 2 → range of f−1 is f−1(x) ≥ 2
Range of f ⟶ Domain of f−1
outputs become inputs
e.g. range of f is f(x) ≥ 5 → domain of f−1 is x ≥ 5
When does an inverse exist?
An inverse function f−1 exists only if f is one-to-one. If f is many-to-one, the “inverse” would have to give multiple outputs for one input — which is forbidden for a function.
Horizontal line testf has an inverse ⟺ every horizontal line crosses the graph of f at most once
Restricting the domain to force one-to-one
Many-to-one functions can be made one-to-one by restricting the domain. The classic example is f(x) = x2: it’s many-to-one over all of ℝ, but if you restrict to x ≥ 0 it becomes one-to-one and the inverse is √x.
Step 1: Set y = f(x) and swapy = (3x + 2)/(x − 1)swap → x = (3y + 2)/(y − 1)Step 2: Multiply both sides by (y − 1)x(y − 1) = 3y + 2xy − x = 3y + 2Step 3: Get y terms on one sidexy − 3y = x + 2y(x − 3) = x + 2y = (x + 2)/(x − 3)f⁻¹(x) = (x + 2)/(x − 3), x ≠ 3check: f(2) = 8/1 = 8; f⁻¹(8) = 10/5 = 2 ✓
WE 3
Find domain and range of an inverse
f(x) = 2x − 5 has domain 0 ≤ x ≤ 10. Find the domain and range of f−1(x).
Step 1: Find range of f (linear, increasing)f(0) = −5, f(10) = 15range of f: −5 ≤ f(x) ≤ 15Step 2: Swap roles for the inversedomain of f⁻¹ = range of frange of f⁻¹ = domain of fdomain of f⁻¹: −5 ≤ x ≤ 15; range: 0 ≤ f⁻¹(x) ≤ 10no need to find an actual formula for f⁻¹ — the swap rule alone gives the answer
WE 4
Restrict a quadratic to make it invertible
Let f(x) = (x + 3)2 − 4. (a) State the largest possible value of m such that f(x), x ≥ m is one-to-one. (b) Find f−1(x) and state its domain.
(a) Find the vertexin completed-square form, vertex is at (−3, −4)to be one-to-one, restrict to x ≥ −3 (right half) or x ≤ −3 (left half)(a) m = −3(b) Swap and rearrangey = (x + 3)² − 4swap → x = (y + 3)² − 4x + 4 = (y + 3)²y + 3 = ±√(x + 4)y = −3 ± √(x + 4)Pick the right sign — range of f⁻¹ = domain of f = x ≥ −3need f⁻¹(x) ≥ −3 → take the + rootrange of f (when x ≥ −3): f(x) ≥ −4 → domain of f⁻¹ is x ≥ −4f⁻¹(x) = −3 + √(x + 4), x ≥ −4check: f(0) = 5; f⁻¹(5) = −3 + 3 = 0 ✓
WE 5
Multi-part inverse question
The function f(x) = (x − 4)2 + 1, x ≥ m, has an inverse.
(a) State the smallest value of m. (b) Find f−1(x). (c) State the domain of f−1(x). (d) Find k such that f(k) = 17.
(a) Vertex at (4, 1) — restrict to x ≥ 4 for one-to-one(a) m = 4(b) Swap and rearrangey = (x − 4)² + 1 → swap: x = (y − 4)² + 1x − 1 = (y − 4)²y − 4 = ±√(x − 1)range of f⁻¹ = domain of f = y ≥ 4 → take + root(b) f⁻¹(x) = 4 + √(x − 1)(c) Range of f when x ≥ 4 starts at 1, increasesrange of f: f(x) ≥ 1 → domain of f⁻¹ is x ≥ 1(c) domain of f⁻¹: x ≥ 1(d) f(k) = 17 ⟺ k = f⁻¹(17)k = 4 + √(17 − 1) = 4 + √16 = 4 + 4(d) k = 8check directly: f(8) = (8−4)² + 1 = 16 + 1 = 17 ✓
WE 6
Verify the inverse property using composition
For f(x) = 5x − 3, find f−1(x) and verify that f(f−1(x)) = x.
Step 1: Find inversey = 5x − 3 → swap: x = 5y − 3y = (x + 3)/5f⁻¹(x) = (x + 3)/5Step 2: Compute f(f⁻¹(x))f(f⁻¹(x)) = f((x + 3)/5)= 5 · (x + 3)/5 − 3= (x + 3) − 3= x ✓f⁻¹(x) = (x + 3)/5; f(f⁻¹(x)) = x as requiredthis composition check is the cleanest way to verify your inverse is correct
💡 Top tips
The swap-and-rearrange method works for almost everything. Write y = f(x), swap x and y, solve for y.
Always state the domain of f−1. It equals the range of the original f, so find that first.
Check by composition. A correct inverse satisfies f(f−1(x)) = x. Plug it through and confirm.
For quadratics under x ≥ h, take the positive square root in the inverse. For x ≤ h, take the negative. Pick based on the required range of f−1.
To solve f(x) = k, just compute x = f−1(k). One step instead of full algebra.
Sketch f(x) on your GDC when you need to find a restricted domain — the vertex or turning point shows you where to cut.
For rational functions like (ax+b)/(cx+d), the algebra is always: multiply both sides by the denominator, expand, group y-terms together, factor.
⚠ Common mistakes
Confusing f−1(x) with 1/f(x). The −1 superscript means inverse, not reciprocal.
Forgetting to state the domain of the inverse. Often worth a mark and easy to miss.
Picking the wrong sign of the square root when inverting a quadratic. Use the range of f−1 to decide.
Trying to invert a many-to-one function without restricting the domain first. x2 over all of ℝ has no inverse.
Forgetting to swap x and y. Just rearranging y = f(x) for x gives the wrong answer — that’s not what an inverse function looks like.
Mixing up domain and range when swapping. Domain of f becomes the range of f−1, not its domain.
Inverting only part of the expression. Every x on the right-hand side has to play a role in the rearrangement.
Inverse functions show up constantly in IB — solving exponential and log equations, inverse trig, and (later) chain-rule integration all hinge on knowing how to invert. The next note covers odd and even functions — special functions whose graphs have symmetry and whose properties make a lot of calculus problems much faster.
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